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Post Graduate Programme

M.Sc. Mathematics (Previous)

Distance Educations

Self Instructional Material

Block:1

Unit-1 Complex Integration Unit-2 Calculus of Residue

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Paper IV Complex Analysis Block -1

Complex Analysis Introduction :

The fundamental idea of extending the real number system by the introduction of complex numbers was first necessitated by the solution of algebraical equations L. Euler (1707-1783) was the first mathematician who introduced the symbol I =√-1 C.F. Gaurs (1777-1855) studied that the algebraic equation with real coefficients has complex roots of the from a+ ib where a and b are real numbers . W.R. Hamilton (1805-1865) also made considerable contribution to the development of the theory of complex number. The BLOCK contains two units. In unit 1 we will discuss complex integration. Under this we will studies the proofs at various theorems based on complex integration like Cauchy’s integral formula. Morera’s theorem, Schwarz lemma etc. in unit 2 we will study about residues and evaluation of integrals using it.

Complex Analysis: Unit-1 Complex integration Cauchy –Gousat theorem Cauchy

integral formula. Higher order derivatives. Morera’s Theorem Cauchy’s inequality and Liouville’s theorem. The Fundamental theorem of algebra. Taylor’s theorem. Maximum modulus principle. Schwarz lemma. Laurent’s series. Lsoluted singularities Menomarphlc Punctions. The argument principle . Rouche’s theorem. Inverse function theorem.

Unit–II Residues. Cauchy’s residue theorem. Evaluation of integrals. Branches of many valued functions with special reference to argz. logzandzn

.

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Unit -1 Complex Integration

Objectives: After reading this unit, your should to able to,

understand the concept of complex integration prove many results based on it understand singularities

STRUCTURE: 1.1 Complex integration 1.2 Basic Definitions

1.2.1 Partition 1.2.2 Jordanarc 1.2.3 Rectifiable Arcs 1.2.4 Contours 1.2.5 Simple and Multiply connected Regional. 1.2.6 Analytic functions

1.3 Cauchy’s –Goursat. Theorem 1.4 Cauchy’s Integral Derivalives 1.5 Higher order Inequality 1.6 Cauchy’s Inequality 1.7 Liouville’s theorem 1.8 Taylor’s theorem 1.9 Zero’s of Analysis functions 1.91 Singularities 1.92 Types of Singularity 1.10 Meromorphic functions 1.11 Argument Principle 1.12 Rouche’s Theorem 1.13 Fundamental theorem of algebra 1.14 Maximum modules principle 1.15 Schwarz lemma 1.16 summary 1.17 Self Assessment Test 1.18 Further Readings

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1.1 Complex integration: In the domain of real variable there are two points of view from which integration may be considered. In the domain of complex variable use start with the definition of an integral as the limit of a sum and latter on establish the connection between differentiation and integration.

1.2 Basic Definitions: 1.2.1 Partitions: Let [a,b] be a closed interval where a,b

are real number. Subdivide the interval [a,b] into n Sub- intervals

[t0,t1], [t1,t2], _ _ _ _ _ _ [tn,tn], by inserting n-1 intermediate. points t1, t2, _ _ _ tn-1, Satisfying the inequality a= t0 < t1 < t2 _ _ _ _ < tn = b Then the set P={ t0, t1 _ _ _ _ tn } is called partition of the interval [a,b]. 1.2.2 Jordan Arc: Suppose a point Z lies on an ace L is

defined by Z=Z(t)= x (t) +iy (t) Where t runs through the interval a ≤ t ≤ b and x (t), y(t) are continuous function of t Then L is called simple or Jordan are if Z(t1)= Z(t2) only when t1 = t2

1.2.3 Rectifiable Arcs: Suppose L denotes a Jordan arc defined by

Z=Z(t)= x (t) + iy (t), a ≤ t ≤ b

Z0

Z1

Z2

Z3

Zn-1

Zn

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Let P={a=t0, t1 , t2 , _ _ _ _ _ tn = b } be any partition of [a,b] let Z0 ,Z1, …...Zn be the points of the are wr responding to their value. Then the length of this polygonal are is;

=

∑

=

∑ | |

If this sum tends to a unique value then it is called rectifiable.

1.2.4 Contour: A contour is a continuous chain of a finite number of regular arc (rectifiable arc)

1.2.5 Simply and Multiply Connected Regions: A region in the argand plane in which every closed curve can be shrunk to a point without passing out of the region is called a simply connected region otherwise it is multiply connected.

1.2.6 Analytic Functions: A function of Z which is one valued and differentiable at every point of a domain D. Save possibly for a finite number of exceptional points is called analytic in the domain D. Those exceptional points are called Singularities.

1.3 Cauchy-Goursat Theorem Statement: If a function f(Z) is analytic and single valued inside and on a simple closed contour C, then

∫ ( )

Proof: For the proof of the theorem use will use two lemmas which are as follows;

Lemma1: if C is a closed contour, then the two integrals

∫

∫

Lemma 2 : Let f(z) be analytic within and on a closed contour C. Then for any given >0, it is always possible to

divide the region inside C into number of meshes either

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0

y

x

complete square Cn or partial Squarer Dn Such that within each mesh there exists a point Z0 such that

| ( ) ( ) ( ) ( )| | |

For every Z within or on the mesh. Now we divide the interior of C into complete squares C1, C2, _ _ _ _ Cm and partial Squares D1, D2, ____ _ _ _ Dn. Now

∫ ( )

∑ ∫ ( ) ∑ ∫ ( )

For the proof of the theorem. We will consider two cases. Case- I : Consider the mesh Cm Then

( ) ( ) ( ) ( ) ( ) ( ) ( )

Where | ( )|

∫ ( )

∫ ,( ) ( ) ( )-

∫ ( ) ( )

From lemma (1) we get

∫ ( ) ( )

as | ( )|

|∫ ( )

| √

Where denotes the length of diagonal of Cm

B

A D

C

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Case II – Consider the irregular sub regions Dn. let Sn denotes the length of the curved part of Dn

Then

|∫ ( ) ( )

| √ ( )

Now (1) becomes

|∫ ( )

| ∑ |∫ ( )

|

∑ ∫ ( )

=

∑ |∫ ( ) ( )

|

∑ |∫ ( ) ( )

|

|∫ ( )

| √ ∑( ) √ ∑

all the terms in the right hand side are constant

|∫ ( )

|

this is only true when

∫ ( )

1.4 Cauchy’s Integral Formula: Statement : let ( ) be analytic within and on a closed

contour C and let any point within C . Then ;

( )

∫

( )

Proof : We define a circle by the equation | | =

where and is the distance from to the neares point of C.

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Consider a function

( ) ( )

∫

( )

∫

( )

where C and are both traversed in the positive sense.

As ( ) is continuous at so by defintion for given

| ( ) ( )| whenever | | -----------(1)

Now

∫

( )

∫

( ) ( )

∫

( )

--------------(2)

Put =

We have

∫

( )

( )

∫

( )

∫

( ) Hence (2) becomes

∫

( )

∫

( ) ( )

( )

or

∫

( )

( )

∫

( ) ( )

---------(3) Since we may choose as small as wee please, we take

. Thus the inequality (1) is satisfied for all points on .

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Hence

|

∫

( ) ( )

| |

∫

( ) ( )

)

|

∫ | ( ) ( )|

∫

Thus

|

∫

( ) ( )

|

−-------------------(4)

Then it follows from (3) and (4) that

|

∫

( )

( )

|

We now observe that is arbitrary and the left hand Side is

independent of it. Thus implies that.

∫

( )

( )

∫

( )

( )

or

∫

( )

1.5 Higher order derivatives: Statement: Let ( ) be analytic within and on the boundary C* of a simply connected region and let be any point within C.

Then derivatives of all orders are analytic and given by

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( )

∫

( )

( )

Proof: We first show that ( ) is analytic inside C. To prove this it is enough to show that ( ) has a differential

coefficient at every point inside C. We have by Cauchy’s

formula for ( ) and ( )

( ) ( )

∫ [

( )

( ) ]

( )

∫

( ) ( )

( ) ( ) ( )

follows that

( ) ( )

∫

( )

( )

∫ ( ) [

( ) ( )

( ) ( )

( ) ]

∫ ( ) [

( )

( ) ( )

]

∫

0 ( ) 1

( ) ( )

( )

Where is the circle | | lying entirely within C. Hence

| ( ) ( )

∫

( )

( )

|

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| |

∫

| | | |

( ) ( )

| ( )|| |

| |

| |

(

⁄ )

by means of arguments parallel to those used in the proof to Cauchy’s formula for ( )

| | .

| |/

Where M is the upper bound of ( ) in D.

Hence when | | the right hand side of (1) will tends to Zero. We then have

( ) ( )

∫

( )

( )

or

( )

∫

( )

( )

Similarly

( )( )

∫

( )

( )

( ) ( )

[∫

( )

( ) ∫

( )

( )

]

∫

( )

6

( )( )

7 ( )

As we have

( )( ) ( )( )

( )

∫

( )

( )

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or

( )( ) ( )

∫

( )

( )

Replacing by

( )

∫

( )

( )

1.6 Cauchy’s Inequality: Statement : let ( ) be analytic within and on a circle C

defined by | | If | ( )| on C, then | ( )( )|

Proof: By Cauchy’s formula for the nth derivative of an analytic function at a point we have

( )

∫

( )

( )

| ( )( )|

∫

| ( )|

| |

| |

∫

| |

| |

---------------(1)

Now we may write the equation of the circle | | as

So that

Thus

| |

Hence (1) will give

| ( )( )|

∫

Thus

| ( )( )|

Remark : If the function ( ) is analytic in every finite region of the Z plane then ( ) is called entire

function. 1.7 Liouvilles’s Theorem: Statement: A bounded entire function is constant.

Proof:- Let a be any arbitrary point of the Z plane. Now ( ) is by hypothesis analytic for | |

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however large R may be Moreover, | ( )| Satisfies

the inequality. | ( )|

On any circle | | where M denotes the upper

bound at | ( )| for z lying in all finite regions of the z plane.

Hence by Cauchy’s inequality we have

| ( )|

As we get ( )

But is an arbitrary point in the Z plane. Thus derivative of

( ) every where hence ( ) is constant. 1.8 Taylor’s Theorem:

Statement: Let ( ) be analytic at all points within a circle

C0 with center Z0 and radius R and Let Z be any point inside C. Then

( ) ( ) ∑( )

( )

Proof: Let Z be any point inside the circle C0 with centre Z0 and radius R. Let | | and Let C be the Circle with

centre Z0 and radius Such that then Z lies inside C. By Cauchy integral formula we have

( )

∫

( )

∫

( )

( ) ( )

∫

( )

( )[

]

∫

( )

( )6

(

)

(

)

7

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∫

( )

( )

∫ ( )

( )

( )

∫

( )

( )( )

Now we will apply Cauchy’s formula and we get

( ) ( ) ( ) ( )

Where

( )

∫

( )

( )( )

Now for the proof of the theorem it is sufficient to show that as

So

| | |( )

∫

( )

( )( )

|

| |

| |∫

| ( )|

| || | | |

∫

( ) | |

( )

(

)

as

( ) ( ) ∫( )

( )

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Remark: when Z0=0 the series is called Maclaurin series Example: Prove that

( ) ( )

( )

+ -----+

Solution : Let ( ) Then by Tuylor’s theorem

( ) ∫( ) ( )( )

( )

( )

( )

( ) ( ) ( )

----- (A)

Now replace Z by 1 in (a)

( )

( )

( ) ‘ ‘ ‘

( ) ( ) ( )

( ) ( ) ( ) ) ( )

( )

( ) ( )

( )

1.9 Zero’s of Analytic functions: A Zero of an analytic function ( )is any value of Z for which ( )vanishes.

1.9.1 Singularities: These points of function ( ) at

which the function ceases to be analytic is called. 1.9.2 Different types of singularities: (a) Isolated Singularity: A point a is said to be an isolated singularity of a function ( ) if ( ) is analytic at each point

in some neighborhood of a excepting at the point a itself.

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(b) Isolated essential Singularity: If there exists no finite value of in such that ( ) ( ) finite non zero

constant then is called an isolated essential singularity

Example: has isolated Singularity at

1.10 Meromorphic Function: To define meromorphic function firstly we will understand the concept of pole. If there exists a positive integer in such that

( ) ( )

Then is called the pole of order of ( ). Meromorphic function are those functions which has poles as its only singularities in the finite part of the plane.

1.11 Argument Principle : Statement: If ( ) is analytic within and on a closed

contour C, having N zeroes inside C but no zeroes on C, then

( )

where ( ) denotes the variation in ( ) as z

moves round C. Proof; Let ( ) is the analytic function with N zeroes inside

C and denotes numbers of poles inside

∫

( )

( )

Put ( ) so that | ( )| and ( ). Then

( ) ( )

( )

( )

∫

Now

∫

, -

Retunes to its original value if z moves once round C. Also,

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∫ , -

( ), as ( ) does not return to its original value us z moves round C and so ( ) is not necessary zero.

( )

As ( ) is analytic i.e. it has no poles inside C hence P=0

( )

1.12 Rouche’s Theorem: Statement: Let ( ) and ( ) be analytic inside and on a

simple closed curve C and Let | ( )| | ( )|

On C, then ( ) and ( ) ( ) have the same number of

zeroes inside C. Proof: As | ( )| | ( )| on C and as | ( )| cannot to negative and hence we get

( )

( ) ( )

( ) ( )

| ( )| | ( )|

But it is a contradiction to the fact that | ( )| | ( )| does

not have a zero on C. Let N and M be the number of zeroes of ( ) and ( ) ( ) respectively inside C.

Then by argument principle

( )

( ) ( )

( ) { ( )

( )}

Now to prove the theorem it is Sufficient to show that

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8 ( )

( )9

Since | ( )| | ( )| C, the transformation

+ ( )

( ) givens point in the plane interior to the circle

with center 1 and radius unity .Then We can write

0 ( )

( )1 must lie between –II/2 and II. This shows

that 0 ( )

( )1 must return to it ‘s original value as Z

describes C.

Since 0 ( )

( )1 cannot increases or decrease by a

multiple of 2 , we conclude that

6 ( )

( )7

Thus we have 2

=> N=M 1.13 Fundamental Theorem of algebra: Statement: Every polynomial of degree n has roots.

Proof: Consider the Polynomial in Z ( )

Where

Clearly has roots and they are all at origin if

Let ( )

( )

Now we define C as the circle with center at the origin and radius R>i then.

| ( )| | | | |

| |

on C and similarly | ( )| (| | | | | |)

| ( )| | ( )| on C if

| | (| | | | | |)

⟩(| | | | | |)

| |

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Then by Rouche theorem we get that ( ) has same

number of Zeroes as ( )

( ) has roots 1.14 Maximum Modulus Principle:

Statement: let ( ) be analytic within and on a simple

closed contour C. Then | ( )| reaches it’s maximum values on C and not inside C, unless ( ) is a constant.

Proof: let us assume that ( ) is analytic within and on a

simple closed contour C ( ) is continuous within and on C

( ) has a maximum value M

Within and on C. Now for the proof of the theorem it is sufficient to show that | ( )| has a maximum value M on the boundary of C not

inside C. On the contrary let ( ) has a maximum value at Z=a

inside C. | ( )| | ( )| ----------------(1)

| ( )| --------------(2)

Now we describe a circle within C with Center at A. ( ) is Z=b inside S’-1

| ( )|

| ( )| ⟩

⟩

|| ( )| | ( )|| | | ( )

as

| ( )| | ( )| || ( )| | ( )||

( ) becoems.

|| ( )| | ( )||

|| ( )| | ( )||

=M-

=M- | ( )| M- inside a circle with center b and radius

Now draw another circle with center at a such that it

passes through b. The are of the circle lies within s.t.

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| ( )| M- on ard QR

and on the remaining arc of we have

| ( )| M

Let is the radius of the circle then =| |

and by Cauchy integral formula

( )

∫

( )

On we have z-a=

( )

∫

( )

∫ ( )

∫ ( )

∫ ( )

| ( )|

∫ ( )

∫ ( )

∫(

⁄ )

∫

(

⁄ )

( )

But

| ( )|

this is a contradiction to our assumption maximum value lies inside C. 1.15 Schwarz lemma

Statement: If ( ) is analytic in a domain | | and

satisfies the conditions | | ( )

Then | ( )| | | and | ( )| Equality holds only if ( ) is a linear transformation

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( ) Where is a real constant.

Proof: Let us consider that ( ) is analytic in the unit disc

| | and hence by taylor’s expansion of ( ) about the origin gives.

( )

Again as ( )

( ) +

We define

( ) ( )

( )

--------(1)

This shows that ( ) defined in (1) has a Singularity at Z=0

which may be removed by defining ( ) ( )

Let z=a be an arbitrary point of the unit disc. Let us choose such that

| |

As | ( )| we get on the circle | | the inequality

given by

| ( )| | ( )|

| |

( )

But by the maximum modulus principle the inequality (2) also holds in the disc | | and hence

| ( )| | ( )|

| |

Letting we get

| ( )| | ( )

|

| ( )| | |

or | ( )| | ( )|

as a is arbitrary we get

| ( )| | | | |

Then by maximum modulus principle we conclude that | ( )| will hold only when

| ( )| ( )

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1.16 Summary: In this block we have understood the concept of complex integration and we have proved a | | of theorems based on it. Some important results are as follows; 1. Let ( ) be a analytic function on and single valued inside

and on a simple closed contour C. then

∫ ( )

2. If C is closed contour then

∫

∫

3. If ( ) be analytic within and on a closed contour C and

let Z0 be any point within C then

( )

∫

( )

4. Every polynomial of degree has roots.

1.17 Self Assessment Test: (a) If ( ) be analytic within and on the boundary C of a

simply connected region and let Z0 be any point within c. Then prove that

( )

∫

( )

( )

(b) Expand and Sin Z in faylor’s series about z=0

(c) Let ( ) be analytic in the ring shaped region between two concentric circles C and C’ with center Z0 and radii R and R’ and Let z be any point of D. Then

( ) ∑ ( )

∑ ( )

Where

∫

( )

( )

∫ (

) ( )

(Laurent’s Theorem)

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(d) Prove that if ( ) is a continuous in a simply connected

domain and Let for every closed contour C in the domain D.

∫ ( )

(Morero’s Theorem) (e) Let a function be analytic at a point z=z0 where

( ) and Let ( ) Then there exists a

of the point in the plane in which the function

( ) has a Unique inverse ( ) and

( )

( )

(Inverse Function theorem)

1.17 Further Readings:

(i) Complex Analysis by Dr. H.K. pathak.

(ii) Function of complex variable by Dr. J.C. Chatuvedi

and Prof. S.J. Seth.

(iii) Complex Analysis by L.V.Anlfor.

(iv) Real and Complex Analysis by Walter Rudin.

(v) Foundations of complex analysis by S. Ponnusumy.

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Unit-2 Calculus of Residue

Objectives: After reading this unit you should able to; Understand the concept of residue . Solve typical problem of integration in a simpler

way. STRUCTURE :

2.1 Concept of Residue 2.1.1 Residue at pole. 2.1.2 Residue at infinity.

2.2 Cauchy Residue theorem. 2.3 Calculations of residue at special cases.

2.3.1 Expression for the residue of ( ) at simple pole z=a 2.3.2 Expression for the residue of ( ) at pole

of order 2.4 Problems based on Cauchy Residue theorem. 2.5 Jordan lemma 2.6 Problems based on Jordan lemma 2.7 Problems having branches of many function. 2.8 Summary 2.9 Self Assessment Test 2.10 Further Readings

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1.1 Concept of Residue: Let us consider a single valued analytic function ( ) then in the neighborhood of an isolated singularity ( ) can

be expanded with the help of Laurent’s series given by

( ) ∑ ( )

∑ ( )

Then the coefficient of B, is called the residue at z=a and

∫ ( )

Where is a circle with center a 2.1.1 Residue of pole: The above stated definition is also known as residue at pole. 2.1.2 Residue at infinity: let ( ) has an isolated

singularity at infinity and C is a closed curve such that ( ) is analytic except at singularity, then clearly C will be very

large and residue will become

∫ ( )

where

integration is taken around C in anti-clockwise direction. 1.2 Cauchy Residue Theorem:

It is the most important theorem which is frequently used in solving the problem of Integration by using the concept of residue. Statement: Let ( ) be single valued and analytic within and on a closed contour C except at a finite number of poles and Let be respectively

the residue of ( ) at these poles then

∫ ( ) (

)

Proof: Let be the circles with centers at

respectively and radii so small that they lie entirely within the closed curve C and do not overlap. Then ( ) is analytic within the region enclosed by the curve C

and these circles. Hence by Cauchy’s theorem for multi-connected region we have

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∫ ( )

∫ ( )

∫ ( )

∫ ( )

--------------------------(1)

But

∫ ( )

∫ ( )

Similarly

∫ ( )

‘ ‘ ‘ ‘ ‘

∫ ( )

Therefore (1) becomes

∫ ( )

( )

2.3 Calculation of Residue: In general we require residue two different situations. 2.3.1 Residue at Simple pole :

( ) ( )

Example :- Find Residue of

at

Solution- we have

( )

( )( )

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Clearly is a simple pole therefore residue at is

( ) ( )

( )

( )( )

( )

2.3.2 Residue at pole of order : If ( ) is an analytic function having as a pole of order

then

( ) ( )

( )

then Residue is given by ( )

( )

2.4 There are many types of problems which can be solved using the concept of Cauchy residue theorem. They are as follows.

2.4.1 Integration Round the Unit Circle. Example: Show that

∫

∫

∫

| |

∫

. /

. /

∫

( )

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∫

( ) . /

∫ ( )

Clearly ( ) has there pole,

out of

which only

lie within C

Therefore Sum of residues at o and

will be

[

( )

( ) . /

( )

. /

( )]

, -

Hence

∫ ( )

∫

Try these questions

1) ∫

) ∫

2.4.2 Integration of the type ∫ ( )

Example: Apply the calculus of residues to prove

∫

Solution : Let us consider

∫

∫ ( )

Where C is the contour consisting of a large semi circle CR by radius R. Then by Cauchy residue theorem

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∫ ( ) ∫

∫

( )

| |

( )

| |

(

)

Hence we get

∫ ( ) ∫

( )

Clearly are the simple poles of ( ) out of which

lies inside C. Therefore residue will be

( )

∫ ( ) (

)

∫

∫

Try these questions

) ∫

( )

) ∫

( )

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2.5 Jordan lemma; This lemma is generally used in solving the integrals of the form

∫ ( )

( )

∫ ( )

( )

Where (i) ( ) and ( ) are polynomials (ii) degree of ( ) exceeds that of ( ) (iii) the equation ( ) has no real

roots. Statement: If (i) ( ) | | , uniformly for

and (ii) ( ) is meromorphic in the upper half

plane, then

∫ ( )

Where CR denotes Semi-circle | | ( )

2.5.1 Problems based on Jordan’s lemma. Example: Apply the calculus of residue to prove that

∫

Solution:

∫

∫ ( )

Where C is a semicircle of very large radius R. Above real . Then by Cauchy residue theorem.

∫ ( ) ∫ ( ) ∫ ( ) ( )

As R by Jordan’s lemma second integral will vanish and

we get

∫ ( ) ( )

----------------(i) as

( )

So, residue at

( ) ( )

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( )

Hence (1) becomes

∫ ( ) (

)

Example: Prove that ∫

( )

using as a contour a

large semicircle in the upper half plane indented at the origin. Solution:

∫

( )

∫ ( )

Where C is the closed contour consisting of the upper half of the large semi-circle by | | and real from –R to R

indented at by a small circle of radius . Now poles of ( ) arc given by ( ) will not lie in C. also is the double pole

Therefore Residue will be ( )

( )

( )

( ) ( )

( )

( )

( )

( )

( )

(

)

.

/

Now by Cauchy residue theorem

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∫ ( ) ∫ ( )

∫ ( )

∫ ( ) ∫ ( ) ( )

4

.

/5

(

)

Also

| | ( )

| |

4

( )

5

Therefore

∫ ( )

Also

( )

( )

( ) ( )

( )

( )

( )

( ) (

)

33 | P a g e

Also

∫ ( )

Taking ( )

∫ ( ) ∫ ( )

(

)

∫

( ) ∫

( )

( )

(

)

∫

( ) ∫

( )

(

)

Equating real parts

∫

(

)

Equating real parts

∫

2.6 Branches of Many- valued functions. If this section we will consider the integrals involving many-valued function. Such as where a is not an integer. In such cases we consider

only those contours whose interiors do not contain any branch points and particulars branches should also be specified. In orders to avoid branch point at origin, we can take double circles contour indented at the center.

Example:1: Prove that

∫

(

Solution: Consider the integral

∫

∫ ( )

34 | P a g e

Contour consisting of a large Semi-circle defined by | |

in the upper half plane and the real from –R to R indented at the origin by a small semi-circle of radius P which avoids the branch point o of the only simple pole of

( ) within C is at Residue at

( ) ( )

.

/

Now we apply Cauchy residue theorem,

∫ ( ) ( )

.

/

OR

∫ ( ) ∫ ( ) ∫ ( )

∫ ( )

.

/

------------------(1) Clearly

|∫ ( )

| ∫ |

|

| |

∫

As R

35 | P a g e

|∫ ( )

|

Also

( )

So

∫

( )

Similarly when ( ) Becomes

∫ ( ) ∫ ( )( ) .

/

Equating real parts we get

∫

( )

∫

2.7 Try these question:

( ) ∫ ( )

( ) ∫

2.8 Summary:

In this section we have studied about Poles. They arc of two types. We have studied about the application of Cauchy

36 | P a g e

residue theorem in solving the typical problems of complex integration.

2.9 Self assessment test: 1. Prove

∫

2. Prove that

∫

( )

√

3.

∫

4. If a>b>0, that prove that

∫

( )( )

4

5

2.10 Further Reading: 1. Complex Analysis by H.K. Pathak.

2. Complex Analysis from Shrivatava.

3. Functions of a complex variable by B.S. Tyagi.

37 | P a g e

FINANCIAL INFORMATION ANALYSIS FOR INTERNAL DECISION MAKERS

ACCOUNTS FOR DECISION MAKING:

3.0 OBJECTIVES

After studying this unit you will be able to:

Understand the concept of marginal costing

Develop an insight about different alternatives available for decision making

Choose the best alternative among the alternatives available

Give an overview on activity based costing

Explain the concept of break even analysis

Describe the concept of cost-volume profit for profit planning

Explain the concept of margin of safety and profit volume ratio in decision making

Understand and analyze the cause of variance between planned and actual results.

3.1 INTRODUCTION

The element of costs can be divided into fixed and variable costs. There are certain costs which

are a combination of fixed and variable costs. These costs are called semi variable costs. It is

necessary to segregate the mixed costs into fixed and variable costs for managerial decisions.

The analysis of cost behaviour is necessary for planning, control and decision making. It means

analysis of variability of each cost element in relation to the level of output. The analysis of

costs plays a vital role in selecting the alternatives available before the management. Costs

could shape alternative opportunities and therefore, it influences and shapes future profits.

Management is not only interested in the historical cost analysis but it is also interested to study

those costs, which are influencing the future operations. After the standard costs have been set,

the next step is to ascertain the actual cost of each element and compare them with the standard

already set. The difference of actual from standard is variance. In this unit you will learn about

the concept of break-even point, cost- volume- profit analysis, margin of safety, profit volume

ratio and their role in decision making. You will also learn about the importance of variance for

effective cost control and decision making.

3.2 MARGINAL COSTING AS A TOOL FOR DECISION MAKING

An important role of cost accounting is to assist in the process of managerial decisions. In this

context profitability of two or more alternative option is compared and such option is selected

which offers maximum profitability along with fulfillment of objectives of the enterprises.

Though there are number of accounting techniques for such comparison and decision making,

marginal costing has their significant role. Marginal costing plays a vital role in decision making

38 | P a g e

.It is a very useful technique in solving various managerial problems and contribution in various

areas of decisions. Few important ones are explained here:

Make or Buy Decision

Lease or Buy

Change in Product Mix

Pricing Decision

Shut –down Decisions.

Exploring New Market

3.2.1 Make or Buy Decision: In this, management has to decide whether a certain product or a

component should be made in the factory itself or bought from outside suppliers. It follows two

types of decision:

Stopping the production of the part and buying it from the market

Stopping the purchase of the component and to produce it in own factory.

Stopping the production of the part and buying it from the market :

A business concern is already making a part or a component which is used in the business. Now

due to some reason, a decision has to be taken whether to bring the part from the market or

should be made in the factory. In case of a decision like the stopping the production of the part

or component and buying it from the market, a comparison of marginal cost of such production

with that of buying price should be made. But in this fixed cost remain fixed.

Marginal Cost < Buying Price ------ Making of a part inside the business is a relevant decision.

Marginal Cost> Buying Cost------Purchasing of a part from the market is a relevant decision.

However opportunity cost may also be taken into consideration while taking such decisions.

Example:

Suppose a component is being manufactured with the help of a machine and 10,000 units at a

cost of Rs 10 per unit.( Rs. 9 variable cost ,Rs1 fixed cost).are produced. The same

component can be bought from the market @ Rs.9.50 per unit.

What will be beneficial to go either for producing the component or buying a component?

What if the machine is released from production and can be hired at an annual rent of Rs

6,000?

If the item to be made: Rs.

Marginal Cost of 10,000 units @ Rs.9 90,000

Buying price of 10,000 units @ Rs. 9.50 95,000

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Savings if „Made‟ 5,000

If the item is bought and machine is let out on hire:

Buying Price 10,000 units @ Rs. 9.50 95,000

Less: Hire of Machine 6,000

89,000

Marginal Cost of Rs. 10,000 @ Rs. 9 90,000

Saving if „bought‟ and „machine is hired‟ 1,000

If we take decision without considering opportunity cost then we can say that it is beneficial

for the company to produce the component. But with the consideration of opportunity cost

decision will totally change then at this it is beneficial for the company to buy the component

from the market.

Stopping the purchase of a component and to produce it in own factory:

Another point to be considered while making make or buy decision may be that a component

thus far being purchased from the market should be purchased or made in the factory or not.

In this case generally some extra arrangements regarding space, labour , machines, raw

material ,rent etc will be required which involves huge investment. Some extra space, staff

requirement is also required which leads to additional cost. Prices paid to the outsiders for

purchasing the component should be compared with the additional cost which will be

incurred in the form of raw material, rent, salaries etc.

Rule Decision

Additional Cost < Buying Price ----- Manufacture the component

Additional Cost> Buying Price------Purchase the component

No compromise in respect of quality .Reliability of regular supply, reputation, financial

position of supplier should be given due consideration. There are large fluctuation in

demands it is better to purchase from outside .But if there is possibility of increasing demand

in future then own production may be preferred because it will lead to low cost.

Example

ABC Ltd. Purchases 20,000 Pens per annum from an outside supplier at Rs. 5 each. The

management feels that these be manufactured and not purchased. The following relevant

information are available:

Material Cost per unit Rs.2

Labour Cost per unit Re.1

Fixed Cost per Pen Re.1

Variable Overheads per Pen 100% of Labour Cost

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You are required to advise whether:

The company should continue to purchase the pens from the outside supplier or should make

them in the factory, and whether the company should accept an order to supply 10,000 pens

to the market at the selling price of Rs. 5 per unit.

Solution:

Calculation of Manufacturing Cost

Particulars Per Pen 20,000 Pen

Rs. Rs.

Variable Cost:

Material 2 40,000

Labour 1 20,000

Variable Overheads 1 20,000

4 80,000

Fixed Cost 1 20,000

5 1,00,000

Decisions: the cost of manufacturing as well as of purchasing per pen is equal. If there is

assurance of regular supply of quality product from the suppliers, the company can continue to

purchase the pens. If there are some problems in this context, the company can decide to make

them in the factory. (II) Yes, the company should accept the order because the marginal cost is

Rs. 4 only, where the order will be at Rs. 5 per pen.

3.2.2 Lease or Buy

The fast changing technology and innovations require replacement of old machineries so as to

modernize the plant and make it up-to-date for the production which is in consonance with the

customers‟ tastes and requirements. Moreover, cost reduction is always the demand of the

industry, particularly in the wake of all time rising prices going to the extent of run-way inflation.

Leasing as an alternative to purchasing the plant, machinery and other equipments, besides land

and building is considered by managements these days. The factors which influence the decision

as to own or lease are: Cost of alternatives, Long-term stability of earnings, Financing, and

Return on Investment

Flexibility

Example:

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In view of increasing cost of operating own fleet of cars, your company is presently considering

two proposals, viz: To hire cars with drivers from an agency @ Rs. 800 per car per month. The

company will bear the cost of petrol, oil and tyres.The executives will be given Rs. 25,000

interest –free loan repayable (in 5 years) to buy their own car. The company will, however,

provide them with free petrol and Rs. 500 per month for maintenance and drivers wages. If the

present cost of a car is Rs. 50,000 and monthly average running is 2,000 kilometers, find out the

most economic way with the help of the following data:

Paisa per km.

Petrol 65

Oil 8

Tyre 7

Repairs 10

Tax and Insurance Rs.560 per year

Drivers‟ wages and bonus Rs. 720 per month

Life of a car 5 years

Resale value at the end of the fifth year @18% per annum Rs. 10,000

Solution:

Comparative Cost of operation

Cost per car per month

Company's own car Hired car Executive car

Rs. Rs. Rs.

Petrol 1,300 1,300 1,300

Oil 160 160 ……

Tyre 140 140 ….

Repairs 200

Tax and insurance 46.67

Wages and Bonus 720

Depreciation 666.67

Hire Charges 800

Maintenance Allowance 500

Interest on Loan 375

Total Cost 3,233.34 2,400 2,175

Cost per km. 1.62 1.2 1.09

* The interest of Rs. 375 relates only to the first year of operation. When annual

instalments of loan

are paid back by the executives,the interest in the subsequent year of operations will

gradually

decrease and thus the cost per km. of operations of Executive cars will be

reduced.

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The cost of opeartion of the Executive cars,being the lowest ,this alternative is

recommended.

3.2.3 Change in Product Mix:

Introducing a New Product Line or Department: It involves two decisions:

Whether a new product or line should be added to the existing production or not.

If it should be introduced, then what should be the model or design or shape of new

product?

For this, the marginal cost of new product in all its possible model should be considered. And

also some additional investment of plant and machinery will likely increase the fixed overheads,

which should also be considered along with marginal costs.

Selecting Optimum Product Mix: When a company is engaged in a number of lines then the

problem arise of selecting the most optimum product mix which would maximize the earnings.

This problem becomes complicated, when one of the factors happens to be the limiting or key

factors. Under such a situation, profitability will be improved only by economizing the scarce

resources .Thus, while deciding a profitable mix of products, Contribution per unit of Key Factor

should be considered. he product giving highest contribution per unit of key factor should be

considered as most desirable product and in this way all products may be assigned ranks ion

order of priority. Selection of products in this way will offer an optimum product-mix at which

the profit will be maximum.

Principles for taking a decision in respect of product-mix are:-

(a) Calculate contribution per unit of key factor

(b) Assign ranks on the basis of highest contribution per unit of key factor.

(c) Available key factor should be utilized in the manufacture of that product which has been

assigned first rank then in the production of product having second rank and so on.

Example:

The following data are available in respect of Product Soap by ABC Co. Ltd:

Rs.

Sales 50,000

Direct Materials 20,000

Direct Labour 10,000

Variable Overheads 5,000

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Fixed Overheads 10,000

The Company now proposes to introduce new product Shampoo so that sales may be increased

by Rs.10, 000.There will be no rise in fixed costs and the estimated variable costs of product

Shampoo are: Rs.

Materials 4,800

Labour 2,200

Overheads 1,400

Advise whether product Shampoo will be profitable or not.

Solution:

Statement of Profit if Shampoo is introduced

Particulars Product Soap

Product Shampoo Total

Rs. Rs. Rs.

Direct Materials 20,000 4,800 24,800

Direct Labour 10,000 2,200 12,200 Variable Overheads 5,000 1,400 6,400

Marginal Costs 35,000 8,400 43,400

Sales 50,000 10,000 60,000

Contribution 15,000 1,600 16,600

Less: Total Fixed Costs 10,000

Net Profit 6,600

Advise: By introducing Shampoo profit has been enhanced by Rs.6,600 .

So, it is profitable to introduce this product.

Illustration:

A company produces two products A and B. The following facts are given regarding them:

A B

Profit contribution per unit Rs 2 Rs.3

Required production hours per unit 1 2

Sales Potential in units 1,500 1,800

Available Production Hours 2,000

Determine the optimum product mix.

Solution:

Particulars A B

Contribution per unit Rs.2 Rs.3

Production hours per unit 1 2

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Contribution per hour Rs.2 Rs.1.50

Because of the limited production hours, product A will be preferred over product B as product A

has greater contribution per hour. The sales potential of A is 1,500 units which will require 1,500

hours. Total available hours are 2,000. Hence, the balance hours can be used in the production of

product B. Product of B will be 250 units @ 2 hours per unit.

Thus, optimum Product Mix=A:B, 1,500units: 250 units.

3.2.4 Pricing Decisions:

In long Run prices should be such as to cover total cost which includes marginal cost and fixed

cost as well as desired profit. And secondly, in competitive markets prices are determined by

market forces. Thus in both the cases Marginal Costing will not play any significant role.

Marginal Cost is helpful in price determination only in short run and monopoly conditions.

Various Aspects to be considered under Pricing Decision are as follows:

Normal Price

Minimum Price

Special Price including dumping

Price changes

Normal Price: The normal price should be such as to maximize the contribution, so that

maximum profit is assured.

Sale Price = Marginal Cost + Contribution (Fixed Cost + Profit)

Minimum Price

Sale Price = Variable Price + Fixed Cost

This price is used:

When there is tough competition.

Price –cut is on war.

When new product is to be introduced in the market for the first time.

In these situations normal price may not be useful .In these situation manufacture faces a problem

of what should be the minimum price. It is obvious that if a manufacturer is not in a position to

earn profit, he will also not incur any loss. As , such ,he would fix a price ,which must cover total

cost at least. Such price is known as Minimum Price.

Special Price:

Sometimes the big concern may face a problem of accepting or rejecting a special offer at a price

which is below at existing price. Similarly, sometimes there may be a possibility of capturing a

45 | P a g e

new market, attracting new customers, patronizing special customers, if goods are sold at a price

below cost. Normally, such offer should be rejected, because total cost will not be recovered.

For this rule should be followed:

Price> Marginal Cost------Accept the proposal

Price< Marginal Cost -----Reject the proposal.

Price Change:

This is needed in order to captalise the market situation. It involves hike in price as well as

reduction in price. In both cases, quantity solids affected depending on the degree of elasticity of

production and in turn affect profit position. A careful analysis is required through the technique

of Marginal Cost. If change is necessary then that course is adopted which leads to maximum

profit or present level profit.

If price –reduction is forced by government ,in such cases adverse affect have to be neutralize by

making suitable adjustments like quantity sold, fixed cost, variable cost, production method and

techniques etc.

Example:

The annual operating capacity of a manufacturing company is 10 lakh direct labour hours and it

is currently operating at 75% of its capacity. The company has recently been approached by a

distribution to buy 1,00,000 units of its product at a special price of Rs. 10.50 per unit .The

standard Cost Sheet is as under: Rs.

Material 6 kg @ Re. 0.40 per kg 2.40

Direct Labour 2 hours @ Rs. 2.50 p.h. 5.00

Overheads:

Variable 2 hours @ Re. 0.75 1.50

Fixed 2 hours @ Rs. 1.20 2.40

11.30

The management of the company is confronted with the following questions:

Would it be profitable to accept the offer?

What would be the minimum price for the offer, if company‟s target is Rs.2,00,000 as

profit on this offer?

Solution:

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Total production capacity is (10,00,000/2) 5,00,000 units and present production

is 3,75,000 units( 5,00,000*75/100)

As present selling price is unknown, decision will be taken on the basis of

contribution of special order.

Statement of contribution on Special Order

Rs.

Sales: 1,00,000units @ rs. 10.50 10,50,000

Less: Marginal Costs:

Materia;ls @ Rs.2.40 p.u. 2,40,000

Direct Labour @ Rs.5.00 p.u. 5,00,000

Variable Overhead @ Rs.1.50 p.u. 1.,50,000

8,90,000

Contribution 1,60,000

Hence, the acceptance of order is profitable.

If company‟s target is Rs. 2,00,000 as profit, the minimum price for the offer will be:

Marginal costs 8,90,000

Desired Profit 2,00,000

Sales desired 10,90,000

Minimum Selling Price =10,90,000/1,00,000 = Rs. 10.90

3.2.5 Shut –Down Decisions:

Shut down decisions may be of two types: Permanent Closure of entire business, and Temporary

Closure of entire business

Permanent Closing Down or Continue: If the project undertaken does not yield the minimum

rate of return expected by the investor it will have to be given up. The decision to continue or

otherwise shut –down will depend upon the cost – benefit analysis of the two alternatives.

Costs items to be considered are:

Setting up Cost;

Effect of Fixed overhead costs;

Packing and storing of plant and equipment costs;

Loss of goodwill and or market;

Retrenchment or lay off compensation to workers;

Items of benefits to be quantified are:

Avoiding operating losses;

47 | P a g e

Saving in Fixed costs;

Saving in repairs and maintenance costs;

Savings in indirect labour costs;

Savings in heat and light costs;

Saving in other indirect costs.

Temporary Closure of the business: When trading activity particularly plant operation is

suspended for a short period; it is known as temporary closure. Such closure is necessitated either

due to depression/recession or due to ensuring off-season. In the former case, the period of

closure will run over the period of recession/depression, while in the latter case, it will cover the

period of off-season. For the period of recession/depression, the trading activity should not be

suspended for a short period till there is contribution on that level of trading activity. But there is

need for re-examination and deep analysis of fixed costs. There may be few items of fixed costs

which can be eliminated or saved by suspending the trading activities. Such fixed costs are

known as „escapable‟ or „avoidable‟ fixed costs. But there are certain items of fixed costs, which

cannot be avoided; there will be incurred after the suspension of activities. These are known as

„inescapable „or „unavoidable‟ fixed costs. Again, additional expenses would have to be incurred

in setting up the plant again after shut-down period. Such fixed costs are known as‟ Special

costs‟. It is significant point to note that inescapable fixed costs are all the time „loss factor‟.

Thus, escapable fixed costs minus special costs, known as net escapable fixed costs is a relevant

factor and amount of contribution should be compared with net escapable fixed costs. This is

done by preparing a statement of marginal costs and contribution for zero level of production and

for possible demanded quantity of production. As regards the level of production to close down

the plant or point of shut –down in terms of quantity, the following formulas may be used:

Shut –down point = Net escapable fixed costs

Contribution per unit

It is also significant to note that a decision regarding temporary closure should not only be based

on cost data; some other economic and social factors may also be considered.

Example:

Mr. Singhania has a sum of Rs.3,00,000 which is invested in a business. He wishes 15% return

on his fund. It is revealed from the present cost data analysis that variable costs of operation are

60% of sales and fixed costs are Rs. 1,50,000 p.a. On the basis of this information, you are

required to find out:

a) Sales Volume to earn 15% return.

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b) Shut- down point of the business, if he would spend RS. 50,000 even if business has to be

closed.

Solution:

(a) Return= 3,00,000 *15/100 = Rs. 45,000

Fixed Cost = Rs. 1,50,000

Total Contribution = Rs. 1,95,000

Contribution percentage or P/V ratio = 100% -60%= 40 %

Required Sales Volume = 1,95,000 *100

40

= Rs. 4,87,500

(b) Net Escapable Fixed Cost = 1,50,000-50,000= Rs. 1,00,000

Shut –Down Point= Net Escapable Fixed Cost

P/V Ratio= = 1,00,000 * 100

40

= Rs. 2,50,000

3.2.6 EXPLORING NEW MARKETS

Whether Indian or foreign, but if new markets is to be explored, the decision rests on a

consideration of the incremental gain resulting there from. However, certain other factors are

also to be considered:

Availability of surplus capacity, If the surplus capacity is not available, the

question of creating additional capacity will have to be separately studied, since it

will involve additional investment.

Maintenance of present sales at current prices, The already existing market

should not be affected at all by the tapping of the fresh market. If in the new

market, the selling price is lower than that prevailing in existing market, it should

be fully ensured that the new reduced selling price outside does not influence the

current higher prices in the present market.

Example:

A manufacturer has planned his level of production at 50% of his total plant capacity of

30,000 units. At 50% of the capacity his expenses are as follows:

a. Direct labour 11,160

b. Direct material 8,280

c. Variable and other manufacturing expenses 3,960

d. Total fixed expenses regardless of production 6,000

The home selling price Rs. 2.00 per unit. Now, the manufacturers receive a trade enquiry

from overseas for 6,000 units at a price of Rs. 1.45 per unit. If you were the manufacturer,

49 | P a g e

would you accept or reject the offer? Support your statement with suitable cost and profit

details.

Solution:

To decide whether it would or would not benefit financially by accepting the offer, the

incremental cost is to be compared with the incremental revenue. The reason is, the fixed has

“sunk” and there is idle capacity to the extent of 50 per cent capacity. As such, only

incremental cost has to be considered and if the incremental revenue exceeds the incremental

cost, the offer is worth accepting. The selling price in overseas market is not going to affect

home selling price.

For calculating incremental cost, we are concerned with variable cost per unit. Assuming

linearity, the variable cost as per question is:

Variable Cost per unit = 23,400

15,000

The above calculation is based on the fact that, at 50% capacity utilization, production is

15,000 units and as per data given in the question, total variable cost would be equal to Rs.

23,400.

Incremental cost per unit = Rs. 1.56

Incremental revenue per unit = Rs. 1.45

Incremental loss per unit = Re. 0.11

Therefore, the offer should not be accepted

3.3 STANDARD COSTING & VARIANCE ANALYSIS

Standard costing is a very important technique of cost control. Every organization

wants to minimize the cost of production and maximize the profits. Standard costing

is such a system which seeks to control the cost of each unit and then it is compared

with actual cost. The difference between the actual and pre-determined costs is

known as variance which need to be analyzed through carefully planned accounting

procedures and then the results are promptly reported to managers.

The term „Standard‟ refers to „a specific measurement‟ or „pre-determined scale or

measurement‟ Webster‟s Dictionary has given several meanings of the word

„Standard‟ but we can prefer to understand it as “ something established as a rule or

basis of comparison in measuring or judging quantity, quality, value, etc.” This

something is not vague but “established as a rule or basis of comparison.” Thus, in

the context of management accounting „standard‟ may be defined as measurable

quantity of material, labor and other elements of cost required in the production of

pre-determined quality, level or technical characteristics.

According to I.C.M.A., LONDON,” Standard cost is defined as a pre-determined

cost, which is calculated from management‟s standards of efficient operation and the

relevant necessary expenditure. It may be used as a basis for price fixing and for cost

control through variance analysis.”

Characteristics or Salient Features of Standard Costing

The salient features of characteristics of standard costing may be summarized as

follows:-

Determination of Standards:- First of all, standards and standard cost of

various elements of cost are ascertained separately.

Computation of Actual:-After the completion of production actual cost

incurred in computed.

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Comparison of Standard and Actual Costs:-The most important feature of

standard costing in the comparison of actual cost incurred with standard cost

specified.

Computation of Variances:-The most important aspect of comparison and

analysis of actual cost and standard cost is to find out variances. These variances are

calculated separately for element of cost.

Ascertainment of Reasons of Variances:-If there is variance between

standard cost and actual cost, then causes of these variances are identified. These

causes may be controllable as well as non-controllable.

Study of Options:-After identifying various causes of variances. Different

options to overcome these variances are studied and points of incidence of these

variances are determined.

Presentation of Report:- After analysis of variances and various options to

deal with a report is prepared for management so that necessary corrective actions

may be taken and variances may be minimized.

Application of Standard Costing

The application of standard costing required the following conditions to be fulfilled:-

A sufficient volume of standard products or components should be produced.

Methods, procedures and materials should be capable of being standardized.

A sufficient number of costs should be capable of being controlled.

OBJECTIVES OF STANDARD COSTING

The following are the main objectives of standard costing:-

Increase in Efficiency and Productivity:-The first objective of standard

costing is to improve the quality and minimize the cost so as to face competition

effectively. In fact, standard costing is a tool of management control with the help of

which efficiency and productivity can be improved and these objectives can be

achieved.

Cost Control:-The purpose of determining standard cost and then to compare

it with actual cost is to make effective.

Determination of Responsibility:- One important objective of standard

costing is to identify the persons or centers responsibility for variances, so that they

may be controlled properly.

Supplement to Budgetary Control:- Standard costing is also adopted to

make budgetary control a success. In fact, management control becomes more

effective if budgetary control and standard costing are introduced simultaneously.

Information to the Management:-To provide important information to

management is also an objectives of standard costing. This costing provides all such

information due to which the production work could not be completed as per pre-

determined plan and standards so that necessary corrective action may be taken at

appropriate time.

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Progressiveness of Management:- One objective of standard costing is to

develop the feeling of looking forward among management personnel and to make

management dynamic and progressive continuously.

ADVANTAGES OF STANDARD COSTING

The various advantages of standard costing are as follows:-

Simplification of Cost Bookkeeping: - It is very simple in comparison to

historical costing. Once the standards are fixed for the product, the records can be

simplified through uniformity which saves the time and money.

Basis for Measuring Operating Performances:-Standards work as

yardsticks for measuring the operating efficiency or inefficiency. For it, the

comparisons are made between standard cost and actual cost.

Cost Reduction and Control:-Standard costing is very useful in cost

reduction and control by eliminating or limiting lost time, spoilage, material, wastage

and lost machine hours.

Helpful In Budgeting:-Standards costing is linked with budgetary control. It

helps in making budgets and planning though budgeting.

Management by Exception:-Standard costing is helpful in applying the

principle of management by exception. Variance analysis brings the inefficient

operations in light and management can focus its attention towards those matters

only.

Prompt Reporting:-In standard costing, reporting is very prompt. Reports

get ready as soon as the operations come to a halt. Moreover, the reports prepared in

it are found to be very simple and improved in nature.

Formulation of Production and Price Policies:-Standard costs represent

long-term estimates, cost and prices. It helps the management in the formulation of

ideal production policy.

Implementing Incentive Schemes:-Standard costing promotes the

implementation of incentive schemes in the organization because every incentive

scheme is based on certain standards which are determined under this system.

Facilities Comparison:-Cost comparison between different products and

departments can be done under standard costing. It also makes possible the

comparison of costs of one period with another.

Promotes Cost Consciousness and Efficiency:- It also promotes cost

consciousness as the employees know that their performances shall be in assessing

manufacturing inefficiencies and fixing responsibilities. This improves the

efficiency of the organization.

LIMITATIONS OF STANDARD COSTING

Through standard costing is an important tool of cost control, it has certain

practical limitations also. These limitations are as follows:-

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Not Appropriate for Small Concerns:-It is not appropriate for small

concerns as the installation of standard costing requires high degree of skill and the

small concerns may not have expert staff for handling or operating the system.

Division of Variances is Difficult:-The exact division of variances into

controllable and uncontrollable variances is a difficult task.

Not suitable for Certain Industries:-This system is not suitable for industries

which produce non-standardized products and for job works which change according

to customer requirements.

Problem of Updating:-There is also a problem of keeping standards up –to-

date. The circumstances keep changing but the standards cannot be changed

frequently.

Unsuitable for Concerns Dealing in Non-Standardized Products:-Standard

costing is not much useful in those concerns where non-standardized products are

produced or production is undertaken according to customer‟s specification. In such

a case it becomes difficult to set up standard for each job separately.

Difficulties in Setting Up Standards:-The process of setting up standards is a

complicated task as it requires technical skill and expertise competence. The time

and motion studies are required to be undertaken for this purpose, which require a lot

of time and money. Moreover, if wrong standards are determined, the whole purpose

of the system would fail.

Not Suitable for Small Firms:-The system of standard costing may not be

suitable for small concerns keeping in view the cost of high degree of skill required

in it.

Difficulty in Fixing Responsibility:-The responsibility for variances is fixed

in the process of standard costing but it is not an easy task. Under many

circumstances variances may arise due to various reasons when it becomes difficult

to fix the specific responsibility. For example, in the case of unfavorable labor

efficiency variance, it is not necessary that workers are inefficient. This situation

may also arise due to inferior quality of raw material or defect in machine.

Changing Business Conditions:-In standard costing standards are fixed under

specified conditions. In case these condition changes, then either standard loose their

suitability or they have to be modified. Thus, standard costing is not suitable in those

concerns where frequent technological changes take place or prices of material

fluctuate frequently.

Need of Budgetary Control:-This system can be effective only when

budgetary control is also adopted in the concern.

Feeing of Dissatisfaction among Employees:-If standards related to labor are

fixed at a high level, it may develop a feeling of dissatisfaction and psychological

frustration among employees.

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VARIANCE ANALYSIS

Variance analysis is the distinctive feature of standard costing. A variance is the

deviation of the „actual‟ from the standard. In standard costing, it means the

differences between the standard cost and the actual cost. Variances depict the

extent to which standard have been achieved or not have been achieved. Variances

are computed to know the deviations and fix the responsibility. It is a very important

managerial tool to interpreting operating results and spotting situations requiring

corrective measures.

MANAGERIAL USES OR VARIANCE ANALYSIS

Measurement of Operational Efficiency:-Efficiency of business activities

can easily be measured on the basis of variances. If variances are unfavorable, the

activity is considered efficient. If it is otherwise the activity is considered inefficient.

Technique of Cost Control:-Variances analysis is a useful managerial

technique from the view of cost control because necessary corrective measures may

be taken after analyzing the cause of unfavorable variances.

Knowledge of Variance Centre:-Variances analysis helps in identifying the

level, activity or departments where variances are occurring so that special attention

may be focused to these activities or departments.

Determination of Responsibility:-On the basis of variance analysis the

persons or departments responsible for variances can be identified which helps in

taking corrections measures.

Measurement of Accuracy of Standards:-If there are no variances or

variance are very limited. It is an indication of accuracy of standards. On the

contrary, permanent existence of variances indicated the inaccuracy of standards.

Relative Measurement of Performances:-The performance of difference of

different department of an enterprise can easily be evaluated in relative terms on the

basis of variance analysis.

Basic of Future Action and Planning:-Analysis of causes of variances,

identification of responsible persons and suggestions of corrective measures provide

a useful base for future action and planning by management.

Cost Consciousness:-Variance analysis and its reporting develops the feeling

of cost consciousness among employees and executives, which proves an effective

motivational tool from the view of management.

CAUSES OF VARIANCES

(a) MATERIAL COST VARIANCE:-Material Cost Variance may arise due to

variance in price or quantity or in both:-

Material Price Variance: - Material price variance may arise due to following

causes:-

Changes in market price of materials

Failure to secure expected discount on purchases

Failure to purchase materials at proper time

Changes in prices of materials due to changes in tax by government

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Purchasing non-standard lots and the consequent reduction in quantity

discount

Increase in transport cost due to small lot or additional transport cost for quick

delivery.

Excessive shrinkage or loss in transit.

Failure to purchase the standard quality of materials.

A material price variance is normally the responsibility of the purchase manager.

However, price variances due to frequent market fluctuations or changes in tax fall in

the category of uncontrollable variances.

Material Usage Variance:-The variance arising due to difference in quantity of

material is called as Material Usage Variance or Material Quantity Variance. This

variance may arise due to the following causes:-

Negligence in the handling and use of materials.

Increase in wastage either due to untrained workers or defective method of

production

Increase in the usage rate of material due to difference in quality

Changes in technique or method of production

More or less yield from materials

Change in material mix, etc.

(b) LABOR COST VARIANCE: - The variances in labor cost may arise due to the

following causes:-

Labor Rate Variance:-

Changes in basic wage rates, it may be either due to change in demand and

supply of worker or due to new wage settlement

Different rates being paid to workers employed to meet seasonal demands or

to get urgent work done

Employment of workers of different grades and wage rates

New workers being paid lower rates from the standard rates

More payment of overtime wages etc.

It may be mentioned that major portion of labor rate variance is beyond the direct

control of management.

Labor Efficiency or Labor Time Variance:-If the actual hours worked are less or

more than standard hours, the difference is known as labor efficiency variances. This

variance may be unfavorable on account of following causes:-

Use of sub-standard materials requiring extra labor

Defective working conditions

Lack of proper supervision of work

Use of defective machinery and equipment

Inadequate training of employees

Discontentment among workers due to unsatisfactory working conditions, etc.

Idle Time Variance:-It will always be unfavorable and the main causes of this

variances are:-

Non-availability of material at proper time

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Gap in work due to defect in machinery

Stoppage of work due to power failure

Closure of work due to strike or lock out

Labor Mix Variance:-If different types or grades of employees are engaged in a

work and practically the actual composition differs from the standard composition,

the variance due to such differences is called as labor mix variance.

3.4 ACTIVITY BASED COSTING

This technique has been of recent origin and primarily concerned with absorption of

overheads (Indirect Costs) in an organization having products that differ in volume

and complexity of production. The crux of activity based costing is in accurately

assigning the overhead cost to the end product. The traditional costing system does

not serve effective purposes of product costing and pricing decision. Activity based

costing is a method of cost attribution to cost units on the basis of benefits received

from indirect activities. Their performance of particular activities and demands made

by these activities on the quantity of resources of organization are linked together so

that the cost of product is arrived at as per the quantum of actual activities performed

to produce a product or service. The reason for such a basis is that products

themselves do not consume resources directly rather several activities are required to

be performed for them, and these activities consume the resources of organization as

driven by cost drivers. Cost centre pay for these resources, depending upon the

number of activities required for a product.

Activity based costing may be defined as a technique which involves identification of

costs with each cost driving activity and making it as the basis for absorption of costs

over different products or jobs.

The Chartered Institute of Management Accountants (CIMA), London, defines

it as “a technique of cost attribution to cost units on the basis of benefits received

from indirect activities, e.g., ordering, setting up, assuring quality.”

3.4.1 Characteristics of ABC

The characteristics of activity based costing can be summarized as follows:-

It increases the number of cost pools used to accumulate overhead costs. The

number of pools depends upon the cost driving activities. Thus, in stead of

accumulated overheads costs in a single company-wise pool or departmental

pools, the costs are accumulated by activities.

It charges overhead costs to different jobs or products in proportion to the costs

driving activities in place of a blanket rate based on direct labor cost or direct

hours or machine hours.

It improves the traceability of the overhead costs which results in more accurate

unit cost data for management.

Identification of cost during activities and their causes not only help in

computation of more accurate cost of a product or a job but also eliminate non-

value added activities. The elimination of non-value added activities would drive

down the cost of the product. This, in fact, is the essence of activity based

costing.

3.4.2 Elements Involved in ABC

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Activity:-An activity can consist of one or more of the tasks associated with one

another to attain an objectives. For example, customer order processing includes

receiving orders from customers, as to its capacity of producing and interaction

with customers regarding delivery time etc.

Activity cost centre (Pool):-The result of identifying the overhead cost to an

activity is called an activity cost pool.

Cost drivers:-These are used to assign costs to products by using an appropriate

measure of resources consumed by each activity.

Process:- When related activities are grouped together, it is known as a process.

Cost Objectives:- are links to the whole of the enterprise.

Non-value Adding Activities:- Certain activities do not contribute anything to the

value of a product, but which are required to be carried out in the organization

because of reasons beyond the control of management.

3.4.3 Steps Involved in ABC

Evaluation of prevalent costing system:- The already existing costing

system should be able to adopt the activity based costing system so that

product costs can be accurately determined and correct pricing decisions

can be taken in a competitive business environment.

Identification activities:- A physical plan of the work place and listing of

pay-rolls can be examined, to begin with, supplemented by holding

interviews with staff. Such an activity analysis can throw light on how the

work spaces have been utilized and how the staff members have spent their

time. After chalking out the different tasks in detail, the prime activity can

be identified. For it, a cost-benefit analysis is required to be performed.

An activity can be a very small job or a combination of several small jobs.

Selection of cost basis:- The immediate past actual cost or average cost of

a specified period may be used under the system.

Determined cost pools:- Cost centre on cost pools should have a

similarity under financial accounting and cost accounting systems in order

to have a comparative utility.

Assigning costs pools:- Cost of resources consumed are to be associated

or allocated (apportioned) to each activity, in order to find out the amount

spent by the enterprise on each of its activity. Methods of direct

attribution , apportionment on a reasonable basis in case of joint costs and

the methods of estimation of certain costs may all be appropriately utilized

assignment of costs to activity cost centre.

Determined activity hierarchies:-The activities may be classified

according to:-

Unit-level activities

Batch-level activities

Product-level activities

Facility-level activities

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Identification outputs:- Outputs are required to be categorically identified

because without it, The basic purpose of applying the system cannot be

meaningful.

Selecting suitable cost drivers:- Interviews should be conducted with

concerned employees to determine the cost drivers for each activity. The

cost drivers should be easily measureable. Past data should be made

available in order to ascertain potential cost drivers. The management can

finally choose a single cost driver or multiple cost drivers for the activities.

Computing Cost Driver Rates:- The cost driver rates are required to be

calculated on a suitable basis.

Identifying cost to products:- The cost driver rates are applied to products. The

aggregate cost can be computed by multiplying the rate of consumption of

resources (cost driver rate) with the number of activities.

Test Run:- The activity cost an all the relevant data used for consumption

purposes should be tested to evaluate and judge the reliability of activity based

costing system adopted.

3.4.4 Activity based Costing versus Traditional Costing

Following are the main differences between activity based costing system and

traditional costing system:-

Effectiveness of purposes:-The objectivities of product costing and pricing

decisions are more effectively served by adopting ABC systems.

Basis of Identification:- Under ABC system, overhead costs are identified to

each major activity in stead of the department as under traditional costing

system. It results in greater number of cost centre under ABC system.

Terms used:-The term „ Cost Drivers‟ is not used under traditional costing

system. Popular terms are basis of allocation or apportionment. Under ABC

system, cost drivers are fewer in number for the purpose of charging

overheads to products.

Methodology:-ABC system uses separate rates for support centre and there is

no reallocation to production centre, as is the case under traditional costing

system.

Thus, ABC is refinement over traditional costing system. This is primarily

because cause and effect relationship is considered under ABC system to

identify support cost objects, which is not the case under traditional costing

system.

Requirement of ABC System: ABC system requires the following:-

Setting up of an information system which could help trace all the costs to cost

objects.

Support from top to bottom because the system involves people at all levels.

Integration of system into financial system. For it, computerization may be

required.

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3.4.5 Reasons for Adoption of ABC System

Manufacturing as well as non-manufacturing industries adopt Activity Based Costing

system because of the following reasons:

More accurate cost estimation:-ABC helps in estimating costs of individual product

or service more accurately, which, in turn, assists in formulating appropriate

marketing and corporate strategy.

More accurate accounting:-ABC improves accuracy of accounting for support

service costs, which occupy an important place in total cost of delivering value of

customers.

More accurate profitable measurement:- Profitability of the product and even

customer-wise profitability can be more accurately measured by tracing the

consumption of resources to each individual product and customer. It is essential as

demand on resources by products and customers differ among products and customers

respectively.

Reduction in cost of data processing:-ABC helps in reducing the cost of data

processing, with consequently brings down the cost of tracking consumption of

resources by variety of activities.

Reduction in cost of wrong decision-making:-ABC induces reduction in costs

associated with poor decisions made to a substantial extent because of rational

apportionment of costs.

The final words of comment over ABC system are that the adoption, implementation

and operation of the system is not an end in itself. The benefits can be derived by

translating the system design and its operation into action-oriented managerial

performance. Ultimately, it amounts to effective cost management for the success of

the system.

3.4.6 Problems with the ABC Approach

Cost of change will be high as everything will have to be worked out from scratch.

It would be difficult to correlate the marginal increase in cost with a particular cost

driver.

Over a period of time, the ABC will tend to standardize the cost of activities related to

a particular product or process. But in practice there will be differences in set-up

time, production run, and meeting a delivery order for the product or process, as well

as for different product.

The ABC system will require a change due to changes associated with new products

and new technology. This will put strain on the costing system and resources due to

certain degree of inbuilt standardization. There exists catch 22 situation in the

implementation of the ABC. Measure of activity performance will change again and

again. A trade-off will be required between the accuracy and time spent on replacing

the existing system with the ABC.

The ABC is at the stage of evolution. Literature on the ABC concept at present is

primarily restricted to the manufacturing environment.

The Activity Based Costing (ABC) has been successfully adopted by many Japanese

Corporations. As a matter of fact, the elimination of “ non-value added activities”

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was the secret of the Japanese capturing a significant market share with limited

products and being considered serious threat to the entire US automobile and

electronics. As a result, now many US Corporations are also increasingly adopting

Activity Based Costing

3.5 COST–VOLUME-PROFIT ANALYSIS

Cost –Volume- Profit Analysis is a logical extension of the concept of marginal

costing, in which cost of production is divided into two parts, i.e., fixed cost and

variable cost. These two cost leads to decrease in cost per unit and increase in profit

per unit with the increase in volume of production. In short, there is negative relation

between cost of production and amount of profit; volume of production and cost of

production and positive relation between volume of production and amount of profit.

The concept of cost –volume-profit analysis is used in the narrower as well as in the

broader sense. In narrower concept, it is concerned with finding out „break –even

point‟, i.e., the point at which there is no profit or no loss. In its broader sense, it is a

technique of management accounting which determines profit, cost and sales value at

different levels of production. It also establishes relationship among these three

factors.

IMPORTANCE OF COST–VOLUME-PROFIT ANALYSIS

It is an important tool in the process of managerial decision and it is extensively

helpful to management in a variety of problems involving planning and control. The

main objectives of such analysis are as follows:

Setting up Flexible Budget: This analysis is helpful in setting up flexible budget

which indicates that what trend of amount of sales and cost of production at different

level of activity will be.

Determination of B.E.P.: The most important objective of cost –volume-profit

analysis is to find out break –even point.

Profit Planning: This analysis is useful in profit planning also because whereas, on

the one hand, we can determine the amount of profits at different levels of activity we

can also determine the volume of sales or production to earn desired profit on the

other hand.

Decision relating to Selection of Alternatives: This analysis helps the management in

taking decision in respect of various alternative proposals, like:

1. Which of the product is more profitable?

2. Whether the firm should accept the proposal of supply of additional products

at a particular price?

3. What is the optimum mix of sales or production, so that profit may become

maximum?

Performance Evaluation for Control: This analysis assists in evaluation of

performance for the purpose of control. On the basis of profit achieved and costs

incurred it can be analyzed that what the role of volume of production and others

factors was in effecting the amount of profit.

Helpful in Price Fixation: This analysis is also helpful in price fixation by studying

the effect of different price structures on cost and profit.

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Allocation of Overhead Costs: This analysis assists in finding out the amount of

overhead costs to be charged to the products at various levels of production because

predetermined overheads rates are related to a selected volume of production.

Analysis of Effect of Changes in Cost: It assists in analyzing the impact of the

fluctuations in fixed cost, material cost, labour cost and overheads etc.

3.5.1 BREAK EVEN ANALYSIS

Break-Even Point (B.E.P.) is that point of sales where the total revenues are equal to

total costs. In other words, it is a point where there is no profit no loss. It is also

known as „No Profit Point „or Zero loss Point.‟ At this point, contribution is equal to

fixed costs. If the sales are more than this level, there will be profit to the firm and if

sales are less than this level, there will be a loss. It is also called as „Equilibrium

Point‟, „Balancing Point‟ or „Critical point‟.

In broader sense, the term break-even analysis refers to the study of relationship

between costs, volume and profit at different levels of sales or production.

DEFINITION:

“ The Break –Even Point is that point of sales volume where total revenues and total

expenses are equal, it is also said as the point of zero profit or zero loss.” --------

Charles T. Horngren

“ The Break-Even Point of a company or unit of a company is that level of sales

income which will equal the sum of its fixed cost and its variable costs.”------Keller

and Ferrara

ASSUMPTIONS OF BREAK EVEN ANALYSIS

1. All elements of costs are segregated into fixed and variable components.

2. There will be no change in general price level.

3. Production and sales are in synchronization

4. Fixed cost remains fixed at all volumes of output.

5. Selling price per unit remains constant.

6. Variable cost remains constant per unit of output.

ADVANTAGES /USES OF BREAK EVEN ANALYSIS

Break –Even Analysis helps to evaluate the profitability of new order.

1. It also helps in determining the sales to earn desired profit.

2. It also helps in estimating margin of safety.

3. Impact of increase or decrease in fixed or variable costs on profit can be

ascertained with break –even analysis.

4. It also helps in make or buy decision.

5. It also helps in determination of optimum sales-mix.

6. It is useful at the time of changing capacity

7. It helps in the calculation of comparative profitability of projects, product

lines, etc.

LIMITATION OF BREAK EVEN ANALYSIS

1. The assumption that variable cost remains constant is not valid assumption

since the prices of material, labour and other expenses do not remain constant.

2. The assumption that the fixed cost remains constant is not valid because the

fixed cost increases after a certain level of production.

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3. The assumption that the selling price remains constant is not valid because it

varies with change in demand and supply of all the products.

4. It may be difficult to divide all costs into fixed and variable costs.

5. The break even analysis assumes a static situation that cannot exist for long

periods of time . For example, it assumes no change in general price level,

selling price, production technology etc. but in practice there is constant

change in these factors as management wants to improve production system

and efficiency.

COMPUTATION OF BREAK EVEN POINT

There are three methods for the computation of B.E.P.

1. Equation Method 2.Contribution Method 3. Graphic Method

(a) EQUATION METHOD: A simple equation is used to calculate the break even

point which also expresses the relationship of the items of income statement.

Sales = Variable Expenses + Fixed Expenses + Profit

For example, selling price is Rs.10 p.u, variable cost is Rs. 6 p.u and fixed cost

Rs.10,000. Assuming x unit is to be sold to break –even, then by applying formula:

10x=6x+10,000

10x-6x=10,000

4x=10,000

X=10,000/4 = 2,500 units.

(b) CONTRIBUTION METHOD: This is the second method to calculate the break

even point. This method involves two basic tools, i.e. „Contribution‟ and „Profit –

Volume Ratio‟.

Contribution: It is the difference of sales and variable cost. It may also be defined as

the excess of sales over variable cost. It contributes towards fixed expenses and profit.

The amount of contribution may be computed as follows:

(A) Total Contribution:

Contribution = Sales-Variable Cost or S-V

Contribution = Fixed Cost + Profit (-Loss)

Contribution= Sales x P/V Ratio

(B) Per Unit Contribution :

Contribution per unit = Sales per unit- Variable Cost per unit

Profit Volume Ratio or P/V Ratio

It is the relationship between contribution and sales and is generally expressed in

terms of percentage. It is also known as „Margin Ratio‟ or „Contribution Ratio‟. It can

be expressed as under:

P/V Ratio = Contribution *100 or C*100

Sales S

= Sales- Variable Cost × 100

Sales

= Fixed Cost + Profit ×100

Sales

= Contribution per unit ×100 OR Spu –Vpu ×100

Sales per unit Spu

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= Change in Profit ×100

Change in Sales

Calculation of Break Even Point

B.E.P. ( Rs.) : It is also known as Sales B.E.P.

= Fixed Cost ×Sales

Contribution

= Fixed Cost × Selling Price p.u.

Contribution p.u

= Fixed Cost

P/V Ratio

= Sales- Margin of Safety

B.E.P.(in units): It is also known as „Break-even Point in Quantity‟ or „output B.E.P.‟

= Fixed Cost

Contribution Per unit

= B.E.P. (Rs.)

Spu

(c ) GRAPHIC METHOD OR BREAK EVEN CHART

Break-even chart is a graphic representation of marginal costing. It presents the

information relating to cost, sales, revenue and profit, etc. It also indicates the break –

even point and the estimated profit and loss. There are three methods of drawing a

break- even chart.

Methods of Drawing a Break- even Chart

First Method: The following steps are followed in it:

1. Volume of production or output or sales is shown on X-axis.

2. Costs and sales revenue are plotted on Y-axis.

3. Fixed cost line is drawn parallel to X-axis. This shows that fixed cost remain

constant at all levels of output.

4. The variable cost and sales value at different levels of output are plotted and

lines are drawn to join their respective points. The variable cost starts above

the fixed cost line.

5. The point where the two lines intersect each other, that point is known as

break –even point.

6. The number of the units that should be produced to reach break-even can be

known by drawing a perpendicular to the X-axis from the break-even point.

7. Similarly sales revenue can be known by drawing a perpendicular from this

point to Y-axis.

8. The area below the break-even point is known as loss area and above it is

known as profit area.

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Second Method: In this method, variable cost is plotted \first and then the fixed cost

line is plotted above the variable cost line. It is drawn parallel to variable cost line. It

represents the total cost of the product. In this method, contribution is represented by

the difference between total revenue line and variable cost line.

3.5.2 Margin of Safety

It is the difference between actual total sales and B.E.P. sales and may be calculated

in rupees, units or even in percentage form as explained below:

M.O.S. in Rupees:

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M.O.S.(Rs.)= Sales (Rs.) – B.E.P (Rs.)

M.O.S.( Rs.)= Profit

P/V Ratio

M.O.S. in Units:

M.O.S. (Units) = Sales (Units) - B.E.P (Units)

M.O.S. (Units) = Profit

Contribution per Unit

M.O.S in Percentage:

= M.O.S × 100

Total Actual Sales

Example: The following information is obtained from XY and Co. for 2013

Rs.

Sales 20,000

Variable Cost 10,000

Fixed Cost 6,000

1. Find the P/V ratio, break-even point and margin of safety at the level.

2. Calculate the effect of:

20% decrease in fixed costs;

10% increase in fixed costs;

10% decrease in variable costs;

10%increase in selling price;

10% increase in selling price together with an increase of fixed overheads by

Rs.1,200

10% decrease in sales price;

10% decrease in sales price accompanied by 10% decrease in variable costs.

Solution:

Profit Volume Ratio = Contribution ×100

Sales

= Sales – Variable Cost × 100

Sales

Break- Even Point (B.E.P) = Fixed Cost × 100

P/V Ratio

Margin of Safety = Profit

P/V Ratio

Or Actual Sales- Sales at B.E.P

(a) At Existing Level :

P/V Ratio = Rs. 20,000 – Rs. 10,000 × 100

Rs. 20,000

= Rs. 10,000 × 100

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Rs. 20,000

= 50%

B.E.P = Rs. 6,000

50%

= Rs. 12,000.

M.O.S = Rs. 20,000- Rs. 12,000

=Rs. 8,000

(b) (i) 20 % decrease in fixed cost

P/V Ratio = 50%

B.E.P. = Rs. 4,800

50%

= Rs. 9,600.

M.O.S = Rs. 5,200

50%

= Rs. 10,400

Or (Rs. 20,000- Rs. 9,600)

(ii) 10 % Increase in Fixed Costs

P/V Ratio = 50%

B.E.P = Rs. 6,600

50%

= Rs. 13,200

M.O.S = Rs. 3,400

50%

= Rs. 6,800

Or (Rs. 20,000 – Rs. 13,200)

(iii) 10% Decrease in Variable Costs

P/V Ratio = Rs 20,000 – Rs 9,000

Rs 20,000

= Rs 11,000 ×100

Rs 20,000

= 55%

B.E.P = Rs. 6,000

55%

= Rs. 10,909.09

M.O.S = Rs. 5,000

55%

= Rs. 9090.91

(iv) 10% Increase in Selling Price

P/V Ratio = Rs. 22,000- Rs. 10,000 ×100

Rs. 22,000

= Rs. 12,000 ×100

Rs. 22,000

= 54.55 % (Approx)

B.E.P = Rs. 6,000

54.55%

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= Rs. 11,000

M.O.S = Rs. 6,000

54.55%

= Rs.11,000

(v) 10% Increase in Selling Price with Increase in Fixed Overhead by Rs. 1,200

P/V Ratio = Rs. 22,000 – Rs 10,000

Rs 22,000

= 54.55%

B.E.P = Rs. 7,200

54.55%

= Rs. 13,200

M.O.S = Rs. 4,800

54.55%

= Rs. 8,800

(vi) 10% Decrease in Sales Price

P/V Ratio = Rs. 18,000 – Rs 10,000 × 100

Rs 18,000

= 44.44 %

B.E.P = Rs. 6,000

44.44 %

= Rs. 13,500

M.O.S = Rs 2,000

44.44%

= Rs. 4,500

(vii) 10% Decrease in Sales with 10% Decrease in Variable Costs

P/V Ratio = Rs. 18,000 – Rs. 9,000 × 100

Rs. 18,000

= 50%

B.E.P = Rs.6,000

50%

= Rs. 12,000

M.O.S = Rs. 3,000

50%

= Rs. 6,000

3.6 CHECK YOUR PROGRESS

1. The break-even point is the point at which:

a. There is no profit no loss;

b. Contribution margin is equal to total fixed cost;

c. Total revenue is equal to total cost;

d. All of the above

2. Cost- Volume profit analysis is based on several assumptions. Which of the

following is not one of these assumptions?

a. The sales-mix of the product is constant;

b. Inventory quantities change during the year;

c. The behavior of both revenues and cost is linear throughout the relevant range;

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d. Factor prices e.g. material prices and wage rate remain unchanged.

3. Standard Costing involves the :

a. Fixation of estimated cost

b. Determination of standard cost

c. Setting of budgeted cost

d. None of these

4. The difference between standard cost and actual cost is known as

a. Variance

b. Profit

c. Differential Cost

d. Marginal Cost

5. An activity-based costing system is one that:

a. traces costs to activities and then to products.

b. traces costs to resources and then to activities.

c. traces activities to costs and then to resources.

d. traces products to activities and then to resources.

6. In Make or Buy decision:

a. Only variable costs are relevant

b. Fixed Cost that can be avoided in future are relevant

c. Fixed Cost that will continue regardless of the decision are relevant

d. Both a and b

7. Relevant Costs are:

a. Standard Costs

b. Controllable Costs

c. Future Costs

d. Historical Costs

8. In deciding whether to manufacture a part or buy it from outside supplier, a cost

that is not relevant to short run decision is:

a. Direct Labour

b. Variable Overheads

c. Fixed overheads that will be avoided if the part is bought from an outside

supplier.

68 | P a g e

d. Fixed overhead that will continue even if the part is bought from an outside

supplier.

3.7 LET US SUM UP

Marginal costing plays a vital role in decision making .It is a very useful technique in

solving various managerial problems and contribution in various areas of decisions.

Few important ones are explained here: Make or Buy Decision, Lease or Buy,

Change in Product Mix, Pricing Decision, Shut –down Decisions and, Exploring

New Market. Standard costing is a very important technique of cost control. Every

organization wants to minimize the cost of production and maximize the profits.

Standard costing is such a system which seeks to control the cost of each unit and

then it is compared with actual cost. The difference between the actual and pre-

determined costs is known as variance which need to be analyzed through carefully

planned accounting procedures and then the results are promptly reported to

managers.

A variance is the deviation of the „actual‟ from the standard. In standard costing, it

means the differences between the standard cost and the actual cost. Variances depict

the extent to which standard have been achieved or not have been achieved.

Variances are computed to know the deviations and fix the responsibility. It is a very

important managerial tool to interpreting operating results and spotting situations

requiring corrective measures. The traditional costing system does not serve effective

purposes of product costing and pricing decision. Activity based costing is a method

of cost attribution to cost units on the basis of benefits received from indirect

activities. Activity based costing may be defined as a technique which involves

identification of costs with each cost driving activity and making it as the basis for

absorption of costs over different products or jobs. Cost –Volume- Profit Analysis is

a logical extension of the concept of marginal costing, in which cost of production is

divided into two parts, i.e., fixed cost and variable cost. These two cost leads to

decrease in cost per unit and increase in profit per unit with the increase in volume of

production. In short, there is negative relation between cost of production and

amount of profit; volume of production and cost of production and positive relation

between volume of production and amount of profit. The concept of cost –volume-

profit analysis is used in the narrower as well as in the broader sense. In narrower

concept, it is concerned with finding out „break –even point‟, i.e., the point at which

there is no profit or no loss. In its broader sense, it is a technique of management

accounting which determines profit, cost and sales value at different levels of

production. It also establishes relationship among these three factors. Break-Even

Point (B.E.P.) is that point of sales where the total revenues are equal to total costs. In

other words, it is a point where there is no profit no loss. It is also known as „No

Profit Point „or Zero loss Point.

3.8 ANSERS TO CHECK YOUR PROGRESS

1 ( a), 2 (b ), 3 (b ), 4 (a ), 5 (a ), 6 (d ), 7 (c ), 8 (c )

Suggested Reading/References

Management Accounting by Aggarwal, Aggarwal and Jain, Ramesh Book

Depot, Jaipur

69 | P a g e

Management Accounting by I M Pandey, Vikas publicationj, Delhi

Management Accounting and Financial Management by D K Goyal, Avichal

Publishing Company, Kala Amb, Himachel Pradesh.

Accounting for Managerial decision by BP Aggarwal and BK Mehta, Sahitya

Bhawan Publication, Agara

Accounting for Managers by Prof. Jawaharlal, Himalaya Publication, Delhi

Accounting for Managers by S. N. Maheshwari, Vikas Publication, Delhi

Financial Accounting by Dr. Mohinder Kumar Gupta, Mahaveer Prakashan,

Delhi

70 | P a g e

Block II - UNIT-IV

SPECTRAL THEORY OF BOUNDED SELF-ADJOINT LINEAR

OPERATORS Introduction

In mathematics, on a finite-dimensional inner product space, a self-adjoint

operator is an operator that is its own adjoint, or, equivalently, one

whose matrix is Hermitian, where a Hermitian matrix is one which is equal to its

own conjugate transpose. By the finite dimensional spectral theorem, such operators

can be associated with an orthonormal basis of the underlying space in which the

operator is represented as a diagonal matrix with entries in the real numbers. In this

Chaptere, we consider generalizations of this concept to operators on Hilbert

spaces of arbitrary dimension.

Self-adjoint operators are used in functional analysis and quantum mechanics. In

quantum mechanics their importance lies in the Dirac–von Neumann formulation of

quantum mechanics, in which physical observables such as

position, momentum, angular momentum and spin are represented by self-adjoint

operators on a Hilbert space. Of particular significance is the Hamiltonian

2

2H V2m

which as an observable corresponds to the total energy of a particle of mass m in a

real potential field V. Differential operators are an important class of unbounded

operators.

Objectives

The structure of self-adjoint operators on infinite-dimensional Hilbert spaces

essentially resembles the finite-dimensional case, that is to say, operators are self-

adjoint if and only if they are unitarily equivalent to real-valued multiplication

operators. With suitable modifications, this result can be extended to possibly

unbounded operators on infinite-dimensional spaces. Since an everywhere defined

self-adjoint operator is necessarily bounded, one needs be more attentive to the

domain issue in the unbounded case. This is explained below in more detail.

For simplicity we will always assume that the Hilbert spaces considered in this

chapter are separable and complex (although most results extend to nonseparable

complex Hilbert spaces).Let H1, H2 be separable Hilbert spaces and A be a linear

operator

A : D(A) ⊂ H1 → H2.

We denote by B(H1,H2) the set of all bounded linear operators from H1into H2 and

write B(H,H) = B(H) for simplicity.We recall that A = B if D(A) = D(B) = D and Ax

= Bx for all x D. Next, let H1 = H2 = H. In this chapter we introduce and explain

spectral properties of Bounded self adjoint Linear operator and and spectral family of

a Bounded self adjoint linear operator with properties. Extension of spectral theorem

to continuous function is main aim.

4.1 Spectral Properties of Bounded Self-Adjoint Linear Operators

Definition 4.1.1. (i) Let T be densely defined in H. Then T* is called the adjoint of T

if,

71 | P a g e

Dom(T*) = {g H | there exists an hg H such that (hg, f) = (g, Tf) for all f

Dom(T)},

T*g = hg.

(ii) An operator A in H is called symmetric if A is densely defined and A A*.

(iii) A densely defined operator B in H is called self-adjoint if B = B*.

(iv) A densely defined operator S in H is called normal if SS* = S*S.

We note that for every self-adjoint operator A in H one has D(A) = H.

For every bounded operator A we will assume D(A) = H unless explicitly stated

otherwise.

Definition 4.1.2. (i) z C lies in the resolvent set of A if (A − zI)−1

exists and is

bounded. The resolvent set of A is denoted by ρ(A).

(ii) If z ρ(A), then (A − zI)−1

is called the resolvent of A at the point z.

(iii) σ(A) = C \ ρ(A) is called the spectrum of A.

We will use the notation, R(z,A) = (A − zI)−1

, z ρ(A).

Remark

1 - (A) = σ(A).

2 - A = A* σ(A) R.

3 - If A is a bounded operator, then σ(A) is a bounded subset of C.

4 - If A is a bounded self-adjoint operator, then σ(A) R is compact.

5 - If A is a bounded self-adjoint operator, then ||A|| = supλσ(A) |λ|.

6 - If A is a self-adjoint operator, then R(z,A) is a normal operator for all z ρ(A).

4.2 Further Spectral Properties of Bounded Self-Adjoint Linear Operators

The spectrum (T) of a bounded self-adjoint linear operator T is real. This important

fact was proved in the last section. We shall now see that the spectrum of such an

operator can be characterized in more detail since it has a number of general

properties which are mathematically interesting and practically important. It is clear

that (T) must be compact, but in the present case we can say more:

Theorem 4.2-1: (Spectrum). The spectrum (T) of a bounded self-adjoint linear

operator T: H H on a complex Hilbert space H lies in the closed interval [m,

M] on the real axis, where

(1) ||x|| 1

m inf Tx, x

, ||x|| 1

M sup Tx, x

Proof: (T) lies on the real axis (by 9.1-3). We show that any real = M + c with c >

0 belongs to the resolvent set (T). For every x 0 and v = ||x||-1

x we have x = ||x||v

and

Tx, x = ||x||2 Tv, v ||x||

2 =

||v|| 1

sup Tv, v

x, xM.

Hence -Tx, x -x, xM, and by the Schwarz inequality we obtain

||Tx|| ||x|| -Tx, x = -(Tx, x) + (x,x) (-M + ) x, x

= c||x||2,

where c = - M > 0 by assumption. Division by ||x|| yields the inequality ||Tx||c||x||.

Hence (T) by 9.1-2. For a real < m the idea of proof is the same.

m and M in (1) are related to the norm of T in an interesting fashion:

Theorem 4.2-2: (Norm). For any bounded self-adjoint linear operator T on a

complex Hilbert space H we have [cf. (1)]

(2) ||T|| = max(|m|, |M|)= ||x|| 1

sup

|Tx, x|.

72 | P a g e

Proof: By the Schwarz inequality,

||x|| 1

sup

|Tx, x| ||x|| 1

sup

||Tx|| ||x|| = ||T||,

that is, K ||T||, where K denotes the expression on the left. We show that ||T||K. If

Tz = 0 for all z of norm 1, then T = 0 (why?) and we are done. Otherwise for any z of

norm 1 such that Tz 0, we set v = ||Tz||1/2

2 and w = ||Tz||-1/2

Tz. Then ||v||2 = ||w||

2 =

||Tz||. We now set y1 = v + w and y2 = v - w. Then by straightforward calculation,

since a number of terms drop out and T is self-adjoint,

Ty1, y, - Ty2, y2 = 2(Tu, w + Tw, v)

(3) =2(Tz,Tz + T2z, z)

= 4||Tz||2.

Now for every y 0 and x = ||y||-1

y we have y = ||y|| x and

|Ty, y| = ||y||2 |Tx, x| ||y||

2

||x|| 1

sup

|T x , x | = K||y||2,

so that by the triangle inequality and straightforward calculation we obtain

|Ty1, y1 - Ty2, y2 |Ty1, y1| + |Ty2, y2|

K(||y1||2 + ||y2||

2)

= 2K(||v||2 + ||w||

2)

= 4K||Tz||.

From this and (3) we see that 4||Tz||2 4K||Tz||. Hence ||Tz||K.

Taking the supremum over all z of norm 1, we obtain ||T||K. Together with K ||T||

this yields (2).

Actually, the bounds for (T) in Theorem 9.2-1 cannot be tightened. This may be

seen from

Theorem 4.2-3: (m and M as spectral values). Let H and T be as in Theorem 4.2-1

and H {0}. then m and M defined in (1) are spectral values of T.

Proof: We show that M (T). By the spectral mapping theorem 7.4-2 the spectrum

of T + kI (k a real constant) is obtained from that of T by a translation, and

M (T) M + k (T + kI).

Hence we may assume 0 m M without loss of generality. Then by the previous

theorem we have

||x|| 1

M sup

Tx, x = ||T||

By the definition of a suprimum there is a sequence (xn) such that

||xn|| = 1, Txn, xn = M - n, n 0, n 0.

Then ||Txn|| ||T|| ||xn|| = ||T|| = M, and since T is self-adjoint,

||Txn – Mxn||2 = Txn - Mxn, Txn - Mxn

= ||Txn||2

- 2M Txn, xn + M2 ||xn||

M2

- 2M(M - n) + M2 = 2Mn 0.

Hence there is no positive c such that

||TMxn|| = ||Txn - Mxn|| c = c ||xn|| (||xn||= 1).

Theorem 4.1-2 now shows that = M cannot belong to the resolvent set of T. Hence

M (T). For = m the proof is similar.

4.3 Positive Operators

4.3.1 Aspects of Positivity

In this subchapter we extend some of the ideas earliar discussion to a more

general context and describe some of the special spectral properties of positive

operators. These were first discovered for n × n matrices with non-negative entries by

73 | P a g e

Perron and Frobenius, but many aspects of the theory can be extended to much more

general level.

When we write B := Lp(X, dx) in this chapter, we usually refer to the space of real-

valued functions. We assume throughout that the measure space satisfies the

assumptions. Sometimes we will consider the corresponding complex space, and

when we need to distinguish between these we do so by adding subscripts, as in BR

and BC.

If X is a countable set and dx is the counting measure we write lp(X) in place of Lp(X,

dx). A number of the theorems have slightly less technical proofs in the discrete case,

because one does not have to worry about null sets and can use pointwise evaluation

of functions.

Later in the chapter we assume that X is a compact metric space, and consider certain

positive one-parameter semigroups acting on C(X).

If f B, the positive and negative parts of f are defined by

f+ := max{f, 0} = 1

2 (|f| + f),

f− := max{−f, 0} = 1

2 (|f| − f).

Note that |f| ≤ |g| implies ||f|| ≤ ||g||. The set B+ of all non-negative f B is a convex

cone, and is closed with respect to the norm and weak topologies of B. An operator A

: B → B is said to be positive, symbolically A ≥ 0, if Af ≥ 0 for all f ≥ 0. We say that

Tt is a positive one-parameter semigroup on B if Tt ≥ 0 for all t ≥ 0.

Lemma 4.3.1 If A is a positive operator acting on B = p

RL (X, dx), then A is bounded

and

||A|| = sup{||Af||/||f|| : f ≥ 0 and f 0}.

Proof: Suppose first that for all n Z+ there exists fn ≥ 0 such that ||fn|| = 1 and ||Afn||

≥ 4n. If we put

n

n

n 1

f : 2 f

then f ≥ 0, ||f|| ≤ 1 and 0 ≤ 2−n

fn ≤ f for all n. Hence 0 ≤ 2−n

Afn ≤ Af, and

2n ≤ 2

−n||Afn|| ≤ ||Af||

for all n. The contradiction implies that there exists c such that ||Af|| ≤ c whenever f ≥

0 and ||f|| = 1. If c is the smallest such constant then c ≤ ||A|| ≤ +∞.

Given f B, the inequality −|f| ≤ f ≤ |f| implies −A|f| ≤ Af ≤ A|f| and hence |Af| ≤ A|f|.

Therefore

||Af|| = || |Af| || ≤ ||A|f| || ≤ c|| |f| || = c||f||.

This implies that ||A|| ≤ c.

In order to study the spectrum of an operator AR acting on BR = p

RL (X, dx), one must

pass to the complexification BC = p

CL (X, dx). The complex-linear operator AC is

defined in the natural way by AC(f + ig) := ARf + iARg. The proof in Theorem 13.1.2

that ||AC|| = ||AR|| is only valid for positive operators. One may also adapt the proof of

Theorem 12.1.1 to Lp(X, dx); this does not require A to be positive, but Problem

13.1.3 shows that it does require p = q.

Theorem 4.3.2 Let 1 ≤ p, q ≤ ∞ and let AR : p

RL (X, dx) → q

RL (X, dx) be a positive

linear operator. Then

|AC(f + ig)| ≤ AR(|f + ig|)

for all f, g p

RL (X, dx). Hence ||AC|| = ||AR||.

74 | P a g e

Proof:. Given R we have

|(ARf) cos () + (ARg) sin()| = |AR(f cos() + g sin())|

≤ AR(|f cos() + g sin()|)

≤ AR(|f + ig|).

Let u, v, w : X → R be functions in the classes of ARf, ARg, AR(|f + ig|). Then we

have shown that

|u(x) cos() + v(x) sin()| ≤ w(x)

for all x not in some null set N(). If n n 1{ } is a countable dense subset of [−, ]

then

|u(x) + iv(x)| = 1 n

sup

|u(x) cos(n) + v(x) sin(n)| ≤ w(x)

for all x not in the null set nn 1

N( )

. This implies the first statement of the

theorem, from which the second follows immediately.

Problem - 1 The following shows that the positivity condition in Theorem 13.1.2 is

necessary if p q. Consider the matrix

1 1A :

1 1

as a bounded operator from l∞ to l

1. Show that ||AR|| = 2 but ||AC|| = 2

3/2.

Problem – 2 Let A be a positive linear operator on Lp(X, dx) where 1 ≤ p < ∞, and let

1/p + 1/q = 1. Use following to prove that

A , A | |,| |

for all complex-valued Lp((X, dx) and L

q((X, dx). Also give the much more

elementary proof available when A has a non-negative integral kernel.

Our next lemma might be regarded as an operator version of the Schwarz inequality.

An operator version of the H older inequality may be proved by the same method.

Lemma 4.3.3 Let 1 ≤ p, q ≤ ∞ and let AR : p

RL (X, dx) → q

RL (X, dx) be a positive

linear operator. Then

|AC(fg)(x)|2 ≤ {AR(|f|

2)(x)}{AR(|g|

2)(x)}

almost everywhere, for all f, g 2p

CL (X, dx).

Proof: If A has a non-negative integral kernel K then

2

2 1/ 2 1/ 2

C gX

| A (f )(x) | {K(x, y) f (y)}{K(x, y) g(y)}dy

2 2

X XK(x, y) | f (y) | dy K(x, y) | g(y) | dy

2 2

R R{A (| f | )(x)}{A (| g | )(x)}

This finishes the proof if X is finite or countable. We deal with the general case by

using an approximation procedure.

If : = {E1, ...,En} is a sequence of disjoint Borel sets with finite measures |Er|, we

define the orthogonal projection P by

r r

n1

r E E

r 1

P f : | E | X f ,X

We then note that the operator PA has the non-negative integral kernel

r r

n1

r E E

r 1

K(x, y) : | E | X (x){A*(X )(y)}

Hence

75 | P a g e

|P(p)(x)|2 ≤ {P(q)(x)}{P(r)(x)}

almost everywhere, where p := AC(fg), q := AR(|f|2) and r := AR(|g|

2). The proof is

completed by choosing a sequence of increasingly fine partitions (n) for which

P(n)(p), P(n)(q) and P(n)(r) converge to p, q and r respectively not only in norm but

also almost everywhere.

The following theorem has a wider scope than is apparent at first sight, because it is

not required that L2(X, dx).

Theorem 4.3.4 Let A be a positive linear operator on L2(X, dx) and let be a

measurable function on X. If (x) > 0 almost everywhere, 0 ≤ A ≤ and 0 ≤ A* ≤

μ then 1/ 2|| A || ( )

Proof:Assume first that L2(X, dx), so that (x)

2dx is a finite measure and L

∞(X)

L2(X,

2dx). We define the unitary operator U : L

2(X,

2dx) → L

2(X, dx) by Uf :=

f. We then observe that B : = U−1

AU is positive and satisfies 0 ≤ B*1 ≤ 1 and 0 ≤

B*1 ≤ μ1. If f L∞(X) then

|(Bf)(x)|2 ≤ B(|f|

2)(x)B(1)(x) ≤ B(|f|

2)(x)

almost everywhere by Lemma 13.1.5. Therefore

2 2 2 2 2

2 2|| Bf || B(| f | ),1 | f | ,B*(1) | f | ,1 || f ||

Since L∞(X) is dense in L

2(X,

2dx) we deduce that ||A|| = ||B|| ≤ (μ)

1/2.

If L2(X, dx) then the assumptions of the theorem have to be interpreted

appropriately. We assume that 0 ≤ A ≤ for all L2(X, dx) that satisfy 0 ≤ ≤

, and similarly for A*. We than define B as before, and observe that 0 ≤ Bf ≤ 1 and 0

≤ B*f ≤ 1 for all f L∞(X) ∩ L

2(X,

2dx) such that 0 ≤ f ≤ 1.

From this point on we work in the weighted L2 space. Let D denote the set of all

bounded functions on X whose supports have finite measure with respect to the

measure (x)2dx. If f D and supp(f) = E then

|(Bf)(x)|2 = |(B(fXE))(x)|

2 ≤ B(|f|

2)(x)B( 2

EX )(x) ≤ B(|f|2)(x)

almost everywhere, If the set F has finite measure then

2 2 2 2

F F| (Bf ) | dx B(| f | ) dx

= B(|f|2), XF

= |f|2, B*XF

≤ μ|f|2, 1

= μ||f||2

Since F is arbitrary subject to having finite measure we deduce that 2 2

2 2|| Bf || || f ||

for all f D, and since D is a dense subspace of L2 we obtain the same bound for all f

L2. Therefore ||A|| = ||B|| ≤ (μ)

1/2.

Corollary 4.3.5 Let A be a positivity preserving2 self-adjoint linear operator on 2

RL

(X, dx) and let L2(X, dx). If (x) > 0 almost everywhere and A = then

||A|| = .

Our next lemma states that the singularity closest to the origin of certain

operatorvalued analytic functions lies on the positive real axis.

Lemma 4.3.6 Suppose that An are positive operators on B := Lp(X, dx) and that for

all z such that |z| < R the series

76 | P a g e

n

n

n 0

A(z) : A z

….(1)

converges in norm to an operator A(z). Suppose also that A(z) may be analytically

continued to the region {z : |z − R| < S}. Then the series (1) is convergent for all z

such that |z| < R + S.

Proof: If 0 ≤ f B and 0 ≤ g B* then the function

F(z) := A(z)f, g

is analytic in

D := {z : |z| < R} {z : |z − R| < S}.

We have

(n) (n)

r RF (R) lim F (r)

m n

m

m n

m!A f ,g R

(m n)!

by a monotone convergence argument that uses the non-negativity of the coefficients.

Moreover

(n) n

n 0

0 F (R)x / n!

for all x such that 0 ≤ x < S by the analyticity of F in {z : |z − R| < S}. Therefore the

series

m m n n

m m

m 0 n 0 m n

m!A f ,g (R x) A f ,g R x

(m n)!n!

of non-negative terms is convergent for 0 ≤ x < S, and the series

m

m

m 0

A f ,g z

has radius of convergence at least (R + S). The same holds for all f B (resp. g B*)

since every element of B (resp. B*) is a linear combination of four elements of B+

(resp. B*+). The proof that (1) is norm convergent for all z such that |z| < R + S is

similar

Theorem 4.3.7 Let A be a positive operator on B and let

r := max{|z| : z Spec(A)}

be its spectral radius. Then r Spec(A).

Proof: If |z| > r then the series

1 n 1 n

n 0

(zI A) z A

is norm convergent. Since the analytic function z → (zI −A)−1

cannot be analytically

continued to any set {z : |z| > r − }, the function must have a singularity at z = r by

Lemma 4.3.6 Therefore r Spec(A).

All of the above ideas can be adapted to the context of one-parameter semigroups.

Semigroups of the following type occur in population growth models and the neutron

diffusion equation. These models are unstable if ||Tt|| increases indefinitely with t. In

such cases one either has a population explosion, or it is prevented by some non-linear

effect.

Lemma 4.3.8 Let the operator Z := −M +A act in B := Lp(X, dx), where M denotes

the operator of multiplication by a measurable function m that is bounded below and

77 | P a g e

A is a bounded, positive operator on B. Then Tt := eZt

is a positive one-parameter

semigroup.

Proof: Putting St := e−Mt

, we note that Z is a bounded perturbation of M, so Tt is a

positive operator for all t ≥ 0 because every term in the perturbation expansion is

positive.

Given f B+ one might interpret f(t, y) := (Ttf)(y) as the local density of some entities

at a site y, which can increase (if m(y) < 0) or decrease (if m(y) > 0) as time passes

without moving from y. Entities at the position y can cause new entities to appear at

the position x at the rate A(x, y) if A has an integral kernel A(x, y). The nth term on

the right-hand side of (11.10) describes the part of the state at time t for which exactly

(n − 1) such creations have taken place.

4.4 Square Roots of a Positive Operator

If T is self-adjoint, then T2 is positive since T

2x, x = Tx, Tx 0. We consider the

converse problem: given a positive operator T, find, a self-adjoint A such that A2 = T.

This suggests the following concept, which will be basic in connection with spectral

representations.

Definition 4.4.1 (Positive square root). Let T : H H be a positive bounded

self-adjoint linear operator on a complex Hilbert space H. Then a bounded self-adjoint

linear operator A is called a square root of T if

(1) A2

= T.

If, in addition, A O, then A is called a positive square root of T and is denoted by

T1/2

exists and is unique:

Theorem 4.4.2: (Positive square root). Every positive bounded self-adjoint linear

operator T: H H on a complex Hilbert space H has a positive square root A,

which is unique. This operator A commutes with every bounded linear operator on H

which commutes with T.

Proof: We proceed in three steps:

(a) We show that if the theorem holds under the additional assumption T I, it also

holds without that assumption.

(b) We obtain the existence of the operator A = T1/2

from Anx Ax, where A0 = 0

and

(2) An+1 = An+ 1

2(T – An

2), n = 0, 1,….,

and we also prove the commutativity stated in the theorem.

(c) We prove uniqueness of the positive square root. The details are as follows.

(a) If T = 0, we can take A = T1/2

= 0. Let T 0. By the Schwarz inequality,

Tx, x) ||Tx|| ||x|| ||T|| ||x||2

Dividing by ||T|| 0 and setting Q = (1/||T||)T, we obtain

Ox, x ||x||2 = Ix, x;

that is, O I. Assuming that Q has a unique positive square root B = Q1/2

, we have

B2= Q and we see that a square root of T = ||T||Q is ||T||

1/2 B because

(||T||1/2

B)2 = ||T||B

2 = ||T||O = T.

Also, it is not difficult to see that the uniqueness of Q1/2

implies uniqueness of the

positive square root of T.

Hence if we can prove the theorem under the additional assumption T I, we are

done.

78 | P a g e

(b) Existence. We consider (2). Since A0 = 0, we have A1 = 1

2T. A2 = T -

1

8T

2, etc.

Each An is a polynomial in T. Hence the An's are self-adjoint and all commute, and

they also commute with every operator that T commutes with. We now prove

(3) An I n = 0, 1,….;

(4) An An+1 n = 0, 1,….;

(5) Anx Ax, A = T1/2

;

(6) ST=TS AS = SA where S is a bounded linear operator on H.

Proof of (3):

We have A0 I. Let n > 0. Since I - An-1 is self-adjoint, (I - An-1)2s 0. Also T I

implies I - T 0. From this and (2) weobtain (3):

0 1

2(I – An-1)

2 +

1

2(I – T)

= I – An-1 - 1

2(T – A

2n-1)

= I - An.

Proof of (4):

We use induction. (2) gives 0 = A0 A1 = 1

2T. We show that An-1An for any fixed n

implies AnAn+1. From (2) we calculate directly

An+1 – An = An + 1

2(T – An

2) – An-1 -

1

2(T – A

2n-1)

= (An – An-1)[I - 1

2(An + An-1)]

Here An - An-1 0 by hypothesis and [….] 0 by (3). Hence An+1 - An 0 by 9.3-1.

Proof of (5):

(An) is monotone by (4) and An I by (3). Hence Theorem 9.3-3. implies the

existence of a bounded self-adjoint linear operator A such that Anx Ax for all x

H. Siace (Anx) converges, (2) gives

An+1x – Anx = 1

2(Tx – An

2x) 0

as n . Hence Tx - A2x = 0 for all x, that is, T = A

2. Also A 0 because 0 =

A0 An by (4), that is, Anx, x 0 for every x H, which implies Ax, x 0 for

every x H, by the continuity of the inner product (cf. 3.2-2).

Proof of (6):

From the line of text before (3) we know that ST = TS implies AnS = SAn, that is,

AnSx = SAnx for all x H. Letting n , we obtain (6).

(c) Uniqueness. Let both A and B be positive square roots of T. Then A2 = B

2 = T.

Also BT=BB2 = B

2B = TB, so that AB = BA by (6). Let x H be arbitrary and y = (A

- B)x. Then (Ay, y) 0 and (By, y) 0 because A 0 and B 0. Using AB = BA and

A2 = B

2, we obtain

Ay, y + By, y = (A+B)y, y = (A2-B

2)x, y = 0.

Hence Ay, y = By, y = 0. Since A 0 and A is self-adjoint, it has itself a positive

square root C, that is, C2 = A and C is self-adjoint. We thus obtain

0 = Ay, y = C2y, y = Cy, Cy = ||Cy||

2

79 | P a g e

and Cy = 0. Also Ay = C2y = C(Cy) = 0. Similarly, By = 0. Hence (A - B)y = 0. Using

y = (A - B)x, we thus have for all x H

||Ax - Bx||2 = (A - B)

2x, x = (A - B)y, x = 0.

This shows that Ax - Bx = 0 for all x H and proves that A = B.

Applications of square roots will be considered in Sec. 9.8. la-deed, square roots will

play a basic role in connection with the spectral representation of bounded self-adjoint

linear operators.

4.5 Projection Operators

The concept of a projection operator P or, briefly, projection P, was ^defined in Sec.

3.3, where a Hilbert space H was represented as the direct sum of a closed subspace Y

and its orthogonal complement Y; thus

H = Y Y

x = y + z (y Y, z Y).

Since the sum is direct, y is unique for any given x H. Hence (1) defines a linear

operator

P : H H

(2)

x y = Px.

P is called an orthogonal projection or projection on H. More specifically, P is called

the projection of H onto Y. Hence a linear operator P : H H is a projection on

H if there is a closed subspace Y of H such that Y is the range of P and Y is the null

space of P and P|Y is the identity operator on Y.

Note in passing that in (1) we can now write

x = y + z = Px + (1 - P)x.

This shows that the projection of H onto Y is I - P.

There is another characterization of a projection on H, which is sometimes used as a

definition:

The results in this section are extension to projection operator

In the following we will assume H+ to be a self-adjoint operator.

By δk l2(Z), k Z we will denote a vector, such that (δk)j = δk,j, j Z.

Theorem 4.5.1 There is a family of projection operators {E+(λ)}λ R corresponding

to the operator H+ and the following representations are valid,

R RI dE ( ) and H dE ( )

Theorem 4.5.2 The following formula is valid,

k, j ,K , j 0 0

R

P ( )P ( )d( ,E ( ) ) (4.1)

In particular, the polynomials P +,j(λ) are orthonormal with respect to the measure

d(δ0, E + (λ)δ0) on R.

Proof: First, note that because of

(H + δj, u) = (δj, H + u) = (H + u)j

= aj − 1uj−1 + ajuj+1 + bjuj

= (aj−1δj−1 + ajδj+1 + bjδj, u), u 2

0l (N0),

H+ acts on each δj as

H+δj = aj−1δj−1 + ajδj+1 + bjδj, j N0,

where we assume δ−1 = 0. Therefore, δj belongs to the domain of any Hn

+,

n N, and analogously to (3.1) we find that

80 | P a g e

δj = P + j(H+)δ0.

Now it is easy to establish (4.1) using

δk,j = (δk, δj)

= (P+,k (H+)δ0, P+,j(H+)δ0)

= (δ0, P+,j(H+)P+,k(H+)δ0)

, j ,k 0 0

R

P ( )P ( )d( ,E ( ) )

Definition 4.5.1 Let σ+(λ) = (δ0, E+(λ)δ0), then dσ+(λ) is called the spectral measure

associated with H+.

Remark 4.5.1 Following the usual conventions we also call dσ+(λ) the spectral

measure of H+ even though this terminology is usually reserved for the operator-

valued spectral measure dE+(λ). (This slight abuse of notation should hardly cause any

confusion.)

Lemma 4.5.3 The set of points of increase of the function σ+(λ) is infinite, that is, for

all polynomials P(λ) L2(R, dσ+(λ)),

2

R

| P( ) | d ( ) 0 if and only if P() = 0

4.6 Further Properties of Projections

Let us start, as indicated, by discussing basic properties of projections. In the first

place we show that projections are always positive operators:

Theorem 4.6-1: (Positivity, norm). For any projection P on a Hilbert space H,

(3) Px, x = ||Px||2

(4) ; P 0

(5) T ||P||1 ||P|| = 1 if P(H) {0}.

Proof. (3) and (4) follow from

Px, x = P2x, x = Px, Px = ||Px||

2 0.

By the Schwarz inequality,

||Px||2 = Px, x ||Px|| ||x||,

so that ||Px||/||x||1 for every x 0, and ||P|| 1. Also ||Px||/||x|| = 1 if x P(H) and x

0. This proves (5).

The product of projections need not be a projection, but we have the basic

Theorem 4.6-2: (Product of projections). In connection with products (composites)

of projections on a Hilbert space H, the following two statements hold.

(a) P = P1P2 is a projection on H if and only if the projections P1 and P2 commute, that

is, P1P2 = P2P1. Then P projects H onto Y = Y1 Y2, where Yj = Pj(H).

(b) Two closed subspaces Y and V of H are orthogonal if and only if the

corresponding projections satisfy PYPv = 0.

Proof: (a) Suppose that P1P2 = P2Pl. Then P is self-adjoint, by Theorem 3.10-4. P is

idempotent since

P2 = (P1P2)(P1P2) = P1

2P1

2 = P1P2 = P.

Hence P is a projection by 9.5-1, and for every x H we have

Px = P1(P2x) = P2(P1x).

Since P1 projects H onto Y1, we must have P1(P2x) Y1. Similarly, P2(P1x) Y2.

Together, Px Y1 Y2 Since x H was arbitrary, this shows that P projects H into

Y = Y1 Y2. Actually, P projects H onto Y. Indeed, if y Y, then y Y1, y Y2,

and

Py = P1P2y = P1y = y.

81 | P a g e

Conversely, if P = P1P2 is a projection denned on H, then P is self-adjoint by 9.5-1,

and P1P2 = P2P1 follows from Theorem 3.10-4.

(b) If Y V, then Y V = {0} and PYPvx = 0 for all x H by part (a), so that PYPv =

0.

Conversely, if PYPv = 0, then for every y Y and v V we obtain

y, v = PYy, Pvv = y, PYPvv = y, 0 = 0.

Hence Y V.

Spectral projections

Let BR denote the set of all Borel subsets of R.

Definition 4.6.1 The family R{P } B of bounded operators in H is called a

projection-valued measure (p.v.m.) of bounded support if the following conditions

(i)–(iv) hold:

(i) PΩ is an orthogonal projection for all Ω BR.

(ii) P = 0, there exist a, b R, a < b such that P(a,b) = I (the bounded support

property).

(iii) If Ω = kk 1

Ωi ∩ Ωj = for i j, then PΩ = s − limN→∞

k

N

k 1P .

(iv) 1 2 1 2

P P P

Next, let A = A* B(H), Ω BR.

Definition 4.6.2 PΩ(A) = χΩ(A) are called the spectral projections of A.

We note that the family {PΩ(A) = χΩ(A)}ΩRB satisfies conditions (i)–(iv) of

Definition 3.1.

Next, consider a p.v.m. {PΩ}ΩRB . Then for any h H, (h, PΩh) is a positive (scalar)

measure since properties (i)–(iv) imply all the necessary properties of a positive

measure. We will use the symbol d(h, Pλh) to denote the integration with respect to

this measure.

By construction, the support of every (h, PΩ(A)h) is a subset of σ(A).

Hence, if we integrate with respect to the measure (h, PΩh), we integrate over σ(A). If

we are dealing with an arbitrary p.v.m. we will denote the support of the

corresponding measure by supp(PΩ).

Theorem 4.6.3 If {PΩ}ΩRB is a p.v.m. and f is a bounded Borel function on

supp(PΩ), then there is a unique operator B, which we will denote by supp(PΩ) f(λ)

dPλ, such that

supp(P )(h,(Bh) f ( )d(h,P h),

h H

Proof: A standard Riesz argument.

Next, we will show that if PΩ(A) is a p.v.m. associated with A, then

(A)f (A) f ( )dP (A)

First, assume f(λ) = χΩ(λ). Then

(A) (A)X ( )d(h,P (A)h) d(h,P (A)h) (h,P (A)h)

= (h, X(A) h)

Hence, holds for all simple functions. Next, approximate any measurable function f(λ)

by a sequence of simple functions to obtain for bounded Borel functions on σ(A).

82 | P a g e

The inverse statement also holds: If we start from any bounded p.v.m. {PΩ}ΩRB and

form A = supp(P )

dP

, then χΩ(A) = PΩ(A) = PΩ. This follows from the fact that for

such an A, the mapping supp(P )

f f ( )dP

forms a functional calculus for A. By

uniqueness of the functional calculus one then gets

supp(P )P (A) X (A) X ( )dP P

Summarizing, one obtains the following result:

Theorem 4.6.4 There is a one-to-one correspondence between bounded self-adjoint

operators A and projection-valued measures {PΩ}ΩRB in H of bounded support given

by

A → {PΩ(A)}ΩRB = {χΩ(A)}Ω

RB ,

RBsupp(P )

{P } A dP

4.7 Spectral Family : We recall from projection operal or(4.5) that our present aim is a representation of

bounded self-adjoint linear operators on a Hilbert space in terms of very simple

operators (projections) whose properties we can readily investigate in order to obtain

information about those more complicated operators. Such a representation will be

called a spectral representation of the operator concerned. A bounded self-ad joint

linear operator T: H H being given, we shall obtain a spectral representation of

T by the use of a suitable family of projections which is called the spectral family

associated with T. In this section we motivate and define the concept of a spectral

family in general, that is, without reference to a given operator T. The association of a

suitable spectral family with a given operator T will be considered separately, in the

next section.

4.8 The spectral family of a bounded selfadjoint Linear operators

The construction of the spectral decomposition for unbounded self-adjoint

operators will be based on the following theorem.

4.8.1 Definition (Spectral family or decomposition of unity). A real spectral family

(or real decomposition of unity) is a one-parameter family = (E)R of projections

E defined on a Hilbert space H (of any dimension) which depends on a real

parameter A and is such that

(1) E E, hence EE = EE = E ( < )

(2) lim E x 0

(3) lim E x 0

(4) 00

E x lim E x E x

(x H).

We see from this definition that a real spectral family can be regarded as a mapping

R B(H, H)

E;

to each A R 'there corresponds a projection E B(H, H), where B(H, H) is the

space of all bounded linear operators from H into H.

is called a spectral family on an interval [a, b] if

(5) E = 0 for < a, E = I for b.

83 | P a g e

Such families will be of particular interest to us since the spectrum of a bounded self-

adjoint linear operator lies in a finite interval on the real line.

Theorem 4.8.1 Assume A = A*. Then there is a measure space (MA, dμA) with μA a finite measure, a

unitary operator UA : H → L2(MA, dμA), and a real-valued function fA on MA which is

finite a.e., such that

(i) ψ D(A) fA(・)(UAψ)(・) L2(MA, dμA).

(ii) If U[D(A)], then (UAAUA−1

)(m) = fA(m)(m).

To prove this theorem we need some additional constructions. First we will prove a

similar result for bounded normal operators.

Definition 4.8.1 Let A be a bounded normal operator in H. Then ψ H is a star-

cyclic vector for A if

0

n m

n,m NLin.span{A (A*) } H .

Lemma 4.8.2 Let A be a bounded normal operator in H with a star-cyclic vector ψ

H. Then there is a measure μA on σ(A), and a unitary operator UA, such that UA : H →

L2(σ(A), dμA) with

(UAAUA-1

f)(λ) = λf(λ).

This equality holds in the sense of equality of elements of L2(σ(A), dμA).

Proof: Introduce n i j

ij iji, j 0P c ,c C,n N

and take any p(・) P .

Define UA by UAp(A)ψ = p.

(1) One can prove that for all x, y H there exists a measure μx,y,A on σ(A) such that

x,y,A(A)

(p(A)x, y) p( )d

, p P

Then

2

, ,A(A)

|| p(A) || (p(A)*p(A) , ) ((pp)(A) , ) p( )p( )d

= 2, ,A

2

L ( (A),d )|| p ||

Next we choose μA = μψ,ψ,A. Since ψ is star-cyclic, UA is densely defined and equation

(1) implies that UA is bounded. Thus, UA can be extended to an isometry

UA : H→ L2(σ(A), dμA).

Since P (σ(A)) is dense in L2(σ(A), dμA), Ran(UA) = L

2(σ(A), dμA) and UA is

invertible. Thus, UA is unitary.

Finally, if p P (σ(A)), then

(UAAUA−1

p)(λ) = (UAAp(A)ψ)(λ) = (UA(λ ・ p)(A)ψ)(λ) =

λp(λ).

By continuity, this can be extended from P (σ(A)) to L2(σ(A), dμA).

Lemma 4.8.3 Let A be a bounded normal operator on a separable Hilbert space H.

Then there is an orthogonal direct sum decomposition H = N

j 1 Hj (N ≤ ∞) such that:

(i) For all j: AHj Hj .

(ii) For all j there exists an xj Hj such that xj is star-cyclic forjHA | .

Proof: Take any h1 0 H. If 1{p(A)h ,p(.) P} = H, then h1 is starcyclic and we

are done. Otherwise, denote H1 = 1{p(A)h ,p(.) P} , take any h2 H1, consider H2 =

84 | P a g e

2{p(A)h ,p(.) P} , etc. Then (i) and (ii) are obvious. To show that {Hj} are

orthogonal one computes

(p(A)hj, q(A)hk) = (q(A)*p(A)hj, hk) = (( qp )(A)hj, hk) = 0, if j k.

Theorem 4.8.4 Let A be a bounded normal operator on a separable Hilbert space H.

Then there is a measure space (MA, dμA) with μA a finite measure, a unitary operator

UA H → L2(MA, dμA), and a bounded continuous function fA on MA, such that

(UAAUA−1

)(λ) = fA(λ)(λ).

Proof: Based on Lemmas 4.8.3 and 4.8.4

Now we return to the principal objective of this section:

Proof of Theorem 4.6.1: Since R(λ, A) is a bounded normal operator, we can apply

Theorem 4.8.4 to (A+i)−1

and get (UA(A+i)−1

UA−1 )(m) = gA(m)(m) for some gA.

Since Ker(A + i)−1

= {0}, then gA 0 μA-a.e., so gA−1

is finite μA-a.e. Define fA(m) =

gA(m)−1

− i.

First, we prove that (i) holds: () Let ψ D(A). Then there exists a H such that

ψ = (A + i)−1 and UAψ = gAUA. Since fg is bounded, one obtains fA(UAψ) L

2(MA,

dμA).

() Let fA(UAψ) L2(MA, dμA). Then UA = (fA + i)UAψ for some H.

Thus, gAUA = gA(fA + i)UAψ and hence ψ = (A + i)−1 D(A).

Next, we show that (ii) holds: Take any ψ D(A). Then ψ = (A + i)−1 for some

H and Aψ = − iψ. Therefore,

(UAAψ)(m) = (UA)(m) − i(UAψ)(m) = (gA(m)−1

− i)(UAψ)(m)

= fA(m)(UAψ)(m).

It remains to show that f is real-valued. We will prove this by contradiction. W.l.o.g.

we suppose that Im(f) > 0 on a set of nonzero measure. Then there exists a bounded

set B {z C | Im(z) > 0} with S = {x R | f(x) B}, μA(S) = 0. Hence, Im((χS,

fχS)) > 0, implying that multiplication by f is not self-adjoint.

Next, we can define functions of an operator A. Let h Bor(R). Then

h(A) = UA−1

Th(fA)UA,

where

A

A

2 2

A A A A

h(f )

h(f ) A

L (M ,d L (M ,d )T :

T h(f (m)) (m)

Using (2), the next theorem follows from the previous facts.

Theorem 4.8.5 Assume A = A*. Then there is a unique map A : Bor(R) → B(H)

such that for all f, g Bor(R) the following statements hold:

(i) A is an algebraic *-homomorphism.

(ii) A (f + g) = A (f) + A (g) (linearity).

(iii) || A (f)||B(H) ≤ ||f|| (continuity)

(iv) If {fn(x)}nN Bor(R), fn(x) n x for all x R, and |fn(x)| ≤ |x| for all n N, then

for any ψ D(A), nlim

A (fn)ψ = Aψ.

(v) If fn(x) n f(x) for all x R and fn(x) are uniformly bounded w.r.t.

(x, n), then A (fn) n A (f) strongly.

Moreover, A has the following additional properties:

(vi) If Aψ = λψ, then (f)ψ = f(λ)ψ. A

85 | P a g e

(vii) If f ≥ 0, then (f) ≥ 0.

Again, formally, (f) = f(A).

4.9 Spectral representation of Bounded self adjaoint Linear operator :

Definition 4.9.1. The family {PΩ}Ω RB of bounded operators in H is called a

projection-valued measure (p.v.m.) if the following conditions (i)–(iv) hold:

(i) PΩ is an orthogonal projection for all ΩBR.

(ii) P = 0, P(−∞,∞) = I.

(iii) If Ω = k i jk 1,

for i j, then PΩ = s

k

N

k 1Nlim P

(iv) 1 2 1 2

P P P .

It is easy to see that {χΩ(A)} is a p.v.m. From now on {PΩ(A)} will always denote

{χΩ(A)}. In analogy to the case of bounded operators we then define g(A) for any g

Bor(R) by

(h,g(A)h) g( )d(h,P (A)h),

h H

where d(h, Pλ(A)h) in (4.3) denotes integration with respect to the measure (h,

PΩ(A)h). One can show that the map g → g(A) coincides with the map g → A (g) in

Theorem 4.6.

At this point we are ready to define g(A) for unbounded functions g. First we

introduce the domain of the operator g(A) as follows:

2D(g(A)) h H | g( ) | d(h,P (A)h

One observes that D(g(A)) = H. Then g(A) is defined by

(h,g(A)h) g( )d(h,P (A)h)

, h D(g(A))

We write symbolically,

(A)g(A) g( )dP (A)

Summarizing, one has the following result:

Theorem 4.9.1 There is a one-to-one correspondence between self-adjoint operators A and

projection-valued measures {PΩ}ΩRB in H given by

A dP

If g is a real-valued Borel function on R, then

g(A) g( )dP (A)

,

2D(g(A)) h H | g( ) | d(h,P (A)h)

is self-adjoint. If g is bounded, g(A) coincides with (g)

4.9.2 Spectral Theorem for Bounded Self-Adjoint Linear Operators. Let T : H

H be a bounded self-adjoint linear operator on a complex Hilbert space H. Then

:

(a) T has the spectral representation

(1) M

m 0T dE

where = (E) is the spectral family associated with T (cf. 9.8-3); the integral is to be

A

A

A

86 | P a g e

understood in the sense of uniform operator convergence

[convergence in the norm on B(H, H)], and for all x, y H,

(1*) Tx, x = M

m 0dw( )

w() = Ex, y

where the integral is an ordinary Riemann-Stieltjes integral (sec. 4.4).

(b) More generally, if p is a polynomial in with real coefficients, say,

p() = nn + n-1

n-1 + …. + 0,

then the operator p(T) defined by

p(T) = nTn + n-1T

n-1 + ……. + 0I

has the spectral representation

(2) M

m 0p(T) p( )dE

and for all x, y H,

(2*) p(T)x, y = M

m 0p( )dw( )

, w() = Ex, y

Theorem 4.9.3: (Properties of p(T)). Let T be as in the previous theorem, and let p,

p1 and p2 be polynomials with real coefficients. Then :

(a) p(T) is self-adjoint.

(b) If p() = p1() + p2(), the p(T) = p1(T) + p2(T).

(c) If p() = p1()p2(), then p(T) = p1(T) = p1(T)p2(T).

(d) If p()0 for all [m, M], then p(T) 0

(e) If p1()p2() for all [m, M], then p1(T) p2(T).

(f) ||p(T)|| J

max

|p()|, where J = [m, M]

(g) If a bounded linear operator commutes with T, it also commutes with p(T).

4.10 Extension of the Spectral Theorem to continuous Functions

Theorem 4.9.1 holds for p(T), where T is bounded self-adjoint linear operator and p is

a polynomial with real coefficients. We want to extend theorem to operators f(T),

where T is as before and f is a continuous real-valued function. Clearly, we must first

define what we mean by f(T).

Theorem 4.10.1 Let T : H H be a bounded self-adjoint linear operator on a

complex Hilbert space H. Let f be a continuous real-valued function on [m, M], where

(1) ||x|| 1

m inf Tx, x

,

as before. Then by the Weierstrass approximation theorem 4.11-5 there is a sequence

of polynomials (pn) with real coefficients such that

(2) pn() f()

Uniformly on [m, M]. Corresponding to it we have a sequence of bounded self-adjoint

linear operators pn(T). By Theorem 9.9-2 (f),

||pn(T) – pr(T)|| J

max

|pn() – pr()|,

where J = [m, M]. Since pn() f(), given any > 0, there is an N such that the

right-hand side is smaller than for all n, r > N. Hence (pn(T)) is Cauchy and has a

limit in B(H, H) since B(H, H) is complete (cf. 2.10-2). We define f(T) to be that

limit; thus

(3) pn(T) f(T).

Or course, to justify this definition of f(T), we must prove that f(T) depends only on f

(and T, of course) but not on the particular choice of a sequence of polynomials

converging to f uniformly.

87 | P a g e

Proof1 Let (np ) be another sequence of polynomials with real coefficients such that

np () f()

uniformly on [m, M]. Thennp (T) f (T) by the previous argument, and we must

show that f (T) = f(T). Clearly,

np () – pn()0, hence

np (T) – pn(T) 0,

again by 4.9-2(f). Consequently, given > 0, there is an N such that for n < N,

|| f (T) - np (T)|| <

3

||np (T) – pn(T)||<

3

||pn(T) – f(T)||< 3

By the triangle inequality it flows that

|| f (T) – f(T)|| || f (T) - np (T)|| + ||

np (T) – pn(T)|| + ||pn(T) – f(T)|| < .

Since > 0 was arbitrary, f (T) – f(T) = 0 and f (T) = f(T).

4.11 Properties of the Spectral Family of a Bounded Self-Adjoint Linear

Operator

It is interesting that the spectral family = (E) of a bounded self-adjoint linear

operator T on a Hilbert space H reflects properties of the spectrum in a striking and

simple fashion. We shall derive results of that kind form the definition of (cf. Sec.

4.8) in combination with the spectral representation in Sec. 4.9 From sec. 4.7 we

know that if H is finite dimensional, the spectral family = (E) has “points of

growth” (discontinuities, jumps) precisely at the eigenvalues of T. In fact

0 0 oE E 0 if and only if 0 is an eigenvlaue of T. It is remarkable, although

perhaps not unexpected, that this property carries over to the infinite dimensional case

:

Theorem 4.11-1: (Eigenvalues). Let T : H H be a bounded self-adjoint linear

operator on a complex Hilbert space H and = (E) the corresponding spectral family.

Then E has a discontinuity at and = 0 (that is, 0 0 oE E ) if and only if

0 is an eigenvalue of T. In this case, the corresponding eigenspace is

(1) N(T - 0I) = (0 0 oE E ) (H).

Proof: 0 is an eigenvalue of T if and only if N(T - 0I) {0}, so that the first

statement of the theorem follows immediately from (1).

Hence it sufficies to prove (1). We write simply

F0 = 0 0 oE E

and prove (1) by first showing that

(2) F0(H) N(T - 0I)

and then

(3) F0(H) N(T - 0I).

Proof of (2) :

Inequality in Sec. 4.8 with = 0 - 1

nand = 0 is

(4) 0 0 0 0 0

1E( ) TE( ) E( )

n

88 | P a g e

where 0 = (0 – 1/n, 0]. We let n . Then E(0) F0, so that (4) yields

0F0 TF0 0F0

Hence TF0 = 0F0, that is, (T - 0I)F0 = 0. This proves (2).

Proof of (3) :

Let x N(T - 0I). We show that then x F0(H), that is, F0x = x since F0 is a

projection.

If 0 [m, M], then 0 (T) by 9.2-1. Hence in this case N(T - 0I) = {0} F0(H)

since F0(H) is a vector space. Let 0 [m, M] By assumption, (T - 0I)x = 0. This

implies (T - 0I)2x = 0, that is by 9.9-1.

b

0a

( )dw( ) 0 w() = Ex, x

where a < m and b > M. Here ( - 0)2 0 and Ex, x is monotone

increasing by 4.7-1. Hence the integral over any subinterval of positive length must be

zero. In particular, for every > 0 we must have

0 0

0

2 2 2

0a a

0 ( ) dw( ) dw( ) E x, x

and

0

0 0

b b2 2 2 2

00 ( ) dw( ) dw( ) Ix, x E x, x

Since > 0, for this and 4.5-2 we obtain

0

E x, x = 0 hence 0

E x = 0

and 0

E x, x = 0 hence 0

x E x = 0

We may thus write

0 0

x (E E )x

If we let 0, we obtain x = F0x because E is a continuous form the

right. This implies (3), as was noted before.

4.12 Assignment

Q 1. It was mentioned in the text that for a self-adjoint linear operator T the inner

product Tx, x is real. What does this imply for matrices ?

Q 2. Let T : H H and W : H H be bounded linear operators on a

complex Hilbert space H. If T is self-adjoint, show that S = W*TW is self-

adjoint.

Q 3. Let T : l2 l

2 be defined by (1, 2,….) (0, 0, 1, 2,….). Is T

bounded ? Self-adjoint ? Find S : l2 l

2 such that T = S

2.

Q 4. What theorem about the eigenvlaues of a Hermitian matrix A = (jk) do we

obtain from Theorem 9.2-1 ?

Q 5. Show that the spectrum of a bounded self-adjoint linear operator on a complex

Hilbert space H {0} is not empty, using one of of the theorems of the present

section.

Q 6. Show that a compact self-adjoint linear operator T : H H on a complex

Hilbert space H {0} has at lest one eigenvalue.

Q 7. Show that a bounded self-adjoint linear operator T on a complex Hilbert space

H is positive if and only if its spectrum consists of nonnegative real values

only. What does this imply for a matrix ?

89 | P a g e

Q 8. Let T : H H and W : H H be bounded linear operators on a

complex Hilbert space H and S W*TW. Show that if T is self-adjoint and

positive, so is S.

Q 9. Let S and T be bounded self-adjoint linear operators on a Hilbert space H. If S

0, show that TST 0.

Q 10. Show that if T 0, then (I + T)-1

exists.

Q 11. Let T be any bounded linear operator on a complex Hilbert space. Show that

the inverse of I + T*T exists.

Q 12. If T : H H and S : H H are bounded linear operators and T is

compact and S*S T*T, show that S is compact.

Q 13. Let T : H H be a bounded linear operator on an infinite dimensional

complex Hilbert space H. If there is a c > 0 such that we have ||Tx|| c||x|| for

all x H, show that T is not compact.

Q 14. Find operators T : R2 R

2 such that T

2 = I, the identity operator. Indicate

which of the square roots is the positive square root of I.

4.13 Check your progress

Q 1. Let T : L2[0, 1] L

2[0, 1] be defined by (Tx)(t) = tx(t). (Cf. 3.1-5) Show

that T is self-adjoint and positive and find its positive square root.

Q 2. Let T : H H be a bounded positive self-adjoint linear operator on a

complex Hilbert space. Using the positive square root of T, show that for all x,

y H,

|Tx, y Tx, x1/2Tx, y

1/2.

Q 3. If S and T are positive bounded self-adjoint linear operators on a complex

Hilbert space H and S2 = T

2, show that S = T.

Q 4. Show that a projection P on a Hilbert space H satisfies

0 P 1.

Under what conditions will (i) P = 0, (ii) P = I?

Q 5. Let Q = S-1

PS : H H, where S and P are bounded and linear. If P is a

projection and S is unitary, show that Q is a projection.

Q 6. If a sum P1 + …. Pk of projections Pj : H H(H a Hilbert space) is a

projection, show that

||P1x||2 + … + ||Pkx||

2 ||x||

2.

Q 7. Show that the difference P = P2 – P1 of two projections on a Hilbert space H is

a projection on H if and only if P1 P2.

Q 8. (Limit of Projections) If (Pn) is a sequence of projections defined on a Hilbert

space H and Pn P, show that P is a projection defined on H.

Q 9. (Invariant subspace) Let T : H H be a bounded linear operator. Then a

subspace Y H is said to be invariant under T if T(Y) Y. Show that a

closed subspace Y of H is invariant under T if and only if Y is invariant

under T*.

Q 10. (Reduction of an operator) A closed subspace Y of a Hilbert space H is said

reduce of linear operator T : H H if T(Y) Y and T(Y) (Y

), that

is, if both Y and Y are invariant under T. (The point is that then the

investigation of T can be facilitated by considering T|Y and Y

T | separately.)

If P1 is the projection of H onto Y and P1T = TP1, show that Y reduces T.

90 | P a g e

Q 11. Prove that T-T = TT

-.

Q 12. Find T+, T

-, (T

2)1/2

and the other square roots of T2 if

2 0

T0 3

Q 13. If in the finite dimensional case a linear operator T is represented by a real

diagonal matrix T , what is the spectrum of T? How do we obtain from T the

matrix (a) T+ (representing T

+), (b) T

- (representing T

-), (c) B (representing

B) ?

Q 14. Verify for the zero operator T = 0; H H.

Q 15. If an operator T : R3 R

3 is represented, with respect to an orthonormal

basis, by a matrix

0 1 0

1 0 0

0 0 1

,

What is corresponding spectral family ? Using the result, verify (1) for this

operator.

Q 16. Find the spectral family of the operator T : l2 l

2 defined by (1, 2,

3,….,) (1/1, 2/2, 3/3….). Find and orthonormal set of eigenvectors.

What from does (1) take in this case ?

4.14 Point for discussion/clarification

91 | P a g e

4.15 Suggested Study material

1. Kelley J. L. (1955) General topology. New Yark Van Nostrand.

2. Kreyszing E (1970) Introductory Mathematical statistics. New York

wieley.

3. Edward R. E. (1965) Functional Analysis. New York.

92 | P a g e

Unit -5

OBJECTIVES: After reading this unit you should be above to;

Define canonical product and prove many results of it Understand the concept of order of an entire function. Understand the meaning of exponent of convergence

and prove a number of important results. Define univalent function.

STRUCTURE:

5.1 Canonical Product 5.1.1 Jensen’s formula 5.1.2 Poisson Jensen formula 5.1.3 Handmard’s there circle theorem 5.2 Order of an entire function 5.2.1 Definition 5.2.2 Theorem 5.2.3 Problems 5.3 Exponent of convergence 5.3.1 Definition 5.3.2 Proposition 5.4 Borel’s Theorem 5.5 Hadamard’s Factorization Theorem. 5.6 The Range of an analytic function 5.6.1 Bloch theorem 5.6.2 Little picard theorem 5.6.3 Montel Caratheodory 5.6.4 Great picrd Theorem 5.7 Univalent Function 5.7.1 Definition 5.7.2 Theorems 5.7.3 Meaning of

5.7.4 Statement of Bieberbach’s Conjecture

5.7.5

theorem

5.8 Summary 5.9 Self assessment problems 5.10 Points of Discussion 5.11 Further Readings

93 | P a g e

5.1 Canonical Product: This product is defined for entire function. Its concept is derived from weiestrass factorization theorem.

Definition: The product of the form

∏(

) 8

(

)

(

)

9

Is called canonical product if it is convergent and represents an entire function with the condition that.

∑

(

)

is converges for all R. Here is called the genus of the canonical product.

Example: Canonical product of sin z is

∏4

5

5.1.1 Jensen’s Formula: Let ( ) be analytic in the closed

disc | | . Assume that ( ) and no zeros of

( ) lie on| | . If are the zeros of ( ) in the open disc | | each repeated as often as its

multiplicity then

| ( )| ∑ (

| |)

∫ | ( )|

Proof; let us define a function

( ) ( )∏

( )

( )

as ( ) is given to be analytic, hence ( ) is also

analytic and hence

( ) on | |

Also ( ) is never zero on an open disc | | for | ( )| | ( )| | |

Let , Then

94 | P a g e

|∏

( )

| |∏

|

∏ ( )

( )

∏|

|

| |

Now we can apply Gauss mean value theorem for | ( )|

| ( )|

∫ | ( )|

( )

From (1)

( ) ( ) ∏(

)

| ( )| | ( )| ∑

| |

Also | ( )| | ( )| | |

| ( )| | ( )|

Putting these values in eq(2), ewe get

| ( )| ∑

| |

∫ | ( )|

| ( )| ∑

| |

∫ | ( )|

5.1.2 Poisson – Jensen Formula : Let ( ) be analytic in

the closed disc | | . Assume that ( ) and no zeros

of ( ) lies on | | . If are the zeros of

95 | P a g e

( ) in the open disc | | , each repeated as often as it’s

multiplicity and ( ) then | ( )|

∑ |

( )|

∫

( ) | ( )|

( )

Proof : Let us consider a function

( ) ( )∏

( )

( )

( ) is analytic in any domain in which f(z) is analytic and

( ) for | | ( ) is analytic and never 0 on an open disc | | for some

Hence | ( )| | ( )| | | Let

and hence

|∏

( )

|

Again as F (z) is analytic and non zero on an open disc |z| < P, P > R. log F (z) is analytic in |z| < P

Real part of log |F (z)| is harmonic so by poisson’s

formula for log |F (z)|,

Log |F (z)| =

∫

( ) | ( )|

( )

d ….. (2)

as log |F (R )| = log |f (R )| on |z| = R

96 | P a g e

again from eq (1),

| ( )| | ( )| ∑ |

( )|

putting these value in eq (2) we get

| ( )|

∑ |

( )|

∫

( ) | ( )|

( )

5.1.3 Hadamard’s Three circles Theorem: Theorem Let f (z) be analytic in the annular region r1 |z| r3, and r1,

r2, r3 if M1, M2, M3 be the max |f (z)| on the three circles |z| = r1, r2, r3 respectively, then

( )

( ) ( )

Proof. Let F (z) = zk f (z), where k is a ral constant to be determined later. Then F (z) is analytic in the annulus r1 |z| r3.

If k is not an integer, then F (z) is a multivalued function, and for convenience, one choose the principal branch. For this we cut the annulus along the negative real axis and we obtain a domain in which the principal branch of this function is analytic. By the maximum modulus principle, the maximum value of |F (z)| is attained on the boundary of the cut annulus. Now consider branch of this function which is

analytic in the part of the annulus for which

arg z

, we see that the principal value cannot attain its

maximum modulus on the cut. Hence the max |F (z)| occurs

97 | P a g e

on one o f the bounding circles. Thus it is shown that when

r1 |z| r3, |F (z)| max {rk

1 M1, rk3 M3}

Hence, on |z| = r2, are must have

rk2 M2 max { rk

1 M1, rk3 M3}

rk2 M2 r

k1 M1 or M2 .

/

M1 …………. (1)

Since k is at our choice, we choose k to our best advantage by making the expressions in the parentheses on the right equal to each other. Thus k is defined by the equation

rk1 M1 = r

k3 M3

whence we have

k log r1 + log M1 = k log r3 + log M3 k [log r1 - log r3] = log M3 - log M1

k = - {log

}/{log

} ………………. (2)

By (1), we have

.

/

- k log

.

/

Now substituting the value of k in the above inequality, we obtain

.

/ .

/ .

/ .

.

/

= .

/ .

/ .

.

/

98 | P a g e

[a log b = ( )

= = ( )

= ]

.

/ .

.

/

.

/ .

.

/

5.2 Order of an entire function :

we have already understood the concept of entire function in the previous unit now we define order of an entire function and for this we must know what is M (r)

M (r) = max {|f (z)| : |z| = r}

5.2.1 Definition : An entire function f (z) is said to be of finite order is

M (r) exp ( ) where

= in f { : M (r) exp ( ) for all larger}

Similarly if M (r) > exp ( ) then the order of f (z) is

considered to be finite

5.2.2 Theorem : Let f be a non constant function Define = in f { ( ) exp ( ) for sufficiently large r}

and

( )

Then =

Proof : For the proof of the theorem we consider two cases Case I : If is finite and > 0 be given then R ( ) > 0

such that M (r) < exp ( ), r R ( ) …….. (1)

Thus for all r > R ( ), in eq (1) we take logar than twice Log log M (r) < ( ) log r

99 | P a g e

( )

Hence

( )

Or

As is arbitrary so as 0

……….. (2)

Case II : If is finite and > 0 be given then

( )

log log M (r) < ( ) log r =

log M (r) < exp ( ( ))

= ( )

log M (r) < exp ( ( ))

in f ( ( ) ( ) for larger)

<

<

Again as us arbitray so an We get ………… (3)

From (2) and (3) we get

100 | P a g e

5.2.3 Problems : Find the order of , Solution let M (r) = | |

So

( )

(| | )

Then we apply L-hospital rule

| |

| |

Hence order of

5.3 Exponent of Convergence :

This is a very common term associated with the sequence of complex numbers.

5.3.1 Definnition : Let { } be a sequence of non-zero

complex numbers such that |z,| |z2| ……….. |zn|

as n . The exponent of convergence of the sequence is define by

= inf { > : ∑

| |

< }

5.3.2 Proposition : The convergence of exponent of a sequence * + is given by

| |

Proof : Let is consider as a finite quantity and hence we

define a series

101 | P a g e

∑

| |

Clearly for < the series is convergent

| |

Let such that

|

| | |

log n - log | | < 0

| |

| | ……………. (1)

Now we consider be any other arbitrary number

exceeding the R.H.S. of (1) ( ) such that

| |

Thus | |

∑

| |

is convergent for

Now from the definitin of exponent of convergence

| |

……………. (2)

102 | P a g e

From (1) and (2) we get

| |

5.4 Barel’s Theorem :

The order of a canonical product is equal to the convergence exponent of its zeros. Proof : Let us consider and as the order and

convergence exponent of a conical product P (z). Then we know that

So to prove the theorem it is sufficient to show that

So let us consider

( ) ( ) 4

5

If we consider P > 0 a, b > 0 Such that | ( )| ( | | )

= ( | | )

If c = a + b, and |w| | and

| ( )| ,( ) -

( | | )

( | | )

On the other hand if |w|

and P > 0

Then

|E (w, p)| exp ( | | )

and if

103 | P a g e

| ( )| ( | | )

again if ½ | | | and p > 0

we get

| ( ) ( | | )|

= | ( ) ( | | )|

Now let us consider {z1 , z2 , z3 ……}

Be the sequence of zeros of the canonical product P (z).

Then by definition h will be the genus and

also by definition of exponent of convergence, we have

{ ∑| |

}

Or

∑| |

If

∑| |

( ) ∏ .

/

We get

| ( )| ∑ . |

|

/

( | | )

Or

|P (z)| ( | | )

Or

104 | P a g e

Hence we get

5.5 Hadamard’s factorization Theorem :

Let f is an entire function of finite order then f has finite genus

or

Let ( ) be an entire function of finite order . Then

( ) ( ) ( ) Where is the order of the (possible) zeros of ( ) at

is a polynomial of degree not exceeding and ( ) is the canonical product associated with the sequence of non-zero zeros of ( ) Proof: By Weierstrass’s factorization theorem, the entire function ( ) can be represented in the form given by

( ) ( ) ( ) Where ( ) is an entire function. Here we have the additional

hypothesis that ( ) is of finite order . This additional

hypothesis is used to show that ( ) is a polynomial. Its is evident that the division of ( )by does not affect either

the hypothesis or the conclusion of the theorem and so it is sufficient to consider the repsrentation

( ) ( ) ( ) ( ) ( )

( )

Therefore | ( )| | ( )

( )|

( ) | ( )

( )| |

( )

( )|

( ) | ( )| | ( )| By the definition of order, it follows that

( ) ( ) For sufficiently large | | and all Therefore

105 | P a g e

| ( )|

Suppose is the convergence exponent of the non-zero zeros of ( ), then

Also by Borel’s theorem, is the order of the cannonical product ( ), and so it follows from theorem II of the

proceeding section . That | ( )| | |

Thus, | ( )|

Form (2) and (3), we obtain

| ( )| | ( )|

Hence by (1). We get

( ( )

Since is large, we conclude from Theorem Ii of the section . That ( ) is a polynomial of degree not exceeding .

So the genus of . This completes the proof.

5.6 Range of an analytic function:

Analytic functions can have a variety of range . Under this article we will prove many important results which given an idea about the range of an analytic function 5.6.1 Bloch’s Theorem: Let be an analytic function in a region containing the closure of the disc * | | + and

satisfying ( ) ( ) . Then there is a disc S⊂D in

which is one-one and such that ( ) contains a disc of

radius

Proof: Let us consider ( ) *| ( )| | | +

( ) ( ) ( ) , - is continuous and ( ) ( ) and ( ) ( )

106 | P a g e

Let * ( ) +

( ) ( )

Also maximum value of ( ) is 1

Now we choose as | | and | ( )| ( ) ( ) ( ) ( )

( ) ( ) | ( )| ( )

( ) Also

| ( )| 4

( )5 | |

[

( )

] (

)

| |

( )

| ( )|

| | ( )

Form (1) and (2)

| ( )| | ( )| | ( )| | ( )|

Now we define

( ) ( ) ( )

Then by Schwarz’s lemma

| ( )| | | * | | +

| ( ) ( )

|

( )

| ( ) ( )| | |

Again if .

/

| |

| ( )| | ( )|

| ( )|

107 | P a g e

So we get that is one-one as we have a result “ Let be

am analytic function in the disc ( ) such that | ( )| | ( )| | ( )| ( ) Now to prove the theorem , it remains to show that ( )

contains a disc of radius

.

So we define

.

/ ( ) ( ) ( )

( ) ( ) ( )

| ( )|

Also

| ( )|

.

/

then we can apply “Let be analytic in ( ) ( ) | ( )| | ( ) |

( ( )) 4

5

and we get

4 .

/5 ( )

.

/

.

/

( ⁄ )

Hence

( ( )

)

{ | ( )|

}

{ ( ) | ( | ( )

}

108 | P a g e

{ ( ) | ( ) |

( )}

* | | + ( )

4 .

/5 ( )

4 .

/5 ( )

( ) ( ( )

)

5.6.2 The little Picard Theorem: To prove this theorem firstly we should know the concept of branch of logarithm. If G is an open connected set in c an is a continuous

function such that ( )

Called a branch of logarithm. Statement: Let be an function that omits two values. Then

is a constant.

Proof: As it is given that omits two values so Let ( ) ( )

( ) ( )

then the function ( ) omits 0 and 1

So we can assume that the given entire function omits two value 0 and 1

( ) ( )

we get a function such that ( ) contains no disc of

radius 1. Now our aim is to prove that is constant, so on the

contrary let is not constant is not constant a point such that ( )

So without the loss of generality we may suppose that

( )

109 | P a g e

Then we use the result “ let be an analytic function in the

region that contains ( ( )) contains a disc of radius

| ( )| ( ( ))contains a disc of radius | ( )|

If we choose R sufficiently large then ( ) does not contain

a contradiction, hence is constant.

5.6.3 Montel Caratheodary Theorem: Statement: Let F be the family of all analytic functions defined in a region G that do not assume the value 0 and 1, then is normal in ( ). Proof: Choose a point in G and keep it fix. Define the

families and by *( | ( )| +

And *( | ( )| + So

So U H. We now show that is normal . in

H(G) and that is normal in ( ) Note that by considering the sequence of constant functions * + in

defined by ( ) we see that .

To show that * | ( )| +11 is normal is H(G) we

shall invoke Montel's theorem; that is, it is just sufficient to

show that is locally bonded.

Let be arbitrary and let be a curve in from

z0 to a. Suppose Do, D1. ...... Dn be discs in G with

centre z0,z1. ..... z, = a, respectively on * + The discs are

so constructed that zk - 1 and zk are in Dk - 1 Dk for

. Assume that ⊂ .

Now Applying Schottky's Theorem to D0, there is a

constant CO such that

| ( )|

We notice that if D0 = B(z0; r) and R > r is such that

110 | P a g e

B(zo; R) C G then by Corollary of Theorem 1 of §124, we

have

| ( )| ( )

whenever is choosen in such a way that .

In particular, f ( ) so that Schottky's Theorem gives

that y is uniformly bounded by a constant C1 on D1.

Continuing this process we have that is uniformly

bounded on D. Since was arbitrary, it follows that is

locally bounded. Hence by Montel's Theorem, is normal in

H(G).

Next, we consider * | ( )| + Then 1/f is

analytic on G because f never vanishes. Also, 11f never assumes the value 1; moreover | ( )| 1.

Hence, if we define * +

then we see that ⊂ and is normal in H(G). It follows that if * +

is a sequence in then there is a subsequence * +

and an analytic function h defined in G

such that

in H(G). Hence, either or never

vanishes. If it is easy to show that

( ) uniformly on

compact subsets of G.

In case never vanishes, we see that is analytic and

it follows that

( ) ( )

uniformly on compact subsets of G.

5.6.4 The Great Picard theorem: Let be an analytic function that has an essential singularity at Then in each neighbourhood of assumes each

111 | P a g e

complex numbers, with one possible exception, an infinite number of times. Proof. Without loss of generality we may suppose that has an essential singularity at z = 0. To effect this we may consider ( ) if necessary.

Suppose, if possible, there is an such that there are two numbers not in * ( ) + we will obtain a contradiction. If ( ) ( ) for all in * + then the function defined by

( ) ( )

omits the values 0 and 1. So we may suppos that ( )

amd ( ) for | | .

Let ( ) * + and define by

( ) .

/

Then each is analytic in , since is analytic and no

assumes the value 0 or 1, Now , by Montel-Caratheodary Theorem, * + is a normal family in ( ).

Let * + be a subsequence of * +

such that

( ) uniformly on .

/ 2 | |

3, where

is either analytic in G or . If is analytic, let

2| ( )|

3 then for e = M>0, such that for

all k , we have

| (

)| |

( )| | ( ) ( )| | ( )|

| ( ) ( )| | ( )|

.

/ |Z| =

Thus

| (

)|

112 | P a g e

For sufficiently large and |Z| =

; hence | ( )|

for sufficiently large and |Z| =

Now, by Maximum Modulus Principle, f is uniformly bounded on concentric annuli about zero. It follows that f is bounded by 2M on a deleted neighbouhood of zero. This shows that z = 0 must be a removable singularity. Therefore, cannot be analytic and so .

Further, we see that if . Then f must have a ple at

zero. Hence we conclude that there is at most one complex number that is never assumed by f. However, if there is a complex number w which is assumed only a finite number of times then by taking a sufficiently small disc, we again arrive at a punctured disc in which f fails to assume two values.

5.7 Univalent Functions : In this section we will study a special type of function 5.7.1 Definition : A function f (z) is said to be univalent in a domain if it is analytic in and assumes no value more

than once in . The condition is , implies ( ) ( ). Points

where ( ) vanishes, ( ) is not univalent. 5.7.2 Theorem 1 : Prove that if w = f (z) is univalent in a domain D then the inverse mapping z = g (w) is univalent in f (D) Proof : As D and f (D) have one-to-one correspondence z is a function of w i.e. z = g (w)

g (w) assume no value in D more than one.

Then by inverse function theorem, g (w) is analytic at each point of f (D)

g (w) is analytic in f (D).

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Theorem 2. A univalent function that maps | | onto | | must be linear. Proof : Consider | |

Then the image of | | under f contain some disk | | Hence if is an essential singularity for f, so by weierstrass

theorem, f (z) comes arbitrary close to in every neighbourhood of .

But this is a contradiction as f is simple is a pole of f is a polynomial and clearly the polynomial must have

degres 1 otherwise ( ) would have atleast one zero. And

hence a univalent function which maps | | onto | | must be linear. 5.7.3 Meaning of : Let us consider U as unit disc and f is

a complex function defined in U if and if ( ) =

( ) ( )

Exists for every Then f is holomorphic in U. Then the

class of all holomorphic function in U is denoted by H (U). The class of all f H (U) which are in one to one in unit disc U and satisfy ( )

5.7.4 Bieberbach Conjecture: This was defined by Bieberback in 1916 and proved by L de Branges in 1984 If and

( ) ∑

then | | for all and ( ) (

)

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5.7.5 ¼ Theorem: Statement: Let and

( ) ∑

then

( )| | ( ) ( ) (

)

Proof (a) We have a result which says if then there exists a such that

( ) ( )

we get such that ( ) ( )

Let G =1/g

( )

√ ( )

√ ∑

√ ∑

( ∑

)

4

5

∑

then as we have |a,|≤1 |a2|≤ 2

Proof (b) Let us consider ( ) we define

( ) ( )

( )

as ( )

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( ), and hence is one to one so by define if

and ( ) ( )

( )

( )

( )

( )

( )

( ) ( ) ( )

( ) ( )

( ) ( ) [ ]

Also, ( ) ( )

( ) ( ), ( ) -

( ),

( )

-

( ),

-

(

)

So that ( ) .

/ It follows that

( ) and ( )

Thus,

Finally, we see that

|

| | | |

| |

|

|

| | |

|

| , | | -

|

|

( )

( ) (

) ( )

This completes the proof.

5.8 Summary: Canonical product is unique

Order of

Order of √

For a finite sequence exponent of convergence is zero for an entire function

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The order of a canonical product is equal to the convergence exponent of its zero.

If f is an entire function of finite order then f has finite genus

If f (z) is an entire function then order of ( ) and f (z) is same

Let f be an entire function that omits two values. Then f is constant.

Univalent function is one-one

5.9 Self assessment problems:

(1) Find the order of sin z,

(2) Use Hadamard’s factorization theorem to show that

∑ 4

5

(3) For each and , 0 < < and 0 < < there is a constant ⊂ ( ) such that if f is an analytic function

defined in some simply connected region containing ( ) that omits the values o and 1 and such that |f (0)| then

| ( )| ( ) for | |

(Schottky’s theorem)

(4) check Z+1/Z is univalent or not.

5.10 Points for Discussion _________________________________________________

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5.11 Further Readings (1) Complex Analysis by Dr. H.K. Pathak (2) Complex Analysis by L.V. Ahlfors. (3) Complex and Real Analysis by walter Rudin.