Outgrowths of Hardy’s Inequality

Wolfgang Arendt, Gisèle Ruiz Goldstein, and Jerome A. Goldstein

1. Introduction and main results

Scaling plays an important role in quantum mechanics and other areas of ap-plied mathematics. When a physicist says that "kinetic energy scales like λ2", shemeans the following. The kinetic energy operator is the negative Laplacian, soconsider the Laplacian ∆ as a selfadjoint operator on the Hilbert space L2

(RN).

For λ > 0 let U (λ) be the normalized scaling operator defined by

(U (λ) f) (x) = λN2 f (λx) , f ∈ L2

(RN), x ∈ RN .

Then U (λ) is unitary on L2(RN)and U (λ)

−1= U

(1λ

). "The Laplacian scales

like λ2" meansU (λ)

−1∆U (λ) = λ2∆

holds for all λ > 0, as is easy to verify.Now we proceed somewhat formally and do not make precise statements about

domains. The operator representing multiplication by 1|x|p scales like λ

p, in thesense that for

(Mpf) (x) = |x|−p f (x) , f ∈ L2(RN), x ∈ RN ,

we haveU (λ)

−1MpU (λ) = λpMp

for all λ > 0. Consequently

(1.1) U (λ)−1

(∆ +

c

|x|2

)U (λ) = λ2

(∆ +

c

|x|2

)holds for all λ > 0 and all c ∈ R. Now let A0

c = ∆ + c|x|2 on D

(A0c

)= C∞c

(RN)

=

D(RN)if N ≥ 5 and D

(A0c

)= C∞c

(RN\ {0}

)= D

(RN\ {0}

)if N ≤ 4. Let Ac

be a selfadjoint extension of A0c for which (1.1) holds on the domain of Ac. By

(1.1) the spectrum of Ac is preserved by multiplication by a positive number. Also,((∆ +

c

|x|2

)ϕ

∣∣∣∣∣ϕ)

= −‖∇ϕ‖2 + c

∫RN

|ϕ (x)|2

|x|2dx

2000 Mathematics Subject Classification. Primary 47A05, 47D08, 47F05, 35B05; Secondary34G10, 47D03, 47D06.

Key words and phrases. Inverse square potential, Hardy’s inequality, Cafferelli-Kohn-Nirenberg inequality, critical constant.

1

2 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

is negative for ϕ 6= 0 and supp(ϕ) shifted far away from the origin, thus, σ (Ac) ⊃(−∞, 0]. Thus σ (Ac) = (−∞, 0] or σ (Ac) = R. If

(A0cϕ|ϕ

)≤ 0 for all ϕ ∈

C∞c(RN\ {0}

), let−Ac be the Friedrichs extension of−A0

c . Then U (λ)−1AcU (λ) =

λ2Ac for λ > 0 and hence σ (Ac) = (−∞, 0].Hardy’s inequality helps to explain how the spectrum of Ac depends on c. The

usual way to express Hardy’s inequality is∫RN|∇u (x)|2 dx ≥

(N − 2

2

)2 ∫RN

|u (x)|2

|x|2dx

for all u ∈ H1loc

(RN)for which the right hand side is finite. The constant

(N−2

2

)2is maximal in all dimensions (including N = 2). Thus, Ac is nonpositive if andonly if c ≤

(N−2

2

)2.

Thus the Cauchy problem∂u

∂t= ∆u+

c

|x|2u x ∈ RN , t ≥ 0

u (x, 0) = f (x) x ∈ RN ,

is well-posed(for f ∈ L2

(RN))if and only if c ≤

(N−2

2

)2, in which case it is

governed by a positive contraction semigroup.It is natural to ask about Lp versions of this result and about other situations

in which scaling plays a role and a critical dimension dependent constant appears.The purpose of this paper is to answer these questions.

Our main result is as follows.

Theorem 1.1. Let β ∈ R, 1 < p <∞, c ∈ R and let

Bu = Bpcβu := ∆u− β

|x|∂u

∂r+

c

|x|2u

with domain D0 = D(RN\ {0}

)if N ≥ 2, D0 = D (0,∞) if N = 1, acting on

Lpβ = Lp(RN , |x|−β dx

)if N ≥ 2 and Lpβ = Lp

((0,∞) , |x|−β dx

)if N = 1;

here ∂u∂r = ∇ · x

|x| is the radial derivative. Then a suitable extension of Bpcβgenerates a (C0) contraction semigroup on Lpβ if

(1.2) c ≤ (N − β − 2)2

(p− 1

p2

)=

(N − β − 2

2

)24

pp′=: K (N, p, β) .

If 2 ≤ p < ∞ and c > K (N, p, β) , then Bpcβ has no quasidissipative extension onLpβ .

Specializing Theorem 1.1 to β = 0 yields

Corollary 1.2. Let 1 < p <∞. Then

A = ∆ +c

|x|2

on D0 = D(RN\ {0}

)if N ≥ 2, D0 = D (0,∞) if N = 1, has an m-dissipative

extension on Lp if

(1.3) c ≤ (N − 2)2

(p− 1

p2

)=: K (N, p) ;

OUTGROWTHS OF HARDY’S INEQUALITY 3

Lp = Lp(RN)if N ≥ 2, Lp = Lp (0,∞) if N = 1. For c > K (N, p) ,

sup{

Re 〈Au, Jp (u)〉 : u ∈ D (A) , ‖u‖p = 1}

=∞

and so A has no quasidississipative extension, provided 2 ≤ p.

This follows from Theorem 1.1 by taking β = 0; here K (N, p) = K (N, p, 0)and Jp is the duality map of Lp. Note that

K (N, 2, β) =

(N − 2− β

2

)2

.

Throughout this paper we consider real vector spaces.

Acknowledgement . The authors are grateful to H. Vogt for several helpfuldiscussions.

2. Hardy’s inequality and the inverse square potential

The aim of this section is to study the semigroup generated by ∆ + c|x|2 for

suitable c. We let D0 = D(RN\ {0}

).

Theorem 2.1. (Hardy’s Inequality)One has (

N − 2

2

)2 ∫ |u|2|x|2

dx ≤∫|∇u|2 dx

for all u ∈ D0. The constant(N−2

2

)2is optimal.

Here the integral is taken over RN if N ≥ 2 and over (0,∞) if N = 1.Theorem 2.1 is well-known. We give a proof of the inequality for the conve-

nience of the reader.

Proof. Let u ∈ D0, u real.a) Let N ≥ 3. Then

u (x)2

= −∫ ∞

1

d

dλu (λx)

2dλ

= −2

∫ ∞1

u (λx) ∇u (λx) · x dλ.

Hence ∫ |u|2|x|2

dx = −2

∫ ∫ ∞1

u (λx)

|x|2∇u (λx) · x dλ dx

= −2

∫ ∞1

∫λ2u (y)

|y|2∇u (y) · y

λ

dy

λNdλ

= −2

∫ ∫ ∞1

λ−N+1 dλu (y)

|y| ∇u (y) · y|y| dy

=2

−N + 2

∫u (y)

|y| ∇u (y) · y|y| dy

≤ 2

N − 2

(∫u (y)

2

|y|2dy

) 12 (∫

|∇u (y)|2 dy

) 12

.

4 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

Hence, (∫u (x)

2

|x|2dx

) 12

≤(

2

N − 2

)(∫|∇u (x)|2 dx

) 12

.

b) For N=1, u ∈ D0,

u (x)2

=

∫ 1

0

d

dλu (λx)

2dλ.

Now use the same arguments as in part a). �

For N=1, D0 is dense in H10 (R\ {0}) =

{u ∈ H1 (R) : u (0) = 0

}, which makes

sense since H10 (R) ⊂ C (R) . By considering even functions in H1

0 (R\ {0}) , we canwrite the one dimensional Hardy inequality as∫ ∞

0

(u′)2dx ≥ 1

4

∫ ∞0

u2

x2dx

for all u ∈ H10 := H1

0 (0,∞) .From now on, integrals (as in (2.1)) will be over RN if N ≥ 2 and over (0,∞)

if N = 1. And for N = 1, D0 will henceforth denote D (0,∞) .If N = 1, then D0 is dense in H1

0 which makes sense, since H10 (0,∞) ⊂

Cb[0,∞). If N ≥ 2, then D0 is dense in H1(RN)(cf. [3, Lemma 2.4]). Con-

sequently, the inequality remains true for all u ∈ H1(RN)if N ≥ 3 and for all

u ∈ H10 (0,∞) if N = 1.

We now sketch why the constant in Hardy’s inequality is optimal. Let x ∈ RN(x > 0 if N = 1) and let r = |x| . For ε > 0, a > 0, b ≥ 0, let

φ (r) =

ε−a−brb if 0 ≤ r ≤ εr−a if ε ≤ r ≤ 12− r if 1 ≤ r ≤ 20 if 2 ≤ r.

Then, letting ψ (x) = φ (r) and using |∇ψ|2 =(φ′)2, we can show that, given δ > 0,∫

|∇ψ|2 <[(

N − 2

2

)2

+ δ

]∫ψ2

|x|2

provided that ε, a, and b are suitably chosen. (We may choose b = 0 for N ≥ 3,

but b > 0 is necessary for N = 1, 2.) In fact, given c >(N−2

2

)2and M > 0, we

may choose ε, a, b so that ∫|∇ψ|2 − c

∫ |ψ|2|x|2∫

ψ2 < −M.

To see how to choose ε, a, b in this way ( in a much more general setting), see[10].

In the following we use the notion of positive linear forms. Given a real Hilbertspace H, a positive form on H is a bilinear mapping a : D (a) × D (a) → R suchthat

a (x, y) = a (y, x) x, y ∈ D (a)

anda (x, x) ≥ 0 x ∈ D (a) .

OUTGROWTHS OF HARDY’S INEQUALITY 5

Here D (a) is a subspace of H, the domain of the form a. The form is called

closed if D (a) is complete for the norm ‖u‖a =(‖u‖2H + a (u, u)

) 12

. The form

is called closable if the continuous extension of the injection D (a) → H to thecompletion D (a) of D (a) is injective. In that case a has a continuous extensiona : D (a)×D (a)→ R that is a positive closed form.

If a is a closed densely defined positive form on H, the associated operator Aon H is defined by

D (A) = {u ∈ D (a) : there is a v ∈ H such that

a (u, ϕ) = (v|ϕ)H for all ϕ ∈ D (a) },Au = v.

The operator A is selfadjoint and form positive. Thus -A generates a (C0) con-traction semigroup

(e−tA

)t≥0

on H.

In the following we let L2 := L2(RN)if N ≥ 2 and L2 = L2 (0,∞) if N = 1.

If H = L2, then the semigroup e−tA is submarkovian (i.e. 0 ≤ f ≤ 1implies 0 ≤ e−tAf ≤ 1) if and only if u ∈ D (a) implies u ∧ 1 ∈ D (a) and

a(u ∧ 1, (u− 1)

+)≥ 0. In that case

∥∥e−tAf∥∥p≤ ‖f‖p for all f ∈ Lp ∩ L2.

Now for c ∈ R, consider the positive form ac on L2 given by

ac (u, v) =

∫∇u · ∇v dx− c

∫u2

|x|2dx,

D (a) = H10 , where H

10 := H1

0

(RN)if N ≥ 2 and H1

0 := H10 (0,∞) if N = 1. We

investigate the question: "For which real c is the form ac closed or closable"?

Proposition 2.2. Let c <(N−2

2

)2. Then the form ac given by D (ac) = H1

0 ,

ac (u, v) =

∫∇u · ∇v dx− c

∫uv

|x|2dx,

is positive and closed.

Proof. Let q = c(N−2

2

)−2< 1. Then by Hardy’s inequality

ac (u) = q

{∫|∇u|2 dx−

(N − 2

2

)2 ∫u2

|x|2dx

}+ (1− q)

∫|∇u|2 dx

≥ (1− q)∫|∇u|2 dx.

Thus, ac (u, u) + ‖u‖2L2 ≥ (1− q) ‖u‖2H1 . This shows that ‖·‖ac is equivalent to‖·‖H1. �

For the critical constant we have the following result.

Proposition 2.3. For cN =(N−2

2

)2, N 6= 2, the form acN is closable.

Proof. Define the symmetric operator B on L2 by D (B) = D0,

Bu = ∆u+ cNu

|x|2.

It is well known that the form b given by b (u, v) = − (Bu|v)L2 , D (b) = D (B) isclosable (and the operator associated with b is called the Friedrichs extension of

6 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

b). Observe that b (u, v) = acN (u, v) for u, v ∈ D0. Since D0 is dense in H10 , it

follows that H10 ⊂ D

(b)and acN = b on H1

0 ×H10 . This shows that acN is closable

and acN = b. �

We next show that acN is not closed. For simplicity we consider only the caseN = 3. Then the critical constant is c3 = 1

4 . We show that

(2.2) D(a 14

)* L6

(R3).

On the other hand, by Sobolev embedding, H10 ⊂ L6. Consequently H1

0 is a proper

subspace of D(a 14

).

Proposition 2.4. The form a 14is not closed and (2.2) holds.

Proof. Let η ∈ D(R3)be a test function such that 0 ≤ η ≤ 1 and η = 1 in

a neighborhood of 0. Let u (x) = |x|−12 η. Then u ∈ L2

(R3), but u /∈ L6

(R3).

Let uε (x) = |x|−12+ε

η. Then uε ∈ H10 and uε −→ u in L2.

Since a 14is closable, it suffi ces to show that a 1

4(uε1 − uε2) −→ 0 as ε1,

ε2 −→ 0. Let δ > 0 such that η (x) = 1 for |x| ≤ δ.It suffi ces to show that

c (ε1, ε2) :=

∫|x|≤δ

{|∇ (uε1 − uε2)|

2 − 1

4

(uε1 − uε2)2

|x|2

}dx −→ 0

as ε1, ε2 −→ 0 + . We compute

4c (ε1, ε2) : =

∫|x|≤δ

4

[(−1

2+ ε1

)|x|ε1 −

(−1

2+ ε2

)|x|ε2

]2

|x|−3dx

−∫|x|≤δ

4(|x|2ε1 − 2 |x|ε1+ε2 + |x|2ε2

)|x|−3

dx

=

∫|x|≤δ

{(−ε1 + ε2

1

)|x|2ε1 + (ε1 + ε2) |x|ε1+ε2

}|x|−3

dx

+

∫|x|≤δ

{−2ε1ε2 |x|ε1+ε2 +

(−ε2 + ε2

2

)|x|2ε2

}|x|−3

dx

=

∫ δ

0

{(−ε1 + ε2

1

)r2ε1−1 + (ε1 + ε2) rε1+ε2−1

}dr

+

∫ δ

0

{−2ε1ε2r

ε1+ε2−1 +(−ε2 + ε2

2

)r2ε2−1

}dr

=(−ε1 + ε2

1

) δ2ε1

2ε1+ δε1+ε2 − 2

ε1ε2

ε1 + ε2δε1+ε2 +

(−ε2 + ε2

2

) δ2ε2

2ε2

= −1

2δ2ε1 +

ε1

2δ2ε1 + δε1+ε2 − 2

ε1ε2

ε1 + ε2δε1+ε2 − 1

2δ2ε2 +

ε2

2δ2ε2

−→ 0

as ε1 ↓ 0, ε2 ↓ 0. �

Next we consider the associated operators and semigroups. In particular, wewill show that the critical constant cN =

(N−2

2

)2is also optimal to associate a

semigroup to the operator ∆ + c|x|2 in a reasonable way.

OUTGROWTHS OF HARDY’S INEQUALITY 7

We denote by −Ac the operator associated with ac for c <(N−2

2

)2and with

ac for c =(N−2

2

)2. Then Ac is given by

(2.3) Acu = ∆u+c

|x|2u

on

D (Ac) =

{u ∈ H1

0 : ∆u+c

|x|2u ∈ L2

}in case c <

(N−2

2

)2, and

D (Ac) =

{u ∈ D (acN ) : ∆u+

c

|x|2u ∈ L2

}in the critical case. Here ∆u+ c

|x|2u is understood in the sense of D′

0. This followsimmediately from the definition of the associated operator.

The semigroup etAc may be described as the limit of the semigroups definedby the cut-off potentials. Denote by ∆p the generator of the Gaussian semigroupon Lp := Lp

(RN)for N ≥ 2 and the Laplacian with Dirichlet boundary conditions

on Lp := Lp (0,∞) if N = 1, 1 ≤ p < ∞ (cf. [8,Theorem 1.4.1]). Then −∆2 isassociated with a0. Let Ac,k = ∆2 + c

(1x2 ∧ k

)for k ∈ N. Then

(2.4) 0 ≤ etAc,k ≤ etAc,k+1 .

If c ≤(N−2

2

)2, then it follows from the monotone convergence theorem for

forms [18, S14] that

(2.5) limk→∞

etAc,k = etAc strongly in L2.

Proposition 2.5. If c >(N−2

2

)2, then

limk→∞

∥∥etAc,k∥∥L(L2)=∞ for t > 0.

Proof. For a selfadjoint operator A on a Hilbert spaceH one has∥∥etA∥∥L(H)

=

ets(A) where s (A) = sup {(Au|u)H : u ∈ D (A) , ‖u‖H = 1} . This follows from thespectral theorem. In our situation

s (Ac,k) = sup

{−∫|∇u|2 dx+ c

∫|u|2

(1

|x|2∧ k)dx : u ∈ D (∆2) , ‖u‖L2 = 1

}.

It follows from the optimality of the constant cN in Hardy’s inequality thats (Ac,k) −→∞ as k −→∞ if c > cN . �

Another way of looking at semigroups associated with the operator ∆ + c|x|2 is

to consider the minimal operator Ac,min on L2 given by

Ac,min (u) = ∆u+c

|x|2u

D (Ac,min) = D0.

8 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

Theorem 2.6. Let N ≥ 5.

a) Let c ≤(N−2

2

)2. Then Ac generates a positive (C0) semigroup on L2 which

is minimal among all positive (C0) semigroups generated by an extension of Ac,min.

b) If c >(N−2

2

)2, then no extension of Ac,min generates a positive (C0) semi-

group on L2.

For the proof we need the following.

Lemma 2.7. Let N ≥ 5. Then D0 is a core of ∆2 (the Laplacian on L2(RN)).

Proof. It is well-known that D(RN)is a core. Let u ∈ D

(RN). Let η ∈

C1 (R) be such that 0 ≤ η ≤ 1, η (r) = 1 for |r| ≥ 2, η (r) = 0 for r ≤ 1, andlet ηk (x) = η (k |x|) . It is easy to see that ηku −→ u and ∆ (ηku) −→ ∆u inL2(RN). �

Proof of Theorem 2.6. Let B be the generator of a positive (C0) semigroup

etB such that Ac,min ⊂ B. Consider the semigroup Tk (t) = et(B−c

(1|x|2 ∧k

))for

c > 0. Then 0 ≤ Tk+1 (t) ≤ Tk (t) . It follows from [2] or [22] that limk→∞ Tk (t) =T∞ (t) exists strongly and defines a (C0) semigroup whose generator we call B∞.Let v ∈ D0. Then there exists k0 such that Bkv = ∆v for all k ≥ k0. Since

Tk (t) v − v =

∫ t

0

Tk (s)Bv ds,

letting k −→∞ we see that

T∞ (t) v − v =

∫ t

0

T∞ (s)Bv ds.

Thus, v ∈ D (B∞) and B∞v = ∆v. Since N ≥ 5, D0 is dense in H2(RN)

=D (∆2) . Thus ∆2 coincides with B∞ on a core. Consequently, ∆2 = B∞. Thus

et∆2 ≤ et(B−c

(1|x|2 ∧k

))for all k. It follows that etAc,k ≤ etB for all k. Now

Proposition 2.5 implies that c ≤(N−2

2

)2, and from (2.5) it follows that etAc ≤ etB

for t ≥ 0. �

By a result of Kalf, Schmincke, Walter and Wüst [12], D0 is a core of Ac iffc ≤

(N−2

2

)2 − 1; see also [18] for dimension N = 5.

3. The Caffarelli-Kohn-Nirenberg inequality and associated semigroups

Let β ∈ R, L2β := L2

(RN , |x|−β dx

)ifN ≥ 2 and L2

β := L2(

(0,∞) , |x|−β dx)

if N = 1. Note that L2β ⊆ L2

loc

(RN\ {0}

)if N ≥ 2 and L2

β ⊆ L2loc (0,∞) if N = 1.

Thus, L2β ⊂ D

′

0 where D0 = D(RN\ {0}

)if N ≥ 2 and D0 = D (0,∞) if N = 1.

We let H100 :=

{u ∈ H1

(RN)

: supp u is compact and 0 /∈ supp u}if N ≥ 2 and

H100 := {u ∈ H1 (0,∞) : supp u is compact, supp u ⊂ (0,∞)} if N = 1.

Consider the unitary operator U : L2β −→ L2

0, u 7−→ |x|− β2 u. It maps H1

00 ontoH1

00. Consider the Dirichlet form

a (u, v) =

∫∇u · ∇v dx

OUTGROWTHS OF HARDY’S INEQUALITY 9

on L20 with domain H

100. We transport the Dirichlet form by U to the space L2

β bydefining

b (u, v) =

∫∇(|x|−

β2 u)· ∇(|x|−

β2 v)dx

= a (Uu,Uv)

for u, v ∈ H100. Then we have

b(u, v) =

∫∇u · ∇v |x|−β dx(3.1)

−{(

N − 2− β2

)2

−(N − 2

2

)2}∫

uv

|x|2|x|−β dx

for all u, v ∈ H100.

Proof of (3.1). Let u, v ∈ H100. Then

b(u, v)

=

∫∇(|x|−

β2 u)· ∇(|x|−

β2 v)dx

=

∫ (−β

2|x|−( β2 )−1 x

|x|u+ |x|−β2 ∇u

)·(−β

2|x|−( β2 )−1 x

|x|v + |x|−β2 ∇v

)dx

=

∫∇u · ∇v |x|−β dx+

β2

4

∫uv |x|−β−2

dx− β

2

∫|x|−β−2

x · ∇ (uv) dx

=

∫∇u · ∇v |x|−β dx−

{(N − 2− β

2

)2

−(N − 2

2

)2}∫

uv

|x|2|x|−β dx

since by integration by parts,

−β2

∫|x|−β−2

x ·∇ (uv) dx = −β2

(β + 2)

∫|x|−β−2

uv dx+β

2N

∫uv

|x|2|x|−β dx.

�

Now by Hardy’s inequality,∫|∇u|2 |x|−β dx−

(N − 2− β

2

)2 ∫u2

|x|2|x|−β dx

= a(|x|−

β2 u, |x|−

β2 u)−(N − 2

2

)2 ∫ (|x|−

β2 u)2

|x|2dx ≥ 0.

Thus, we have proved the following.

Theorem 3.1. One has

(CKN)(N − 2− β

2

)2 ∫u2

|x|2|x|−β dx ≤

∫|∇u|2 |x|−β dx

for all u, v ∈ H100, with optimal constant.

10 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

This is the Caffarelli-Kohn-Nirenberg inequality [7] which we have deducedfrom Hardy’s inequality by similarity. Note that the case of N = 2 is includedhere.

Now for c ∈ R we consider the form bc on L2β given by

bc (u, v) =

∫∇u · ∇v |x|−β dx− c

∫uv |x|−β dx

with domain D (bc) = H100. Then by Theorem 3.1 the form bc is positive iff c ≤(

N−2−β2

)2

.

Theorem 3.2. Let β ∈ R and let c ≤(N−2−β

2

)2

. Then the form bc is closable

and D0 is a form core. Let −Bc be the operator on L2β associated with bc. Then

D0 ⊂ D (Bc) and

(3.2) Bcu = ∆u− β

|x|∂u

∂r+ c

u

|x|2

for all u ∈ D0. The operator Bc generates a positive (C0) semigroup on L2β which

is submarkovian iff c ≤ 0.

Proof. Letting δ = −(N−2

2

)2+(N−2−β

2

)2

we have, by (3.1) , bδ = b and

more generally,

(3.3) ac−δ (Uu,Uv) = bc (u, v)

for all u, v ∈ H100. Since UD0 = D0, it follows from the results of Section 2 that bc

is closable and D0 is a form core. It follows from (3.3) that

Bc = U−1Ac−δU

and so

etBc = U−1etAc−δU

for t ≥ 0. In particular, etBc ≥ 0.Let u ∈ D0, v = ∆u− β

|x|∂u∂r + c u

|x|2 . Integration by parts shows that

bc (u, ϕ) = −∫vϕ |x|−β dx

for all ϕ ∈ D0. Hence, u ∈ Bc and Bcu = v.Finally we show that etBc is submarkovian iff c ≤ 0. Let u ∈ H1

00. It followsfrom [9, p.152] that (u ∧ 1) sgn u ∈ H1

00 and

Dj ((|u| ∧ 1) sgn u) = Dju 1|{|u|<1}

where sgn u = u|u| 1{u6=0}. Thus

b0 ((1 ∧ |u|) sgn u, (1 ∧ |u|) sgn u)

=

∫|∇u|2 1|{|u|<1} |x|−β dx ≤ b0 (u, u) .

It follows from [17, Theorem 2.6, Theorem 2.14] that(etB0

)t≥0

is submarkovian.

OUTGROWTHS OF HARDY’S INEQUALITY 11

If c > 0, then the semigroup etBc is not submarkovian. In fact, let u ∈ D0

be such that (u− 1)+ 6= 0. Then Dj (u− 1)

+= Dju 1{u>1} and Dj (u ∧ 1) =

Dju · 1{u<1}. Thus

bc

(u ∧ 1, (u− 1)

+)

=

∫∇ (u ∧ 1) · ∇

((u− 1)

+)

−c∫

(u ∧ 1) (u− 1)+

|x|2|x|−β dx

= −c∫

(u− 1)+

|x|2|x|−β dx < 0.

Thus etBc is not submarkovian by [17, Corollary 2.17]. �

Next we want to describe the domain of bc. We denote by

H1β :=

{u ∈ L2

β : Dju ∈ L2β , j = 1, ..., N

}the weighted Sobolev space where Dju is understood in the sense of D

′

0. Then H1β

is a Hilbert space for the scalar product

(u|v)H1β

:= (u|v)L2β+

∫∇u · ∇v |x|−β dx.

D(bc)is the completion of D0 with respect to the norm(

‖u‖2L2β

+ b (u, u)

) 12

= ‖u‖H1β.

Thus, D(bc)is the closure H1

β0 of D0 in H1β . It follows from Fatou’s Lemma

that (CKN) remains true for all u ∈ H1β0. As in the proof of Proposition 2.2, one

deduces that for 0 < c <(N−2−β

2

)2

, one has D(bc)

= H1β0 and

bc (u, v) =

∫∇u · ∇v |x|−β dx− c

∫ |u|2|x|2|x|−β dx.

We summarize these facts as

Proposition 3.3. One has D(b)

= H1β0 and

(CKN)(N − 2− β

2

)2 ∫u2

|x|2|x|−β dx ≤

∫|∇u|2 |x|−β dx

for all u ∈ H1β0. If 0 ≤ c ≤

(N−2−β

2

)2

, then D(bc)

= H1β0.

For N ≥ 3 and β 6= N − 2, we have the following

Proposition 3.4. If N ≥ 3 and β 6= N − 2, then H1β = |x|

β2 H1 = H1

β0.

Proof. Let u ∈ H1, v = |x|β2 u. Then v ∈ L2

β and Djv = β2 |x|

β2−1 xj

|x|u +

|x|β2 Dju in D

′

0. It follows from Hardy’s inequality that |x|β2−1

u ∈ L2β . Hence

Djv ∈ L2β and v ∈ H1

β .

12 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

Conversely, let v ∈H1β , u = |x|−

β2 v. Then u ∈ L2 andDju = −β2 |x|

− β2−1 xj|x|v+

|x|−β2 Djv. It follows from (CKN) that |x|−

β2−1

v ∈ L2. Since v ∈ H1β , one has

|x|−β2 Djv ∈ L2. Thus, Dju ∈ H1.

We have shown that H1β = |x|

β2 H1. The above computation also shows that

‖v‖H1βand

∥∥∥|x|− β2 v∥∥∥H1

are equivalent norms on H1β . Since |x|−

β2 D0 = D0 and

since D0 is dense in H1 it follows that D0 is also dense in H1β . �

Next we consider realizations of the semigroup(etBc

)on Lpβ := Lp

(RN , |x|−β dx

)if N ≥ 2 and Lpβ := Lp

((0,∞) , |x|−β dx

)if N = 1. Here p ∈ [1,∞) will have to

be restricted to a certain interval containing 2.Let us write B = B0. Since

(etB)t≥0

is submarkovian, there exists a consistentfamily (

etB(p))t≥0

of (C0) semigroups on Lpβ , 1 ≤ p < ∞ with B(2) = B. Consider the boundedpotentials

Vk (x) = inf

{1

|x|2, k

}.

Then et(B(p)+cVk) is a (C0) semigroup on Lpβ for all c > 0, 1 ≤ p <∞ . Moreover,

0 ≤ et(B(p)+cVk) ≤ et(B

(p)+cVk+1).

Now let c <(N−2−β

2

)2

. Let 1 ≤ p− (c) < 2 < p+ (c) ≤ ∞ be such that

[p− (c) , p+ (c)] =

{p ∈ [1,∞] : c ≤ 4

pp′

(N − 2− β

2

)2}.

Note that 1 < p− (c) < p+ (c) < ∞ if c > 0, and p− (c) = 1, p+ (c) = ∞ if c ≤ 0.Observe that p ∈ [p− (c) , p+ (c)] iff p′ ∈ [p− (c) , p+ (c)]. By Bc,min we denote theoperator given by D (Bc,min) = D0 ,

Bc,minu = ∆u− β

|x|∂u

∂r+

c

|x|2u

which acts on all Lpβ , 1 ≤ p <∞.

Theorem 3.5. Let p ∈ [p− (c) , p+ (c)], p < ∞. Then Tp,c (t) := strong limk−→∞

et(B(p)+cVk)exists in L

(Lpβ

)and defines a (C0) semigroup Tp,c on L

pβ . Its generator

B(p)c is an extension of Bc,min. In addition B

(2)c = Bc.

It clearly follows from Theorem 3.5 that the semigroup Tp,c are consistent, i.e.,

Tp,c (t) f = Tq,c (t) f

for all t ≥ 0, f ∈ Lpβ ∩Lqβ , p− (c) ≤ p, q ≤ p+ (c) . For the proof of Theorem 3.5 we

will use the following simple argument for increasing semigroups referring to Voigt[22], [23] for further information and developments.

OUTGROWTHS OF HARDY’S INEQUALITY 13

Proposition 3.6. Let Tk be (C0) contraction semigroups on Lp, 1 ≤ p <∞ ,such that 0 ≤ Tk (t) ≤ Tk+1 (t) . Then T (t) f = lim

k−→∞Tk (t) f exists for all f ∈ Lp,

t ≥ 0, and defines a (C0) semigroup T.

Proof. The strong limit exists by the Beppo Levi Theorem. Then T (t) ∈L (Lp) and T (t+ s) = T (t)T (s) for t, s ≥ 0. It remains to prove strong continuity.Let tn ↓ 0, 0 ≤ f ∈ Lp. We have to show that fn := T (tn) f −→ f as n −→ ∞.Let gn := T1 (tn) f. Then 0 ≤ gn ≤ fn and gn −→ f as n −→ ∞. Moreover,‖gn‖Lp ≤ ‖f‖Lp .

a) Let p = 1. Then∫(fn − gn) dx+

∫gn dx =

∫fn dx ≤ ‖f‖L1.

Since (fn − gn)≥ 0 and∫gn dx −→ ‖f‖L1. , it follows that ‖fn − gn‖L1 =

∫(fn − gn)

dx −→ 0 as n −→∞. Since gn −→ f in L1, also fn −→ f in L1.b) Let 1 < p < ∞. It suffi ces to show that each subsequence of (fn) has a

subsequence converging to f in Lp. Since Lp is reflexive, we may assume that fnconverges weakly to a function h ∈ Lp (consider a subsequence otherwise). Sincegn ≤ fn and gn −→ f it follows that f ≤ h. Hence ‖f‖Lp ≤ ‖h‖Lp . Since Lpisuniformly convex, this implies that fn converges strongly to h. It follows that‖h‖Lp ≤ ‖f‖Lp . Since f ≤ h, this implies that f = h. �

Proof of Theorem 3.5. Let p ∈ [p− (c) , p+ (c)], i.e. c ≤ 4pp′

(N−2−β

2

)2

.

We show that B(p) + cVk is dissipative for all k ∈ N. Let u ∈ D(B(p)

). We have

to show that ⟨B(p)u+ cVku, |u|p−1

sgn u⟩≤ 0.

By a result of Liskevich-Semenov [17, Theorem 3.9] one has u |u|p2−1 ∈ D

(b)and⟨

B(p)u, |u|p−1sgn u

⟩≤ − 4

pp′b(u |u|

p2−1

, u |u|p2−1).

Hence by (CKN) in Proposition 3.3,⟨B(p)u, |u|p−1

sgn u⟩≤ − 4

pp′

(N − 2− β

2

)2 ∫ |u|p|x|2|x|−β dx.

Since

c⟨Vku, |u|p−1

sgn u⟩

= c

∫|x|2≥ 1

k

u

|x|2|u|p−1

sgn u |x|−β dx

= c

∫|x|2≥ 1

k

|u|p

|x|2|x|−β dx,

and the claim follows. By Proposition 3.6, there exists a (C0) semigroup Tc,p on

Lpβ such that Tc,p (t) = limk−→∞

et(B(p)+cVk) strongly. Denote the generator of Tc,p

by B(p)c . Let u ∈ D0. Then there exists k0 ∈ N such that u (x) = 0 if |x|2 ≤ 1

k0.

Hence (B(p) + cVk

)u =

(B(p) + c

1

|x|2

)u = Bc,minu =: v

for all k ≥ k0.

14 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

Observe that

et(B(p)+cVk)u− u =

∫ t

0

es(B(p)+cVk)v ds

for all k ≥ k0. Passing to the limit as k −→∞, we obtain

Tc,p (t)u− u =

∫ t

0

Tc,p (s) v ds (t ≥ 0) .

This implies that u ∈ D(B

(p)c

)and B(p)

c u = v. We have shown that Bc,min ⊂ B(p)c .

It follows from the convergence theorem for an increasing sequence of closed positiveforms [18,Theorem S.14, p.373] that B(2)

c = Bc. �

Finally, we describe the behavior of B(p)c by rescaling. For λ > 0, the operator

Uλ given by

(Uλf) (x) = λN−βp f (λx)

is an isometric isomorphism on Lpβ which is unitary if p = 2. Moreover, UλH100 =

H100 and

(3.4) bc (Uλu, Uλv) = λ2bc (u, v)

for all u, v ∈ H100 (as one easily shows). It follows that UλD

(bc)

= D(bc)and that

(3.4) remains valid for all u, v ∈ D(bc). This implies that

(3.5) UλB−1cλ = λ2Bc (λ > 0)

by the definition of the associated operator. This implies that

(3.6) UλetB(2)c U−1

λ = etλ2B(2)

c (t > 0)

for all λ > 0. By consistency it follows that

(3.7) UλetB(p)c U−1

λ = etλ2B(p)

c (t > 0)

for all λ > 0, p ∈ [p− (c) , p+ (c)], p <∞, whenever c <(N−2−β

2

)2

.

From these rescaling properties we deduce the following result on the spectrum.

Proposition 3.7. (a) For p = 2 one has σ (Bc) = (−∞, 0] for all c ≤(N−2−β

2

)2

.

(b) Let c <(N−2−β

2

)2

. Then σ(B

(p)c

)∩R = (−∞, 0], for all p ∈ [p− (c) , p+ (c)],

p <∞.

Proof. (a) It follows from (3.5) that σ (Bc) = λ2σ (Bc) for all λ > 0. Sinceσ (Bc) 6= ∅, the claim follows.

(b) Let p ∈ [p− (c) , p+ (c)], p < ∞. We first show that σ(B

(p)c

)6= ∅.

In fact, if σ(B

(p)c

)= ∅, then it follows from Weis’Theorem [4, Theorem 5.3.6]

that ω(B

(p)c

)= −∞, where ω denotes the growth bound (or type) of the semi-

group generated by B(p)c . But then it follows from the Riesz-Thorin Theorem that

ω(B

(2)c

)= −∞, which is impossible by (a) .

OUTGROWTHS OF HARDY’S INEQUALITY 15

The spectral bound

s(B(p)c

)= sup

{Reµ : µ ∈ σ

(B(p)c

)}

is finite. It follows from (3.7) and (a) that s(B

(p)c

)= 0 and (−∞, 0] ⊂ σ

(B

(p)c

).

�

Remark 3.8. We refer to [21], [20], [16] and [1] for results on p-independenceof the spectra of consistent semigroups. In general the spectrum of the generatorof a symmetric submarkovian semigroup may depend on p ∈ [1,∞). An interestingexample is the Neumann Laplacian on irregular domains (see Kunstmann [14]).

Finally, we show that the constant 4pp′

(N−2−β

2

)2

in Theorem 3.5 is optimal.

For β = 0 this result has been proved by Vogt [21, Example 3.31].

Theorem 3.9. Let 2 ≤ p <∞, 1p + 1

p′ = 1, and let c > 4pp′

(N−2−β

2

)2

. Then

the operator Bc,min is not quasidissipative. Thus no extension of Bc,min generatesa (C0) semigroup T on Lpβ satisfying ‖T (t)‖L(Lpβ) ≤ eωt for t ≥ 0 and for some

ω ∈ R.

Recall that an operator A on a Banach space X is dissipative iff

‖x‖ ≤ ‖x− tAx‖ x ∈ D (A) , t ≥ 0.

An operator A is quasidissipative iff A− ωI is dissipative for some ω ∈ R. Let U :X −→ X be an isometric isomorphism. Then this characterization of dissipativityshows that A is dissipative iff UAU−1 is dissipative. If A is dissipative, then alsoλA is dissipative for all λ > 0, and A− ω is dissipative for all ω ≥ 0. Conversely,let ω ≥ 0 and assume that λA− ω is dissipative for all λ > 0. Then it follows thatA is dissipative.

Proof of Theorem 3.9. Let ω ≥ 0. Assume that Bc,min − ω is dissipativein Lpβ . Let λ > 0, Uλ = const ·Uλ where the constant is chosen in such a way thatUλ is an isometric isomorphism on Lpβ . Then λ

2Bc,min−ω = Uλ (Bc,min − ω) Uλ−1

is dissipative. Since λ > 0 is arbitrary, it follows that Bc,min is dissipative.Thus, we merely have to prove that Bc,min is not dissipative in L

pβ . Let us

assume that Bc,min is dissipative. Let u ∈ D0. It follows that

(3.8) 0 ≥⟨

∆u− β

|x|∂u

∂r+ c

u

|x|2, |u|p−1

sgn u

⟩.

16 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

We compute, since p ≥ 2,⟨∆u, |u|p−1

sgn u⟩

=

∫∆u(|u|p−1

sgn u)|x|−β dx

= −∫|∇u|2 (p− 1) |u|p−2 |x|−β dx

−∫ N∑

j=1

Dju |u|p−1sgn u · (−β) |x|−β−1 xj

|x| dx

= − (p− 1)

∫|∇u|2 |x|p−2 |x|−β dx

+

∫β

|x|∂u

∂rsgn u |u|p−1 |x|−β dx.

Thus it follows from (3.8) that

c

∫ |u|p|x|2|x|−β dx =

⟨cu

|x|2, |u|p−1

sgn u

⟩

≤ (p− 1)

∫|∇u|2 |u|p−2 |x|−β dx.

In order to rewrite the last expression we compute

∇ |u|p2 =

p

2|u|

p2−1∇u u|u| ,

hence ∣∣∣∇ |u| p2 ∣∣∣2 =p2

4|u|p−2 |∇u|2 .

So we obtain

c

∫ |u|p|x|2

|x|−β dx ≤ 4

p2(p− 1)

∫ ∣∣∣∇ |u| p2 ∣∣∣2 |x|−β dx.

Hence

(3.9) c

∫ |u|2|x|2

|x|−β dx ≤ 4

pp′

∫ ∣∣∣∇u p2 ∣∣∣2 |x|−β dx

for u ∈ D0+. By replacing u by λε ∗ u where {λε}ε>0 is a Friedrichs mollifier, wededuce that (3.9) remains true for all u ∈ H1

00∩L∞+ . Here we again use that p ≥ 2.Let r = p

2 . We define

ϕn (s) = s ∧ nsr (s ≥ 0) , un = ϕn ◦ u.

Then u1rn ∈ H1

00. So (3.9) implies

(3.10) c

∫ |un|2|x|2

|x|−β dx ≤ 4

pp′

∫|∇un|2 |x|−β dx.

Letting n −→∞, since |∇un|2 ≤ p2

4 |∇u|2 for all n ∈ N, we deduce that

(3.11) c

∫ |u|2|x|2

|x|−β dx ≤ 4

pp′

∫|∇u|2 |x|−β dx

for all u ∈ H100 ∩ L∞+ . Applying (3.11) to u ∧ n instead of u and letting n −→ ∞,

we see that (3.11) remains true for all u ∈ H100+. Finally, applying (3.11) to |u|

OUTGROWTHS OF HARDY’S INEQUALITY 17

instead of u it follows that (3.11) holds for all u ∈ H100. Now Theorem 3.1 implies

c ≤ 4pp′

(N−2−β

2

)2

. �

We remark that even though for 0 < c <(N−2−β

2

)2

the interval [p− (c) , p+ (c)]

is optimal for obtaining quasicontractive (equivalently, contractive) extrapolatingsemigroups to Lpβ for all p ∈ [p− (c) , p+ (c)] , one might still have extensions whichare not quasicontractive. In fact, for β = 0, it is shown by Vogt [21, Example

3.31] that for p ∈(

NN−2p− (c) , N

N−2p+ (c)), and N ≥ 3, such an extrapolation

semigroup on Lp(RN)exists.

References

[1] W. Arendt, Gaussian estimates and interpolation of the spectrum in Lp, Differential andIntegral Equations, 7 (1994) 1153-1168.

[2] W. Arendt and C. Batty, Abosrption semigroups and Dirichlet boundary conditions, Math.Ann. 295 (1993), 427-448.

[3] W. Arendt and C. Batty, Exponential stability of a diffusion equation with absorption, Dif-ferential and Integral Equations, 6 (1993), 1009-1024.

[4] W. Arendt, C. Batty, M. Hieber, and F. Neubrander, Vector-valued Laplace Transforms andCauchy Problems, Birkhäuser, Basel 2001.

[5] P. Baras and J. Goldstein, The heat equation with a singular potential, Trans. Amer. Math.Soc., 284 (1984), 121-139.

[6] P. Baras and J. Goldstein, Remarks on the inverse square potential in quantum mechanics,North Holland Math. Stud., 92 (1984), 31-35.

[7] L. Caffarelli, R. V. Kohn and L. Nirenberg, First order interpolation inequality with weights,Compositio Math. 53 (1984), 259-275.

[8] E. B. Davies, Heat Kernels and Spectral Theory, Oxford University Press 1989.[9] D. Gilbarg and N. S. Trudinger, Elliptic Partial Diff erential Equations of Second Order,

Springer, Berlin 1998.[10] G. R. Goldstein, J. A. Goldstein and I. Kombe, The power potential and nonexistence of

positive solutions, Proc. 2004 Cortona Conf., to appear.[11] J. A. Goldstein and I. Kombe, Instantaneous blow-up, Contemporary Math 327 (2003),

141-150.[12] H. Kalf, U.-W. Schmincke, J. Walter and R. Wüst, On the spectral theory of Schrödinger and

Dirac operators with strongly singular potentials, Lecture Notes in Math., Vol. 448, Springer,Berlin (1975), 182-226.

[13] H. Kalf and J. Walter, Strongly singular potentials and essential self-adjointness of singularelliptic operators in C∞

0

(RN\ {0}

), J. Functional Analysis 10 (1972), 114-130.

[14] P. C. Kunstmann, Lp-spectral properties of the Neumann Laplacian on horns, comets, andstars, Math. Z. 242 (2002), 183-201.

[15] V. Liskevich and Yu. Semenov, Some inequalities for submarkovian generators and theirapplications to the perturbation theory, Proc. Amer. Math. Soc. 119 (1993), 1171-1177.

[16] V. Liskevich, A. Sobol and H. Vogt, On the Lp−theory of C0−semigroups associated withsecond order elliptic operators, II, J. Functional Analysis 193 (2002), 55-76.

[17] E.M. Ouhabaz, Analysis of Heat Equations on Domains, LMS Monograph Series 31, Prince-ton University Press, Princeton 2004.

[18] M. Reed and B. Simon, Methods of Modern Mathematical Physics Vol. I Functional Analysis,2nd edition, Academic Press, London 1980.

[19] M. Reed and B. Simon, Methods of Modern Mathematical Physics Vol. II Fourier Analysis,Self-Adjointness, Academic Press, London 1975.

[20] Z. Sobol and H. Vogt, On the Lp−theory of C0−semigroups associated with second orderelliptic operators, I, J. Functional Analysis 193 (2002), 24-54.

[21] H. Vogt, Lp−properties of second order elliptic diff erential equations, PhD thesis, Dresden2001.

18 WOLFGANG ARENDT, GISÈLE RUIZ GOLDSTEIN, AND JEROME A. GOLDSTEIN

[22] J. Voigt, Absorption semigroups, their generators and Schrödinger semigroups, J. FunctionalAnalysis 67 (1986), 167-205.

[23] J. Voigt, Absorption semigroups, J. Operator Theory 20 (1988), 117-131.[24] J. Voigt, One-parameter semigroups acting simultaneoously on diff erent Lp−spaces, Bull.

Soc. Roy. Sci. Liège 61 (1992), 465-470.

Department of Mathematics, University of Ulm, Ulm, GermanyE-mail address : arendt@mathematik.uni-ulm.de

Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152E-mail address : ggoldste@memphis.edu

Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152E-mail address : jgoldste@memphis.edu