Cylindrical Symmetry of Extremals of a Hardy-Sobolev Inequality by G. Mancini1 and K. Sandeep2
1
Introduction
In this note we discuss symmetry properties of the extremals of a Hardy-Sobolev inequality introduced by Gabriella Tarantello and Marino Badiale. Let n ≥ 3, 2 ≤ k ≤ n, IRn = Rk × IRn−k , x = (x0 x00 ) ∈ IRk × IRn−k , 0 ≤ σ ≤ 1 and 2n pσ = n−2+2σ . Hardy Sobolev inequality [1]. There exists a constant C > 0 such that the inequality
Z
IRn
pσ
|u| |x0 |σpσ
2 pσ
≤C
Z
|∇u|2
(1.1)
IRn
holds for all u ∈ D 1,2 (IRn ). We will refer to (1.1) as a (HS) inequality. More general inequalities of type (1.1), in case k = n, have been considered by Caffarelli-Kohn-Nirenberg ([3]). We refer to ([5]) for several results and references concerning best constants and corresponding extremal functions. When k = n, (1.1) was thoroughly investigated in [7]. Extremal functions for (1.1) in this case have been computed by Lieb ([8]), when σ < 1 (if σ = 1, (1.1) is the classic Hardy inequality which is known not to have extremal funtions). So, our interest is for the case k < n. It has been shown in [2] that the best constant in (1.1) is achieved if σ < 1: if S = S(σ) :=
sup u∈D 1,2 (IRn ) |∇u|2 =1
R
Z
IRn
pσ
|u| |x0 |σpσ
2 pσ
(1.2)
then S is attained at some u ∈ D 1,2 (IRn ). It satisfies the Euler-Lagrange equation Z
n
IR
pσ |u|pσ −2 u ϕ = S2 0 σp σ |x |
Z
IR
∇u ∇ϕ ∀ϕ ∈ D 1,2 (IRn )
(1.3)
n
1 Dipartimentodi Mathematica, Universita di Roma Tre, Via S.Leonardo Murialdo 1, Roma,Italy. e-mail:
[email protected] 2 TIFR Centre, IISc Campus, P.B. 1234, Bangalore - 560 012, India. e-mail:
[email protected]
1
We are interested in symmetry properties of extremal functions. If k = n, they are radial functions (this is not always the case for more general CKN inequalities, see ([5] and [6]). However, if k < n, it is clear, from the Euler Lagrange equation that they cannot be radial. Still, since the inequality (1.1) is symmetric in the variables x0 and x00 separately, one expects the extremals to be symmetric in these variables. We prove this is infact the case. THEOREM 1.1 Extremals of the Hardy-Sobolev inequality (1.1) are cylindrically symmetric,
i.e, if the supremum in (1.2) is attained at u ∈ D 1,2 (IRn ), then (i) (ii)
u(., x00 ) is a radially symmetric decreasing function in IR k for all x00 ∈ IRn−k . There exists x000 ∈ IRn−k such that u(x0 , . + x000 ) is a radially symmetric decreasing
function in IRn−k for all x0 ∈ IRk , x0 6= 0. Remark. Extremals are clearly smooth away from the subspace x0 = 0. See ([1]) for more subtle regularity results. and [?] as required. Don’t forget to give each section and subsection a unique label (see Sect. 2).
2
An equivalent weighted Hardy-Littlewood-Sobolev inequality
In this section we prove that the (HS) inequality (1.1) is equivalent to a doubly weighted Hardy Littlewood Sobolev inequality (wHLS). Moreover, the corresponding sharp constants are the same and we establish a relation between corresponding extremal functions. We first derive from (1.1) a doubly weighted Hardy-Littlewood Sobolev inequality. We will use the following notations: G = (n−2)w1n |x|n−2 will denote the fundamental solutions of −∆ in IRn , and 2n 1 1 2n qγ : = , ( + = 1) (2.1) ∀γ ∈ [0, 1], pγ : = n − 2 + 2γ n + 2 − 2γ p γ qγ THEOREM 2.1 Let σ, ρ ∈ [0, 1]. Then Z Z
IR
n
Rn
1 1 h(x) f (y) G(x − y) 0 ρ dydx ≤ S(σ) 2 S(ρ) 2 khkqσ kf kqρ 0 σ |x | |y |
for all h ∈ L (IR ), f ∈ Lqρ (IRn ) where S(σ), S(ρ) are as in (1.2). qσ
n
2
(2.2)
Remark. When k = n, more general weighted HLS inequalities are known (see Lieb [8]): if 0 < δ < n, 0 ≤ σ < nq , 0 ≤ ρ < nr , p1 + 1q = 1t + 1r = p1 + 1t + σ+ρ−δ = 1, then n
n
h(x) 1 f (y) dydx ≤ Ckhkp kf kt σ n−δ |x| |x − y| |y|ρ
1 p
+
Z Z n
IR IR
Notice that the constraint
1 t
+
σ+ρ−δ n
(2.3)
= 1, which is implied by the scale invariance of
(2.3), is of course satisfied (when δ = 2) by p = qσ , t = qρ in (2.1). Proof of Theorem 2.1 Since, for γ ∈ [0, 1], Hγ g :=
g(x) , |x0 |γ
2n
is in L n+2 (IRn ) ∀g ∈ C0 (IRn ), by
classic (HLS) inequality (i.e. (2.3) with σ = ρ = 0, δ = 1) G ∗ Hγ g is in D 1,2 (IRn ) and Z
ϕ Hγ g =
Z
∇ϕ ∇(G ∗ Hγ g) ∀ϕ ∈ D 1,2 (IRn ), g ∈ C0 (IRn ), γ ∈ [0, 1]
(2.4)
Thus, taking γ = ρ, g = f ∈ C0 , ϕ = G ∗ Hρ h in (2.4), we have
R
|∇(G ∗ Hρ f )|2 = ≤ kf kqρ
R
Hρ f (G ∗ Hρ f ) ≤ Z
(G ∗ Hρ f )pρ |x0 |ρpρ
!
1 pρ
q
≤ kf kqρ S(ρ)k∇(G ∗ Hρ f )k2
and hence q
(2.5)
k∇(G ∗ Hρ f )k2 ≤ kf kqρ S(ρ) Now, let h, f ∈ C0 (IRn ). By Holder and (HS) inequalities, we see that Z
(G ∗ Hρ f ) Hσ h ≤ khkqσ
Z
(G ∗ Hρ f )pσ |x0 |σpσ
!
1 pσ
≤ khkqσ
q
S(σ)k∇(G ∗ Hρ f )k2
This, jointly with (2.5), gives (2.2) for h, f ∈ C0 (IRn ). A density argument using Fatou’s Lemma gives the result.
COROLLARY 2.1 f → Hρ (G ∗ Hσ f ) is a continuous linear map from Lqσ into Lpρ ,
f → G ∗ Hσ f is a continuous linear map from Lqσ into D 1,2 (IRn ) and −∆(G ∗ Hσ f ) = Hσ f ∀f ∈ Lqσ in the weak sense.
3
Proof of Corollary 2.1. The first statement is just a rewriting of (2.2). In particular, if we 2n choose ρ = 0 we see that fj → f in Lqσ ⇒ G ∗ Hσ fj → G ∗ Hσ f in L n−2 . Also, from (2.5), k∇(G ∗ Hσ f )k2 ≤ c kf kqσ , and hence G ∗ Hσ fj converge to some u ¯ in D 1,2 (IRn ) and also in R R R 2n ¯ ∈ D 1,2 (IRn ). In particular, ϕHσ fj = ∇(G∗Hσ fj )∇ϕ → ∇(G∗ L n−2 . Thus G∗Hσ f = u Hσ f )∇ϕ. By (HS), ϕ ∈ D 1,2 (IRn ) ⇒ Hσ ϕ ∈ Lpσ and hence R R and hence ∇(G ∗ Hσ f )∇ϕ = ϕ Hσ f ∀ϕ ∈ D 1,2 (IRn )
R
ϕ Hσ f j =
R
f j Hσ ϕ →
R
Hσ f ϕ
Now, if we set
Z
ˆ Sˆ = S(σ) := sup kf kqσ =1
n
n
IR ×IR
f (y) f (x) G(x − y) dx dy, 0 σ |x | |y0|σ
(2.6)
( 2.2) yields Sˆ ≤ S. We have in fact equality: ˆ S(σ) = S(σ) and S is achieved at some u iff Sˆ is
THEOREM 2.2 Let 0 ≤ σ < 1 Then upσ −1
achieved at f =
|x0 |σ(pσ −1)
Proof of Theorem 2.2. Since σ < 1, (1.1) possesses an extremal function u. Since |u| is extremal as well, and hence solves (1.3), it turns out to be positive away from {x0 = 0} by the maximum principle, i.e. u does not change sign, and we can assume u ≥ 0. After normalization, we have Z
|u|pσ = 1, |x0 |σpσ
pσ −1
|u|pσ −1 ϕ = S |x0 |σpσ
Z
IRn
Z
∇u ∇ϕ ∀ϕ ∈ D 1,2 (IRn )
(2.7)
IRn
pσ
If f : = |x0u|σ(pσ −1) , then f qσ = |xu0 |σpσ = 1. In particular, by Corollary 2.1 we have that G ∗ Hσ f ∈ D 1,2 (IRn ), and hence, using (2.7), R
S
But Hσ f =
Z
R
u Hσ f = S
|u|pσ −1 |x0 |σpσ
Z
∇(G ∗ Hσ f ) ∇u =
Z
IRn
and hence S =
R
|u|pσ −1 G ∗ Hσ f |x0 |σpσ
R Hσ f G ∗ Hσ f ≤ Sˆ because f qσ = 1.
pσ −1 This proves that S = Sˆ and that f = |x0u|σ(pσ −1) is an extremal function for (2.6) Conversely, let f be an extremal function for the (wHLS) quotient (2.6). Clearly, f cannot change sign, so we can assume f ≥ 0. We have:
kf kqσ = 1 and
Z
n
IR ×IR
n
f (x) g(y) G(x − y) = Sˆ 0 σ |x | |y0|σ 4
Z
f qσ −1 g ∀g ∈ Lqσ
Let u : = f qσ −1 |x0 |σ so that rewrites Z
IRn
upσ −1 |x0 |σ(pσ −1)
= f ∈ Lqσ . The Euler-Lagrange equation for f
upσ −1 g(x) G(x − y) = Sˆ |x0 |σ |y0|σpσ
Z
u g ∀g ∈ Lqσ |x0 |σ
ˆ = G ∗ Hσ f and hence, by Corolary 2.1, u ∈ D 1,2 (IRn ) In particular, we find that Su R σ upσ −1 ˆ ˆ R |∇u|2 = R u0 pσp and −S∆u = Hσ f = |x = f qσ = 1 i.e. u is an extremal 0 |σpσ . Thus S |x | σ function for (HS).
3
Rearrangement inequalities and Proof of Theorem 1.1.
The main tool we use to prove Theorem 1.1 is a weighted version of the Riesz’s rearrangement inequality. First we recall the usual Riesz’s rearrangement inequality. For a proof and more details we refer to [9]. Definition. f : IRn → IR is said to be vanishing at ∞ if it satisfies |{|f | > t}| < ∞ for all t > 0, where |.| denotes the Lebesgue measure. For an f which vanishes at infinity we denote by f ∗ its symmetric decreasing rearrangement. Lemma 3.1 (Riesz’s rearrangement inequality). Let f, g, h be three non negative functions on IRn , vanishing at ∞, thus Z Z n
IR IR
n
f (x)g(x − y)h(y)dxdy ≤
Z Z n
IR IR
f ∗ (x)g ∗ (x − y)h∗ (y)dxdy
(3.1)
n
Moreover if g is strictly symmetric decreasing, then equality in the above inequality holds if and only if f (x) = f ∗ (x − x0 ) and h(x) = h∗ (x − x0 ) for some common x0 ∈ IRn . For our purpose we need a weighted form of the above rearrangement inequality. Lemma 3.2. Let f, g, h be three nonnegative functions on IRn , vanishing at ∞ and a ≥ 0, then we have Z Z Z Z h(y) h∗ (y) f (x) f ∗ (x) ∗ g(x − y) dxdy ≤ g (x − y) dxdy (3.2) |x|a |y|a |x|a |y|a n n n n IR IR
IR IR
5
Moreover if g is strictly symmetric decreasing, then equality holds if and only if f = f ∗ and h = h∗ . 1
Proof: For s > 0, let us denote by Bs the ball in IRn with centre at zero and radius ( 1s ) a . With this notation we have Z∞ 1 = χBs (x) ds |x|a 0
Using the above representation of R
R
IRn IRn
f (x) g(x |x|a
− y) h(y) dxdy = |y|a ≤ ≤ =
1 , |x|a
Fubini-Tonelli and Lemma 3.1 we get
R∞ R∞ R
(
R
0 0 IRn IRn R∞ R∞ R R
(
0 0 IRn IRn R∞ R∞ R R
(
n
n
(f χBs )(x)g(x − y)(hχBt )(y)dydx)dsdt (f χBs )∗ (x)g ∗ (x − y)(hχBt )∗ (y)dydx)dsdt (f ∗ χBs )(x)g ∗ (x − y)(h∗ χBt )(y))dsdt
0 0 IR IR R R f ∗ (x) ∗ g (x |x|a IRn IRn
Here we used the fact that (f χBs )∗ ≤ f ∗ χBs ,
∗
− y) h|y|(y) a dxdy (3.3)
∀s > 0, and this follow since:
{(f χBs )∗ > t} ⊆ {f ∗ χBs > t} for all t ({(f χBs )∗ > t} = {f χBs > t}∗ = ({f > t} ∩ Bs )∗ ⊆ {f > t}∗ ∩ Bs = {f ∗ > t} ∩ Bs = {f ∗ χBs > t}) and hence ∗
(f χBs ) (x) =
Z∞
χ{(f χBs )∗ >t} (x)dt ≤
Z∞
χ{f ∗ χBs >t} (x)dt = (f ∗ χBs )(x)
0
0
This proves our weighted rearrangement inequality. Converse. Let g be strictly symmetric decreasing. Assume that we have equality in (3.2), for an f and h. We want to show that f = f ∗ and h = h∗ . It is clear from (3.3) that equality implies for almost every s > 0, ∞ R 0
R
R
(f χBs )(x)g(x − y)(hχBt )(y)dydx dt
IRn IRn R∞ R R
= (
!
0 IRn IRn
(f χBs )∗ (x)g ∗ (x − y)(hχBt )∗ (y)dydx)dt.
6
Let us fix an s for which the above equality holds say s0 . Then again we have: R
R
n IRn IR R
=
(f χBs0 )(x)g(x − y)(hχBt )(y)dydx R
IRn IRn
(3.4)
(f χBs0 )∗ (x)g ∗ (x − y)(hχBt )∗ (y)dydx
for almost all t > 0. Therefore from Lemma 3.1 we get: There exists x0 ∈ IRn such that (f χBs0 ) and (hχBt ) are symmetrically decreasing with respect to x0 . We claim that x0 = 0, unless f ≡ 0 or h ≡ 0. Suppose x0 6= 0. Let t > 0 be such that (3.4) holds for this t and
1 1 t
a
< |x0 |. Then
(hχBt ) is symmetric decreasing with respect to x0 and (hχBt ) is zero near x0 ⇒ hχBt ≡ 0 ⇒ h ≡ 0 in Bt . Now varying t we get h ≡ 0 in the ball B(0, |x0 |) := {x : |x| < |x0 |}. Again, since (hχBt ) is symmetrically decreasing with respect to x0 for almost all t > 0 and h ≡ 0 in the ball with center at zero and radius |x0 |, we get h ≡ 0 in IRn . This proves that x0 = 0. ie.(f χBs0 ) and (hχBt ) are radially decreasing functions for a.e. t ∈ (0, ∞) ⇒ h is radially decreasing in IRn and f is radially decreasing in Bs0 . Varying s0 , we get f = f ∗ and h = h∗ . This proves lemma 3.2. Proof of Theorem 1.1 In view of Theorem 2.2 it is enough to prove that the extremals of (2.7) have the required symmetry properties. Let f ∈ Lqσ (IRn ) be such that Z
f qσ = 1 and
IRn
f (y) f (x) G(x − y) dydx = S |x0 |σ |y 0|σ
Z Z
IRn IRn
Let f ∗0 denote the function obtained by taking the k-dim rearrangement of f (., x00 ) for all 0 x00 ∈ IRn−k . i.e f ∗ (., x00 ) = f (., x00 )∗ , where ∗ denotes the rearrangement of functions in IRk . Then, from the properties of symmetrization, Lemma 3.2 and the fact that f is an extremal R 0 for (2.7), we get (f ∗ )qσ = 1 and IRn
S =
R R
Rn
=
RIR
n
f (x) G(x |x0 |σ
R
(
f (y) − y) |y 0 |σ dydx
R R
IRn−k IRn−k IRk IRk
≤ =
R
R
(
R R
f (x0 ,x00 ) G(x0 |x0 |σ 0
f ∗ (x0 ,x00 ) G(x0 |x0 |σ
IRn−k IRn−k IRk IRk R R f ∗0 (x) G(x − |x0 |σ IRn IRn
∗0
0
− y 0 , x00 − y 00 ) f
y) f|y0(y) dydx |σ
≤ S
7
00
− y 0 , x00 − y 00 ) f (y|y0,y|σ ) ) ∗0 (y 0 ,y 00 )
|y 0 |σ
)
Hence we get equality in all the steps and therefore from Lemma 3.2, we obtain u(., x00 ) is symmetrically decreasing for almost all x00 ∈ IRn−k . Now let us prove the symmetry in the x00 variable. Let v be the function obtained by taking the (n − k) dim symmetrization of f (x0 , .) for a.e.x0 ∈ IRk . i.e.v(x0 , .) = (f (x0 , .))∗ , where ∗ denotes the rearrangement in IRn−k . Then we have from Lemma 3.1
≤
k k IR R IRR
IRk IRk
=
R
=
R
IRn IRn
⇒ ⇒
IRn
v qσ = 1 and
R R 1 1 ( f (x0 , x00 )G(x0 − y 0 , x00 − y 00 )f (y 0 , y 00 )) |x0 |σ |y|σ n−k n−k IR R IR R 1 1 ( v(x0 , x00 )G(x0 − y 0 , x00 − y 00 )v(y 0 , y 00 )) 0 σ σ |x | |y| n−k n−k IR IR v(x) v(y) G(x − y) ≤S 0 σ |x | |y 0 |σ
R R
S =
R
we have equality in the second step k for a.e. xR0 ∈ IR we have R 1 R f ((x0 , x00 )G(x0 − y 0 , x00 − y 00 )f (y 0 , y 00 ))dy 0 ( |y 0 |σ
k IR R
IR
k
n−k IRn−k R IR R 1 ( |y 0 |σ IRn−k IRn−k
v(x0 , x00 )G(x0 − y 0 , x00 − y 00 )v(y 0 , y 00 ))dy 0 .
Fix such an x0 ∈ IRk , say x00 . Then from the above equality we have R
R
f (x00 , x00 )G(x00 − y 0 , x00 − y 00 )u(y 0 , y 00 )dx00 dy 00
IRn−kRIRn−kR
=
IR
n−k
IR
n−k
v(x00 , x00 )G(x00 − y 0 , x00 − y 00 )v(y 0 , y 00 )dx00 dy 00
holds for a.e y 0 ∈ IRk . Since, G(x0 , .) is strictly symmetric decreasing in IRn−k we get from Lemma 3.1, that f (x00 , .) and f (y 0 , .) are symmetrically decreasing with respect to a common point in IRn−k for almost all y 0 ∈ IRk , ie, there exists x000 ∈ IRn−k such that x00 → f (x0 , x00 +x000 ) is symmetrically decreasing in IRn−k . This completes the proof of Theorem 1.1 .
After this work was completed, we got to know a preprint by S. Sechi, D. Smets and M. Willem (0 Remarks on a Sobolev- Hardy inequality0 ) where the Authors prove, among other things, that the infimum S = S(σ) is the same when taken on the class of D 1,2 functions u(y, z) which are radially symmetric and decreasing in y, z separately, thus obtaining existence of cylindrically symmetric minimizers. What we prove in fact is that all minimizers enjoy (up to tranlation in the autonomous variable) these properties. This is a basic information while looking for uniqueness. Our approach should help to work out uniqueness exploiting the presence of 0 competing symmetries 0 This is part of work in progress.
8
Acknowledgement. This work was done while the second author was visiting University of Rome Tre under the MURSAT Young Scientists programme. He wishes to thank the department for the hospitality . The second author also acknowledges the partial support he received from the Kanwal Rekhi scholarship of the TIFR Endowment fund ,during the course of this work.
References [1] Badiale M. and E. Serra ,Critical nonlinear elliptic equations with singularities and cylindrical symmetry, preprint SISSA, 2002 [2] Marino Badiale, and Gabriella Tarantello ,A Sobolev-Hardy inequality with applications to a nonlinear elliptic equation arising in astrophysics, Arch. Rational Mech. Anal. 163 (2002), 259-293. [3] Caffarelli L., R. Kohn and L. Nirenberg , First order interpolation inequalities with weights, Compositio Math. 53 (1984), 259-275 [4] Caldiroli P., R. Musina, On the existence of extremal functions for a weighted Sobolev embedding with critical exponent, Calc. Var. 8 (1999) , 365-387 [5] Catrina F., Z.Q. Wang, On the Caffarelli-Kohn-Nirenberg inequalites: sharp constants, existence (and nonexistence) and symmetry of extremal functions, C.P.A.M., 54 (2001) no. 2, 229-258 [6] Felli V., M. Schneider, Perturbation results of critical elliptic equations of CaffarelliKohn-Nirenberg type, preprint SISSA, 2002 [7] Glaser, V., A. Martin, H. Grosse, W. Thirring, A family of optimal conditions for absence of bound states in a potential, in Studies in Mathematical Physics, E.H. Lieb, B. Simon, A.S. Wightman , eds., Princeton University Press, (1976) , 169-194 [8] Elliott H Lieb, Sharp constants in the Hardy-Littlewood-Sobolev and related inequalities, Ann. of Math. 118 (1983), 349-374. [9] Elliott H Lieb, and Michael Loss Analysis, Graduate Studies in Mathematics ,14 American Mathematical Society, Providence, RI, 1997.
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