SHARP AFFINE Lp SOBOLEV INEQUALITIES

Erwin Lutwak, Deane Yang and Gaoyong Zhang

0. Introduction In this paper we prove a sharp affine Lp Sobolev inequality for functions on Rn . The new inequality is significantly stronger than (and directly implies) the classical sharp Lp Sobolev inequality of Aubin [A2] and Talenti [T], even though it uses only the vector space structure and standard Lebesgue measure on Rn . For the new inequality, no inner product, norm, or conformal structure is needed at all. In other words, the inequality is invariant under all affine transformations of Rn . That such an inequality exists is surprising because the classical sharp Lp Sobolev inequality relies strongly on the Euclidean geometric structure of Rn , especially on the isoperimetric inequality. Zhang [Z] formulated and proved the sharp affine L1 Sobolev inequality and established its equivalence to an L1 affine isoperimetric inequality that is also proved in [Z]. He also showed that the affine L1 Sobolev inequality is stronger than the classical L1 Sobolev inequality. The L1 Sobolev inequality is known to be equivalent to the isoperimetric inequality (see, for example, [F], [FF], [M], [BZ], [O], and [SY]). The geometry behind the sharp Lp Sobolev inequality is also the isoperimetric inequality. For the affine Sobolev inequalities the situation is quite different. The geometric inequality and the critical tools used to establish the affine L1 Sobolev inequality are not strong enough to enable us to establish the affine Lp Sobolev inequality for p > 1. A new geometric inequality and new tools are needed. The inequality needed is an affine Lp affine isoperimetric inequality recently established by the authors in [LYZ1] (see Campi and Gronchi [CG] for a recent alternate approach). We will also need the solution of an Lp extension of the classical Minkowski problem obtained in [L2]. It is crucial to observe that while the geometric core of the classical Lp Sobolev inequality (i.e., the isoperimetric inequality) is the same for all p, the geometric inequality (i.e., the affine Lp isoperimetric inequality) behind the new affine Lp Sobolev inequality is different for different p. Let Rn denote n–dimensional Euclidean space; throughout we will assume that n ≥ 2. Let H 1,p (Rn ) denote the usual Sobolev space of real-valued functions of Rn with Lp partial derivatives. Research supported, in part, by NSF Grant DMS–0104363 Typeset by AMS-TEX

1

2

ERWIN LUTWAK, DEANE YANG AND GAOYONG ZHANG

The classical sharp Lp Sobolev inequality of Aubin [A2] and Talenti [T] states that if f ∈ H 1,p (Rn ), with real p satisfying 1 < p < n, and if q is given by 1q = p1 − n1 , then µZ ¶ p1 p |∇f | dx ≥ c0 kf kq , (0.1) Rn

where |∇f | is the Euclidean norm of the gradient of f , while kf kq is the usual Lq norm of f in Rn , and ¢1− p1 £ ¤1 1¡ ωn Γ( np )Γ(n + 1 − np )/Γ(n) n , c0 = n p n−p p−1 where ωn is the n-dimensional volume enclosed by the unit sphere S n−1 in Rn . Generalizations and related problems have been much studied (see, e.g., [AL], [BL], [Be], [BP], [BH], [BN], [CL], [CC], [D], [HV], [HS], [LZ], [Lie], [Y], [Z], and the references therein). Since the case p = 1 of the sharp affine Lp Sobolev inequality was settled in [Z], in this paper we will focus exclusively on the case p > 1. The basic concept behind our new inequality is a Banach space that we will associate with each function in H 1,p (Rn ). The critical observation here is that this association is affine in nature. For real p ≥ 1, we associate with each f ∈ H 1,p (Rn ) a Banach norm k · kf,p on Rn . For v ∈ S n−1 define µZ ¶ p1 µZ ¶ p1 p p = |v · ∇f (x)| dx , kvkf,p = kDv f kp = |Dv f (x)| dx Rn

Rn

where Dv f is the directional derivative of f in the direction v. The integral on the right immediately provides the extension of k · kf,p from S n−1 to Rn . Now (Rn , k · kf,p ) is the n-dimensional Banach space that we shall associate with f . Its unit ball Bp (f ) = {v ∈ Rn : kvkf,p ≤ 1} is a symmetric convex body in Rn and our new inequality states that the volume of this unit ball, |Bp (f )|, can be bounded from above by the reciprocal of the ordinary Lq -norm of f . Specifically, we have: Theorem 1. Suppose p ∈ (1, n) and q is given by

1 q

=

1 p

− n1 . If f ∈ H 1,p (Rn ), then

|Bp (f )|1/n ≤ c1 /kf kq ,

(0.2)

where the best possible c1 is given by ¡ p−1 ¢1− p1 ¡ ¢ n1 ¡ √πΓ( n+p ¢ p1 Γ(n) 2 ) c1 = n−p p+1 n Γ( n )Γ(n+1− n ) nΓ( )Γ( ) p

p

2

2

and equality is attained when p

n

f (x) = (a + |A(x − x0 )| p−1 )1− p , with A ∈ GL(n), real a > 0, and x0 ∈ Rn . Since the volume of the symmetric convex body Bp (f ) is obviously given by Z 1 kDv f k−n |Bp (f )| = p dv, n S n−1 we can rewrite our main theorem as the following affine Lp Sobolev inequality:

SHARP AFFINE Lp SOBOLEV INEQUALITIES 1 q

Theorem 10 . Suppose p ∈ (1, n) and q is given by

=

1 p

3

− n1 . If f ∈ H 1,p (Rn ), then

¶−1/n

µZ S n−1

kDv f k−n p dv

≥ c2 kf kq ,

(0.3)

where the best possible c2 is given by ¡ ¢1− p1 ¡ Γ( np )Γ(n+1− np ) ¢ n1 ¡ nΓ( n2 )Γ( p+1 ¢1 2 ) p √ c2 = n−p n+p p−1 Γ(n+1) ) πΓ( 2

and equality is attained when p

n

f (x) = (a + |A(x − x0 )| p−1 )1− p , with A ∈ GL(n), real a > 0, and x0 ∈ Rn . Using the obvious fact that Z Z 1 1 −kDv f kp e dv = kDv f k−n p dv, n! Rn n S n−1 we can in turn rewrite Theorem 10 as: Theorem 100 . Suppose p ∈ (1, n) and q is given by µZ −kDv f kp

e

1 q

=

1 p

− n1 . If f ∈ H 1,p (Rn ), then

¶− n1 dv

Rn

≥ c3 kf kq ,

(0.4)

where the best possible c3 is given by ¢1− p1 ¡ Γ( np )Γ(n+1− np ) ¢ n1 ¡ nΓ( n2 )Γ( p+1 ¢1 ¡ 2 ) p √ , c3 = n−p n+p p−1 Γ(n)Γ(n+1) πΓ( ) 2

and equality is attained when p

n

f (x) = (a + |A(x − x0 )| p−1 )1− p , with A ∈ GL(n), real a > 0, and x0 ∈ Rn . Observe that inequality (0.4), and thus also inequality (0.3), is invariant under affine transformations of Rn , while the Lp Sobolev inequality (0.1) is invariant only under rigid motions. That the affine Lp Sobolev inequality (0.3) or (0.4) is stronger than the classical Lp Sobolev inequality (0.1) follows directly from the H¨older inequality, as will be shown in Section 7. We also note that the affine L2 Sobolev inequality and the classical L2 Sobolev inequality are equivalent under an affine transformation since the L2 Banach norm k · kf,2 is Euclidean. In Section 8, we present an application of the affine Lp Sobolev inequality to information theory. For a random vector X in a finite dimensional Banach space that is associated to a function f , we prove a sharp inequality that gives the best lower bound of the moments of X with respect to the Banach norm in terms of the λ-Ren´ yi entropy of X and the Lq norm of f . Additional applications will be given in a forthcoming paper.

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ERWIN LUTWAK, DEANE YANG AND GAOYONG ZHANG

1. Background For quick reference we list some facts about convex bodies. See [G], [S] and [Th] for additional details. A convex body is a compact convex set in Rn with nonempty interior. In this paper it will always be assumed that a convex body contains the origin in its interior. A convex body K is uniquely determined by its support function h(K, · ) = hK : Rn → (0, ∞), defined for v ∈ Rn by hK (v) = max{v · x : x ∈ K}, where v · x denotes the usual inner product of v and x in Rn . The n-dimensional volume of K will be denoted by V (K) or |K|. For real p ≥ 1, convex bodies K, L and real ε > 0, the Minkowski-Firey Lp combination, K +p ε·L, is the convex body whose support function is given h(K +p ε·L, ·)p = h(K, ·)p + εh(L, · )p . The Lp -mixed volume Vp (K, L) of convex bodies K and L is defined by Vp (K, L) =

V (K +p ε · L) − V (K) p lim+ . n ε→0 ε

Note that Vp (Q, Q) = V (Q) for each convex body Q. It was shown in [L2] that there exists a unique finite positive Borel measure Sp (K, · ) on S n−1 such that Z 1 Vp (K, Q) = hQ (v)p dSp (K, v), (1.1) n S n−1 for each convex body Q. The measure Sp (K, · ) is called the Lp -surface area measure of K. The measure S1 (K, · ) = SK is the classical surface area measure of K. It was shown in [L2] that the measure Sp (K, · ) is absolutely continuous with respect to SK and the Radon-Nikodym derivative dSp (K, · ) = h1−p K . dSK If boundary, ∂K, of K is C 2 with positive curvature, then the Radon-Nikodym derivative of SK with respect to the Lebesgue measure on S n−1 is the reciprocal of the Gauss curvature of ∂K (when viewed as a function of the outer normals of ∂K). A compact domain is the closure of a bounded open set. For compact domains M1 , M2 and real λ1 , λ1 ≥ 0, the Minkowski linear combination λ1 M1 +λ2 M2 is defined by λ1 M1 + λ2 M2 = {λ1 x1 + λ2 x2 : x1 ∈ M1 and x2 ∈ M2 }. The Brunn-Minkowski inequality states that if M1 , M2 are compact domains in Rn and λ1 , λ1 ≥ 0, then V (λ1 M1 + λ2 M2 )1/n ≥ λ1 V (M1 )1/n + λ2 V (M2 )1/n ,

SHARP AFFINE Lp SOBOLEV INEQUALITIES

5

where V denotes n-dimensional volume. If M is a compact domain and K is a convex body in Rn define the mixed volume, V1 (M, K), of M and K by V (M + εK) − V (M ) . ε ε→0+ We shall require the following Minkowski mixed volume inequality for compact domains: If M is a compact domain in Rn and K is a convex body in Rn , then nV1 (M, K) = lim inf

V1 (M, K)n ≥ V (M )n−1 V (K).

(1.2)

Note that (1.2) follows immediately from the definition of mixed volumes and the Brunn-Minkowski inequality: 1 V (M + εK) − V (M ) V1 (M, K) = lim inf + n ε→0 ε 1/n [V (M ) + εV (K)1/n ]n − V (M ) 1 ≥ lim n ε→0+ ε = V (M )

n−1 n

1

V (K) n .

We will also require the following integral representation: Suppose M is a compact domain with C 1 boundary, ∂M . Then, if K is a convex body in Rn , Z 1 V1 (M, K) = hK (ν(x))dSM (x), (1.3) n ∂M where ν(x) denotes the exterior unit normal at x ∈ ∂M , and dSM (x) is the surface area element at x ∈ ∂M . Identity (1.3) can be found, e.g., in [Z]. 2. Affine Lp isoperimetric inequalities We require an Lp -affine isoperimetric inequality that was first proved in [LYZ1] (see Campi and Gronchi [CG] for an alternative proof and generalizations). This inequality is one of the key ingredients in the proof of Theorem 1. Special cases of this new inequality and their relations to other affine isoperimetric inequalities can be found in, e.g., [L1] and [Le]. While the Lp mixed volume Vp ( · , · ) has been defined only for compact convex sets that contain the origin in their interiors, a simple continuity argument allows us to extend the definition of the Lp -mixed volume Vp (K, L) to the case where K is a compact convex set that contains the origin in its interior and L is a compact convex set that contains the origin in its relative interior. For u ∈ S n−1 , let u ¯ denote the line segment connecting the points −u/2 to u/2. Note that from (1.1) we have Z 1 Vp (K, u ¯) = p |v · u|p dSp (K, v), (2.1) 2 n S n−1 for each u ∈ S n−1 . Let c4 be defined by



c4 =

πΓ( n+p 2 ) n Γ( 2 +1)Γ( p+1 2 )

.

The following affine isoperimetric inequality was established in [LYZ1] and will be critical in establishing the affine Lp Sobolev inequality:

6

ERWIN LUTWAK, DEANE YANG AND GAOYONG ZHANG

Theorem 2.1. If real p ≥ 1 and K is a convex body in Rn that contains the origin in its interior, then µZ ¶ np n−p p −n Vp (K, v¯) p dv V (K) n ≤ 2p−1 n1+ n c4 , (2.2) S n−1

with equality if and only if K is an ellipsoid. As an aside, we note that the actual inequality presented in [LYZ1] relates the volume of a convex body to that of its polar Lp projection body. However, the polar coordinate formula for volume quickly shows the equivalence of (2.2) and the polar Lp projection inequality that was established in [LYZ1]. 3. The Lp Minkowski problem We shall construct a family of convex bodies from a given function by using the solution to the even Lp -Minkowski problem. This will allow us to use the affine isoperimetric inequality (2.2) to establish Theorem 1. A Borel measure on S n−1 is said to be even if for each Borel set ω ⊂ S n−1 the measure of ω and −ω = {−x : x ∈ ω} are equal. In [L2], the following solution to the even case of the Lp -Minkowski problem is given: Theorem 3.1. Suppose µ is an even positive measure on S n−1 that is not supported on a great hypersphere of S n−1 . Then for real p ≥ 1 such that p 6= n there exists a unique origin-symmetric convex body K in Rn whose Lp -surface area measure is µ; i.e., µ = Sp (K, · ). We now define for functions (rather than bodies) the notions of Lp mixed volumes. Suppose f ∈ H 1,p (Rn ) ∩ C ∞ (Rn ). For each real t > 0, define the level set, [f ]t = {x ∈ Rn : |f (x)| ≥ t}. Then [f ]t is compact for each t > 0. By Sard’s theorem, for almost all t > 0, the boundary of the level set ∂[f ]t is a C 1 submanifold with everywhere nonzero normal vector ∇f . Abbreviate the surface area element of ∂[f ]t by dSt . If Q is a compact convex set that contains the origin in its relative interior, then define the Lp -mixed volume Vp (f, t, Q), by Z 1 Vp (f, t, Q) = hQ (ν(x))p |∇f (x)|p−1 dSt (x), (3.1) n ∂[f ]t where ν(x) = ∇f (x)/|∇f (x)| is the outer unit normal at x ∈ ∂[f ]t . In particular, when Q is the line segment joining the points −v/2 and v/2, we have Z 1 Vp (f, t, v¯) = p |v · ∇f (x)|p |∇f (x)|−1 dSt (x), (3.2) 2 n ∂[f ]t for each v ∈ S n−1 . The following lemma shows that, for each fixed real p ≥ 1, there is a natural way to associate a family of convex bodies with a given function.

SHARP AFFINE Lp SOBOLEV INEQUALITIES

7

Lemma 3.2. If f ∈ H 1,p (Rn ) ∩ C ∞ (Rn ), then for almost every t > 0, there exists an origin-symmetric convex body Kt whose volume is given by V (Kt ) = Vp (f, t, Kt )

(3.3a)

Vp (f, t, v¯) = Vp (Kt , v¯).

(3.3b)

and such that for all v ∈ S n−1 ,

Proof. Define the positive Borel measure µt on S n−1 by Z Z g(v)dµt (v) = g(ν(x))|∇f (x)|p−1 dSt (x), S n−1

(3.4)

∂[f ]t

for each continuous g : S n−1 → R. Define the even Borel measure µ∗t on S n−1 by letting µ∗t (ω) = 12 µt (ω) + 12 µt (−ω), for each Borel ω ⊂ S n−1 . Obviously, for each continuous even g : S n−1 → R, Z Z ∗ g(v) dµt (v) = g(v) dµt (v). S n−1

(3.5)

S n−1

From (3.5), (3.4), and the fact that [f ]t has non-empty interior, it follows that for each u ∈ S n−1 , Z

Z |u ·

S n−1

v|dµ∗t (v)

Z

= S n−1

|u · ν(x)||∇f (x)|p−1 dSt (x) > 0.

|u · v|dµt (v) = ∂[f ]t

Hence, the measure µ∗t is not supported on any great hypersphere of S n−1 . By Theorem 3.1, there exists a unique origin-symmetric convex body Kt so that dµ∗t = dSp (Kt , · ) = h1−p Kt dSKt .

(3.6)

To see that for each origin-symmetric convex body Q, Vp (Kt , Q) = Vp (f, t, Q),

(3.7)

note that from (1.1) and (3.6), (3.5), (3.4) and definition (3.1), it follows that Z 1 Vp (Kt , Q) = hQ (v)p dµ∗t (v) n S n−1 Z 1 = hQ (v)p dµt (v) n S n−1 Z 1 = hQ (ν(x))p |∇f (x)|p−1 dSt (x) n ∂[f ]t = Vp (f, t, Q). Now (3.7) and a continuity argument immediately yields (3.3b). To get (3.3a) take Q = Kt in (3.7) and recall that Vp (Q, Q) = V (Q) ¤

8

ERWIN LUTWAK, DEANE YANG AND GAOYONG ZHANG

4. An integral inequality The following well-known (see, e.g., [A1]) consequence of Bliss’ inequality [B] will be needed. For the sake of completeness we include an elementary proof that uses techniques similar to ones used in [CNV]. Lemma 4.1. Let f be a nonnegative differentiable function in (0, ∞), q = 1 < p < n. If the integrals exist, then µZ ∞ ¶ p1 µZ ∞ ¶ q1 0 p n−1 q n−1 |f (x)| x dx ≥ c5 f (x) x dx , 0

and

0

where c5 = n

np n−p ,

1 q

1 1− p ( n−p ) p−1

h i n1 1 n n Γ( p )Γ(n + 1 − p )/Γ(n) = (nωn )− n c0 ,

p

n

Equality holds if f (x) = (ax p−1 + b)1− p , with a, b > 0. Proof. It suffices to prove the inequality for a nonnegative compactly supported smooth function f satisfying Z ∞

f (x)q xn−1 dx = 1.

0

Let f0 : (0, ∞) → [0, ∞) be a continuous function that is supported on a bounded interval [0, R) for some R > 0 and that satisfies Z ∞ Z ∞ ∗ q n−1 f0 (x) x dx = f0 (x)q xp +n−1 dx = 1, where p∗ =

p p−1 .

0

0

Define y : [0, ∞) → [0, R] by Z x Z y(x) q n−1 f (s) s ds = f0 (t)q tn−1 dt. 0

0

It follows that

q q 1 y f0 (y)q− n y n−1 y 0 = f (x)q− n xn−1 [( )n−1 y 0 ] n x q 1 y q− n n−1 ≤ f (x) x ((n − 1) + y 0 ) n x q 1 = f (x)q− n (xn−1 y)0 . n Equality in the inequality holds if and only if y = λx, λ > 0. Integration by parts and the H¨older inequality, will give Z ∞ Z ∞ q q q q− n n−1 0 f (x) (x y) dx = −(q − ) f (x)q− n −1 f 0 (x)xn−1 y dx n 0 0 Z ∞ q q f (x)q− n −1 |f 0 (x)|xn−1 y dx ≤ (q − ) n 0 µZ ∞ ¶ p1∗ µZ ∞ ¶ p1 q p∗ q n−1 0 p n−1 y f x dx ≤ (q − ) |f | x dx n 0 0 µZ ∞ ¶ p1 ¶ p1∗ µZ ∞ q q p∗ +n−1 = (q − ) f0 y |f 0 |p xn−1 dx dy . n 0 0

(4.1)

(4.2)

SHARP AFFINE Lp SOBOLEV INEQUALITIES

9

By (4.1) and (4.2), µZ



0 p n−1

|f | x 0

¶ p1 Z n(n − p) ∞ q−q/n n−1 dx ≥ f0 y dy. p(n − 1) 0

(4.3)

Since (4.3) holds for any compactly supported function f0 , it holds for any positive n 1 continuous function. Moreover, equality holds for (4.3) if y = λx and f 0 = βf n−p y p−1 for some constant β. Integrating this gives the extremal function. In particular, the p n desired inequality follows by setting f0 (y) = (ay p−1 + b)1− p , where a and b are chosen so that f0 satisfies the required normalizations. ¤ 5. A lemma about rearrangements For f : Rn → R and real t > 0, let [f ]t = {x ∈ Rn : |f (x)| ≥ t}, denote the level sets of f . We always assume that our functions are such that the level sets [f ]t are compact for all t > 0. The decreasing rearrangement, f¯, of f : Rn → R is defined by f¯(x) = inf{t > 0 : V ([f ]t ) < ωn |x|n }, where ωn |x|n is the n-dimensional volume of the ball of radius |x| in Rn . The set [f¯]t = {x ∈ Rn : f¯(x) ≥ t} is a dilate of the unit ball, B = {x ∈ Rn : |x| ≤ 1}, and its volume is equal to V ([f ]t ); i.e., V ([f¯]t ) = V ([f ]t ). The functions f and f¯ are equimeasurable, and therefore for all q ≥ 1 kf kq = kf¯kq .

(5.1)

Note that since f¯(x) depends only on the magnitude, |x|, of x (and not on the direction of x), there exists an increasing function fˆ : (0, ∞) → R defined by f¯(x) = fˆ(1/|x|).

(5.2)

Observe that provided f is sufficiently smooth, f¯(x) = t implies (by definition of f¯) that V ([f ]t ) = ωn |x|n , or equivalently, fˆ(1/|x|) = t implies V ([f ]t ) = ωn |x|n . The following is needed in the proof of the main theorem:

10

ERWIN LUTWAK, DEANE YANG AND GAOYONG ZHANG

Lemma 5.1. If f ∈ H 1,p (Rn ) ∩ C ∞ (Rn ), then Z



V ([f ]t )

p(n−1) n

0

1−p

(−V ([f ]t ))

dt =

p−n n

ωn

0

and

Z kf¯kqq

Z

n1−p



= nωn



(fˆ0 (s))p s2p−n−1 ds,

(5.3)

0

fˆ(s)q s−n−1 ds.

(5.4)

0

Proof. Let t = fˆ(s). Since fˆ(1/|x|) = t implies V ([f ]t ) = ωn |x|n , V ([f ]t ) = s−n ωn , and hence −V 0 ([f ]t ) = ns−n−1

ds ωn . dt

It follows that µ V ([f ]t )

p(n−1) n

0

(−V ([f ]t ))

1−p

1−p 2p−n−1

dt = n

s

ds dt

¶−p

p 1− n

ds ωn

,

which gives (5.3). To get (5.4) simply rewrite the defining integral for kf¯kqq in polar coordinates. ¤ 6. Affine Lp Sobolev inequalities 1 q

Theorem 10 . Suppose p ∈ (0, 1), and q is given by µZ

=

1 p

− n1 . If f ∈ H 1,p (Rn ), then

¶−1/n

S n−1

kDv f k−n p dv

≥ c2 kf kq ,

(6.1)

where the optimal c2 is given by c2 =

¡ n−p ¢1− p1 ¡ Γ( np )Γ(n+1− np ) ¢ n1 ¡ nΓ( n2 )Γ( p+1 ¢1 2 ) p p−1

Γ(n+1)



πΓ( n+p 2 )

1

1

= (2/nc4 ) p (nωn )− n c0

and equality is attained when p

n

f (x) = (a + |A(x − x0 )| p−1 )1− p , with A ∈ GL(n), real a > 0 and x0 ∈ Rn . Proof. It suffices to prove the inequality for compactly supported f ∈ C ∞ (Rn ). For t > 0, consider the level sets of f , [f ]t = {x ∈ Rn : |f (x)| ≥ t}.

SHARP AFFINE Lp SOBOLEV INEQUALITIES

11

By Sard’s theorem, for almost all t > 0 the boundary, ∂[f ]t , of the level set is a C 1 submanifold which has everywhere nonzero normal vector ∇f . Let dSt denote the surface area element of ∂[f ]t . For t > 0, let Kt be the convex body constructed from f in Lemma 3.2. We first need Z ∞

kDv f kpp = 2p n

Vp (Kt , v¯)dt.

(6.2)

0

To see this simply note that by rewriting the integral, using (3.2), and then using (3.3b) we have Z kDv f kpp

|v · ∇f (x)|p dx n ZR∞ Z = |v · ∇f (x)|p |∇f (x)|−1 dSt (x) dt =

0

Z

= 2p n

∂[f ]t ∞

Vp (f, t, v¯)dt Z

0

p



=2 n

Vp (Kt , v¯)dt. 0

We need the fact that V (Kt )n−p ≥ V ([f ]t )(n−1)p (−n−1 V 0 ([f ]t ))n(1−p) .

(6.3)

To see this, note that from (3.3a), definition (3.1), the H¨older inequality, definition (3.1) again, the extended Minkowski mixed volume inequality (1.2), and the co-area formula, we have V (Kt )

n−p np

1

1

= V (Kt )− n Vp (f, t, Kt ) p à Z ! p1 1 1 = V (Kt )− n hKt (ν(x))p |∇f (x)|p−1 dSt (x) n ∂[f ]t ÃZ ! 1−p Z p 1

1

≥ n− p V (Kt )− n

|∇f |−1 dSt ∂[f ]t

∂[f ]t

ÃZ =n

1 1− p

! 1−p p

1 −n

|∇f |−1 dSt

V (Kt )

V1 ([f ]t , Kt )

∂[f ]t

ÃZ 1

≥ n1− p

! 1−p p |∇f |−1 dSt

V ([f ]t )

∂[f ]t

=n

1 1− p

(−V 0 ([f ]t ))

hKt (ν(x))dSt (x)

1−p p

V ([f ]t )

n−1 n

.

n−1 n

12

ERWIN LUTWAK, DEANE YANG AND GAOYONG ZHANG

To complete the proof, observe that from (6.2), the Minkowski inequality for integrals, the affine inequality (2.2), (6.3), (5.3), (5.4), and (5.1), ÃZ µZ ¶− np µZ ∞ ¶− np !− np kDv f k−n = 2p n Vp (Kt , v¯) dt dv p dv S n−1

S n−1

Z



≥ 2p n 0

2 ≥ c4 np/n

µZ −n p

Z 0 p

=

2n

Vp (Kt , v¯)

S n−1 ∞

V (Kt )

2np−1− n ≥ c4 p −n

0

Z

n−p n

ωn

n−p n



V ([f ]t )

¶− np dv

dt

dt

(n−1)p n

(−V 0 ([f ]t ))1−p dt

0

Z

c4



(fˆ0 (s))p s2p−n−1 ds

0

2 c(fˆ)p kf¯kpq ≥ nc4 2 = c(fˆ)p kf kpq , nc4 where

µZ c(fˆ) =



¶ p1 µZ 0 p 2p−n−1 ˆ (f (s)) s ds

0



¶− q1 q −n−1 ˆ . f (s) s ds

0

Make the substitution t = 1/s and then define the function g by g(t) = fˆ(1/t), to get µZ ∞ ¶ p1 µZ ∞ ¶− q1 0 p n−1 q n−1 c(fˆ) = |g (t)| t dt g(t) t dt , 0

0

(recall that fˆ is increasing and thus g 0 is always negative). Lemma 4.1 gives, c(fˆ) ≥ c5 and this proves the desired inequality. ¤ Remark. The affine Lp -Sobolev inequality (0.4) implies the affine Lp -isoperimetric inequality (2.2). This can be seen by taking ¡ p ¢1− n p f (x) = 1 + ρK (x) 1−p , where ρK (x) = max{λ ≥ 0 : λx ∈ K} denotes the radial function of K. A simple calculation shows that kDv f kpp = c6 Vp (K, v¯), where c6 = n2p

³

n−p p−1

´p−1

Γ( np )Γ(n + 1 − np )/Γ(n).

Therefore, (2.2) is one of the consequences of the new inequality (0.4).

SHARP AFFINE Lp SOBOLEV INEQUALITIES

13

7. The Lp and affine Lp Sobolev inequalities We will show that the new affine Lp Sobolev inequality is indeed stronger (and directly implies) the sharp Lp Sobolev inequality. First observe that µ ¶− np Z Z 1 2 −n kDv f kp dv |∇f (x)|p dx. (7.1) ≤ nωn S n−1 nc4 Rn To see this, note that from the H¨older inequality and Fubini’s theorem we have µ

1 nωn

Z S n−1

kDv f k−n p dv

¶− np ≤ = = = =

Z 1 kDv f kpp dv nωn S n−1 Z Z 1 |v · ∇f (x)|p dxdv nωn S n−1 Rn Z Z 1 |v · ∇f (x)|p dvdx nωn Rn S n−1 Z Z 1 p |u0 · v| dv |∇f (x)|p dx, nωn S n−1 n R Z 2 |∇f (x)|p dx, nc4 Rn

where u0 is any fixed unit vector. Now combine (7.1) with the affine Lp Sobolev inequality (6.1) to get: µZ p

¶ p1

|∇f | dx Rn

µ 1/p

≥ [nc4 /2]

1 nωn

Z S n−1

kDv f k−n p dv

¶− n1 ≥ c0 kf kq .

(7.2)

From the equality conditions in the H¨older inequality it follows that equality in the left inequality in (7.2) occurs precisely when f is such that kDv f kp is independent of v ∈ S n−1 . 8. An application to information theory In this section we use Theorem 1 to prove a moment-entropy inequality for a Banach space–valued random variable X. Let X be a random vector in Rn with probability density g. Given λ > 0, the λ-Renyi entropy power Nλ (X) is defined by ( λ kgkλ1−λ λ 6= 1 Nλ (X) = R e− g log g λ = 1. Observe that N1 (X) is the Shannon entropy power of X, and lim Nλ (X) = N1 (X).

λ→1

14

ERWIN LUTWAK, DEANE YANG AND GAOYONG ZHANG

A random vector X in Rn with density function g is said to have finite rth -moment, r > 0, if Z |x|r g(x)dx < ∞. Rn

If X is a random vector in Rn with finite rth -moment, and K is an origin-symmetric convex body in Rn , then the dual mixed volume V˜−r (X, K) was defined in [LYZ2] by n+r V˜−r (X, K) = n

Z Rn

kxkrK g(x) dx,

(8.1)

where g is the density function of X and k·kK is the norm of the n-dimensional Banach space whose unit ball is K; i.e., for x ∈ Rn kxkK = min{λ > 0 : x ∈ λK}. The following is a special case of the dual Minkowski inequality established in [LYZ2]. n Lemma 8.1. Suppose r > 0 and λ > n+r . If K is an origin-symmetric convex body n n in R and X is a random vector in R with finite rth -moment, then 1 1 V˜−r (X, K) r ≥ c7 [Nλ (X)/|K| ] n ,

where the best constant c7 is given by ³ 1 ´ 1 ³ n ´− r ³ ¡n 1 ¢´− n1 n(1−λ) n(1−λ) λ− n+r n n  1 − rλ   1−λ r B r , 1−λ − r   ¡n ¢− n1 n+r r1 c7 = Γ ( ) + 1 re r   1  ´ n(1−λ) ³ λ− n ´− r1 ³ ³  ¡ n λ ¢´− n1  n n+r 1 + n(λ−1) B rλ λ−1 r r , λ−1

λ < 1, λ = 1, λ > 1.

Given a function f ∈ H 1,p (Rn ), let k · kf,p denote the associated Banach norm defined in the Introduction. If X is a random vector in the Banach space (Rn , k · kf,p ), the rth moment of X is E(kXkrf,p ). The following theorem gives a sharp lower bound of the moment E(kXkrf,p ) in terms of the Renyi entropy power Nλ (X) and the Lq norm kf kq . np Theorem 8.2. Suppose 1 ≤ p < n, q = n−p , r > 0, and λ > n th and X is a random vector in R with finite r -moment, then r

E(kXkrf,p ) ≥ c8 Nλ (X) n kf krq , where the best constant c8 = ncr7 c−r 1 /(n + r).

n n+r .

If f ∈ H 1,p (Rn )

SHARP AFFINE Lp SOBOLEV INEQUALITIES

15

Proof. Let Bp (f ) denote the unit ball associated with the norm k · kf,p . Let g be the density function of X. Note that inequality (0.3) holds when p = 1 (see [Z]), and thus inequality (0.2) holds when p = 1. From (8.1), Lemma 8.1, and (0.2), we have Z r E(kXkf,p ) = kxkrf,p g(x) dx Rn

n ˜ V−r (X, Bp (f )) n+r r n r ≥ c7 [Nλ (X)/|Bp (f )| ] n n+r r n r n n ≥ c7 [Nλ (X)c−n 1 kf kq ] n+r r = c8 [Nλ (X)kf knq ] n . ¤ =

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16 [H]

[HV] [He] [HS]

[Le] [LZ] [Lie] [L1] [L2] [LO] [LYZ1] [LYZ2] [LZ] [M] [O] [P] [S] [SY] [T] [Th] [Y] [Z]

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