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# Analyse Master Cours de Francis Clarke

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20 pages
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Niveau: Supérieur, Master
Chapter 6 Lebesgue spaces Analyse Master 1 : Cours de Francis Clarke (2011) The spaces LP(?) play an important role in many applications of functional analy- sis. We focus upon them in this chapter. First, we examine a geometric property of the norm that turns out to have a surprising consequence. 6.1 Uniform convexity 6.1 Definition. A normed space X is uniformly convex if it satisfies the following property: ? ? > 0, ? ? > 0 such that x ? B, y ? B, x? y> ? =? x+ y 2 < 1?? . In geometric terms, this is a way of saying that the unit ball is curved. The property depends upon the choice of the norm on X , even among equivalent norms, as one can see even in R2. 6.2 Exercise. The following three norms on R2 are equivalent: (x,y)1 = |x |+ |y |, (x,y)2 = |(x,y)| = |x |2 + |y |2 1/2, (x,y)∞ = max |x |, |y | . 97

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Sujets

##### Universal quantification

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Chapter 6 Lebesgue spaces
Analyse Master 1 : Cours de Francis Clarke (2011)
The spaces L P ( ) play an important role in many applications of functional analy-sis. We focus upon them in this chapter. First, we examine a geometric property of the norm that turns out to have a surprising consequence.
6.1 Uniform convexity
6.1 Deﬁnition. A normed space X is uniformly convex if it satisﬁes the following property: ε > 0 , δ > 0 such that x B , y B , ￿ x y ￿ > ε = ￿ x 2 + y ￿ < 1 δ . In geometric terms, this is a way of saying that the unit ball is curved. The property depends upon the choice of the norm on X , even among equivalent norms, as one can see even in R 2 . 6.2 Exercise. The following three norms on R 2 are equivalent: ￿ ( x , y ) ￿ 1 = | x | + | y | , ￿ ( x , y ) ￿ 2 = | ( x , y ) | = | x | 2 + | y | 2 1 / 2 , ￿ ( x , y ) ￿ = max | x | , | y | .
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98 Cours de Francis Clarke : Lebesgue spaces Which ones make R 2 a uniformly convex normed space?  Despite the fact that uniform convexity is a norm-dependent property, the very exis-tence of such a norm yields an important property of the underlying space, one that does not depend on the choice of equivalent norm. 6.3 Theorem. (Milman) Any uniformly convex Banach space is reﬂexive. Proof. Let θ X ∗∗ satisfy ￿ θ ￿ ∗∗ = 1, and ﬁx any ε > 0. We shall prove the exis-tence of x B such that ￿ Jx θ ￿ ∗∗ ￿ ε . Since JB is closed in X ∗∗ (see Prop. ?? ), this implies JB = B ∗∗ , and consequently that JX = X ∗∗ , so that X is reﬂexive. Let δ correspond to ε as in the deﬁnition of uniform convexity. We choose ζ X , ￿ ζ ￿ = 1, such that θ , ζ > 1 δ / 2, and we set V = θ X ∗∗ : θ θ , ζ < δ / 2 , which is an open neighborhood of θ in the topology σ X ∗∗ , X . By Goldstine’s lemma (see the proof of Theorem 5.45), V intersects JB : there exists x B such that ζ , x θ , ζ = Jx θ , ζ < δ / 2 . We claim that ￿ Jx θ ￿ ∗∗ ￿ ε . We reason from the absurd, by supposing that θ W , where W is the complement in X ∗∗ of the set Jx + ε B ∗∗ . Since Jx + ε B ∗∗ is closed in σ X ∗∗ , X , W is open in σ X ∗∗ , X . Thus V W is an open neighborhood of θ in this topology. By Goldstine’s lemma, there exists y B such that Jy V W . Thus we have | ζ , y  −  θ , ζ | < δ / 2 by deﬁnition of V . We calculate 1 δ / 2 < θ , ζ = 21   θ , ζ ζ , y   + 12   θ , ζ ζ , x   + 12 ζ , x + y < δ / 4 + δ / 4 + ￿ x + y ￿ / 2 . It follows that ￿ x + y ￿ / 2 > 1 δ , whence ￿ x y ￿ ￿ ε (from uniform convexity). However, Jy W yields ε < ￿ Jy Jx ￿ = ￿ y x ￿ , a contradiction which completes the proof.  The theorem asserts that if a Banach space admits an equivalent norm for which it is uniformly convex, then it is reﬂexive (a property that doesn’t depend on the choice of equivalent norm). There exist reﬂexive spaces, however, which fail to admit an equivalent norm that is uniformly convex; thus the existence of such a norm is not a necessary condition for reﬂexivity. But it is a useful sufﬁcient condition. 6.4 Theorem. If 1 < p < , the Banach space L p ( ) is reﬂexive.
6.1 Uniform convexity 99 Proof. We treat ﬁrst the case 2 ￿ p < . 1 Then, we claim, L p ( ) is uniformly convex, and therefore reﬂexive by Theorem 6.3. The function θ ( t ) = t 2 + 1 p / 2 t p 1 is increasing on [ 0 , ) , which implies, by writing θ ( 0 ) ￿ θ ( α / β ) , the inequality α p + β p ￿ α 2 + β 2 p / 2 α , β 0 . Let a , b R and take α = | a + b | / 2 , β = | a b | / 2; we ﬁnd a + b p a b p ￿ a + 2 b 2 + a 2 b 2 p / 2 = a 2 2 + b 2 2 p / 2 ￿ a 2 p + b 2 p 2 + 2 (we have used the convexity of the function t → | t | p / 2 , which holds because p 2). This yields Clarkson’s inequality : ￿ f 2 + g ￿ Lp p + ￿ f g p p ￿ 12 ￿ f ￿ pL p + ￿ g ￿ pL p f , g L p ( ) . 2 ￿ L Fix ε > 0, and let f , g in the unit ball of L p ( ) satisfy ￿ f g ￿ L p > ε . From Clark-son’s inequality we deduce + p f g < ￿ 2 ￿ L p 1 2 ε p = ￿ f + 2 g ￿ L p < 1 δ , where δ = 1 1 2 ε p 1 / p . This completes the proof of the case p 2. We now prove that L p ( ) is reﬂexive for 1 < p < 2. Let q = p , and consider the operator T : L p ( ) ( L q ( )) deﬁned as follows: for u L p ( ) , the effect of Tu on L q ( ) is given by Tu , g = u ( x ) g ( x ) dx g L q ( ) . Then we have (see Exer. 1.29) ￿ Tu ￿ ( L q ( )) = ￿ u ￿ L p ( ) . Thus T is an isometry between L p ( ) and a closed subspace of ( L q ( )) (since L p ( ) is complete). Now q > 2, so L q ( ) is reﬂexive (by the case of the theorem proved above); thus, its dual ( L q ( )) is reﬂexive (Prop. 5.41). Then T L p ( ) , as a closed subspace, is reﬂexive (Exer. 5.47), and therefore L p ( ) is as well (Exer. 5.40).  6.5 Corollary. The spaces AC p [ a , b ] are reﬂexive for 1 < p < . 1 WefollowBr´ezis,th´eor`emeIV.10.
100 Cours de Francis Clarke : Lebesgue spaces 6.6 Exercise. Let Λ : [ 0 , 1 ] × R × R R be continuous and bounded below, and such that, for almost every t [ 0 , 1 ] , the function ( x , v ) → Λ ( t , x , v ) is convex (see Example 2.30). We suppose in addition that, for certain numbers r > 1 , α > 0 and β , we have Λ ( t , x , v ) α | v | r + β ( t , x , v ) [ 0 , 1 ] × R × R . Fix x 0 , x 1 R , and consider the problem (P) of minimizing 1 f ( x ) = 0 Λ t , x ( t ) , x ( t ) dt over the functions x AC r [ 0 , 1 ] which satisfy x ( 0 ) = x 0 , x ( 1 ) = x 1 . Prove that (P) admits a solution.  6.7 Theorem. (Riesz) For 1 < p < , the dual space of L p ( ) is isometric to L q ( ) , where q is the conjugate exponent of p . More precisely, each ζ of the dual admits a function g L q ( ) (necessarily unique) such that ζ , f = f ( x ) g ( x ) dx f L p ( ) . We then have ￿ ζ ￿ = ￿ g ￿ q . Proof. We deﬁne T : L q ( ) ( L p ( )) via T g , f = f ( x ) g ( x ) dx f L p ( ) . We then have ￿ T g ￿ ( L p ( )) = ￿ g ￿ q (see Exer. 1.29). It sufﬁces to prove that T is surjective. Since T ( L q ( )) is closed (as the image of a Banach space under an isometry), it sufﬁces to prove that T L q ( ) is dense in ( L p ( )) . To prove this, it sufﬁces in turn to prove that θ L p ( ) ∗∗ , θ , T g = 0 g L q ( ) = θ = 0 . We proceed to establish this now. Since L p ( ) is reﬂexive, there exists f L p ( ) such that θ = J f . Then θ , T g = 0 = J f , T g = T g , f = f ( x ) g ( x ) dx g L q ( ) . We discover f = 0, by taking g = f p 2 f , whence θ = 0.  6.8 Exercise. Characterize the dual of AC p [ a , b ] for 1 < p < .  6.9 Theorem. The dual of L 1 ( ) is isometric to L ( ) . More precisely, ζ belongs to L 1 ( ) if and only if there exists z L ( ) (necessarily unique) such that
101
6.1 Uniform convexity ζ , f = ( x ) f L 1 ( ) . z ( x ) dx f When this holds we have ￿ z ￿ L ( ) = ￿ ζ ￿ ( L 1 ( )) . Proof. That any z L ( ) engenders an element ζ in the dual is clear; that any ζ so generated satisﬁes ￿ ζ ￿ ( L 1 ( ) = ￿ z ￿ L ( ) is easy to show. We consider now the converse. By considering B ( 0 , k ) and letting k , the proof can be reduced to the case in which is bounded. Let ζ L 1 ( ) . Any f L 2 ( ) then belongs to L 1 ( ) , by Ho¨ lder’s inequality, and we have: ζ , f ￿ ￿ ζ ￿ ￿ f ￿ L 1 ( ) ￿ ￿ ζ ￿ meas ( ) 1 / 2 ￿ f ￿ L 2 ( ) . It follows that ζ can be viewed as an element of L 2 ( ) . According to Theorem 6.7, there is a (unique) function z in L 2 ( ) such that (using the preceding inequal-ity) ζ , f = z ( x ) f ( x ) dx ￿ ￿ ζ ￿ ￿ f ￿ L 1 ( ) f L 2 ( ) . Thus, for any f L ( ) L 2 ( ) , we have (by rewriting): ￿ ζ ￿ f ( x ) z ( x ) f ( x ) dx 0 . It follows from this (see Theorem 6.20) that, for almost every x , we have: ￿ ζ ￿ f z ( x ) f 0 f R , which implies z ( x ) ￿ ￿ ζ ￿ a.e. Thus z belongs to L ( ) , and satisﬁes ζ , f = z ( x ) f ( x ) dx f L ( ) . Given any f L 1 ( ) , there is a sequence f i L ( ) such that ￿ f f i ￿ L 1 ( ) 0; for instance, let f i ( x ) = f ( x ) if f ( x ) ￿ i , and 0 otherwise. We have, by the above ζ , f i = z ( x ) f i ( x ) dx i 1 . Passing to the limit, we obtain the same conclusion for f ; it follows that z represents ζ on L 1 ( ) , as we wished to show.  It can be shown (for example, by the method of the next exercise) that L 1 ( ) is not reﬂexive; this implies that L ( ) is not reﬂexive either (in view of Theorem 6.9). One can also show that L p ( ) is separable if and only if 1 ￿ p < .
102 Cours de Francis Clarke : Lebesgue spaces 6.10 Exercise. Let X be a Banach space. By deﬁnition, we have, for ζ X , ￿ ζ ￿ = sup   ζ , x : x X , ￿ x ￿ ￿ 1 . Prove that the supremum is attained if X is reﬂexive. Show that the supremum is not attained in general, by taking X = L 1 ( 0 , 1 ) , and by considering the function which equals 1 2 n on the interval 2 n 1 , 2 n , n = 0 , 1 , 2 . . . . Deduce that L 1 ( 0 , 1 ) is not reﬂexive.  Certain useful compactness properties do hold in L 1 ( ) and L ( ) , despite the fact that these spaces fail to be reﬂexive. The next two exercises are cases in point. 6.11 Exercise. Let f n be a bounded sequence in L ( ) . Prove the existence of a subsequence f n i and f L ( ) such that g L 1 ( ) = g ( x ) f n i ( x ) dx g ( x ) f ( x ) dx .  6.12 Exercise. Let k ( ) L 1 ( ) , where is an open subset of R n . We set X = L ( ) equipped with the weak topology σ L ( ) , L 1 ( ) , Y = L 1 ( ) equipped with the weak topology σ L 1 ( ) , L ( ) . a) We deﬁne a linear functional Λ : X Y by Λ g = kg . Prove that Λ is continuous. We now consider the set K = f L 1 ( ) : f ( x ) ￿ k ( x ) , x a.e. . b) Prove that K is weakly compact in L 1 ( ) . c) Prove that K is sequentially weakly compact in L 1 ( ) . For each x , let F ( x ) be a closed convex set in X satisfying F ( x ) ￿ k ( x ) . d) Prove that the set Φ = f L 1 ( ) : f ( x ) F ( x ) , x a.e. is sequentially weakly compact in L 1 ( ) .
6.2 Measurable multifunctions

Recall that a multifunction Γ from R m to R n refers to a mapping from R m to the subsets of R n ; thus, for each x R m , Γ ( x ) is a subset of R n , possibly the empty set. A recurrent issue in future developments will be the possibility of ﬁnding a
6.2 Measurable multifunctions 103 measurable selection of Γ ; that is, a measurable function γ : R m R n such that γ ( x ) belongs to Γ ( x ) for almost all x for which Γ ( x ) = 0/ . A major ingredient in the analysis is the following concept. Measurable multifunctions. The multifunction Γ is called measurable provided that the set Γ 1 ( V ) = x R m : Γ ( x ) V = /0 is (Lebesgue) measurable for every open subset V of R n . It is easy to see that we obtain an equivalent deﬁnition by taking closed sets V in the deﬁnition (or compact sets, or compact convex sets), and that the deﬁnition corresponds to the familiar concept of measurable function when Γ is of the form Γ ( x ) = { γ ( x ) } . The domain dom Γ of Γ is deﬁned as follows: dom Γ = x R m : Γ ( x ) = 0/ . When Γ is measurable, then dom Γ is a measurable set. As in the case of a func-tion, redeﬁning Γ on a set of measure zero does not affect its measurability, so in discussing measurable multifunctions we deal implicitly with equivalence classes, as we do with Lebesgue spaces. 6.13 Exercise. Let u : R m R n and r : R m R + be measurable, and let W be a measurable subset of R n . Prove that the multifunction x → W + B u ( x ) , r ( x ) is measurable.  It is not hard to show that if γ i is a sequence of measurable functions, then the multifunction Γ ( x ) = γ i ( x ) : i 1 is measurable. The following shows that closed-valued measurable multifunctions are essentially generated this way. 6.14 Theorem. Let Γ be closed-valued and measurable. Then there exists a count-able family γ i : dom Γ R n of measurable functions such that Γ ( x ) = cl γ i ( x ) : i 1 , x dom Γ a.e. Proof. Let = dom Γ . We begin by noting that, for any u in R n , the function s d Γ ( s ) ( u ) is measurable on , where d Γ ( s ) is as usual the Euclidean distance function, since we have s : d Γ ( s ) ( u ) < r = s : Γ ( s ) B ( u , r ) = /0 . Now let u i be a countable dense subset of R n , and deﬁne a function f 0 : R n as follows: f 0 ( s ) = the ﬁrst u i such that d Γ ( s ) ( u i ) ￿ 1. Lemma. The functions s f 0 ( s ) and s d Γ ( s ) f 0 ( s ) are measurable.
104 Cours de Francis Clarke : Lebesgue spaces To see this, observe that f 0 assumes countably many values, and that we have, for each i 1: s : f 0 ( s ) = u i = ij = 11 s : d Γ ( s ) ( u j ) > 1 s : d Γ ( s ) ( u i ) ￿ 1 . This implies that f 0 is measurable. To complete the proof of the lemma, we need only note s : d Γ ( s ) f 0 ( s ) > r = j 1 s : f 0 ( s ) = u j s : d Γ ( s ) ( u j ) > r   . We pursue the process begun above by deﬁning for each integer i a function f i + 1 such that f i + 1 ( s ) is the ﬁrst u j for which both the following hold: u j f i ( s ) ￿ 23 d Γ ( s ) f i ( s ) , d Γ ( s ) ( u j ) ￿ 23 d Γ ( s ) f i ( s ) . It follows much as above that each f i is measurable. Furthermore, we deduce the inequalities d Γ ( s ) f i + 1 ( s ) ￿ 32 i d Γ ( s ) f 0 ( s ) ￿ 23 i , together with f i + 1 ( s ) f i ( s ) ￿ 23 i + 1 . It follows that f i ( s ) is a Cauchy se-quence converging for each s to a value which we denote by γ 0 ( s ) , and that γ 0 ( x ) Γ ( x ) for almost every x . We now deﬁne a new multifunction Γ i , j for every pair of positive integers i , j , as follows: /0if x / Γ i , j ( x ) = Γ ( x ) B ( u i , 1 / j ) if x and Γ ( x ) B ( u i , 1 / j ) = 0/ γ 0 ( x ) otherwise . 1 For any closed subset V of R n , the set Γ i , j ( V ) is given by x : Γ ( x ) V B ( u i , 1 / j ) = /0 x : Γ ( x ) B ( u i , 1 / j ) = 0/ γ 0 1 ( V ) . It follows that Γ i , j is measurable, as well as closed-valued and nonempty on . By the argument above (applied to Γ i , j rather than Γ ), there exists a measurable function γ i , j such that γ i , j ( x ) Γ i , j ( x ) , x a.e. We claim that the countable collection γ i , j satisﬁes the conclusion of the theorem. To see this, let S i , j be the null set of points x for which the inclusion γ i , j ( x ) Γ ( x ) fails. Now let x \ [ i , j S i , j ] , and ﬁx any γ Γ ( x ) . There exists a sequence u i k in { u i } and an increasing sequence of integers j k such that | u i k γ | < 1 / j k . Then we have γ i k , j k ( x ) B ( u i k , 1 / j k ) = | γ i k , j k ( x ) u i k | < 1 / j k = | γ i k , j k ( x ) γ | < 2 / j k .