Plan 2 Outline: Basic definitions and facts Relation between L p , - PowerPoint PPT Presentation

Function spaces L p Updated April 22, 2020

Plan 2 Outline: Basic definitions and facts Relation between L p , interpolation of norms Uniform convexity of L p with 1 ă p ă 8 Projections on closed convex sets Peculiarities of L p -spaces with 0 ď p ă 1 L 2 -space as inner product space

Definition 3 Definition Given measure space p X , F , µ q , for p P p 0, 8q let ż ! ) L p p X , F , µ q : “ | f | p d µ ă 8 f : X Ñ R : F -measurable ^ . For p : “ 0 we set L 0 p X F , µ q : “ � ( f : X Ñ R : F -measurable and for p : “ 8 we let ! f P L 0 : ˘) L 8 p X F , µ q : “ ` D M P r 0, 8q : µ p| f | ą M q “ 0 q Alternative notations: L p , L p p µ q or L p p X q When X is countable and µ is the counting measure: ℓ p p X q

Vector space 4 Lemma For each p P r 0, 8s and each a , b P R , f , g P L p ñ af ` bg P L p . In particular, L p is a vector space over R . Integrability condition: use | f ` g | p ď 2 p | f | p ` 2 p | g | p

L p -norm 5 Lemma For f P L 0 and for p P r 1, 8q let ´ ż ¯ 1 { p | f | p d µ } f } p : “ and for p : “ 8 set � ( } f } 8 : “ inf M P r 0, 8q : µ p| f | ą M q “ 0 Then } ¨ } p is a seminorm on L p such that } f } p “ 0 ñ f “ 0 µ -a.e. In particular, } ¨ } is the norm on tr f s : f P L p u where r f s : “ t g P L p : g “ f µ -a.e. u and }r f s} p : “ } f } p .

Proof 6 Homogeneity checked directly. For triangle inequality, we proceed by showing: Young’s inequality: @ α P r 0, 1 s @ x , y P r 0, 8q : x α y 1 ´ α ď α x ` p 1 ´ α q y older’s inequality: @ f , g P L 0 @ p , q P r 1, 8s : H¨ p ` 1 1 ż q “ 1 ñ | fg | d µ ď } f } p } g } q . Minkowski’s inequality: @ p P r 1, 8q @ f , g P L p : } f ` g } p ď } f } p ` } g } p The case of p “ 8 is checked directly or via limits using @ f P L 0 : p Ñ8 } f } p “ } f } 8 lim

Banach space structure 7 Theorem (Completeness of L p -spaces) For each p P r 1, 8s and any t f n u n ě 1 Ď L p that is Cauchy in the sense N Ñ8 sup lim } f n ´ f m } p “ 0 m , n ě N there exists f P L p such that n Ñ8 } f n ´ f } p “ 0. lim Moreover, we then have } f } p “ lim n Ñ8 } f n } p where the limit on the right exists.

Proof 8 Assume p ă 8 , pick t n k u k ě 1 with sup m , n ě n k } f n ´ f m } p ď 2 ´ k . Define n ÿ ÿ F n : “ | f n k ` 1 ´ f n k | ^ F : “ | f n k ` 1 ´ f n k | k “ 1 k ě 1 Then } F n } p ď 1 and, by Fatou’s lemma, } F } p ď 1. So F ă 8 µ -a.e. and so ÿ ` ˘ f : “ f n 1 ` f n k ` 1 ´ f n k k ě 1 absolutely convergent µ -a.e. Telescoping: f n k Ñ f µ -a.e. and so, by Fatou again, } f n k } p Ñ } f } p and } f n j ´ f n k } p ď 2 ´ k } f ´ f n k } p ď lim sup j Ñ8 This gives f n Ñ f in L p . The case p “ 8 checked directly.

Two corollaries 9 Corollary If f n Ñ f in L p p µ q then there exists a subsequence t n k u k ě 1 such that | f n k | P L p p µ q sup k ě 1 and Ý Ñ f n k k Ñ8 f µ -a.e. Corollary For each p P r 1, 8s , the space tr f s : f P L p u endowed with norm } ¨ } p is a complete normed vector space a.k.a. Banach space.

Relations between L p -spaces 10 Lemma Let L p “ L p p X , F , µ q . (1) If µ is finite then p ÞÑ L p is non-increasing, @ p , r P r 0, 8s : p ă r ñ L r Ď L p (2) If X is countable and µ is the counting measure, then p ÞÑ ℓ p p X q is non-decreasing, @ p , r P r 0, 8s : p ă r ñ ℓ p p X q Ď ℓ r p X q Neither of these hold in general measure spaces but we have @ p , q , r P p 0, 8s : q ă p ă r ñ L p Ď L q ` L r where L q ` L r : “ t f ` g : f P L q ^ g P L r u .

Proof 11 If µ finite, } f } q ď µ p X q 1 { q ´ 1 { p } f } p @ p ą q ą 0: so p ÞÑ L p is non-increasing. For X countable and µ the counting measure, } f } r ď } f } 1 ´ p { r } f } p { r @ r ą p ą 0: 8 p so p ÞÑ ℓ p p X q is non-decreasing. For general case: f “ f 1 t| f |ě 1 u ` f 1 t| f |ď 1 u and use above for the parts.

Interpolation of p -norms 12 Lemma For any q , p , r P p 0, 8s with q ă p ă r we have } f } p ď } f } α q } f } 1 ´ α r where α P p 0, 1 q is the unique number such that 1 p “ α q ` 1 ´ α . r In particular, L q X L r Ď L p . Proof: ż ż | f | α p | f | p 1 ´ α q p d µ ď } f α p } q α p } f p 1 ´ α q p } | f | p d µ “ r p 1 ´ α q p

Uniform convexity 13 Definition A normed linear space V is uniformly convex if for each ǫ ą 0 there is δ ą 0 such that for all x , y P V with } x } “ 1 “ } y } , › x ` y › › } x ´ y } ě ǫ ñ › ď 1 ´ δ › › 2 Says that unit ball is strictly convex and “uniformly curved”. Fails in L 1 and L 8 , but: Theorem (Clarkson 1936) For each p P p 1, 8q , the normed linear space L p is uniformly convex. Proof based on Clarkson’s inequalities. We will follow the proof by O. Hanner (1956)

Hanner’s inequalities 14 Lemma For all p P r 1, 2 s and all f , g P L p , we have ˘ p ` ˇ p } f ` g } p p ` } f ´ g } p ` ˇ ˇ p ě } f } p ` } g } p ˇ } f } p ´ } g } p and ˘ p ` ˇ p } f } p p ` } g } p 2 p ` ˘ ` ˇ ˇ ě } f ` g } p ` } f ´ g } p ˇ } f ` g } p ´ } f ´ g } p p The inequalities are reversed for p P r 2, 8q . Note: 2nd inequality obtained from 1st by subs f Ñ f ` g ^ g Ñ f ´ g Generalize parallogram law for p “ 2

Proof: main steps 15 Suppose p P r 1, 2 s (for p P r 2, 8q all inequalities reversed). Key identity: @ t P r 0, 1 s @ u , v P R : | u ` v | p ` | u ´ v | p ě a p t q| u | p ` b p t q| v | p where a p t q : “ p 1 ` t q p ´ 1 ` p 1 ´ t q p ´ 1 and p 1 ` t q p ´ 1 ´ p 1 ´ t q p ´ 1 ‰ t 1 ´ p “ b p t q : “ Proved by calculus. Apply to u “ f p x q and v “ g p x q and integrate: } f ` g } p p ` } f ´ g } p p ě a p t q} f } p p ` b p t q} g } p p As a p t q ` b p t q t p “ p 1 ` t q p ` p 1 ´ t q p , setting t : “ } g } p {} f } p we then get the result.

Proof of uniform convexity 16 Suppose p P r 2, 8q . For } f } p “ } g } p “ 1, the 1st inequality gives 1 ´ } f ´ g } p ˙ 1 { p ˆ › f ` g › › p p ď › › 2 p 2 › so claim holds with δ : “ 1 ´ p 1 ´ p ǫ { 2 q p q 1 { p For p P r 1, 2 s assume WLOG } f ` g } p ě } f ´ g } p . The 2nd inequality gives ´ } f ´ g } p 2 p ` 1 ě } f ` g } p ¯ p h } f ` g } p where h p x q : “ p 1 ` x q p ` p 1 ´ x q p . Calculus shows h p x q ě 2 ` c p p q x 2 where c p p q : “ p p p ´ 1 q This gives › f ` g 1 ` c p p q ¯ ´ 1 › › ´ 8 } f ´ g } 2 p ď › › p 2 › and claim follows with δ : “ 1 ´ p 1 ´ 1 8 c p p q ǫ 2 q ´ 1 .

Projections on closed convex sets 17 Proposition (Projection on closed convex subsets) Let p P p 1, 8q and let C Ď L p be non-empty and convex, i.e., @ f , g P C @ α P r 0, 1 s : α f ` p 1 ´ α q g P C and closed (i.e., with all L p -Cauchy sequences in C convergent in C). Let f P L p � C. Then C contains an element that is closest to f, D h P C : } f ´ h } p “ inf g P C } f ´ g } p

Proof of Proposition 18 By shift suppose f “ 0 and C Ď L p closed convex with 0 R C . Let t h n u n ě 1 be s.t. n Ñ8 } h n } p “ inf lim g P C } g } p Convexity of C ensures } h n } p ` } h m } p › h n ` h m › › ě p ě inf g P C } g } p › › 2 2 › So › h n ` h m › › lim sup p “ inf g P C } g } p › › 2 › N Ñ8 m , n ě N For normalized functions h 1 n : “ h n {} h n } p this implies › h 1 n ` h 1 › › m N Ñ8 sup lim p “ 1 › › 2 › m , n ě N Uniform convexity: t h 1 n u n ě 1 Cauchy and thus convergent in L p . Then: t h n u n ě 1 convergent to some h P C , } h } p “ inf g P C } g } p .

L p -spaces with 0 ă p ă 1 19 Lemma (Reverse Minkowski inequality) For each p P p 0, 1 q , we have @ f , g P L p : › › › | f | ` | g | p ě } f } p ` } g } p › The inequality is strict whenever µ p f ¨ g ‰ 0 q ą 0 . Proof: As x ÞÑ x p concave, so for a , b ě 0, t P p 0, 1 q : ta b ¯ p p a ` b q p “ ´ ě t 1 ´ p a p ` p 1 ´ t q 1 ´ t b p t ` p 1 ´ t q 1 ´ t True for t “ 0, 1 by continuity, strict if a , b ą 0 and t P p 0, 1 q . Now set a : “ | f p x q| , b : “ | g p x q| and integrate to get }| f | ` | g |} p p ě t 1 ´ p } f } p p ` p 1 ´ t q 1 ´ p } g } p Finally, take t : “ } f } p {p} f } p ` } g } p q .

L p (still) a complete metric space 20 Lemma For p P p 0, 1 q and f , g P L p p µ q , define ż ̺ p p f , g q : “ } f ´ g } p | f ´ g | p d µ p “ Then ̺ p is a pseudometric with ̺ p p f , g q “ 0 ô f “ g µ -a.e. In particular, ̺ p is a metric on the factor space tr f s : f P L p u and the resulting metric space is complete. Proof: Triangle inequality because p a ` b q p ď a p ` b p @ a , b ě 0: Completness: same argument as for p ě 1.

Convexity curse 21 Ball in L p , 0 ă p ă 1, not convex! Proposition Let p P p 0, 1 q and consider the L p space over p X , F , µ q such that for each δ ą 0 there are disjoint t A n u n ě 1 Ď F such that ď X “ A n ^ sup µ p A n q ă δ n ě 1 n ě 1 Then every open, convex subset of p L p , ̺ p q is either empty or L p itself Convexity = algebra, open = topology so: L p cannot be remetrized to a locally-convex vector space (let alone Banach space)