Plan 2 Outline: Continuous linear operators on normed spaces - PowerPoint PPT Presentation

Continuous linear operators on L p Updated May 6, 2020

Plan 2 Outline: Continuous linear operators on normed spaces Explicit examples: shift on R d Fourier transform on R d Convolution Some inequalities

Definition 3 Definition (Continuous/bounded linear operators) Let X and Y be normed linear spaces over R . A map T : X Ñ Y is said to be linear if @ a , b P R @ x , y P X : T p ax ` by q “ aTx ` bTy Moreover, T is said to be (1) continuous if x n Ñ x in X implies Tx n Ñ Tx in Y . (2) bounded if D c P r 0, 8q @ x P X : } Tx } Y ď c } x } X .

Operator norm 4 Lemma A linear map T between normed vector spaces X and Y is continuous if and only if it is bounded. Lemma (Operator norm) L p X , Y q is a linear space over R with addition and scalar multiplication defined by p aT 1 ` bT 2 q x “ aT 1 x ` bT 2 x. Setting } Tx } Y } T } : “ sup } x } X x P X � t 0 u defines a norm on L p X , Y q .

Example 1: Shift on R d 5 Lemma (Shift on R d ) For each a P R d and each f : R d Ñ R let T a f be defined by p T a f qp x q : “ f p x ` a q For each p P r 1, 8s , the map T a is a continuous linear operator imaging L p onto L p . Moreover, T a is an isometry and } T a } L p Ñ L p “ 1 . Moreover, @ f P L p : a ÞÑ T a f is continuous in norm topology

Proof of Lemma 6 Linearity & isometry clear (measure preserving property enough) For continuity of a ÞÑ T a f , need @ f P L p @ ǫ ą 0 D g ǫ P C c p R d q : } f ´ g ǫ } p ă ǫ . From } T a f ´ T a g ǫ } p “ } f ´ g ǫ } p we get } T a f ´ f } p ď 2 } f ´ g ǫ } p ` } T a g ǫ ´ g ǫ } p . Continuity of g ǫ : a Ñ 0 } T a g ǫ ´ g ǫ } p “ 0 lim General a handled by T a ` b “ T a T b .

Example 2: Fourier transform over L 1 7 Lemma For each C -valued f P L 1 p R d , L p R d q , λ q define ż p e 2 π i k ¨ x f p x q d x f p k q : “ f defines a continuous linear map L 1 Ñ L 8 with Then Tf : “ p } T } L 1 Ñ L 8 “ 1 p f is called the Fourier transform of f Note: NOT claiming that f ÞÑ p f is an isometry!

Proof of Lemma 8 f P L 1 + Fubini-Tonelli: p f well defined and } p f } 8 ď } f } 1 so } T } L 1 Ñ L 8 ď 1. Linearity immediate. For f p x q : “ e ´ π | x | 2 we get ż p e 2 π i k ¨ x ´ π | x | 2 d x “ e ´ π | k | 2 f p k q “ and so } p f a } 8 “ 1 “ } f a } 1 .

Note on continuity 9 Q: Is the image all of L 8 ? A: Not even close; only continuous functions! Lemma For each f P L 1 , the function p f is uniformly continuous on R d Proof: ż ˇ ˇ | e 2 π i x ¨ a ´ 1 || f p x q| d x ˇp f p k ` a q ´ p ˇ ď f p k q Dominated Convergence: vanishes as a Ñ 0 uniformly in k .

Fourier inversion 10 Lemma (Inverse Fourier transform) For f P L 1 denote f _ p k q : “ p f p´ k q . Then ´ ¯ f P L 1 ñ f q _ “ f ^ y @ f P L 1 : p p p p f _ q “ f Proof: Key identity ż ż @ f , g P L 1 : p f p f g d λ “ g d λ Iterates to ż p ż g P L 1 ñ @ f , g P L 1 : p p f p f , p p f g d λ “ g d λ For special choice of g “ e ´ a π | x ` z | 2 . . .

Proof of Fourier inversion 11 . . . we get p g p y q “ e ´ a π | y ´ z | 2 . p and so f P L 1 with p f P L 1 implies ż ż p p f q _ p y q e ´ a π | y ´ z | 2 d y “ f p y q e ´ a π | y ´ z | 2 d y @ a ą 0: Suffices to find a n Ñ 8 such that for h “ f , p p f q _ ż ˇ ˇ n Ñ8 a d { 2 ˇ h p¨ ` t q ´ h p¨q ˇ e ´ a n π | t | 2 d t “ 0, lim λ -a.e. n For h : “ p p f q _ , which is bounded and continuous, this follows by Dominate Convergence. For h : “ f we integrate over λ to get ż } T t {? a n f ´ f } 1 e ´ π | t | 2 d t By continuity of a ÞÑ T a f this tends to zero in L 1 . Choosing a subsequence gives convergence λ -a.e.

Fourier transform on other L p ’s? 12 Q: How does Fourier transform behave on other L p spaces? Lemma (Parseval-Plancherel identity) We have @ f P L 1 X L 2 : } p f } 2 “ } f } 2 In particular, @ f , g P L 1 X L 2 : x p f , g y “ x f , g _ y M.-A. Parseval (late 18th century) M. Plancherel (1910)

Proof of Parseval-Plancherel identity 13 Pick ǫ ą 0 and compute: ż ¡ | p f p k q| 2 e ´ ǫπ | k | 2 d k “ f p x q f p y q e 2 π i k ¨p x ´ y q´ ǫπ | k | 2 d x d y d k Integrate over k and substitute y “ x ` z to write RHS as ij f p x q f p x ` z ? ǫ q ǫ ´ d { 2 e ´ π | z | 2 { ǫ d x d z ş ǫ ´ d { 2 e ´ π | z | 2 { ǫ d z “ 1, this can be written as Since ż ij f p x qp f p x ` z ? ǫ q ´ f p x qq e ´ πǫ | z | 2 d x d z | f p x q| 2 d x ` Bound 2nd term using Cauchy-Schwarz by ż } T z ? ǫ f ´ f } 2 e ´ π | z | 2 d z } f } 2 Now use continuity and boundedness of a ÞÑ } T a f ´ f } 2 .

Fourier transform as isometric bijection on L 2 14 Theorem (Plancherel’s theorem) There exists a unique continuous linear map T : L 2 Ñ L 2 such that @ f P L 1 X L 2 : Tf “ p f . In addition, T is an isometry and } T } L 2 Ñ L 2 “ 1 . Moreover, T is also surjective and the inverse map T ´ 1 is, for f P L 1 X L 2 given by T ´ 1 f “ f _ . In addition, @ f , g P L 2 : x Tf , g y “ x f , T ´ 1 g y Note: Not claiming that integral defining p f converges for f P L 2 ! (It does not if f R L 1 .) Same about f _ .

Proof of Parseval’s Theorem 15 As L 1 X L 2 is dense in L 2 , the map T is densely defined. Boundedness: T extends uniquely to closure of L 1 X L 2 , i.e., to L 2 . Remains to be an isometry. NTS: T is surjective. First: Ran p T q is closed for if t f n u n ě 1 Ď Ran p T q is Cauchy then so is t T ´ 1 f n u n ě 1 . Completeness of L 2 then gives T ´ 1 f n Ñ h and so f n Ñ Th . Hence Ran p T q is closed. Next we prove that Ran p T q is all of L 2 . If not there is f P L 2 such that x f , g y “ 0 for all g P Ran p T q . Hence, x f , p h y “ 0 for all h P L 1 X L 2 . Then . . .

Proof of Parseval’s Theorem continued ... 16 . . . for h p x q : “ e ´ 2 π i x ¨ z ´ π | x | 2 { a , for which p h p x q “ a d { 2 e ´ a π | x ´ z | 2 , ż f p x q a d { 2 e ´ a π | x ´ z | 2 d x “ 0 @ a ą 0: Change of variables recasts this as ż f p z ` t {? a q ´ f p z q e ´ π | t | 2 d t “ ´ f p z q @ a ą 0: We claim that the LHS tends to zero as a Ñ 0 for λ -a.e. z . For this put absolute value and multiply by e ´ π | z | 2 inside, integrate over z and use Cauchy-Schwarz to bound the result by ż } T t {? a f ´ f } 2 e ´ π | t | 2 d t So LHS above vanishes in L 1 as a Ñ 0. Choosing subsequence, it vanishes λ -a.e. So f “ 0 and Ran p T q “ L 2 . Q: Are there extensions to L p for p ‰ 1, 2? A: Yes! Needs interpolation.

Example 3: Convolution 17 Lemma For any f , g P L 1 p R d , B p R d q , λ q set ż f ‹ g p x q : “ f p x ´ y q g p y q d y Then f ‹ g P L 1 and f ‹ g “ g ‹ f. Moreover, for each g P L 1 , the map T g f : “ g ‹ f defines a continuous linear operator T g : L 1 Ñ L 1 with } T g } L 1 Ñ L 1 ď } g } 1 Proof: Tonelli gives ij ˇ ˇ ˇ ˇ ˇ ˇ f p x ´ y q ˇ g p y q ˇ d x d y “ } f } 1 } g } 1 . This shows existence of f ‹ g . Change of var’s: f ‹ g “ g ‹ f .

Fourier transform: convolution Ø product 18 Lemma We have @ f , g P L 1 : f ‹ g “ p y f p g . Proof: ż ´ ż ¯ y e 2 π i k ¨ x d x f ‹ g p k q “ f p x ´ y q g p y q d y ż ´ ż ¯ f p x ´ y q e 2 π i k ¨p x ´ y q d x g p y q e 2 π i k ¨ y d y “ “ p f p k q p g p k q Q: Are there other maps that do this?

Banach algebra L 1 19 Definition A Banach algebra is a complete normed linear space V over R or C that admits a product ‹ : V ˆ V Ñ V that is associative, distributive around addition, commutes with scalar multiplication and obeys @ x , v P V : } x ‹ y } ď } x }} y } The Banach algebra is commutative if @ x , y P V : x ‹ y “ y ‹ x . Examples: R , C , R n , C n , C p X q Lemma L 1 endowed with the product f ‹ g given by the convolution is a commutative Banach algebra.

Homomorphism on Banach algebra 20 Definition (Homomorphism) A map φ : V Ñ R (or C ) is a homomorphism if φ is linear, φ p ax ` by q “ a φ p x q ` b φ p y q and, multiplication-preserving, φ p x ‹ y q “ φ p x q φ p y q . Lemma Let φ be a homomorphism of Banach algebra V . Then φ is continuous and, in fact } φ } ď 1 . Proof: Suppose x P V obeys | φ p x q| ą } x } . Set z : “ φ p x q ´ 1 x . Then } z } ă 1 and } z n } ď } z } n . Define y : “ ř n ě 1 z n (converges in norm). Then y “ z ` z ‹ y and so φ p y q “ φ p z q ` φ p z q φ p y q “ 1 ` φ p y q a contradiction!

All homomorphisms achieved by Fourier 21 Proposition Let φ be a homomorphism on the Banach algebra L 1 p R d , B p R d q , λ q (of either R or C -valued functions). Then ż D k P R d @ f P L 1 : e 2 π i k ¨ x f p x q d x φ p f q “ ş If L 1 is over R , then k : “ 0 so φ p f q “ f p x q d x.

Proof of Proposition 22 Let φ : L 1 Ñ C be a homomorphism. Continuous so φ P p L 1 q ‹ : ż D g P L 8 @ f P L 1 : φ p f q “ f p x q g p x q d x Applying to convolution f ‹ h gives ż ´ż ¯ φ p f q φ p h q “ φ p f ‹ h q “ h p y q f p x ´ y q g p x q d x d y This implies ż λ -a.e. y P R d φ p f q g p y q “ f p x ´ y q g p x q d x , RHS continuous by continuity of the shift in L 1 so g admits a continuous version. Plugging for φ p f q then shows ż ż @ y P R d : g p y q f p x q g p x q d x “ f p x q g p x ` y q d x and so @ x , y P R d : g p x ` y q “ g p x q g p y q All solutions then of form g p x q “ e i t ¨ x (or g “ 1 over R ).