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

Continuous linear operators on Lp

Updated May 6, 2020

Plan

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Outline: Continuous linear operators on normed spaces Explicit examples: shift on Rd Fourier transform on Rd Convolution Some inequalities

Definition

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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 : Tpax ` byq “ aTx ` bTy Moreover, T is said to be (1) continuous if xn Ñ x in X implies Txn Ñ Tx in Y. (2) bounded if Dc P r0, 8q @x P X : }Tx}Y ď c}x}X .

Operator norm

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Lemma A linear map T between normed vector spaces X and Y is continuous if and only if it is bounded. Lemma (Operator norm) LpX , Yq is a linear space over R with addition and scalar multiplication defined by paT1 ` bT2qx “ aT1x ` bT2x. Setting }T} :“ sup

xPX t0u

}Tx}Y }x}X defines a norm on LpX , Yq.

Example 1: Shift on Rd

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Lemma (Shift on Rd) For each a P Rd and each f : Rd Ñ R let Taf be defined by pTafqpxq :“ fpx ` aq For each p P r1, 8s, the map Ta is a continuous linear operator imaging Lp onto Lp. Moreover, Ta is an isometry and }Ta}LpÑLp “ 1. Moreover, @f P Lp : a ÞÑ Taf is continuous in norm topology

Proof of Lemma

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Linearity & isometry clear (measure preserving property enough) For continuity of a ÞÑ Taf, need @f P Lp @ǫ ą 0 Dgǫ P CcpRdq: }f ´ gǫ}p ă ǫ. From }Taf ´ Tagǫ}p “ }f ´ gǫ}p we get }Taf ´ f}p ď 2}f ´ gǫ}p ` }Tagǫ ´ gǫ}p. Continuity of gǫ: lim

aÑ0 }Tagǫ ´ gǫ}p “ 0

General a handled by Ta`b “ TaTb.

Example 2: Fourier transform over L1

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Lemma For each C-valued f P L1pRd, LpRdq, λq define p fpkq :“ ż e2πik¨x fpxqdx Then Tf :“ p f defines a continuous linear map L1 Ñ L8 with }T}L1ÑL8 “ 1 p f is called the Fourier transform of f Note: NOT claiming that f ÞÑ p f is an isometry!

Proof of Lemma

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f P L1 + Fubini-Tonelli: p f well defined and }p f}8 ď }f}1 so }T}L1ÑL8 ď 1. Linearity immediate. For fpxq :“ e´π|x|2 we get p fpkq “ ż e2πik¨x´π|x|2dx “ e´π|k|2 and so }p fa}8 “ 1 “ }fa}1.

Note on continuity

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Q: Is the image all of L8? A: Not even close; only continuous functions! Lemma For each f P L1, the function p f is uniformly continuous on Rd Proof: ˇ ˇp fpk ` aq ´p fpkq ˇ ˇ ď ż |e2πix¨a ´ 1||fpxq|dx Dominated Convergence: vanishes as a Ñ 0 uniformly in k.

Fourier inversion

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Lemma (Inverse Fourier transform) For f P L1 denote f _pkq :“ p fp´kq. Then @f P L1 : p f P L1 ñ ´ pp fq_ “ f ^ y pf _q “ f ¯ Proof: Key identity @f, g P L1 : ż p f g dλ “ ż f p g dλ Iterates to @f, g P L1 : p f,p g P L1 ñ ż p p f g dλ “ ż f p p g dλ For special choice of g “ e´aπ|x`z|2 . . .

Proof of Fourier inversion

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. . . we get p p gpyq “ e´aπ|y´z|2. and so f P L1 with p f P L1 implies @a ą 0: ż pp fq_pyqe´aπ|y´z|2dy “ ż fpyqe´aπ|y´z|2dy Suffices to find an Ñ 8 such that for h “ f, pp fq_ lim

nÑ8 ad{2 n

ż ˇ ˇhp¨ ` tq ´ hp¨q ˇ ˇe´anπ|t|2dt “ 0, λ-a.e. For h :“ pp fq_, which is bounded and continuous, this follows by Dominate Convergence. For h :“ f we integrate over λ to get ż }Tt{?an f ´ f}1e´π|t|2dt By continuity of a ÞÑ Taf this tends to zero in L1. Choosing a subsequence gives convergence λ-a.e.

Fourier transform on other Lp’s?

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Q: How does Fourier transform behave on other Lp spaces? Lemma (Parseval-Plancherel identity) We have @f P L1 X L2 : }p f}2 “ }f}2 In particular, @f, g P L1 X L2 : xp f, gy “ xf, g_y M.-A. Parseval (late 18th century)

• M. Plancherel (1910)

Proof of Parseval-Plancherel identity

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Pick ǫ ą 0 and compute: ż |p fpkq|2e´ǫπ|k|2dk “ ¡ fpxqfpyq e2πik¨px´yq´ǫπ|k|2dx dy dk Integrate over k and substitute y “ x ` z to write RHS as ij fpxqfpx ` z?ǫq ǫ´d{2e´π|z|2{ǫ dx dz Since ş ǫ´d{2e´π|z|2{ǫdz “ 1, this can be written as ż |fpxq|2dx ` ij fpxqpfpx ` z?ǫq ´ fpxqq e´πǫ|z|2 dx dz Bound 2nd term using Cauchy-Schwarz by }f}2 ż }Tz?ǫ f ´ f}2 e´π|z|2 dz Now use continuity and boundedness of a ÞÑ }Taf ´ f}2.

Fourier transform as isometric bijection on L2

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Theorem (Plancherel’s theorem) There exists a unique continuous linear map T: L2 Ñ L2 such that @f P L1 X L2 : Tf “ p f. In addition, T is an isometry and }T}L2ÑL2 “ 1. Moreover, T is also surjective and the inverse map T´1 is, for f P L1 X L2 given by T´1f “ f _. In addition, @f, g P L2 : xTf, gy “ xf, T´1gy Note: Not claiming that integral defining p f converges for f P L2! (It does not if f R L1.) Same about f _.

Proof of Parseval’s Theorem

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As L1 X L2 is dense in L2, the map T is densely defined. Boundedness: T extends uniquely to closure of L1 X L2, i.e., to L2. Remains to be an isometry. NTS: T is surjective. First: RanpTq is closed for if tfnuně1 Ď RanpTq is Cauchy then so is tT´1fnuně1. Completeness of L2 then gives T´1fn Ñ h and so fn Ñ Th. Hence RanpTq is closed. Next we prove that RanpTq is all of L2. If not there is f P L2 such that xf, gy “ 0 for all g P RanpTq. Hence, xf,p hy “ 0 for all h P L1 X L2. Then . . .

Proof of Parseval’s Theorem continued ...

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. . . for hpxq :“ e´2πix¨z´π|x|2{a, for which p hpxq “ ad{2e´aπ|x´z|2, @a ą 0: ż fpxq ad{2e´aπ|x´z|2dx “ 0 Change of variables recasts this as @a ą 0: ż fpz ` t{?aq ´ fpzq e´π|t|2dt “ ´fpzq 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

• ver z and use Cauchy-Schwarz to bound the result by

ż }Tt{?a f ´ f}2 e´π|t|2dt So LHS above vanishes in L1 as a Ñ 0. Choosing subsequence, it vanishes λ-a.e. So f “ 0 and RanpTq “ L2. Q: Are there extensions to Lp for p ‰ 1, 2? A: Yes! Needs interpolation.

Example 3: Convolution

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Lemma For any f, g P L1pRd, BpRdq, λq set f ‹ gpxq :“ ż fpx ´ yqgpyq dy Then f ‹ g P L1 and f ‹ g “ g ‹ f. Moreover, for each g P L1, the map Tgf :“ g ‹ f defines a continuous linear operator Tg : L1 Ñ L1 with }Tg}L1ÑL1 ď }g}1 Proof: Tonelli gives ij ˇ ˇfpx ´ yq ˇ ˇ ˇ ˇgpyq ˇ ˇ dxdy “ }f}1 }g}1. This shows existence of f ‹ g. Change of var’s: f ‹ g “ g ‹ f.

Fourier transform: convolution Ø product

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Lemma We have @f, g P L1 : y f ‹ g “ p f p g. Proof: y f ‹ gpkq “ ż ´ ż fpx ´ yqgpyqdy ¯ e2πik¨xdx “ ż ´ ż fpx ´ yqe2πik¨px´yqdx ¯ gpyqe2πik¨ydy “ p fpkqp gpkq Q: Are there other maps that do this?

Banach algebra L1

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Definition A Banach algebra is a complete normed linear space V over R

• r 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, Rn, Cn, CpXq Lemma L1 endowed with the product f ‹ g given by the convolution is a commutative Banach algebra.

Homomorphism on Banach algebra

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Definition (Homomorphism) A map φ: V Ñ R (or C) is a homomorphism if φ is linear, φpax ` byq “ aφpxq ` bφpyq and, multiplication-preserving, φpx ‹ yq “ φpxqφpyq. 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 |φpxq| ą }x}. Set z :“ φpxq´1x. Then }z} ă 1 and }zn} ď }z}n. Define y :“ ř

ně1 zn (converges in

norm). Then y “ z ` z ‹ y and so φpyq “ φpzq ` φpzqφpyq “ 1 ` φpyq a contradiction!

All homomorphisms achieved by Fourier

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Proposition Let φ be a homomorphism on the Banach algebra L1pRd, BpRdq, λq (of either R or C-valued functions). Then Dk P Rd @f P L1 : φpfq “ ż e2πik¨xfpxqdx If L1 is over R, then k :“ 0 so φpfq “ ş fpxqdx.

Proof of Proposition

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Let φ: L1 Ñ C be a homomorphism. Continuous so φ P pL1q‹: Dg P L8 @f P L1 : φpfq “ ż fpxqgpxqdx Applying to convolution f ‹ h gives φpfqφphq “ φpf ‹ hq “ ż hpyq ´ż fpx ´ yqgpxqdx ¯ dy This implies φpfqgpyq “ ż fpx ´ yqgpxqdx, λ-a.e.y P Rd RHS continuous by continuity of the shift in L1 so g admits a continuous version. Plugging for φpfq then shows @y P Rd : gpyq ż fpxqgpxqdx “ ż fpxqgpx ` yqdx and so @x, y P Rd : gpx ` yq “ gpxqgpyq All solutions then of form gpxq “ eit¨x (or g “ 1 over R).

Convolution beyond L1

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Q: Does f ÞÑ f ‹ g extend to f, g P Lp? Proposition (Young’s inequality) Let 1 ď p ď 8. Then @f P L1 and @g P Lp the integral defining f ‹ gpxq converges for λ-a.e. x P Rd. Moreover, f ‹ g P Lp and }f ‹ g}p ď }f}1 }g}p In particular, for each f P L1, the linear operator Tfg :“ f ‹ g maps Lp Ñ Lp continuously with }Tf}LpÑLp ď }f}1

Minkowski’s inequality for integrals

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Lemma Let pX, F, µq and pY, G, νq be σ-finite measure spaces. Then for all p P r1, 8q and all F b G-measurable f : X ˆ Y Ñ r0, 8q, ˆż ´ż fpx, yqµpdxq ¯p νpdyq ˙1{p ď ż ´ż fpx, yqpνpdyq ¯1{p µpdxq Proof: WLOG p ą 1. Write C for RHS of claim. By H¨

• lder

ˇ ˇ ˇ ˇ ż ´ż fpx, yqµpdxq ¯ gpyqνpdyq ˇ ˇ ˇ ˇ ď ż ´ż fpx, yq|gpyq|νpdyq ¯ µpdxq ď ż ´ż fpx, yqpνpdyq ¯1{p }g}Lqpνqµpdxq So hpyq :“ ş fpx, yqµpdxq obeys |φgphq| ď C}g}q. For g :“ |h|p´2h we have φgphq “ }h}p}g}q and so }h}p ď C.

Proof of Young’s inequality

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For p “ 8 checked directly so assume 1 ď p ă 8. Then ˆż ´ż ˇ ˇgpx ´ yq ˇ ˇˇ ˇfpyq ˇ ˇdy ¯p dx ˙1{p ď ż ´ż ˇ ˇgpx ´ yq ˇ ˇpdx ¯1{pˇ ˇfpyq ˇ ˇdy “ }g}p }f}1. So integral defining f ‹ gpxq converges absolutely for λ-a.e. x P Rd and }f ‹ g}p ď }f}1}g}p This now gives }Tf}LpÑLp ď }f}1. Q: What if f, g R L1? A: Will address using interpolation.

Schur’s test

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Convolution is an example of an integral operator Tfpxq :“ ż Kpx, yqfpyqµpdyq where K: X ˆ Y Ñ R is the kernel. Proposition (Schur’s test) Given 1 ď p ď 8 and two σ-finite measure spaces pX, F, µq and pY, G, νq, let K: X ˆ Y Ñ R be F b G-measurable and such that DC P p0, 8q: }Kpx, ¨q}L1pνq ď C for µ-a.e. x P X and Dr C P p0, 8q: }Kp¨, yq}L1pµq ď r C for ν-a.e. y P Y. The integral defining Tfpxq then converges absolutely for µ-a.e. x P X and T maps Lppνq Ñ Lppµq continuously with }T}LppνqÑLppµq ď C1´1{pr C1{p.

Proof of Schur’s test

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Assume first 1 ă p ă 8 and let q obey p´1 ` q´1 “ 1. H¨

• lder:

ż ˇ ˇKpx, yq ˇ ˇ ˇ ˇfpyq ˇ ˇνpdyq ď ´ż ˇ ˇKpx, yq ˇ ˇνpdyq ¯1{q´ż ˇ ˇKpx, yq ˇ ˇ ˇ ˇfpyq ˇ ˇpνpdyq ¯1{p The first term is bounded by C1{q. Raising both sides to power p and integrating with respect to µ then gives ż ´ż ˇ ˇKpx, yq ˇ ˇ ˇ ˇfpyq ˇ ˇνpdyq ¯p µpdxq ď Cp{qr C ż ˇ ˇfpyq ˇ ˇpνpdyq. The integral thus converges and }Tf}p ď C1´1{pr C1{p}f}p. The cases p “ 1 and p “ 8 are checked directly.

Singular integral kernels

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Note: Schur’s test yields operators Lp Ñ Lp which allows iterations etc Permits us to treat kernels with ˇ ˇKpx, yq ˇ ˇ ď c |x ´ y|d´ǫ (1)

• n bounded subsets of Rd or those with

ˇ ˇKpx, yq ˇ ˇ ď c 1 ` |x ´ y|d`ǫ (2)

• n all of Rd. The case ǫ “ 0 open; requires interpolation and
• ther techniques (Calderon-Zygmund).