Ralph Chill Harmonic analysis – in Banach spaces – January 15, 2014 c by R. Chill Contents 1 Bochner-Lebesgue and Bochner-Sobolev spaces ................ 1 1.1 The Bochner integral . .1 1.2 Bochner-Lebesgue spaces . .5 1.3 The convolution . .7 1.4 Bochner-Sobolev spaces . 11 2 The Fourier transform ......................................... 15 2.1 The Fourier transform in L1 ................................ 15 2.2 The Fourier transform on L2 ............................... 21 2.3 The Fourier transform on ................................ 22 S 2.4 The Fourier transform on 0 ............................... 24 2.5 Elliptic and parabolic equationsS in RN ...................... 29 2.6 The Marcinkiewicz multiplier theorem . 29 3 Singular integrals ............................................. 31 3.1 The Marcinkiewicz interpolation theorem . 31 3.2 The Hardy-Littlewood maximal operator . 34 3.3 The Rubio de Francia extrapolation theorem . 44 3.4 Calderon-Zygmund operators . 51 3.5 The Hilbert transform . 58 3.6 Covering and decomposition . 63 References ........................................................ 65 Index ............................................................. 67 v Chapter 1 Bochner-Lebesgue and Bochner-Sobolev spaces 1.1 The Bochner integral Let X and Y be Banach spaces, and let (Ω, ;µ) be a measure space. A A function f : Ω X is called step function, if there exists a sequence (An) ! ⊆ of mutually disjoint measurable sets and a sequence (xn) X such that A P ⊆ f = 1A xn. A function f : Ω X is called mesurable, if there exists a n n ! sequence ( fn) of step functions fn : Ω X such that fn f pointwise µ- almost everywhere. ! ! Remark 1.1. Note that there may be a difference to the definition of mesura- bility of scalar valued functions. Measurability of a function is here depend- ing on the measure µ. However, if the measure space (Ω, ;µ) is complete in the sense that µ(A) = 0 and B A implies B , then theA above definition of measurability and the classical⊆ definition of2 measurability A coincide. Note that one may always consider complete measure spaces. Lemma 1.2. If f : Ω X is measurable, then f : Ω R is measurable. More generally, if f : Ω X! is measurable and if g :kXk Y! is continuous, then g f : Ω Y is measurable.! ! ◦ ! Proof. This is an easy consequence of the definition of measurability and the continuity of g. Note that in particular the norm : X R is continuous. k · k ! Lemma 1.3. If f : Ω X and g : Ω K are measurable, then f g : Ω X is ! ! ! measurable. Similarly, if f : Ω X and g : Ω X0 are measurable, then g; f X0;X : Ω K is measurable. ! ! h i ! Proof. For the proof it suffices to use the definition of measurability and to show that the (duality) product of two step functions is again a step function. This is, however, straightforward. Theorem 1.4 (Pettis). A function f : Ω X is measurable if and only if x0; f is measurable for every x X (we say that! f is weakly measurable) andh if therei 0 2 0 1 2 1 Bochner-Lebesgue and Bochner-Sobolev spaces exists a µ-null set N such that f (Ω N) is separable (we say that f is almost separably valued). 2 A n For the following proof of Pettis’ theorem, see Hille &Phillips [Hille and Phillips (1957)]. Proof. Sufficiency. Assume that f is measurable. Then f is weakly measurable by Lemma 1.2. Moreover, by definition, there exists a sequence ( fn) of test functions and a µ-null set N such that 2 A fn(t) f (t) for all t Ω N: ! 2 n Hence, [ f (Ω N) fn(Ω): n ⊆ n Since for every step function fn the range is countable, the set on the right- hand side of this inclusion is separable, and hence f is almost separably valued. Necessity. Assume that f is weakly measurable and almost separably val- ued. We first show that f is measurable. By assumption, there exists a k k µ-null set and a sequence (xn) in X such that D := xn : n N is dense in f (Ω N). By the Hahn-Banach theorem, there existsf a sequence2 g (x ) in X n n0 ∗ such that xn0 = 1 and xn0 ;xn = xn . Since f is weakly measurable, xn0 ; f is measurablek k for everyhn. Asi a consequence,k k sup x ; f is measurable.jh Butij n jh n0 ij supn xn0 ; f = f on Ω N by the choice of the sequence (xn0 ) and the density of D jhin theij f (Ωk k N). Sincen our measure space (Ω, ;µ) is supposed to be complete, we obtainn that f is measurable. In a similarA way, one shows that k k f x is measurable for every x X, and in particular for x = xn. k −Nowk fix m N and define 2 2 Am1 := f x1 inf f xk ; fk − k ≤ 1 k m k − kg ≤ ≤ Am2 := f x2 inf f xk Am1; fk − k ≤ 1 k m k − kg n ≤ ≤ Am3 := f x3 inf f xk (Am1 Am2); fk − k ≤ 1 k m k − kg n [ ≤ ≤ : : : : m 1 [− Amm := f xm inf f xk ( Amk): fk − k ≤ 1 k m k − kg n ≤ ≤ k=1 Then (Amn)1 n m is a family of measurable, mutually disjoint sets such that Sm ≤ ≤ 1 n=1 Amn = Ω. Define 1 We are grateful to Anton Claußnitzer for the definition of the sets Amn and the functions fm. 1.1 The Bochner integral 3 Xm fm := 1Amn xn: n=1 Then ( fm) is a sequence of step functions, ( fm f )m is decreasing pointwise everywhere, and since D is dense in f (Ω kN),− k n lim fm(t) f (t) = 0 for every t Ω N: m !1 k − k 2 n that is, fm f µ-almost everywhere. As a consequence, f is measurable. ! Corollary 1.5. If ( fn) is a sequence of measurable functions Ω X such that fn f pointwise µ-almost everywhere, then f is measurable. ! ! Proof. We assume that this corollary is known in the scalar case, that is, when X = K. By Pettis’s theorem (Theorem 1.4), for all n there exists a µ-null set Nn 2 A such that fn(Ω Nn) is separable. Moreover there exists a µ-null set N0 Ω n S 2 such that fn(t) f (t) for all t Ω N0. Let N := n 0 Nn; as a countable union of µ-null sets, N!is a µ-null set.2 n ≥ Then f (restricted to Ω N) is the pointwise limit everywhere of the se- n quence ( fn). In particular f is weakly measurable. Moreover, f (Ω N) is separable since n [ f (Ω N) fn(Ω N); n ⊆ n n and since fn(Ω N) is separable. The claim follows from Pettis’ theorem. n R A measurable function f : Ω X is called integrable if f dµ < . ! Ω k k 1 P Lemma 1.6. For every integrable step function f : Ω X, f = 1A xn the series P ! n n n xnµ(An) converges absolutely and its limit is independent of the representation of f . P Proof. Let f = 1A xn be an integrable step function. The sets (An) are n n ⊆ A mutually disjoint and (xn) X. Then ⊆ X Z xn µ(An) = f dµ < : n k k Ω k k 1 P Let f : Ω X be an integrable step function, f = n 1An xn. We define the Bochner integral! (for integrable step functions) by Z X f dµ := xn µ(An): Ω n Lemma 1.7. (a) For every integrable function f : Ω X there exists a sequence ! ( fn) of integrable step functions Ω X such that fn f and fn f pointwise ! k k ≤ k k ! 4 1 Bochner-Lebesgue and Bochner-Sobolev spaces µ-almost everywhere. (b) Let f : Ω X be integrable. Let ( fn) be a sequence of integrable step functions ! such that fn f and fn f pointwise µ-almost everywhere. Then k k ≤ k k ! Z x := lim fn dµ exists n !1 Ω and Z x f dµ. k k ≤ Ω k k Proof. (a) Let f : Ω X be integrable. Then f : Ω R is integrable. ! k k ! Therefore there exists a sequence (gn) of integrable step functions such that 0 gn f and gn f pointwise µ-almost everywhere. ≤ ≤ k k ! k k Since f is measurable, there exists a sequence ( f˜n) of step functions Ω X ! such that f˜n f pointwise µ-almost everywhere. Put ! f˜n gn f := : n 1 f˜n + k k n (b) For every integrable step function g : Ω X one has ! Z Z g dµ g dµ. Ω ≤ Ω k k Hence, for every n, m Z Z fn fm dµ fn fm dµ, Ω − ≤ Ω k − k R and by Lebesgue’s dominated convergence theorem the sequence ( fn dµ) R Ω is a Cauchy sequence. When we put x = limn fn dµ then !1 Ω Z Z x liminf fn dµ = f dµ. n k k ≤ !1 Ω k k Ω k k Let f : Ω X be integrable. We define the Bochner integral ! Z Z f dµ := lim fn dµ, n Ω !1 Ω where ( fn) is a sequence of step functions Ω X such that fn f and ! k k ≤ k k fn f pointwise µ-almost everywhere. The definition of the Bochner integral ! for integrable functions is independent of the choice of the sequence ( fn) of step functions, by Lemma 1.7. Moreover, if f is a step function, then this definition of the Bochner integral and the previous definition coincide.