Fourier Analysis. Dr. Felix Schwenninger University of Hamburg

Transcription

1 Fourier Analysis Dr. Felix Schwenninger University of Hamburg

2 his is a preliminary (and still incomplete) version of the things we have encountered in the lectures so far. For example, the motivation presented in Lecture is not yet included here. here may and will still be numerous typos and misprints and the file will regularly be updated. Furthermore, there may be proofs missing that were actually presented in the lectures. herefore these notes can not serve as a alternative for attending the lectures, but should rather help in preparing the exercise sheets. Please also observe that the numbering of theorems, lemmata etc. may slightly differ from the ones I used on the blackboard. Summarizing this script should be seen as supplementary support in order to follow the lectures and do the exercise sheets. his lecture notes as well as the exercise sheets can be found at Hamburg, July 208

3 CHAPER 0 Introduction. What is Fourier Analysis? Example. (hresholding and Fourier series periodic, square-integrable functions). 2. Historical Remarks Example 2. (Fourier s investigation on heat flow). Example 2.2 (A reminder on functional analysis on Hilbert spaces). 3. he main questions and goals Q: Does the Fourier series of a function f converge to f pointwise? 4. Prerequisites and Literature In this course we will strongly rely on concepts from measure theory (e.g. Lebesgue integration, L p spaces,..), functional analysis (e.g. bounded linear operators on Banach spaces, uniform boundedness principle, Hahn-Banach, closed graph,..) it should go without saying that fundamentals from analysis (Bachelor courses including Analysis III) will be required too. If any of the mentioned keywords do not ring a bell you better consult relevant literature, such as for example ao s book An Epsilon of Room, I: Real Analysis, available as online version (see ) as well as in the library. Alternatively you may consider looking into Analysis III scripts from past courses at the University of Hamburg. Further reading: Among the vast literature on Fourier Analysis, the books by Y. Katznelson ( An Introduction to Harmonic Analysis, Cambridge University Press) and L. Grafakos ( Classical Fourier Analysis, Springer Graduate exts in Mathematics) shall be mentioned here. In the first part of the course we will loosely follow the corresponding part of Katznelson s book.

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5 CHAPER Fourier Series his chapter deals with the interplay of 2π-periodic functions f : R C and associated (formal) objects (0.) a n e int n Z with which we aim to represent the functions f. Sometimes it will be convenient to view such functions as defined on the torus = R/2πZ or upon the isomorphism t e it on {z C: z = }. Furthermore, we can identify 2π-periodic functions on R with functions defined on [0, 2π). We will make use of the later identification in the following without specifying. Note that this particularly means that for f : R C Lebesgue measurable (and integrable on [0, 2π), 2π f(t) dt = f(x) dλ(x), 0 where λ denotes the Lebesgue measure. he presented theory generalizes to any other (positive) period by obvious transformations (in the literature often the -periodic functions are discussed).. Definitions and basics Definition.. An object of the form (??) is called a trigonometric series, and in particular trigonometric polynomial if a n = 0 for almost all n. Having in mind the goal to represent a function f by a trigonometric series, that is, we would like to link a (preferably unique) trigonometric series to f, f n a n e int, (this notation is, for the moment, nothing else than formal) and recalling the following fundamental identity for k Z 2π { (.) e ikt 2π k = 0 dt = 0 k 0 we (formally) derive (.2) a n = 2π 0 2π 0 f(t)e int dt. If f was a trigonometric polynomial (or more generally, if the trigonometric sum converges uniformly), then this formula is indeed justified. From now on and unless stated otherwise, we will assume that functions f are in L (), in which case the integral in (??) exists. 3

6 4. FOURIER SERIES Definition.2. For a 2π-periodic function f L (R), the sequence (a n ) n Z defined by (??) is called the Fourier coefficients of f and we denote it by ˆf(n) = a n. he trigonometric series S(f) = a n e int = (t a n e int ) n Z n Z is called the Fourier series of f and we also use the notation f n Z a ne int. Given a (formal) trigonometric series S of the form (??), we call it a Fourier series if there exists an f L such that S is the Fourier series of f. Remark.3. rigonometric polynomials take in some sense over the role of usual polynomials but on the unit sphere. More generally, trigonometric series take over the role of power series. In that sense, Fourier series can be compared to aylor series of a function. However, while aylor series exploit the local behavior of a function, Fourier series encode the global behavior, as we will see in this course. he latter fact makes them very powerful on one hand, but far more difficult in study as we shall see. From the analysis lectures we recall the following fundamental result For f C([a, b]; C) there exists a sequence of polynomials (p n ) n N (with complex coefficients) such that lim n f p n,[a,b] = 0. In particular, this yields heorem.4 (Weierstrass approximation theorem). he trigonometric polynomials lie dense in C(). We remark that there are several ways to prove Weierstass theorem, among which is a modern approach via the more general Stone Weierstrass theorem which we now have in mind when we take this theorem for granted. However, we will shortly face a constructive proof based on Fourier series. heorem.5 (Basic facts on Fourier series). he following holds. () f ˆf is a linear, injective mapping from L () to C Z (2) f ˆf bounded from L () to l (Z), the space of bounded, complex-valued (double) sequences, i.e. ˆf := sup ˆf(n) n Z 2π f L (). (3) for any s R, τs ˆf = (s e is ˆf(s)), where (τ s f)(t) = f(t + s), (Modulation) Proof. he linearity follows directly form (??). We will present a proof of the injectivity (and shall see another one in the Exercise classes). Let us assume that ˆf = 0. By linearity it suffices to show that that f = 0. Again by linearity and the definition of ˆf, we have that (.3) 2π 0 f(t)g(t) dt = 0 for all trigonometric polynomials g 2 and hence for all continuous functions g by Weierstrass theorem. o conclude that f = 0 we have several alternatives: In these lectures, one may have only seen the real version of this theorem. he complex version however follows from that case by applying it to real- and imaginary part of the considered continuous function. 2 If f is continuous, we can conclude that f = 0 by Weierstrass theorem. For more general f L (), we need a more refined argument

7 . DEFINIIONS AND BASICS 5 (α) If we can show that for any measurable set E [0, 2π] it holds that f(t) dt = 0, then E it is not hard to see that f = 0. In fact, if f 0, then either Rf 0 or If 0 (in the L sense). Let us assume that there exists ε such that E = {t [0, 2π]: Rf(t) > ε} has positive measure the other cases follow analogously. hen R f(t) dt > ελ(e) E which yields a contradiction. In order to show that f(t) dt = 0 for any measurable E set E, recall Lusin s theorem 3 Let h : [a, b] C be a measurable function. hen for any δ > 0 there exists a continuous function h δ : [a, b] C such λ({t [a, b]: h(t) h n (t) 0}) < δ 4 with h δ h Apply this with h = χ E and set g n := h 2 n and conclude that lim n g n (t) = g(t) for almost every t [a, b] 5. As g n, dominated convergence together with (??) yields 2π 2π 0 = lim f(s)g n (s) ds = f(s)g(s) ds = f(s) ds. n 0 0 (β) For any function g in L () there exists a sequence of continuous functions that converge to g in weak sense (see e.g. Ex. 2.3), and consequently 2π 0 f(t)g(t) ds = 0 g L (). Using the fact that L () is isometrically isomorph to the dual space of X = L () (Riesz Representation theorem), this implies that f, g X,X = 0 for all g X. Hence, a consequence of the Hahn-Banach theorem yields f = 0. heorem.6 (discrete Riemann Lebesgue). f ˆf maps L () to c 0 (Z), where c 0 (Z) denotes the space of sequences that converge to 0 (as n ± ). Proof. We present a sketch of the proof (the details of each step are left for the exercises). his follows the routine prove for smooth + functional analysis First, one shows (sees) that for any trigonometric polynomial p it holds that ˆp c 0 (Z). As a second step, use that f ˆf is bounded from L () to l (Z), heorem?? and conclude the assertion by density of the trigonometric polynomials (Weierstrass heorem) and density of continuous functions in L (). Remark.7 (a warning). One may be tempted to think that Weierstrass theorem, heorem??, already gives an indication on functions that can safely be approximated by Fourier series: the continuous functions on, as Fourier himself conjectured. his however turns out to be ( very ) wrong as we shall see. In a similar, but different, spirit, we convince ourselves that the approximating sequence in Weierstrass theorem is in general not given by the partial sums of a aylor series. E 3 See for instance ao s book (the condition on the supremum follows easily from the proof) mentioned in the introduction. 4 where the essential supremum on left hand side of the inequality is allowed to be. 5 here, one may use that λ({t: h(t) hn(t) 0 for infinitely many n N}) = 0 by Borel-Cantelli

8 6. FOURIER SERIES 2. Convolutions On our way to understand the relation between f and the series S(f) In the following let Ω be either R or and consider L p = L p (Ω) = L p (Ω, λ) where dλ(s) = ds refers to the corresponding Lebesgue measure. In the present chapter we shall be interested in the case Ω =, but later on the case Ω = R will be needed. Definition 2. (Convolution). Let f, g L (Ω), then the convolution f g of f and g is defined as (f g)(t) = f(t s)g(s) ds for t Ω such that the integral exists. Ω Formally one can define the convolution for more general functions f, g only requiring that f g exists for a.e. t Ω. In the case of L (Ω) this is indeed guaranteed as the following result shows. heorem 2.2. he convolution is a bilinear mapping from L (Ω) L (Ω) L (Ω) that satisfies for all f, g, h L (Ω): (2.) f g L f L g L f g = g f f (g h) = (f g) h (commutativity) (associativity). Proof. Bilinearity easily follows from linearity of the integral. All other assertions are based on applying Fubini s theorem. Let us show that is well-defined from L (Ω) L (Ω) to L (Ω) and (??). he mapping I(t, s) = f(t s)g(s) is clearly measurable on Ω Ω since f and g are measurable. Furthermore, t I(t, s) L (Ω) for a.e. s Ω because f L (Ω) and by I(t, s) dt ds = g(s) f(t s) dt ds = g L (Ω) f L (Ω), Ω Ω Ω Ω where we have used the translation invariance of the Lebesgue measure on our choices for Ω, we get that s I(, s) L (Ω) is in L (Ω). hus, Fubini s theorem (or Fubini-onelli), yields that t (f g)(t) = I(t, s) ds is integrable and Ω f g L (Ω) = I(t, s) ds dt Ω Ω I(t, s) ds dt = I(t, s) dt ds = g L (Ω) f L (Ω). Ω Ω Inequality (??) is only a special instance of the following result. Ω Ω heorem 2.3 (Young s inequality (for convolutions)). Let p + q = + r and f L p (Ω), g L q. hen f g L r (Ω) and f g L r f L p g L q. Proof. See Exercise 5.4. for p, q, r [, ] Corollary 2.4 (Minkowski s inequality). For any g L (Ω), the mapping f f g is bounded on L p (Ω) with norm less or equal to. For a separate proof of Minkowski s inequality see Ex. 2.4.

9 3. APPROXIMAE IDENIIES 7 Remark 2.5. We have seen convolutions on L p spaces defined on R and with the Lebesgue measure. It is important to realize that the latter are (additive) groups with a translationinvariant measure. In fact, this are the crucial properties needed to define a convolution. We remark that this is also possible for more general locally compact topological groups and their, roughly speaking, translation-invariant measures. heorem 2.6 (Convolution and Fourier coefficients). Let f, g L (), n Z. hen f g = 2π ˆf ĝ, where the multiplication on the right has to be understood pointwise, i.e. f g(n) = 2π ˆf(n)ĝ(n). Furthermore, (f e in )(t) = 2π ˆf(n)e int. Proof. By definition of the convolution and the Fourier coefficients, 2π f g(n) = f(t s)g(s) ds e int dt = f(t s)e in(t s) e ins g(s) ds dt. Using Fubini (or Fubini-onelli)), we can interchange the order of integration, and also using that translation-invariance of the the Lebesgue measure on, we get f(t s)e in(t s) e ins g(s) ds dt = e ins g(s) f(t s)e in(t s) dt ds = (2π) 2 ĝ(n) ˆf(n). he second claim follows either by a similar calculation or by recalling that êin = e n, where e n denotes the n-th canonical basis vector in c 0 (Z), and applying the first part of this theorem: hus f e in = ˆf(n)e n and by injectivity of the linear map f ˆf, heorem??, we deduce (f e in )(t) = ˆf(n)e int. 3. Approximate identities Definition 3. (Approximate identity). A sequence (k n ) n N C() is called approximate identity if () k n(s) ds = for all n N, (2) sup n N k n L () < 2π δ (3) lim n k δ n (s) ds = 0 for all δ (0, π). If k n 0 we say that the approximate identity is positive. Note that the requirement that the k n are continuous is made for technical reasons here and actually can be dropped. he name approximate identity is justified by the following proposition. Recall the Riemann integral for Banach space valued functions. Proposition 3.2. Let X be a Banach space, φ : X continuous and (k n ) n be an approximate identity. hen for any t k n (s)φ(s) ds = φ(0). lim n Proof. By the properties of an approximating identity we have k n (s)φ(s) ds φ(0) = k n (s)[φ(s) φ(0)] ds. o show that the right-hand side equals zero, we split the integral 2π δ k n (s)[φ(s) φ(0)] ds = k n (s)[φ(s) φ(0)] ds + δ [0,δ] [2π δ,2π] k n (s)[φ(s) φ(0)] ds

10 8. FOURIER SERIES with δ (0, π). Since φ(2π) = φ(0) and since k n L is uniformly bounded in n, the second term on the right-hand-side can be made arbitrarily small as δ goes to 0 (independent of n). On the other hand, 2π δ 2π δ k n (s)[φ(s) φ(0)] ds 2 sup φ(s) X k n (s) ds 0 δ for any δ as n by the definition of an approximate identity. Proposition 3.3 (Strong continuity). Let X be one of the following Banach spaces k C k () with norm f C k () = f (j), L p () with p [, ). hen for s, the operator is linear and isometric. Furthermore, j=0 s τ s : X X, f f s := f( s) f X : s τ s f is continuous. he latter property is called strong continuity of τ. he assertion of the theorem remains true if is replaced by R and C() is replaced by C 0 (R). Proof. We only give a sketch see the exercises for details. hat τ s is well-defined, linear and isometric is seen directly (also using the translation invariance of the Lebesgue measure). Hence in particular τ s L(X) = for all s. Let f C k (). hen f (j) is uniformly continuous on and hence lim t s f (j) s δ (t) f (j) t (s) = 0 for j {,.., k}. For X = L p (), p [, ), recall that (the set) C() lies dense in X. Since (the Banach space) C() is continuously embedded in X, s τ s f is continuous for f C() by what we have already shown. he continuity for general f X now follows by approximation with continuous functions and the fact that s τ s is uniformly bounded (triangle inequality). he case where is replaced by R can be shown similarly, using compactly supported continuous functions when approximating. he previous result motivates the following definition. Definition 3.4. A Banach space X that is continuously embedded in L (), i.e. C > 0 with X C L (), is called homogenous (Banach) space if () for f X it follows that f s = f( s) X with f X = f s X and (2) the shift group τ s is strongly continuous on X. heorem 3.5 (Approximation in homogenous spaces). Let (k n ) n N be an approximate identity and X be a homogeneous Banach space. hen for any f X it holds that f = lim k n (s)f s ds = lim k n f, n n where the limits as well as the integral exist in X. Proof. he first identity follows directly from Prop.??. In fact, s φ(s) = f s is continuous from to X since X is a homogenous Banach space and φ(0) = f 0 = f. For the second identity, we will show that for k C() B f := k(s)f s ds = k f =: B 2 f

11 4. HE FEJÉR KERNEL AND DIRICHLE KERNEL 9 for all f L (). If this holds, then the second identity follows as X is continuously embedded in L () and therefore, the integral B f exists as limit of Riemann sums in both L and X and coincides. It is easy to see that B, B 2 are bounded linear operators on L (). Hence, it suffices to show that they coincide on a dense subspace. For t let C t : C() C, f f(t) denote the point evaluation operator at t. Also note that the integral B f exists as limit of Riemann sums in C() as τ s is strongly continuous on C(). Since C t is linear and bounded, we have that (B f)(t) = C t B f = C t (k(s)f s ) ds = (k f)(t). his theorem has several consequence concerning approximation of elements in homogenous Banach spaces and in particular in the context of trigonometric series. 4. he Fejér kernel and Dirichlet kernel In the following we would like to apply the abstract result derived in the previous section to specific approximate identities. Recalling that thus (4.) ( one is tempted to believe that n k= n (e ik f)(t) = ˆf(k)e int 2π, e ik f)(t) = 2π (4.2) D n (t) = 2π n k= n n k= n e ikt ˆf(k)e ikt is an approximate identity. If this was true, then heorem?? would imply that the (partial sums of the) Fourier series n (D n f)(t) = ˆf(k)e ikt k= n of f X converge to f in X for any homogenous Banach space X. Example 4. (he Dirichlet kernel). he sequence k n defined by (??) is called the Dirichlet kernel and is not an approximate identity. In fact, D n(s) ds =, but D n L () is not uniformly bounded in n N. More precisely, using the identity D n (s) = 2π one can show that D n L () log n, see Exercises. sin((n + 2 )t) sin t, 2 Not only that (D n ) n fails to be an approximate identity, it can moreover be shown that the assertion of heorem?? does not hold with k n = D n at least not in the case X = L () and X = C(). Motivated by convergence properties for sequences we may instead hope for some better properties of the arithmetic mean of D n Motivated by the fact that the convergence of a sequence (a n ) n N0 may improve upon con- ( sidering the sequence of arithmetic means instead the following kernel. a 0+a +..+a n n+ ) n N 0, see Ex. 2.0, we introduce

12 0. FOURIER SERIES Definition 4.2 (Fejér kernel). he sequence (F n ) n N0 C() defined by F n (t) = n D k (t), t, n + where (D k ) k is defined in Ex.??, is called the Fejér kernel. For f L (), we call F n f = n (D k f) n + k=0 the n-th Fejér mean (or Césaro mean ) of f. By (??), the n-th Fejér mean is nothing else than the arithmetic mean of the first n + partial sums of the Fourier series of f. Proposition 4.3. he Fejér kernel (F n ) n N0 defined in Def.?? is an approximate identity, additionally satisfying for all n N 0, t and δ (0, π), k=0 F n (t) 0, F n (t) = F n ( t), lim n F n L (δ,2π δ) = 0. Furthermore, the following identities hold. (4.3) F n (t) = n ( k 2π n + k=0 ) e int = 2π(n + ) sin((n + ) t 2 ) sin t. 2 Proof. See Ex. 2. (first show (??) from which the rest follows rather directly). heorem 4.4 (Approximation by Fejér kernel). Let X be a homogeneous Banach space and f X. hen (F n f) converges to f in X as n. Proof. his follows from heorem?? and since (F n ) n is an approximate identity, Proposition??. Corollary 4.5. he set of trigonometric polynomials rig() lies dense in any homogeneous Banach space X in the sense that the closure of rig() X in X equals X. Moreover, for any f X, the sequence n ( f n (t) = k ) ˆf(k)e int n + k=0 converges to f in the X norm. In particular, the trigonometric polynomials lie dense in C k () and L p () for any p [, ), k N 0. Proof. Note that F n f = f n by Proposition?? and heorem??. he rest follows from heorem??. Corollary?? provides a constructive proof of Weierstrass theorem. However, we should be aware of the fact that the proof of heorem?? relied on the fact that C() lies dense in L () which then should not be argued by Weierstrass theorem, but more directly (e.g. use that the simple functions are dense in L and apply Lusin s theorem). heorem 4.6. For X = L () or X = C() there exists a function f X such that the partial sums D n f of the Fourier series of f do not converge in X. Moreover, there exists f C() and t 0 such that (D n f)(t 0 ) does not converge. Proof. his is proved in Ex. 2.5 by an abstract argument employing the uniform boundedness principle.

13 5. CONVERGENCE OF FOURIER SERIES IN HOMOGENEOUS BANACH SPACES Remark 4.7. Although the proof of heorem was based on an abstract existence argument, it is possible to construct such f C() explicitly, see e.g. [?,?]. heorems?? and?? leave us with two main questions/goals. (I) Do the Fejér means (F n f) n converge pointwise almost everywhere for any f L ()? (II) Do the partial sums (D n f) n of a Fourier series converge in L p () for any f L p ()? We remark that we could have asked the second question for more general homogeneous Banach spaces X other than C() or L (), but here will restrict to the important class of L p spaces only. Let us first approach (I): heorem 4.8 (Pointwise convergence of Fejér means). Let f L (), t 0 and assume that there exists L t0 C such that h (4.4) lim h 0 + h 2 (f(t 0 + s) + f(t 0 s)) L t0 ds = 0. hen lim n (F n f)(t) = L t0. Proof. 0 Remark 4.9. Note that the condition (??) is in particular satisfied if (4.5) lim s (f(t 0 + s) + f(t 0 s)) = L t0 and hence especially for any point t 0 if f is in C(). We will see that heorem?? even implies that the Fejér means F n f of a function f L () converge pointwise almost everywhere to f. he proof is postponed since we will apply some deeper techniques. heorem 4.0. Let f L (), then lim n (F n f)(t) = f(t) for almost every t. 5. Convergence of Fourier series in homogeneous Banach spaces We have already seen that for the spaces X = C() and X = L () the partial sums of a Fourier series (i.e. the sequence of Dirichlet means) does not converge in the respective norms, i.e. D n f f in X for n. his failure was seen by relating convergence with uniform boundedness of the operator sequence (M Dn ) n N defined by (5.) M Dn : X X, f D n f. Let us begin this section with showing that this is actually a manifestation of an argument for general spaces. In the following the phrase the Fourier series of f converges to f in X always refers to the convergence of the partial sums D n f. heorem 5.. Let X be a homogeneous Banach space. hen the Fourier series of all f X converges to f in X if and only if the operators (M Dn ) n N defined by (??) are uniformly bounded. Proof. he sufficiency ( ) is a direct consequence of the uniform boundedness principle. In fact, for f X the sequence (M Dn f) n N is bounded in X, i.e. sup n N M Dn f X < since M Dn f converges for n by assumption. hus, by the uniform boundedness principle (recalling that M Dn : X X is a bounded operator on X for every n N, see Ex. 2.4), we conclude that (M Dn ) n N is bounded in B(X, X), the space of bounded linear operators on X.

14 2. FOURIER SERIES Conversely, assume that C := sup n N M Dn <. In order to show that M Dn f f as n, we first consider the case when f is a trigonometric polynomial and show the general case by density of these polynomials in X and that C <. As we have seen in Exercise 2.4, the norm of S n = M Dn as operator on X can be bounded by D n L (), but this inequality may be strict. Example 5.2. We can apply heorem?? to the Dirichlet mean in X = L 2. From Exercise 2.4 (d), we know that S n L 2 () L 2 () 2π D n l (Z) = and hence the partial sums of the Fourier series for L 2 functions converge in L 2 (something we knew already from the functional analysis class). In contrast to this example (and X = C() or X = L ()), for general homogeneous spaces X it is difficult to calculate (or estimate) S n X X in order to conclude on the convergence of the Fourier series. For that reason, we will reformulate this property once more. 5.. Conjugate trigonometric series. Definition 5.3 (Conjugate trigonometric series). Given a trigonometric series n Z a ne int, we call i sgn(n)a n e int n Z the conjugate (trigonometric) series 6. In the following we shall be particularly interested in conjugate series of Fourier series. Let us first study the question if Fourier series are invariant under conjugation, i.e. whether the conjugate series of a Fourier series is again a Fourier series. Recall that we say that a trigonometric series is a Fourier series if its coefficients are the Fourier coefficients of some function in L (). his question looks harmless, but, as we shall see shortly, has a negative answer. However, functions f L () for which an affirmative answer can be given are e.g. f A() = {f L (): ˆf l (Z)} (why?). Proposition 5.4. () Let (a n ) n Z c 0 (Z) be a given sequence such that a n = a n, a n 0 and 2 (a n + a n+ ) a n for all n N 0. hen there exists f L () nonnegative such that ˆf(n) = a n for all n Z. (2) Let f L () be such that ˆf(n) = ˆf( n) 0 for all n N 0. hen l (N). ( ) ˆf(n) n n N Proof. (): From the assumption, (a n a n+ ) 0 and moreover, n(a n a n+ ) 0 as n (this follows from n N 0 (a n a n+ ) = a 0 ). Define f(t) = 2π n(a n + a n+ 2a n )F n (t) n= for t. o see that the series converges in L (), note that by sup n N F n L () <, it suffices to see that k n= n(a n + a n+ 2a n ) converges for k. he latter follows easily from the identity k n(a n + a n+ 2a n ) = a 0 a k k(a k a k+ ). n= 6 Recall that without further specifications these series are only formally defined by their coefficients sequence.

15 5. CONVERGENCE OF FOURIER SERIES IN HOMOGENEOUS BANACH SPACES 3 his also shows that the limit as k equals a 0. It remains to show that ˆf(m) = a m for all m Z. By linearity and boundedness of f ˆf we have ˆf(m) = n(a n + a n+ 2a n )2π F n (m). n= Since F n (m) = 0 for m n and 2π F n (m) = ( m n ) for m < n, this yields ˆf(m) = n(a n + a n+ 2a n )( m n ). n= m + Using the above derived convergence properties of (a n ) it is not hard to see that the last sum equals a m (index shift!). (2): see Ex. 3.(3). Corollary 5.5. here exists a Fourier series such that its conjugate series is not a Fourier series. More precisely, the following holds: Let a n > 0 for all n Z such that a nn n Z =. hen i n Z a nsgn(n)e int is not a Fourier series. Proof. Let a n = log(n) for n 2 and 0 else. Definition 5.6 (Homogeneous spaces invariant under conjugation). We say that a homogeneous Banach space X is closed under conjugation if for any f X, it follows that there exists f X with ˆ f(m) = isgn(m) ˆf(m) m Z. Note that the property of being invariant under conjugation is non-trivial: In particular, the homogeneous space L () does not possess it by Corollary??. On the other hand, examples of a space which are invariant under conjugation are given by X = L 2 () or X = A() as can be seen easily. Proposition 5.7. Let X be a homogeneous Banach space. equivalent. he following assertions are () X is invariant under conjugation, (2) f f is a well-defined, bounded linear operator from X to X. (3) f P + f = 2 ˆf(0) + 2 (f + i f) is a well-defined, bounded linear operator from X to X. Note that (P + f)(m) = ˆf(m) for m 0, and (P + f)(m) = 0 for m < 0. Proof. We show () = (2) = (3) = (). he first implication is a consequence of the closed graph theorem, by which it suffices to show that the mapping f f is closed. By (), the mapping is well-defined and obviously linear. Let f n, f, g X, n N, be such that f n f and f n g in X as n. o show that f = g, consider the Fourier coefficients. Since X L, we have that lim f n (m) = ˆf(m) and lim f n (m) = ĝ(m) n n for all m Z, by heorem??. On the other hand, we have that fn (m) = isgn(m) f n (m) by definition. herefore, ˆ f = ĝ and thus f = g by uniqueness of the Fourier coefficients, heorem??. hus, by the closed graph theorem (X is a Banach space), the mapping f f is bounded. (2) = (3): his is clear since f ˆf(0) and f f are bounded operators on X and therefore P + is bounded as sum of bounded operators.

16 4. FOURIER SERIES (3) = (): his follows by f = i(2p + f ˆf(0) f) and since the right-hand side is well-defined for f X. In fact, homogeneous spaces which are invariant under conjugation characterize the property that the partial sums of Fourier series converge: heorem 5.8. Let X be a homogeneous Banach space such that f e in f is an isometry from X to X 7 for any n Z. hen the following assertions are equivalent. () For all f X the Fourier series converges to f in X, (2) X is invariant under conjugation. Proof. By heorem?? and Proposition?? it suffices to show that M Dn are uniformly bounded operators on X if and only if P + is a bounded operator on X. From the definition of the Dirichlet kernel and heorem?? we get the following dentity for f X, (5.2) 2n k=0 ˆf(k)e ikt = e int (D n (fe in ))(t) = e int (M Dn (fe in )) For n Z let J n be the operator defined by (J n f)(t) = e int f(t). Clearly, J n = (J n ) and, by assumption, J n f X = f X and hence particularly J n =. Let f be a trigonometric polynomial. hen for sufficiently large n, P + f = 2n k=0 ˆf(k)e ikt = J n M Dn J n f, where we used (??). hus P + f X M Dn f X. If we assume that C = sup n N M Dn X X <, then it follows that P + f X C f X for any trigonometric polynomial and thus P + is bounded on X by density, Corollary??. Conversely, if P + is bounded on X, then by (J n P + J n J n+ P + J n )f = M Dn f for any f X, we have that M Dn 2 P + for all n N. heorem 5.9 (Convergence of Fourier series in L p, Riesz 927). Let p (, ). hen for every f X = L p () the partial sums D n f of the Fourier series of f converge to f in X as n. Sketch of the proof. he proof of heorem?? is split into the following steps () Show heorem?? for the special case that p = 2k with k N, (Lemma??) (2) Show that this implies the assertion for p 2 (interpolation) (3) Show the case p (, 2) (duality). Lemma 5.0. he assertion of heorem?? holds for p = 2k with k N. Proof. As we have seen in Proposition??, boundedness of the conjugation f f on X = L p characterizes the convergence of the Fourier series. Hence, it suffices to show that there exists some constant C 2k > 0 such that f L 2k C 2k f L 2k for all trigonometric polynomials f, since the latter set is dense in X, Corollary??. For a trigonometric polynomial f, P + f = 2 ( ˆf(0)+f +i f) = N ˆf(j)e ik, j=0 for sufficiently large N. If we additionally assume that ˆf(0) = 0, then the latter identity implies using the elementary identity (??) that (f(t) + i f(t)) 2k dt = 0. 7 i.e. t e int f(t) X and e in f X = f X for all f X.

17 5. CONVERGENCE OF FOURIER SERIES IN HOMOGENEOUS BANACH SPACES 5 From this identity we will now extract an estimate for f L 2k. By expanding the binomial, we derive 2k ( 2k )i 2k j f(t) j f(t) 2k j dt = 0. j j=0 Let us for the moment further assume that f and f are real-valued. hus, taking real parts gives k j= and thus by Hölder s inequality Dividing by f 2k L yields that 2k ( 2k )( ) k j f(t) 2j f(t) 2k 2j dt = ( ) k 2k f 2k 2j k ( ) 2k 2k f L f(t) 2j f(t) 2k 2j dt 2k 2j j= k ( ) 2k f 2j 2k 2j f. 2j L 2k L 2k j= ( f ) 2k L 2k f L 2k k j= ( ) ( 2k f L 2k 2j f L 2k ) 2k 2j for all f 0 satisfying the listed assumptions. In other words, x = f L 2k the polynomial p(x) = x 2k k j= f L 2k satisfies p(x) 0 for ( 2k 2j) x 2k 2j. By the mean value theorem and the fact that p(0) < 0, we conclude that p has at least one zero and we can hence denote by C 2k the largest zero of p. It follows that f L 2k C 2k f L 2k for all real-valued trigonometric polynomials f such that f is also real-valued and ˆf(0) = 0. It remains to remove the made conditions on the trigonometric polynomials f. If f is such that ˆf(0) = 0, then we can decompose f by f = P + iq where P (t) = N j= N 2 ( ˆf(j) + ˆf( j) ) } {{ } c j e ijt, Q(t) = N j= N ( ) ˆf(j) ˆf( j) e ijt, 2i and where N is sufficiently large. hen P (t) = N j= 2Rc j cos(jt) and Q(t) = N j= 2Ic j sin(jt) are real-valued. herefore, by linearity of f f and by what we have already shown, (5.3) f L 2k = P + i Q L 2k C 2k ( P L 2k + Q L 2k) 2C 2k f L 2k. where the last inequality follows by max{ P (t), Q(t) } ( P (t) + Q(t) ) 2 = f(t) for t. Finally, we remove the assumption that ˆf(0) = 0: For a general trigonometric polynomial we have by linearity and the fact that g = 0 if g is constant, f L 2k = (f ˆf(0)) + ˆf(0) L 2k = (f ˆf(0)) L 2k 2C 2k f ˆf(0) L 2k

18 6. FOURIER SERIES where we used (??). By heorem?? and Hölder s inequality, f ˆf(0) L 2k f L 2k + ˆf(0) (2π) 2k f L 2k + f L (2π) 2k f L 2k + f L 2k(2π) k 2 f L 2k. hus, altogether, f L 2k 4C 2k f L 2k for trigonometric polynomials f and hence for all f L 2k. Recall that a function f : C C is called entire if there exists a sequence (a n ) n N0 of complex numbers such that f(z) = n=0 a nz n for all z C. For entire functions one has the following maximum modulus principle: (5.4) Ω C open, bounded = sup f(z) = max f(z), z Ω z Ω where Ω denotes the boundary of Ω, see Exercise 5.2 for a proof. We even need version of the maximum modulus principe for unbounded domains, which is covered by the following result. Lemma 5. (Hadamard s hree Lines lemma for entire functions). Let H be entire and assume that H grows at most exponentially on the vertical strip S = {z C: Rz (0, )}, i.e. there exists c, c 2 > 0 such that H(z) c e c2i(z) for all z S. Further suppose that C j := sup Rz=j H(z) < for j {0, }. hen 8 H(z) C θ 0 C θ if Rz = θ. Proof. Define G by G(z) = H(z)(C z 0 C z ). Since C z 0 C z = C θ 0 C θ for Rz = θ, it thus remains to show that G(z) for all z S. As we cannot apply the maximum modulus principle to the unbounded set S directly, we first assume that G(x + iy) 0 as y uniformly in x (0, ). Hence there exists y 0 > 0 such that G(z) < for z S with I(z) > y 0. Applying the maximum modulus principle, (??), to G and Ω = {z S : Iz < y 0 }, yields that G(z) for z Ω since it follows from the assumption on H that G(z) when either Rz = 0 or Rz =. ogether this shows that G(z) for all z S. In the general case when H is only assumed to grow at most exponentially, we consider G n defined by G n (z) = G(z)e z2 n which converges to G(z) for any z S as n. Since e z2 n = e n ( (Iz)2 +(Rz) 2 ), it follows that for every fixed n, G n (z) goes to 0 uniformly in S as I(z) and thus satisfies the assumption made in the first proof step. herefore, G n (z) for all n N and thus G(z) for all z S. Recall that for a measure space (Ω, µ) a function f : Ω C is called simple if it is of the form n f = f jχ Aj j= with n N, f j C and measurable, disjoint sets A j Ω. We say that a simple function f has finite measure support if µ(a j) < for j {,..., n}. Note that for p [, ), the set of simple functions with finite measure support M simple,finite (Ω, µ) lies dense in any L p (Ω, µ) if the measure space is σ-finite. 8 Note that this statement actually holds more generally for functions f which are analytic on S and continuous on the closure S.

19 5. CONVERGENCE OF FOURIER SERIES IN HOMOGENEOUS BANACH SPACES 7 heorem 5.2 (Riesz horin interpolation theorem). Let (Ω i, µ i ), i =, 2 be σ-finite measure spaces and let p i, q i [, ]. Let be an operator mapping the space of simple functions Ω C with finite measure support M simple,finite (Ω, µ) to L q (Ω 2, µ 2 ) L q2 (Ω 2, µ 2 ) such that (5.5) f L q i (Ω2) C i f L p i (Ω) f M simple,finite (Ω, µ). hen there exists a constant C p,q > 0 such that where (5.6) f L q (Ω 2) C p,q f L p (Ω ) f M simple,finite (Ω, µ), p = θ + θ, p p 2 q = θ + θ, θ (0, ), q q 2 and C p,q C θ C θ 2. In particular, if p <, then extends to a bounded operatore from L p (Ω, µ ) to L q (Ω 2, µ 2 ). Proof. In the following we will abbreviate L p = L p (Ω ) and L q = L q (Ω 2 ) as the space Ω i is clear from the context. Recall that q denotes the Hölder conjugate of q, i.e. q + q =. By a Hahn-Banach argument and duality of L p spaces, it suffices to show that ( f)g dµ 2 C p,q Ω 2 for simple functions f : Ω C and g : Ω 2 C with finite measure support and with f L p, g L q = (why?!, check this in particular for limiting cases p, q = ). Our goal is to apply Hadamard s three lines lemma. Let us factorize f and g as follows f = e i arg(f) F θ F θ 2, g = e i arg(g) G θ G θ 2, with 9 F i, G i non-negative and such that F i L pi, G i L q i with norms Fi L p i = G i L q i = this can be achieved by setting F i = f p q p q i and G i = g i. Now define h : C C by h(z) = (e i arg(f) F z F2 z )e i arg(g) G z G z 2 dµ 2, Ω 2 which is well-defined and entire since F i and G i are simple functions with finite measure support: More precisely, this by the fact that there exists pairwise disjoint measurable sets A j Ω, and B k Ω 2, and a j, b j, c k, d k > 0, for j =,.., m, k =,.., l such that h(z) = j,k a z j b z j c z k d z k Ω 2 (e i arg(f) χ Aj )e i arg (g) χ Bk dµ 2. o apply Lemma??, it remains to check that h is bounded in modulus along the lines ir and + ir. Let us first check that F z F2 is L p and G is G is 2 L q. his holds with norms equal to since F is F2 is = F and G is G is 2 = G (since F i, G i are nonnegative). Hence, for s R, h(is) = (e i arg(f) F is F2 is )e i arg(g) G is G is 2 dµ 2 Ω 2 (e i arg(f) F is F2 is ) L q G is G is 2 L q C F is F2 is ) L p G is G is 2 L q = C 9 here, arg denotes a principal branch of the argument and we set e i arg (x) = if x = 0.

20 8. FOURIER SERIES where the first estimate follows from Hölder s inequality and the second is clear by the assumed boundedness of. Analogously, we get h( + is) C 2 for all s R. hus by Lemma??, h(z) C θ C2 θ for Rz = θ. Setting z = θ, the assertion follows since h(θ) = ( f)g dµ 2. Ω 2 Proof of heorem??. Let k N. We apply the Riesz horin heorem to the operators f = f defined from L 2k () to L 2k+2 () and from L 2k+2 () to L 2k+2 (). By Lemma??, is bounded on both spaces and hence heorem?? yields that : L p () L p () is bounded as well for all p (2k, 2k + 2). Since this holds for arbitrary k N, we conclude that L p is invariant under conjugation for all p 2 and thus the assertion of the heorem holds for such p by Proposition??. he final step that the statement also holds for p (, 2) now follows by duality: For that consider the dual operator : (L q ()) (L q ()) of the conjugation operator on L q () for q > 2. By duality of L q -spaces and the definition of the dual operator, this means that f, g L q,l q = f, g L q,l q f Lq (), g L q (), where q is the Hölder conjugate of q, i.e. q + q f = f, 2πi N m= N ˆf(m)ĝ(m)sgn(m) = 2π =. Let us consider f, g rig(). hen by N m= N f(m)ĝ(m), and therefore, f(m) = isgn(m) ˆf(m) = f(m) for all m { N,.., N} where N is chosen sufficiently large. By density of the trigonometric polynomials it follows that f = f. Since with also is a bounded operator, we hence conclude that f f = f is bounded on L q. Since this holds for all q (2, ), the assertion follows. We give another application of the Riesz horin interpolation theorem, which comes natural knowing that f ˆf is a bounded operator from both L () to l (Z) and L 2 () l 2 (Z). heorem 5.3 (Hausdorff Young). For p [, 2] and q [2, ] defined by = p + q, the operator f ˆf is bounded from L p () to l q (Z) with operator norm equal to (2π) 2 q (2π) q. Proof. he claim holds for p = and p = 2 by heorem?? and Hilbert space theory respectively. hus we can apply heorem?? for the σ-finite measure spaces (, λ) and (Z, ) refers to the counting measure and get that f ˆf is bounded from L p () to l q (Z) for all pairs (p, q) of the form p = θ + θ 2, q = θ + θ, θ (0, ). 2 Hence, p = θ 2 and q = 2 θ, which gives the assertion.

21 CHAPER 2 he Fourier tranform Note: In the following L p will unless state otherwise refer to L p (R) = L p (R, λ) where λ refers to the Lebesgue measure (on the Lebesgue measurable sets). Motivated by the question How to represent a non-periodic function on R in an analogous way as with Fourier series? we are lead to a continuous transformation of a function f in contrast to the Fourier series which is discrete.. he Fourier transform and the Schwartz class Definition. (Fourier transform on L ). For f L (R) the function F(f) : R C defined by F(f)(s) = f(s)e ist ds, s R is called Fourier transform of f. In the following we use the operators, ω R \ {0}, g L (R), R τ w f = f( + ω), m g f = fg, Rf = f( ), D ω f = f(w ) defined for f L p (R) for any p [, ]. Proposition.2 (Basics of the Fourier transform). Let ω R \ {0}. () F : L (R) L (R) is linear and bounded (2) he range ran F of F lies in C 0 (R) = {f C(R): lim x ± f(x) = 0}, that is F C 0 (R) for all f L (R) (Riemann Lebesgue) (3) For e iω (s) := e iωs we have Fm e iω = τ ω F (Modulation) Fτ ω = m eiω F FR = RF F = RF FD ω = m / ω D ω F (ranslation) (Reflection) (Conjugation) considered on L (R). (4) F(f g) = F(f) F(g) (Convolution heorem) (5) for any f C k (R) such that f (l) L (R) for all l = 0,.., k, F(f (k) ) = m (is) kf(f), (Dilation) where is refers to the function s is. Conversely, if x x l f(x) L (R) for all l = 0,.., k, [F(f)] (k) = F(m ( is) kf) 9

22 20 2. HE FOURIER RANFORM Proof. See Exercise 6.. Remark.3. Compared to the definition of the Fourier series for a given function in L (), it is not clear how to define the Fourier transform of a general function in L p (R) for p > since the integral defining F may not converge absolutely then: Recall that L p (R) L (R) since λ(r) =! his shows already a crucial technical difficulty when dealing with the Fourier transform of a function. However, as we shall see shortly, it is possible to define the Fourier transform on L 2 (R) in a consistent way. In fact, one may naively think that this is possible if the integral R eist is allowed to converge conditionally, which is not meaningful in terms of the Lebesgue integral however. In the following we will use the notation α f = α f x α = f (α) for α N 0 and a α-times differentiable function f : R C. Definition.4 (he Schwartz class). A function f C (R) is defined to be in the Schwartz class S(R) if α, β N 0 : ρ α,β (f) := sup x α β f(x) <. x R Note that (x e x2 ) and any compactly supported C (R)-function is in S(R). Further note that it is easily seen that a function f C (R) lies in S(R) if and only if N, β N 0 C N,β > 0 x R : f (β) (x) C( + x ) N. If f S(R), the property follows by the binomial formula. Conversely, we have x α f (β) (x) C x α which is bounded on R for N α. Let us introduce the following type of topology on (+ x ) N the Schwartz space. Definition.5 (Convergence in the Schwartz space). Let (f n ) n N S(R) and f S(R). S We say that the sequence (f n ) converges to f in S(R), f n f, if α, β N : ρ α,β (f f n ) = sup x α (f f n ) (β) (x) 0, as n. x R NEUER EIL, Benedikt. Am besten die Definitionen des Schwartzraums wiederholen. Remark.6. One may wonder how the sequential convergence defined in Def.?? relates to a topology on the space S(R). In fact, this convergence can be shown to be equivalent to convergence of a sequence in the following metric defined on S(R) d(f, g) = j N 2 j ρ αj,β j (f g) + ρ αj,β j (f g), where j (α j, β j ) is a bijection from N N 0 N 0. See Ex In other words, the local convex topological vector space defined by defined by the seminorms ρ α,β, α, β N 0 as the initial topology for which ρ α,β is continuous is metrizable. Moreover, it can be shown that the above metric even yields a complete space. Proposition.7 (Properties of S(R)). () he inclusions C c (R) S(R) L p (R) are dense in the L p-norm for p [, ). (2) S(R) is continuously embedded in L p (R) for all p [, ]. S Here continuously embedded means that the identity is sequentially continuous, i.e. fn f implies that L f p n f. However, by Remark??, sequential continuity coincides with continuity between the topological spaces.

23 . HE FOURIER RANSFORM AND HE SCHWARZ CLASS 2 (3) For f, g S(R) we have that fg and f g are elements of S(R) as well as α f S(R) with α (f g) = α f g = f α g. Proof. See Ex for (), (2) and (3) without the part on α f. he fact that for f S(R) also α f S(R) simply follows by definition of S(R). o see that α commutes with the convolution consider α = : Since s t f(t s)g(s) is dominated by an integrable function of the form s C( + t s ) 2 g, we conclude that (f g) = f g, which, by symmetry also equals f g by (a consequence of) dominated convergence. he rest follows by induction. Definition.8 (he inverse Fourier transform). For f L (R) the function F (f) defined by F (f)(s) = 2π F(f)( s) = f(s)e ist ds 2π R is called inverse Fourier transfrom of f. Let us show that the latter notion is meaningful. Lemma.9. Let f, g, h S(R). hen () R ff(g) dλ = (Ff)g dλ. R (2) F(F (f)) = f = F (F(f)) (3) f, h = 2π F(f), F(h) (4) Ff, F h = f, h (5) f L2 (R) = 2π F(f) L 2 (R) = 2π F f L2 (R) (Parseval s identity) where f, h = f(s)g(s) ds. R Proof. Assertion () follows by Fubini s theorem. he idea to prove (2) is as follows: Choose a function h L (R) with the following properties h S(R), 2π F(h( 2π ) = h, h(0) =. he existence of such a function will be shown in the exercises (hint: h(x) = ae bx2 for suitable a, b R). he idea is to represent f by an approximate identity, f = lim λ R f(s)h λ(t s) ds and then use () in order to find to prove the equality in (2). By exercise 6.2 and the assumptions on h, h λ = λh(λ ), λ > 0, defines an approximate identity. Hence, by a version of heorem??, we have that f(t) = lim λ (f h λ )(t). On the other hand motivated by idea to choose g such that F(g) = h λ (t ) = λrτ t D λ h we find g(t) = 2π Rm e it D h, where we used the 2πλ basic properties of the Fourier transform and that F(h) = 2πh. By (), we now have (f h λ )(t) = f(s)f(g)(s) ds = (F(f)(s))g(s) ds R R = (F(f)(s))(Rm e it D h)(s) ds 2π λ R = (F(f)(s))e its h( s 2π R 2πλ ) ds Since h(0) =, the last integral converges to 2π R (F(f)(s))eits ds for λ, by dominated convergence. Altogether, we thus have f(t) = (F Ff)(t) for all t R. he other identity follows by applying what we have shown to Rf = f( ). Assertion (3) and (4) follows from () and (2) by setting g = F (h) and noting that F (h) = F(h). Assertion (5) follows direcltly from (3). 2π

24 22 2. HE FOURIER RANFORM heorem.0 (he Fourier transform on S(R)). he Fourier transform is a well-defined linear mapping from S(R) to S(R). Moreover, () F is continuous and bijective on S(R) (with in the natural topology on S(R)) (2) (2π) 2 F is isometric on S(R) with respect to the L 2 -norm, i.e. f S(R) : F(f) L2 (R) = (2π) 2 f L2 (R) hus F is also continuous with continuous inverse on the S(R) equipped with the L 2 - norm. he inverse F on S(R) is given by F. Proof. By Lemma??(2), F is bijective on S(R) with inverse F = F. For the continuity in () see Ex. 6.2 and use the analogous argument for the continuity of F. Assertion (2) follows from Lemma??(5). 2. he Fourier transform on L 2 heorem 2. (he Fourier transform on L 2 ). he definition of the Fourier transform F can be extended to functions in L 2 (R) such that (2π) 2 F is an isometric isomorphism from L 2 (R) to L 2 (R) 2. For functions f in L (R) L 2 (R), this abstract extension coincides pointwise almost everywhere with the action of F. Proof. Let F : L 2 L 2 denote the unique bounded operator which extends the operator F : S(R) S(R) such that F L 2 = F note that we use the density of S(R) in L 2 here. By heorem??, we even have that f L 2 (R) : F(f) L2 (R) = (2π) 2 f L 2 (R) hus (2π) 2 F is an isometric isomorphism. Let f L L 2. It remains to show that F(f) = Ff pointwise almost everywhere. Note that this is not clear a-priori as F is only defined as operator on L 2 (R) and Ff is defined as the L 2 -limit of the Cauchy sequence Ff n where (f n ) n N is a sequence in S(R) which converges to f L 2 (R) in the L 2 -norm, i.e. Ff n Ff L 2 = 0 3. Let g n = fχ { x n} and let f n C (R) with support in [0, n] such that g n f n L 2 (R) = g n f n L2 ([0,n]) < n. hen f f n L2 (R) 0 as n and also for fixed s R, (Ff Ff n )(s) f f n L (R) f g n L (R) + g n f n L (0,n) By definition of g n, it is easy to see that the term f g n L (R) goes to 0 as n since f L. he second term g n f n L (0,n), can be estimated using Cauchy-Schwarz so that we get g n f n L (0,n) n g n f n L 2 (0,n) n 0, (n ). hus Ff n (s) Ff(s) for any s R. On the other hand, we know that L 2 -convergence always implies the existence of a subsequence which converges pointwise almost everywhere (see Ex..0) therefore, Ff nk Ff pointwise almost everywhere. hus, Ff(s) = Ff(s) for a.e. s R. Remark 2.2. Note that isometric isomorphisms on Hilbert spaces are nothing else than unitary operators the latter being defined as operators : X X such that = = I. 2 Note that we identify the operator F with its extension defined on L 2. 3 As we know from Ex..0, L 2 -convergence need not imply pointwise convergence.

25 3. EMPERED DISRIBUIONS 23 heorem 2.3 (Hausdorff Young inequality for the Fourier transform). Let p (, 2) and let q denote the Hölder conjugate of p. he Fourier transform F can be uniquely extended to a bounded operator from L p (R) to L q (R) with where C p = (2π) p. Ff L q C p f L p f L p (R), Proof. his is an application of Riesz horin s theorem since F act as (extension of a) bounded operator on the spaces L L, Ff L f L L 2 L 2, Ff L 2 (2π) 2 f L 2, see the basics of the Fourier transform and heorem??. his thus implies that f L p C p f L q for simple functions f with finite measure support and p, q, θ linked through (??). he constant C p can be computed from C p = θ (2π) θ 2 where θ is given through p = θ + θ 2. hus C = (2π) θ 2 = (2π) p. By density of the simple functions with finite measure support in L p, p <, the assertion follows. 3. empered distributions Definition 3.. Any linear functional u : S(R) C which is continuous with respect to sequential convergence on S(R), i.e. f n S f = u(fn f) 0, as n is called tempered distribution. he set of all tempered distributions is denoted by S (R). Proposition 3.2. A linear functional 4 u : S(R) C is a tempered distribution if and only if there exists k, m N 0 and C > 0 such that u(f) C ρ α,β (f) β m,α k 4 Here linear functional simply refers to a linear mapping that maps to the field of complex numbers.