TOPICS IN HARMONIC ANALYSIS

Transcription

1 TOPICS IN HARMONIC ANALYSIS T.KOMOROWSKI Contents. Fourier Series 3.. Definition 3.2. Why do we bother about Fourier Series? 3.3. Convolution of two functions 3.4. Dirichlet kernel 4.5. The question about the convergence of the Fourier series 4.6. Regularity of a function and the asymptotics of its Fourier coefficients 7.7. Some remarks on the pointwise convergence The L 2 -convergence 2.9. Sobolev spaces 3.. Filters* 7 2. The maximal function. Pointwise convergence Applications of the maximal inequality Lebesgue density theorem Almost everywhere convergence of Fejér sums Interpolation theorems Marcinkiewicz real interpolation theorem Applications Riesz-Thorin complex interpolation theorem Applications Fourier transform on R d The mollifiers and the Schwartz class of functions The definition and basic properties of the Fourier transform The definition of the Fourier transform on L 2 (R d ) Singular operators Examples The Calderon-Zygmund decomposition of a function Calderon-Zygmund theorem on singular operators Applications 5 7. The Sobolev spaces. Elliptic partial differential equations The definition Bessel potentials Proof of Theorem Some useful inequalities Non-increasing rearrangement of a function The proof of Theorem The Sobolev embeddings The Dirichlet problem for a second order elliptic equation More on the Fourier transform. Paley-Wiener theorems H p spaces on a half-plane Finite speed of propagation for solutions of strongly hiperbolic p.d.e-s 84 Date: May 23, 25. File: lecture-final.tex.

2 2 T.KOMOROWSKI References 89

3 HARMONIC ANALYSIS 3. Fourier Series.. Definition. We assume that T := R/ with identification relation x y iff x y = 2πn, n Z. Let f L (T) we let ˆf(n) := f(x)e inx dx T for all n Z. The measure on T is dx := (2π) dx, where dx is the standard Lebesgue measure on R. The formal series (needs not even be convergent) ˆf(n)e inx shall be called the Fourier series of f. Basic properties: P ) T n e inx dx = δ n P 2) if f is a trigonometric polynomial, i.e. f(x) = then a n = ˆf(n). P 3) if f L (T) n N a n e inx ˆf(n) f L..2. Why do we bother about Fourier Series? Consider a periodic, initial value problem (Cauchy problem) for the heat eqt. t u(t, x) = xxu(t, 2 x), t >, x R u(, x) = f(x), where f(x + 2π) = f(x), x R. How can we solve it? Look for the solutions of the form u(t, x) = e kt+inx Note, k = n 2 and if we require u to be 2π-periodic then n Z. The superposition principle yields that u(t, x) = n a(n)e n2 t+inx is a solution to the heat eqt. The initial condition yields f(x) = n a(n)e inx. So the best candidates for the coefficients a(n) are the Fourier ones ˆf(n) Convolution of two functions. Suppose that f, g L (T). We define f g(x) := f(x y)g(y)dy. T Proposition.. P ) For any f, g L (T) we have f g is defined as an element of L (T). P 2) We have f g = g f P 3) f g(k) = ˆf(k)ĝ(k) P 4) f g L (T) f L (T) g L (T), in fact for nonnegative f, g we have equality. P 5) f g L (T) f L (T) g L (T). P 6) For any f L (T) and g C(T) we have f g C(T).

4 4 T.KOMOROWSKI Proof. Exercise..4. Dirichlet kernel. Consider the following trigonometric polynomial D n (x) := e ikx. k n It is called the Dirichlet kernel (of degree n). We introduce also the Fejer kernel (of degree n) Proposition.2. We have A n (x) := n + n D n (x). k= D n (x) = sin[(n + 2 )x] sin x, 2 In addition, we also have A n = n + S n (x; f) := k n ( ) sin[(n + )x/2] 2. sin x 2 ˆf(k)e ikx = f D n (x), F n (x; f) := n S k (x; f) = f A n (x), k= Exercise. Prove the above proposition..5. The question about the convergence of the Fourier series. Does the Fourier series converge? If so, in what sense?.5.. Uniform and the L convergences. Theorem.3. (Fejér) Suppose that f C(T). Then lim F n(f) f =. n + Proof. The theorem follows from the following lemma. Lemma.4. We have (.) A n (x), (.2) for any δ > T A n (x)dx =, (.3) lim sup A n (x) =. n + x δ

5 HARMONIC ANALYSIS 5 Proof. Exercise. Note, from (.2) it follows that for any x T and δ > F n (x; f) f(x) = A n (y)[f(x y) f(x)]dy T A n (y)[f(x y) f(x)]dy [ y δ] + A n (y)[f(x y) f(x)]dy [ y <δ] 2 f A n (y)dy + ω f (δ), [ y δ] where ω f (δ) = sup x y <δ f(x) f(y) is the modulus of continuity of a given function f. Hence, using (.2) lim sup F n (f) f ω f (δ) n + for arbitrary δ > and the conclusion of the theorem follows. Remark. The argument above in fact shows that if we have any family of functions A α L (T), indexed by some parameter α, such that (.4) A α (x), (.5) A α (x)dx =, for any δ > (.6) lim sup A α (x) =. α x δ then Note that we also have T lim α A α f f =, f C(T). (.7) lim α A α f f L (T) = for f L (T). Indeed, for any ε > there exists g C(T) such that f g L write then < ε we can A α f f L A α (f g) L (T) + A α g g L + f g L (T) A α L (T) f g L + A α g g L }{{} (T) + f g L (T) = 2ε + A α f g L (T). Letting δ + we obtain lim sup A α f f L (T) 2ε. δ + Since ε > has been arbitrary we have shown (.7). We shall return to the issue of a.s. convergence and the L p convergence later on...

6 6 T.KOMOROWSKI Example. (Poisson kernel). For any r (, ) we have P r (x) := n r n e inx = r2 re ix 2 and the limit taken here corresponds to r. Exercise. The kernel satisfies the assumptions made in the remark. We have therefore for any f C(T) r n ˆf(n)e inx = P r f(x) f(x), n as r uniformly on T and in L (T) if f L (T). Example 2. (Mollifiers). Another important example is provided by the following construction. Suppose that c (, π) and g : T R is such that ) g, 2) g(x) = for x c > and 3) T gdx = (density). Define a function on R by taking 2π-periodic extension of g. We use the same symbol for its notation. Let (.8) g δ (x) := δg(x/δ), x T. This family satisfies conditions (.4) (.6) with δ +. Then for any f C(T) we have lim g δ f f = and for any f L (T) we have lim g δ f f L δ + δ + (T) =. Corollary.5. (Weierstrass) The set of trigonometric polynomials is dense in C(T). The set of polynomials is dense in C[, ]. Proof. The first part is obvious in light of Theorem.3. As for the other one suppose that f C[, ]. Then (.9) g(x) := f(cos x) belongs to C(T). For any ε > one can find a trig. polynomial g(x) := a n e inx n N such that g g < ε. But, of course also g( x) satisfies this estimate and so does Choose h(x) := N 2 [ g(x) + g( x)] = (a n + a n ) cos nx. }{{} n= =b n h(x) = N b n cos(n arccos x). n= Exercise. Prove that cos (n arccos x), x [, ] is an n-th degree polynomial for a given n. It is called a Chebyshev polynomial and is denoted by T n (x).

7 HARMONIC ANALYSIS 7 Finally observe that h is real valued if f is so. Indeed, then g given by (.9) is real valued and, in consequence a n = g(x)e ixn dx = a n so b n = a n + a n = 2Re a n R. T Corollary.6. (Uniqueness of the Fourier coefficients) Suppose that ˆf(n) = for all n and f L (T). Then, f =. The above means in particular that if two functions f, f 2 L (T) have the same Fourier coefficients then, they are equal. Proof. Suppose first that f C(T). Note that F n (f) =. According to Fejér theorem f is the limit of F n (f), so f =. If f L (T) then for a mollifier g C(T) we have f f g δ L (T) as δ +. But f g δ (n) = f(n)ĝ δ (n) = so f g δ = (note that f g δ C(T)), thus f =..6. Regularity of a function and the asymptotics of its Fourier coefficients. Proposition.7. Let m be an integer. Suppose that f C m (T). Then, (.) for all k Z. f (m) (k) = (ik) m ˆf(k) Proof. Exercise. Suppose that f C (T). We have ˆf(n) = f(x)e inx dx parts = n Hence, More generally, if f C k (T) one gets T ˆf(n) f L (T), n. n T f (x)e inx dx. (.) ˆf(n) f (k) L (T) n k, n. Definition. We say that f Lip α (T) for α (, ), if M > such that f(x) f(y) M x y α. The infimum of M-s satisfying the above condition shall be denoted by f Lipα. Proposition.8. Suppose that f Lip α (T) for some α (, ). Then ˆf(n) f Lip α 2 n α, n Z.

8 8 T.KOMOROWSKI Proof. Note that hence, ˆf(n) = 2π T f(x)e inx dx = f(x)e iπ inx dx = ( f x + π ) e inx dx 2π T 2π T n ˆf(n) = 4π T [ ( f(x) f x + π )] e inx dx n f Lip α 2n α. Question: Can we characterize the condition f Lip α (T) in terms of the asymptotic behavior of ˆf(n)? The answer to this question is given in Theorem.25 below. Theorem.9. (Riemann-Lebesgue) Suppose that f L (T). Then lim ˆf(n) =. n + Proof. If f C (T) the theorem is obvious. Suppose that f L (T). For any ε > one can find g C (T) such that f g L < ε. We have therefore Hence, ˆf(n) ˆf(n) ĝ(n) + ĝ(n) ε 2π + ĝ(n). lim sup ˆf(n) ε n + 2π thus it equals to. One can observe that the more regular a function coefficients the faster its Fourier coefficients decay at infinity. In fact, we can easily conclude the following. Theorem.. Suppose that f C +α (T). Then, lim S n(f) f =. n + Proof. From the previous estimates we get that there exists a constant C >, for which ˆf(n) C, n. n +α Using Weierstrass convergence criterion for function series we conclude that S n (f) g. But Hence g = f. F n (f) = n n S k (f) f. k=

9 HARMONIC ANALYSIS 9.7. Some remarks on the pointwise convergence. Is it true that for f C(T) given x T we have S n (x; f) f(x), as n +? The answer is in the negative. Exercise. Show that there exist < C < C 2 < + such that (.2) C log n D n (x) dx C 2 log n, n 2. T Define s n : C(T) R as s n (f) := S n (; f). It is a bounded linear functional. In fact s n (f) D n (x)f(x) dx C 2 log n f. Choosing we would find that T, D n (x) > f n :=, D n (x) <, s n (f) = T D n (x) dx C log n. This is not entirely correct because f C(T), but that can be easily remedied by taking f n = f except close to zeros of D n (x) and then extrapolate e.g. linearly. We can achieve easily f n and s n (f n ) C log n. 2 Note that we have lim n + s n = +. Hence by the Banach-Steinhaus theorem there must be f C(T) for which lim sup n + s n (f) = +, or equivalently lim sup n + S n (; f) = +. In fact the set of those f-s is residual, i.e. its complement is contained in a union of closed and nowhere dense sets. Remark. It has been shown by Carleson that if f C(T) then S n (x; f) f(x) for x T \ N, where m (N) =. Let ω f (x; δ) := sup[ f(y) f(x) : y x δ] be the modulus of continuity of f at x. As an application of the Lebesgue Jordan theorem we obtain immediately the following. Theorem.. (Dini Theorem) Suppose that for some δ > δ Then, lim n + S n (x; f) = f(x), x T. ω f (x; δ) dδ < +. δ Proof. With no loss of generality assume that x = (otherwise we consider f(y) := f(x+y)). We have sin [(n + /2)y] S n (; f) f() = D n f() f() = [f(y) f()]dy. sin(y/2) Let g(y) := T f(y) f(). sin(y/2)

10 T.KOMOROWSKI We have g L (T) for g(y) dy = Using Theorem.9 we obtain T lim [S n(; f) f()] = n + T f(y) f() dy C sin(y/2) T lim n + T ω f (; y) dy < +. y sin [(n + /2)y] [f(y) f()] =. sin(y/2) Theorem.2. (Dirichlet-Jordan Theorem) If f BV (T) (a function of bounded variation) then S n (x; f) /2(f(x ) + f(x+)) for x T. Proof. Let Sf() := /2(f( ) + f(+)). We have Let We need to show that S n (; f) = 2π (.3) T λ g := 2π as λ +. We have however π π g(y) := π π T λ g = 2π sin [(n + /2)y] f(y)dy. sin(y/2) f(y) sin(y/2). y/2 sin(y(2λ) ) g(y)dy Sg(), y/2 πλ πλ sin(y/2) g (λy) dy y/2 We divide the above integral into two integrals: πλ and πλ. Let Si(y) := π + y sin(z/2) dz. z/2 We have Si() =. Since g( ) is of bounded variation on R we can use integration by parts formula and obtain (.4) 2π πλ sin(y/2) [g (λy) g(+)]dy parts y/2 = G(y) πλ + y= 2π πλ Si(y)d[g (λy) g(+)], where G(y) := [g (λy) g(+)]si(y). Since lim y + Si(y) = and [g (λy) g(+)] is bounded we conclude that lim y + G(y) =. On the other hand the second term on the right hand side of (.4) equals π Si(yλ )dg (y), 2π

11 HARMONIC ANALYSIS which tends to, as λ +, by virtue of the Lebesgue dominated convergence theorem. We have shown therefore that (.5) lim λ + 2π Similarly, we prove that (.6) lim λ + 2π πλ πλ Combining (.5) with (.6) we conclude (.3). sin(y/2) [g (λy) g(+)]dy =. y/2 sin(y/2) [g (λy) g( )]dy =. y/2 Example. Gibbs phenomenon., x [nπ, (n + )π), n even f(x) =, x [nπ, (n + )π), n odd Calculate!" " Figure. Function f(x) ˆf(n) = (.7) S 2N+ (x; f) = n N, n even 2 iπn, n odd. 2e inx iπn = 4 π N n= sin[(2n + )x]. 2n + Calculate the maximum of S 2N+ (x; f). Note that S 2N+(x; f) = 4 N cos[(2n + )x]. π n=

12 2 T.KOMOROWSKI Since we get 2 cos[(2n + )x] sin x = sin[2(n + )x] sin(2nx) S 2N+(x; f) = 2 sin[2(n + )x]. π sin x We have S 2N+ (x N; f) =, where x N = π/[2(n + )]. It is easy to see that the function S 2N+ (x; f) atains a local maximum at x N. In fact, a moment s reflection allows us to conclude that the maximum is global on (, π). To calculate the value of the maximum note that from (.7) we get ( ) 2 N sin 2n+ 2(N+) S 2N+ (x N ; f) = (N + )π 2n+ 2 π sin x dx.7... π 2(N+) x n= Figure 2. Gibb s phenomenon.8. The L 2 -convergence. Suppose that f L 2 (T). The space is Hilbert when equipped with the scalar product (f, g) L 2 (T) = fḡdx. Note that e n (x) := e inx are orthonormal. S n (f) is the orthogonal projection onto the subspace of n-th degree orthogonal polynomials, i.e. f S n (f) e k, k =, ±,..., ±n. As a result we get f 2 L 2 (T) = f S n(f) 2 L 2 (T) + S n(f) 2 L 2 (T) = f S n (f) 2 L 2 (T) + ˆf(k) 2. Also, for m n T k n (.8) S m (f) S n (f) 2 L 2 (T) = m< k n ˆf(k) 2.

13 Corollary.3. We have n ˆf(n) 2 < +. HARMONIC ANALYSIS 3 Theorem.4. We have lim n + f S n (f) L 2 (T) = for any f L 2 (T). Proof. Using (.8) we conclude that S n (f) converges in L 2 (T). Denote its limit by g. However, the above means that F n (f) converges also to g in L 2 (T). According to (.7) we obtain F n (f) f, so f = g. Corollary.5. lim n + f F n (f) L 2 (T) =. Corollary.6. (Parseval s identity) (.9) f 2 L 2 (T) = n ˆf(n) 2. Questions:. Can we claim lim n + f S n (f) L p (T) =, or at least lim n + f F n (f) L p (T) = for f L p (T) and p [, + )? 2. How about the a.s. convergence? 3. Is there a similar duality relation to (.9) for other L p spaces?.9. Sobolev spaces. According to (.) for any f C (T) and a positive integer k we have ˆf(n) C/n k for all n. For a given s R we can define H s (T) - the Sobolev space of order s as the completion of C (T) in the norm (.2) f 2 H s (T) := n ( + n 2 ) s ˆf(n) 2. Let h s (Z) be the space of all complex valued sequences a = (a n ) that satisfy a 2 h s (Z) := n ( + n 2 ) s a n 2. It is clear that h s (Z) is a Hilbert space, when equipped with the scalar product a, b H s (T) := n ( + n 2 ) s a nˆb n, a = (a n ), b = (b n ) h s (Z). Define U : C (T) h s (Z) by Uf := ( ˆf(n)), f C (T). Proposition.7. For any s R mapping U extends to an isometric isomorphism between H s (T) and h s (Z). Exercise. Prove the above proposition. Thanks to the above result, we can identify the element f H s (T) with the corresponding sequence Uf = ( ˆf(n)). To mark that identification we shall write f = ˆf(n)e n inx. Proposition.7 implies in particular that H s (T) is a Hilbert space, when equipped with the scalar product f, g H s (T) := ( + n 2 ) s ˆf(n)ĝ (n), f, g H s (T). n For s we can identify H s (T) with a subspace of L 2 (T) consisting of those elements f = ˆf(n)e n inx for which the norm in (.2) is finite. In the particular case s = we have H (T) = L 2 (T).

14 4 T.KOMOROWSKI Since h s (Z) h s (Z) for s s, using the aforementioned identification between the elements of H s (T) and h s (Z) we can write In fact more is true. H s (T) H s (T), for s s. Proposition.8. Suppose that s s. Then, mapping J : C (T) H s (T) given by Jf := f, f C (T) extends to a compact linear injection. Proof. The fact that J extends to a bounded operator is trivial. Exercise. Show that the extended operator is one-to-one. Only the compactness part requires a proof. For that it suffices to show that K := J ( [f C (T) : f H s (T) ] ) is pre-compact in H s (T). Observe that ẽ n := ( + n 2 ) s /2 e n, e n (x) := e inx, n Z is an orthonormal base in H s (T). Denote by Π N the orthogonal projection in H s (T) onto span {ẽ n, n N}. To show pre-compactnes of K it suffices to prove that it is bounded and for any ɛ > there exists N > such that Π N Jf H s (T) < ɛ, f C (T), f H s (T), see e.g. Corollary IV. 5, p. 26 of [3]. The first fact is trivial, due to the fact that K is contained in the unit ball in H s (T). As for the second, choose an arbitrary ɛ >. Then Π N Jf f, ẽ n H s (T)ẽn = ˆf(n)e n hence n N n N Π N Jf 2 H s (T) = ( + n 2 ) s ˆf(n) 2 ( + n 2 ) s s ( + n 2 ) s ˆf(n) 2 n N ( + N 2 ) s s f 2 H s (T) ( + N 2 ) s s < ɛ n N for a sufficiently large N, as s < s. Suppose that s = k is a positive integer. In light of Parseval s identity and Theorem.7 we conclude that (I + D) k f L 2 (T) = f H s (T), f C k (T). Here I is the identity operator and Df(x) := f (x) is the differentiation operator on C k (T) In particular, we conclude that for any such k we have C k (T) H k (T) L 2 (T). Proposition.9. Given s R and a positive integer k, operator D k extends from C (T) to a bounded linear operator from H s (T) into H s k (T). We call it a generalized differentiation operator of order k.

15 Proof. According to Proposition.7 for any f C (T) we have HARMONIC ANALYSIS 5 D m f(n) = (in) m ˆf(n), n Z. Hence D m f 2 H s m (T) = n ( + n 2 ) s m n 2m ˆf(n) 2 n ( + n 2 ) s ˆf(n) 2 f 2 H s m (T). The conclusion of the proposition follows from density of C (T) in H s (T). Consider now a duality pairing mapping (.2) C (T) C (T) (f, g) f, g L 2 (T) = n ˆf(n)ĝ (n) C. In what follows we denote by (H s (T)) the dual to H s (T). The relationship between H s (T) and (H s (T)) can be expressed as follows. Proposition.2. For any s R the duality pairing map (.2) extends continuously to a mapping from H s (T) H s (T) into C. This extension, denoted by, L 2 (T), satisfies a generalized Parseval equality (.22) f, g L 2 (T) = n ˆf(n)ĝ (n), (f, g) H s (T) H s (T). The mapping Ω : H s (T) (H s (T)) g, g L 2 (T) is an isometric isomorphism between H s (T) and (H s (T)). Proof. From the Cauchy-Schwartz inequality we get (.23) f, g L 2 (T) f H s (T) g H s (T), f, g C (T). This proves that the mapping defined in (.2) is uniformly continuous on C (T) C (T). The possibility of its continuous extension to H s (T) H s (T) follows immediately from density of C (T) in both H s (T) and H s (T). Equality (.22) is a direct consequence of (.2). For any g H s (T) we have Ω(g) = sup{ f, g L 2 (T), f H s (T)= = } { = sup ˆf(n)ĝ (n), } ( + n 2 ) s ˆf(n) 2 = = g H s (T), n n which proves that Ω is an isomorphic embedding. To prove that it is onto suppose that Φ is a linear functional on H s (T). Since H s (T) is Hilbert, according to Riesz representation theorem, there exists g H s (T) such that Φ(f) = f, g H s (T), Define g = n ĝ(n)einx as the element corresponding to f H s (T). ĝ(n) := ( + n 2 ) s ĝ (n), n Z.

16 6 T.KOMOROWSKI One can easily check that ( + n 2 ) s ĝ(n) 2 = g 2 H s (T) < +, n therefore g H s (T). From (.22) Ω(g)(f) = ˆf(n)ĝ (n) = ( + n 2 ) s ˆf(n)ĝ (n) = Φ(f), f H s (T). n n Given α (, ) denote by C k,α (T) the subset of C k (T) made of such functions that f (k) Lip α (T). We let also C,α (T) := Lip α (T). It is a Banach space when equipped with the norm f k,α := f + f (k) + f (k) Lipα. For k = we omit f (k) from the definition. Recall also that, given s R symbols [s] and (s) := s [s] denote the integer and fractional parts of s. Theorem.2. Suppose that s > /2. Let α := (s /2). Then for any α (, α ) there exists a constant C > such that (.24) f [s /2],α C f H s (T), f C (T). Any f H s (T) has therefore its representative in C [s /2],α (T). Proof. The conclusion about the existence of an appropriate representative follows from the fact that C [s /2],α (T) is complete in the norm [s /2],α. Only (.24) requires a proof. To simplify the notation we write a := ( + a 2 ) /2, a R. Also, we prove the result for s such that [s /2] =. The general case can be handled in a similar way by considering f ([s /2]) instead of f. We have (.25) f(x) = e ixn ˆf(n) n s ˆf(n) n s. n n Using Cauchy-Schwartz inequality we obtain that the right hand side is estimated by ( ) /2 ( ) /2 (.26) f n 2s ˆf(n) 2 n 2s C f H s (T), f C (T). n n Furthermore, we have f(x + h) f(x) = [exp {ihn} ] e ixn ˆf(n) (.27) n n s exp {ihn} ˆf(n) n s. Since e ix x and e ix 2 for x R we conclude that n e ix 2 α x α

17 HARMONIC ANALYSIS 7 and the utmost right hand side of (.27) can be estimated by C n n s+α h α ˆf(n) n s C h α ( n n 2s+2α ) /2 ( n ˆf(n) 2 n 2s ) /2, by virtue of the Cauchy-Schwartz inequality. Note that both series on the right hand side converge, due to the fact that 2α 2s < and the definition of the norm f H s (T). We have shown therefore that (.28) f Lipα C f H s (T), f C (T). Corollary.22. For any positive integer k the space H k (T) can be continuously embedded into C k,α (T) for an arbitrary α (, /2). Theorem.23. (Poincare inequality) We have (.29) f 2 L 2 (T) f (x) 2 dx + f(x)dx T T 2, f H (T). Proof. Suppose first that f C (T). By Parseval s identity (.9) and formula (.) we can write f (x) 2 dx = n 2 ˆf(n) 2 = f 2 L 2 (T) + ( n 2 ) ˆf(n) 2 T n n f 2 L 2 (T) 2 f(x)dx. T Inequality extends to the entire H (T) by the density argument... Filters*. Fourier coefficients of D n (x):, k > n ˆD(k) =, k n. Fourier coefficients of A n (x): Â(k) =, k > n n+ k n+, k n. Exercise. Suppose that m n we let A n,m (x) := k m  n (k)e ikx.

18 8 T.KOMOROWSKI!n n Figure 3. Filter D n (x)!n n Then, Figure 4. Filter A n (x) A n,m = n m n + D m + m + n + A m. In consequence (see the first example in Section.7), (.3) A n,m L C n m n + log m + m + n +, which implies that (.3) sup A n,m L C log n. m n Theorem.24. (Bernstein) Suppose that f is a trigonometric polynomial of degree n then f n f.

19 HARMONIC ANALYSIS 9 Proof. We shall present a simple proof with the constant 2n in place of n. The proof with the proper constant n is a bit more involved, see []. We have f = K n + f + Kn f, where, k > 2n, or k < ˆK n + (k) = k, k n, 2n k, n < k 2n, ˆK n (k) = ˆK + n ( k). We show the estimate for K + n f for the other case is analogous. Note that K + n (x) = ne inx A n (x) so by virtue of Proposition. P 5) we obtain K + n f K + n L f and K n + L = n A + n L = n. Let V n := 2A 2n A n, n, V := A. Let W n := V 2 n V 2 n, n, and W := V. Let m 2 n+ and W n,m = A m W n. Using (.3) we conclude that for m 2 n+ (.32) W n,m L A m L W n L 3. 2!(2n!)!(n!) n! 2n! Exercise. Show that for f C(T) Figure 5. Filters 2A 2n (x) and A n (x) (.33) f = n W n f = lim n + V 2 n f and the convergence is uniform. The Hölder class L α := [f : C > f(x + h) f(x) Ch α ], when α (, ).

20 2 T.KOMOROWSKI! 2 n! 2 n! 2 n! 2 n Figure 6. Filter V 2 n(x) n n! n!2 n+ n!! 2! n+ Figure 7. Filter W n (x) Theorem.25. Suppose that α (, ). Then, the necessary and sufficient condition for f L α is that (.34) C > n : W n f C2 nα. Proof. Suppose that (.34) holds. Then, the series in (.33) converges uniformly and g = n W n f C(T). On the other hand we have ĝ(k) = so, in fact, f = g. Suppose also that lim ˆV 2 n(k) ˆf(k) = ˆf(k) n + (.35) h [2 N, 2 N ). We can write f(x + h) f(x) n W n f(x + h) W n f(x)

21 HARMONIC ANALYSIS 2 W n f(x + h) W n f(x) + W n f(x + h) W n f(x). N n N<n In the case when n > N we use (.34) and obtain that the second sum can be bounded from above by W n f(x + h) W n f(x) C 2 nα C 2 Nα Chα. n>n n>n As for the first sum we use Bernstein s theorem.24 to estimate by W n f(x + h) W n f(x) (W n f) h 2 n+2 W n f h C2 n( α) h and, in consequence, N n (.8) C2 (n N)( α) h α W n f(x + h) W n f(x) Ch α C N n 2 n( α) h α Ch α. N n 2 (n N)( α) Thus, f L α. Conversely, suppose that f L α, then f(x + h) f(x) Ch α. Note that W n f(x) = 2 n k 2 n+ Ŵ n (k) ˆf(k)e ikx 2 (eikh + e ikh ) 2 (e ikh + e ikh ). Exercise. Suppose that ˆf() =, λ k = λ k for all k > and g = ikx λ k ˆf(k)e Hint: Show that k N N 2 (.36) g = λ N S N f + (λ N λ N )NF N f + (λ k + λ k+2 2λ k+ )(k + )F k f. Prove first (.36) holds for f such that ˆf(k) = for all k <. In general case decompose first f = f + + f, where f + = ˆf(k)e N k> ikx and f = ˆf(k)e N k< ikx. Let λ k := [2 (e ikh + e ikh )] and M n f(x) = Ŵ n (k) ˆf(k)e ikx [2 (e ikh +e ikh )] = W n [f(x+h)+f(x h) 2f(x)]. 2 n k 2 n+ Using formula (.36) we can obtain (.37) W n f(x) = λ 2 n+m n f(x) + (λ 2 n+ λ 2 n+ )2 n+ A 2 n+ M n f(x) 2 n+ 2 + (λ k + λ k+2 2λ k+ )(k + )A k M n f(x). k=2 n Exercise. Show that for any complex valued function F : [a, b] C that belongs to C 2 [a, b] we have k= (.38) F (x + h ) F (x) sup F h

22 22 T.KOMOROWSKI (.39) F (x + h ) + F (x h ) 2F (x) 2 sup F h 2 for any x h, x, x + h [a, b], h >. Using (.39), with F (s) = [2 (e i(k++s)h + e i(k++s)h )], h =, we can estimate by (.4) W n f(x) λ 2 n+ M n f(x) +2 n+ h sup 2 2 (e i(2n+ +s)h + e i(2n+ +s)h ) 2 A 2 n+ M n f(x) s [,] 2n+ 2 +2h 2 Lemma.26. For any m k=2 n (k + ) sup s [,] [4 2 (e i(2n+ +s)h + e i(2n+ +s)h ) 3 ] + sup [2 2 (e i(2n+ +s)h + e i(2n+ +s)h ) 2 ]} A k M n f(x), s [,] 2 (e i(k+s)h + e i(k+s)h ) m (kh) 2m. Proof. Exercise. Since A k M n f = W k,n [f( + h) + f( h) 2f( )], according to (.32), we have (.4) A k M n f 2 sup W n,m L f( + h) f( ) Ch α m k Let h = 2 n. Note that (.42) 2 (e i(2n+ +s)2 n + e i(2n+ +s)2 n ) = 4 sin 2 { 2 ( + s )} 2n+ for s [, ] Taking all of the above into account we get from (.4) (for h = 2 n ) C > (.43) W n f(x) C W n L f( + 2 n ) + f( 2 n ) 2f( ) +2 n+ h sup [2 2 (e i(2n+ +s)2 n + e i(2n+ +s)2 n ) 2 ] A 2 n+ M n f s [,] 2n+ 2 +2h 2 k=2 n (k + ){ sup s [,] [4 2 (e i(2n+ +s)2 n + e i(2n+ +s)2 n ) 3 ] + sup [2 2 (e i(2n+ +s)2 n + e i(2n+ +s)2 n ) 2 ]} A k M n f, s [,] Lemma.26 2 C n+ 2 nα + A 2 2 n+ M n f + C2 2(n+) k=2 n (k + ) A k M n f (.4) C 2 n+ 2 nα + 2 C2 nα + C2 n+ 2 nα (k + ) k 3 C 2 nα. k=2 n Remark. The above result holds also for α =. But in this case one has to modify a little the definition of the space Lip := [f : C > f(x + h) + f(x h) 2f(x) Ch 2 ], see [].

23 HARMONIC ANALYSIS 23 Remark 2. Combining (.4) and (.43) we obtain [ ( W n f C W n f + ) ( 2 n+ + f ) 2 n+ 2f ( )]. Exercise. Use the above remark to prove that if f Lip α then C > n : W n f C 2 n(+α). In fact, for any k if f (k) Lip α then C C > n : W n f 2 n(k+α). Exercise. The converse to the above is also true. Namely, if C C > n : W n f 2 n(k+α). then f (k) Lip α. Hint: Use the fact that f (k) Lip α to conclude C > n : W n f (k) C 2 nα. then use Theorem The maximal function. Pointwise convergence. Our main objective in this section is to show the following. Theorem 2.. Suppose that f L (T). Then lim F n(f) = f, n + We start with a quite useful covering lemma, due to Vitali. It concerns intervals (cubes) in R d. By an open cube centered at c R d and the length 2s > we understand I = d j= (c j s, c j + s). For a > denote by ai := d j= (c j as, c j + as). Its volume equals ai = a d I. Lemma 2.2. Suppose that K R d is a compact set and {I α } is its covering with open cubes. Then there exists a finite family of cubes {I αi } i=,...,n such that i) I αi I αj =, if i j, ii) K N i= 3I α i. Proof. Since K is compact there is a finite subfamily F of {I α } that covers K. Choose I α as the cube of F that has the largest volume. In case of a tie choose any one of those intervals with the largest volume. Among the cubes of F choose a subfamily F 2 of cubes disjoint with I α. If it is non-empty pick the cube of F 2 that has the largest volume, etc. Suppose that N is the largest natural such that F N. In the described above way we construct a family of cubes {I αi } i=,...,n that clearly satisfies i). We show that it must also satisfy ii). Suppose that x K then there is I F such that x I. If I = I α there is nothing to prove. If otherwise, then either x F 2, or I I α. In the latter case though we must have I I α so a.e.

24 24 T.KOMOROWSKI x 3I α. In the first case we repeat the above procedure replacing F with F 2. After a finite number of steps we convince ourselves that ii) must hold. Exercise. Was the assumption that K is compact important? Substantiate your answer. For any cube I denote by I(c) its center. Definition 2.3. ( Maximal function of Hardy Littlewood) Suppose that f L loc (Rd ). For any x R d define Mf(x) := sup f(y) dy. x=i(c) I I Theorem 2.4. ( The maximal inequality of Hardy-Littlewood) For any f L (R d ) and λ > we have (2.) m d (Mf > λ) 2 3d f(y) dy. λ i= i= I i [ f >λ/2] Proof. Let E λ := [Mf > λ]. Suppose that K E λ is compact. For any x K there is I x such that x = I x (c) and f(y) dy > λi x. I x Clearly {I x } x K covers K. According to Lemma 2.2 one can find {I i } a finite subfamily of disjoint cubes such that K N i= 3I i. We can write therefore that N N m d (K) 3 d I i 3d 3 d f(y) dy f(y) dy. λ λ Since K was chosen arbitrarily we conclude, from Radon property of the Lebesgue measure, that the following inequality holds (2.2) m d (Mf > λ) = sup K E λ m d (K) 3d λ f L (R). Finally we show how (2.2) implies (2.). Define f(x), if f(x) > λ 2, (2.3) f(x) :=, if f(x) λ 2. Note that R d therefore We have f(x) f(x) + λ 2 Mf(x) M f(x) + λ 2. m d (Mf > λ) m d ( M f > λ 2 ) 2 3d λ R d f(y) dy.

25 HARMONIC ANALYSIS 25 The last inequality follows from (2.2). Using the definition of f we conclude (2.). and Remark. Suppose that supp f B R () for some R >. Then, for x 2R we have I = Mf(x) d (x j 2 x, x j + 2 x ) B R () j= (2 x ) d f(y) dy = I (2 x ) d f L (R d ). This implies that Mf L (R d ). In fact one can show that if f L (R d ) does not identically vanish then we have Mf L (R d ). One can prove, see p. 3 of [7], that there exists C > such that Mf(x) C, x. x d However we can bound the L loc norm of Mf. Recall that supp f B R() for some R >. We have then Mfdx m d (B R ()) + (Mf ) + dx B R () = m d (B R ()) + + B R () m d (Mf λ)dλ m d (B R ()) d + dλ λ [f λ/2] f dy = m d (B R ()) d f dy 2 f dλ λ m d(b R ()) d (log 2 + log + f ) f dy. Remark 2. The corresponding version of the maximal inequality on the torus T d := T }. {{.. T } can be proved in the same way as in the case of R d. We have the following d times estimate. Theorem 2.5. ( The maximal inequality of Hardy-Littlewood) For any f L (T d ) and λ > we have (2.4) m d (Mf > λ) 2 3d f(y) dy λ and (2.5) m d (Mf > λ) 3d λ [ f >λ/2] T d f(y) dy

26 26 T.KOMOROWSKI 3. Applications of the maximal inequality 3.. Lebesgue density theorem. Suppose that f L (R). Let also (3.) F (x) := We shall prove the following. Theorem 3.. We have x f(y)dy. F (x + h) F (x) (3.2) lim = f(x). h h for a.e. x. Note that when f C c (R) the above result is obviously true. For any f L (R) there exists a sequence (f n ) C c (R) such that lim n + f f n L (R) =. Denote by F n the antiderivative corresponding to f n by (3.). Of course (3.2) would follow if we could write F (x + h) F (x) (3.3) lim = lim h h h F n (x + h) F n (x) = lim lim n + h lim n + F n (x + h) F n (x) h = lim n + f n(x) = f(x). h The question is whether we could interchange the order of taking the limits lim and h The proof of Theorem 3.. Let c (, 2] and (3.4) S c f(x; h) := sup I I I f(y) f(x) dy, lim. n + where the supremum is taken over all subintervals I [x h, x + h] such that I ch. It suffices only to show that (3.5) lim h + S cf(x; h) = for a.e. x. Let We show that [ E n (f) := x : lim sup S c f(x; h) h + n (3.6) m(e n (f)) = for any n, which of course implies (3.5). Note however that for any g C c (R) we have E n (f) = E n (f g). Choose ε > arbitrary and find g so that f g L (R) < ε. For any function u L (R d ) and I as in the definition of we have (3.7) I I u(y) u(x) dy I (3.8) 2 Mu(x) + u(x) c and in consequence for any c (, 2] we have I u(y) dy + u(x) ch ]. x+h (3.9) S c u(x; h) 2 Mu(x) + u(x), x R and h >. c x h u(y) dy + u(x)

27 HARMONIC ANALYSIS 27 Using the above estimate, with u := f g, we can write ( (3.) m(e n (f)) = m(e n (f g)) m M(f g) c ) ( + m f g > ). 4n 2n Using the maximal inequality, see (2.2), we obtain that the first term on the right hand side of (3.) can be estimated by 2n c f g L (R) < 2n c ε. The second term can be estimated by Chebyshev s inequality by Summarizing we showed that 2n f g L 2 (R) < 2nε. ( m(e n (f)) < 2n 6 + ) ε c for any ε > and n. Thus (3.6) follows Almost everywhere convergence of Fejér sums. Theorem 3.2. Suppose that f L (T). Then lim F n(f) = f, n + a.e. Proof. Again it suffices only to prove that (3.) lim A N (x y) f(y) f(x) dy = N + T for a.e. x. Let [ E n (f) := x T : lim sup A N (x y) f(y) f(x) dy ]. N + T n As before we need to show that (3.2) m(e n (f)) =, n. To this purpose we show the following. Lemma 3.3. There exists a constant C (, + ) such that Proof. We have A N f(x) CMf(x), x T, f L (T), N. (3.3) A N (x) (N + ) π2 4 π 2 (N + )x 2 Exercise. Show the above estimate. Hint: Prove that 2 π sin x x, x π 2 and sin x x, x R., x T.

28 28 T.KOMOROWSKI For k and N let We can write A N f(x) (3.3) (N + ) π2 4 T Z k := [ y : 2 k ] 2k+ x y. N + N + A N (x y) f(y) dy = + x y /(N+) k:2 k π(n+) f(y) dy + π2 2 Mf(x) + π2 π2 2 Mf(x) + π2 k:2 k π(n+) and the conclusion of the lemma follows. x y /(N+) Z k A N (x y) f(y) dy k:2 k π(n+) k:2 k π(n+) N + 2 2k N + 2 2k A N (x y) f(y) dy Zk π 2 (N + )(x y) 2 f(y) dy Z k f(y) dy [ x y 2 k+ /(N+)] π2 2 Mf(x) + 8π2 Mf(x) 2 k k f(y) dy Define (3.4) F (x; f) := sup F N (x; f). N Corollary 3.4. There exists constants C, c (, + ) such that (3.5) m[f (f) > λ] C f dx, f L (T), λ >. λ [ f >cλ] The remainder of the proof is exactly the same as in the previous case, as we can L approximate f by elements g C(T), for which Theorem.3 holds. Exercise. Finish the proof of Theorem Interpolation theorems 4.. Marcinkiewicz real interpolation theorem. We start with the following simple observation. Lemma 4.. Suppose that (X, Σ, µ) is a measure space and F : X [, + ] is a measurable function. Then, for any p (4.) f p dµ = p + λ p µ[f λ]dλ.

29 Proof. We have hence, p + + = p Fubini HARMONIC ANALYSIS 29 µ[f λ] = λ p µ[f λ]dλ = p [F λ] dµ + λ p [F λ] dλ dµ = p λ p [F λ] dλdµ F λ p dλ dµ = F p dµ. Denote by L (L (X, Σ, µ)) the space of all measurable functions f : X [, + ]. Suppose that p < p 2 < + and T : L p + L p 2 L is a subadditive operator, i.e. (4.2) T (f + g) T f + T g, f, g L p + L p 2. Definition 4.2. We say that the operator T is of type (p, q) if there exists + > C > such that T f L q C f L p f L p. The infimum of those C for which the above inequality holds shall be denoted by T p,q. Definition 4.3. Suppose that p + and q +. We say that the operator T is of weak (p, q) type if a) in case when q < + there exists a constant C > such that (4.3) µ[ T f λ] C f q L p λ q, f L p. Define the infimum of such C-s as T w p,q. b) in case when q = : the operator is of type (p, ). Exercise. We define as L p w the Lorentz space that consists of all functions f that satisfy (4.4) µ[ f λ] C f p L p λ p, λ > and some C >. i) Prove that L p L p w, ii) Is it true that L w(, ) \ L (, )? iii) Define f w L p as the infimum of C-s such that (4.4) holds. Can you compare it with the standard L p norm? iv) Check that f + g w L 2( f w L + g w L ). v) The triangle inequality though needs not to hold. Consider the example f = x, g = x on L w(, ). Exercise 2. Prove that for p p p 2 we have L p (X, Σ, µ) L p (X, Σ, µ) + L p 2 (X, Σ, µ).

30 3 T.KOMOROWSKI The following result holds. Theorem 4.4. (Marcinkiewicz interpolation theorem) Suppose that p < p 2 < + and T is a subadditive operator acting on L p + L p 2 is of both weakly (p, p ) and (p 2, p 2 ) types. Assume also that p (p, p 2 ) and p = θ + θ p p 2 for some θ (, ). Then, we have T (L p (L p + L p 2 )) L p and T f L p C f p. The constant C depends only on T w p,p, T w p 2,p 2 p p, p 2 p. that Proof. For any λ > set f λ := f [ f λ] and f (λ) := f [ f >λ]. We have µ[ T f 2λ] µ[ T (f λ ) λ] + µ[ T (f (λ) ) λ] Since the operator T is both weak (p, p ) and (p 2, p 2 ) types we can write that the right hand side of the above inequality can be further estimated from above by ( ) ( ) C λ p f p dµ + C 2 [ f >λ] λ p f p 2 dµ. 2 [ f λ] We obtain therefore + ( + ) λ p µ[ T f 2λ]dλ C λ p p dλ f p dµ [ f >λ] ( + ) +C 2 λ p p2 dλ f p 2 dµ [ f λ] f + C f p dµ λ p p dλ + C 2 f p 2 dµ λ p p2 dλ f C f p dµ + C 2 f p dµ p p p 2 p and the theorem follows from Lemma 4.. Remark. When T is linear it extends continuously from Π p := L p (L p + L p 2 ) to the entire L p. Indeed, according to the above theorem we have T f T g L p = T (f g) L p C f g L p thus, T is continuous and allows for extension to the L p closure of Π p, which is L p. A similar conclusion can be reached for T f if T is only sublinear, i.e. T (αf + βg) α T (f) + β T (g), f, g L p, α, β R. Exercise. Theorem 4.4 extends also to the case when p 2 = +

31 HARMONIC ANALYSIS Applications L p bounds on the maximal function. For any p (, + ) there exists a constant C p (, + ) such that Mf L p C f L p, for all f L p (R). For the proof it suffices only to note that M(f + g) Mf + Mg and that Hardy-Littlewood inequality guarantees that M is of weak (, ) type. Of course, directly from the definition we conclude that it is also of (, ) type so the assertion follows from the Marcinkiewicz interpolation theorem L p convergence of Fejér sums. As a direct consequence of (3.5) and the Marcinkiewicz interpolation theorem we obtain that for any p (, + ) there exists a constant C p (, + ) such that (4.5) F (f) L p (T) C f L p (T), for all f L p (T), where F has been defined in (3.4). Let ε > be arbitrary. For any f L p (T) there exists g C(T) such that f g L p (T) < ε. We can write therefore F N (f) f L p (T) F N (f g) (f g) L p (T) + F N (g) g L p (T) F N (f g) L p (T) + f g L p (T) + F N (g) g F (f g) L p (T) + ε + F N (g) g. Using (4.5) and Theorem.3 we get Combining the above with (.7) we conclude. Theorem 4.5. For any p [, + ) we have lim sup F N (f) f L p (T) (C + )ε. N + lim F N(f) f L N + p (T) =, f L p (T) Riesz-Thorin complex interpolation theorem. We have the following maximum principle. Theorem 4.6. (Hadamard s three-lines theorem) Suppose that Ω := [z : a < Re z < b] for some reals a < b and f : Ω C is a continuous function, holomorphic on Ω. Assume also that there exist α < π(b a) and A < + such that (4.6) f(z) exp {A exp(α Im z )}, z Ω and (4.7) f(a + iy) M, f(b + iy) M, y R. Then, (4.8) f(z) M, z Ω.

32 32 T.KOMOROWSKI Example. Note that one cannot improve on the value of α. Take a = π/2, b = π/2 and f(z) = exp(e iz ). We have f( π/2 + iy) = exp(e iπ/2 e y ) = exp( ie y ) and f( π/2 + iy) = and f(π/2 + iy) =. On the other hand, f(iy) = exp(e y ) + when y. The proof of Theorem 4.6. Let g(z) := M f((a + b)/2 + (b a)π z). We have g : Ω C, where Ω := [z : π/2 < Re z < π/2], (4.9) g( π/2 + iy), g(π/2 + iy), y R and (4.) g(z) exp {A exp(α Im z )}, z Ω for some α < and A < +. With no loss of generality we can assume that α (, ). It suffices only to show that (4.) g(z), z Ω. For any ε > and β (α, ) we let Denote z = x + iy where x, y R. Note that h ε (z) := exp{ 2ε cos(βz)}. e βy + e βy Re(e iβz + e iβz ) = (e βy + e βy ) cos(βx) δ(e βy + e βy ), with δ := cos(βπ/2) >, since β (, ). Thus (4.2) exp{ 2ε cosh(βy)} h ε (z) exp{ 2εδ cosh(βy)} <, z Ω. hence g(z)h ε (z) for z Ω. Since β > α there exists R > such that Therefore for y R. Hence, for any R R we have Ae α y 2εδ cosh(βy) <, y R. { } g(z)h ε (z) exp Ae α y 2εδ cosh(βy) g(z)h ε (z), z R, where R := [ π/2, π] [ R, R]. Thus, by virtue of the maximum principle for the modulus of an analytic function we conclude that g(z)h ε (z) for all z R and R R. As a result we conclude that g(z)h ε (z) for all z Ω and ɛ >. From this and (4.2) we obtain g(z) h ε (z) = exp{2ε cosh(βy)} for all z Ω and ɛ >.

33 HARMONIC ANALYSIS 33 Letting ε + we conclude (4.), which as we have already observed, implies the assertion of the theorem. The second result gives more information about the behavior of the modulus of a holomorphic function. Theorem 4.7. Suppose that Ω := [z : a < Re z < b] and f : Ω C satisfies the assumptions of Theorem 4.6. For z = x + iy, such that x, y R we let Then, M(x) := sup[ f(x + iy) : y R]. (4.3) [M(x)] b a [M(a)] b x [M(b)] x a, x [a, b]. Remark. The above means that x log M(x) is a convex function. The proof of Theorem 4.7. For t R consider g(z) := f(z)e tz. The function satisfies the assumptions of Theorem 4.6. As a consequence of Hadamard s three-line theorem we conclude that M(x)e tx max{m(a)e ta, M(b)e tb }. Hence (4.4) M(x) max{m(a)e t(a x), M(b)e t(b x) }, t R. We optimize the right hand side of (4.4) over t and find that the smallest value is attained when M(a)e t(a x) = M(b)e t(b x), or and M(a) M(b) = et(b a) (4.5) M(x) [M(a)] (b x)/(b a) [M(b)] (x a)/(b a). Suppose that T : L p (X, Σ, µ) L (X, Σ, µ). Define This quantity can be finite or not. T p,q := sup{ T f L q : f L p }. Theorem 4.8. (Riesz-Thorin) Suppose that p < p 2 +, q < q 2 + and T : L p +L p 2 L is a linear operator such that it is of both (p, q ) and (p 2, q 2 ) types. Then, for any p [p, p 2 ] such that (4.6) p = θ p + θ p 2 for some θ [, ] the operator T extends continuously to an operator of (p, q) type, where (4.7) In addition, q = θ + θ. q q 2 (4.8) T p,q T θ p,q T θ p 2,q 2

34 34 T.KOMOROWSKI Proof. Let We can write (4.9) T p,q = = sup sup f L p= g L q = q + q =. sup T fgdµ sup sup sup f L p=,f> g L q =,g> φ = ψ = = sup For any f, g > define u(z; f, g, φ, ψ) := sup sup f L =,f> g L =,g> φ = ψ = sup T (fφ)gψdµ T (f /p φ)g /q ψdµ. T (f ( z)/p +z/p 2 φ)g [( z)/p +z/p 2 ] ψdµ. The function is holomorphic on the strip [ Re z ]. Let M(x; f, g, φ, ψ) := sup u(x + iy; f, g, φ, ψ), x [, ]. y R For p, q and θ as in (4.6) and (4.7) we can write (4.2) T p,q = sup sup sup f L =,f> g L =,g> φ = ψ = sup M(θ; f, g, φ, ψ). Each function u(z; f, g, φ, ψ) satisfies the hypotheses of Theorem 4.6. Hence we obtain that M(x; f, g, φ, ψ) M x (; f, g, φ, ψ)m x (; f, g, φ, ψ), x [, ]. Taking the supremum on both sides over φ, ψ, f, g, as in (4.2), we conclude estimate (4.8). Remark. The above result can be also worded as follows. Theorem 4.9. Suppose that T is defined on D a dense subset of all L r, r [p, p 2 ] and satisfies T f L q i M i f L p i, i =, 2, f D and (p, q, θ) are as in (4.6) and (4.7). Then, T f L q M θ M θ 2 f L p, f D. Proof. The proof follows from an application of Theorem 4.8 and obvious estimates T pi,q i M i, i =, 2.

35 4.4. Applications. HARMONIC ANALYSIS Young s inequality. Suppose that f L p (R d ), g L q (R d ), p, q, /p+/q = /r for some r then f g L r (R d ) and (4.2) f g L r f L p g L q. Suppose that T (f) := f g for f, g S(R d ). In the same way as in Proposition., P 4) we have T (f) L f L g L and obviously T (f) L f L g L. Using Riesz-Thorin interpolation theorem we obtain T (f) L p f L p g L for any p [, + ] and all f L p. Let S(g) := f g. We have From Hölder inequality so again by interpolation if we obtain where q = θ and (4.2) follows. S(g) L p f L p g L. S(g) L f L p g L p r = θ + θ p = θ p S(g) L r f L p g L q, + θ ( p = θ + θ ) = θ p p = p + r Markov operator with an invariant measure. Suppose that (X, Σ) is a measurable space and B(X) is the space of bounded measurable functions. An operator P : B(X) B(X) is called Markov if P = and P f, if f. A probability measure µ on (X, Σ) is called invariant if P fdµ = fdµ, f B(X). We show that P extends to a contraction on any L p (µ). This is trivially true for p = for f L f f L so f L P f f L and in consequence P f L f L. On the other hand, for any f we have f f f so P f P f and P f dµ P f dµ inv. = f dµ. Hence, P f L f L. Using Riesz-Thorin interpolation theorem we have P f L p f L p for any p [, + ].

36 36 T.KOMOROWSKI Hausdorff-Young inequality. Recall that the Fourier coefficients are given by ˆf(n) = e inx f(x)dx. T So, denoting by ˆf the respective complex number sequence ( ˆf(n)) n Z we obtain ˆf l f L (T). On the other hand, Parseval identity shows that ˆf l 2 = f L 2 (T). Interpolating between these two estimates with the help of the Riesz-Thorin theorem we obtain ( ) /p ˆf l p = ˆf(n) p f L p (T) where /p + /p =. n Weakly singular operators. Suppose that T : L [, ] L [, ] is given by where α [, ). Note that On the other hand, T f L T f L T f(x) := ( sup x [,] ( dx f(y)dy x y α, ) dy x y α f L. ) f(y) dy x y α f L dx x α. Interpolating between these inequalities we obtain that T extends to a bounded operator between T : L p [, ] L p [, ] for any p [, + ]. 5. Fourier transform on R d. Throughout this section all the L p spaces are complex. 5.. The mollifiers and the Schwartz class of functions. Definition 5.. The Schwartz class of functions S(R d ) is defined as the space that contains all functions f : R d C that are of C (R d ) class and sup( + x 2 ) n D k f(x) < + for any n and multi-index k = (k,..., k d ). Here D k f(x) := k... k d d f(x). Definition 5.2. The convolution of functions on R d. Suppose that f, g L (R d ). Define f g L (R d ) by f g(x) := f(x y)g(y)dy. R d

37 HARMONIC ANALYSIS 37 Proposition 5.3. C ) For any f, g L (R d ) the convolution f g is defined as an element of L (R d ). C 2) We have f g = g f C 3) f g L (R d ) f L (R d ) g L (R d ), in fact for nonnegative f, g we have equality. C 4) f g L (R d ) f L (R d ) g L (R d ). C 5) For any f L (R d ) and g C b (R d ) L (R d ) we have f g C b (R d ) L (R d ). Exercise. Prove the above proposition. Suppose that φ Cc (R d ) and φ, R φ(x)dx =. For any δ > define φ d δ (x) := δ d φ(x/δ). Let T δ f(x) := φ δ f(x), f L loc (Rd ). Exercise. Prove that i) T δ f C (R d ) for any f L (R d ), ii) T δ f C c (R d ) for any f L (R d ) that is compactly supported. Exercise. Verify that i) T δ f L f L for any f L (R d ), ii) T δ f L f L for any f L (R d ), iii) conclude from i) and ii) that T δ f L p f L p for any f L p (R d ) and p, iv) using the argument contained in the proof of Lemma.4 conclude that for any f C c (R d ) we have T δ f f, as δ + uniformly on compact sets, v) from iii) and iv) show that (5.) lim δ + T δf f L p = for any f L p (R d ), where p [, + ). Exercise. Show that Cc (R d ) (thus also S(R d )) is dense in any L p (R d ), p [, + ) The definition and basic properties of the Fourier transform. Definition 5.4. Suppose that f L (R d ). The Fourier transform of the function is defined as ˆf(ξ) := e ix ξ f(x)dx, ξ R d. Sometimes we shall also denote the Fourier transform by F(f). R Basic properties. Here we assume always that the minimal assumptions for the existence of the respective Fourier transforms are satisfied. Denote also by j p (x) := x p, x R d, p =,..., d. Then, we have (F) (F2) (F3) Corollary 5.5. F(S) S. ˆf L (R d ) f L (R d ), F(D k f) = i k j k... jk d ˆf, d F(j k... jk d d f) = ik D k ˆf,

38 38 T.KOMOROWSKI Proof. Suppose that f S. Note that for any l, l Z, a multi-index k = (k,..., k d ) we have ( + ξ 2 ) l D k (F3) ˆf(ξ) = ( i) k ( + ξ 2 ) l F(j k... jk d d f)(ξ) (F2) = ( i) k F((I ) l (j k... jk d d f))(ξ). Here, k = d j= k j, := d p= 2 p and, as a consequence of (F), we obtain sup[( + ξ 2 ) l D k ˆf(ξ) ] (I ) l (j k... jk d d f) L (R d ) < +. ξ Exercise. Show (F) (F3). (F4) ˆf C (R d ), where C (R d ) denotes the class of functions g : R d continuous and satisfy lim x + g(x) =. C that are Proof. According to Corollary 5.5 the above result holds for any f S(R d ). Suppose that f L (R d ). Choose an arbitrary ε >. There exists g S(R d ) such that f g L < ε. Note that ˆf(ξ) ĝ(ξ) f g L < ε, so ˆf is a limit, uniform on compacts, of continuous functions. Thus, it must be continuous. In addition, lim sup ξ + ˆf(ξ) ĝ(ξ) = lim sup ˆf(ξ) ε, ξ + therefore, since ε > was arbitrary, we must have ˆf C (R d ). Theorem 5.6. (F 5) The inversion formula. Suppose that ˆf L (R d ) then there exists a version of f that belongs to C (R d ) and satisfies (5.2) f(x) = (2π) d e ix ξ ˆf(ξ)dξ. R d Proof. Since ˆf L (R d ) we can write that the right hand side of (5.2) equals ( ) (5.3) (2π) d e iy ξ f(y)dy dξ = ( ) R d (2π) d e i(x y) ξ f(y)dy dξ R d R d e ix ξ Since the integral in parentheses equals e ix ξ ˆf(ξ) it belongs to L (R d ) so using the Lebesgue dominated convergence theorem we can write that the right hand side of (5.3) equals } ( ) (5.4) (2π) d lim exp { ξ 2 e i(x y) ξ f(y)dy dξ λ + R d λ R d ( } ) Fubini = (2π) d lim f(y) exp { ξ 2 e i(x y) ξ dξ dy λ + R d R d λ Lemma 5.7. For any a R and λ > we have (5.5) exp { ξ2 λ R } e iaξ dξ = πλ exp R d } { λa2. 4

39 HARMONIC ANALYSIS 39 Exercise. Prove the above lemma. Hint: Use the contour integration. Using (5.5) we can write that the right hand side of (5.4) equals ( ) λ d/2 } λ y x 2 lim f(y) exp { dy. λ + 4π R d 4 The remaining part of the proof of the inversion formula can be concluded from the following two facts, that are left out as exercises. Exercise 2. Verify that the family e λ (x) := ( ) λ d/2 } exp { λ x 2 4π 4 (the so called heat kernels) satisfies the assumptions of the remark after Lemma.4. Exercise 3. Applying an analogue of the argument used to show Lemma.4 prove that lim e λ f f L λ + (R d ) = for any f L (R d ). Corollary 5.8. Suppose that ˆf. Then, f. Theorem 5.9. (F 6) (The Fourier transform of a convolution) For any f, g L (R d ) we have (5.6) f g(ξ) = ˆf(ξ)ĝ(ξ). Proof. Note that the left hand side of (5.6) equals ( ) ( ) e iξ x f(x y)g(y)dy dx Fubini = e iξ y g(y) e iξ (x y) f(x y)dy dx R d R d R d R d ( ) = e iξ y g(y) e iξ (y) f(y)dy dx = ˆf(ξ)ĝ(ξ). R d R d (F7) Fourier transform of a space translation. Suppose that a R d and T a f(x) := f(x a), f L (R d ). Then Exercise: prove (F7). F(T a f)(ξ) = e ia ξ ˆf(ξ).

40 4 T.KOMOROWSKI 5.3. The definition of the Fourier transform on L 2 (R d ). Theorem 5.. (F8) Suppose that f, g S(R d ) then (5.7) f(x)g (x)dx = R d (2π) d Proof. The right hand side of (5.7) equals ( ) (2π) d e iξ x f(x)dx R d R d ĝ (ξ)dξ Fubini = (F5) = R d R d f(x)g (x)dx. ˆf(ξ)ĝ (ξ)dξ. ( ) (2π) d f(x)dx e iξ x ĝ (ξ)dξ R d R d Corollary 5.. (F9) The Plancherel identity. In the special case when f = g then (5.8) f(x) 2 dx = R d (2π) d ˆf(ξ) 2 dξ. R d As a consequence of the above equality we conclude that F extends continuously to a unitary map F : L 2 (R d, dx) L 2 (R d, dξ), where dξ := (2π) d dξ. Exercise. Note that for any f L (R d ) L 2 (R d ) we have F(f) = F(f). For the above reason we shall unify the notation and call F(f) the Fourier transform of f L 2 (R d ). Corollary 5.2. Both formulas (5.7) and (5.8) hold for all f, g L 2 (R d ). 6.. Examples. 6. Singular operators Hilbert transform. We have shown that f T f given by T f(x) := x y α f(y)dy is a bounded operator on any L p [, ], p [, + ] when α (, ). One could inquire whether the above holds for α =. Note that we cannot even define then T f for f so we have to proceed with some degree of caution. First, we assume some symmetry that allows to claim that T f = for f =const. To simplify matters assume that f : R R. Let (6.) T f(x) = + f(y)dy x y. Even here we should be a bit careful since we still have not defined the meaning of the integral on the right hand side of (6.). We could use for instance the principal value, i.e. + f(y)dy f(y)dy (6.2) T f(x) = p.v. := lim x y ε + x y, y x ε when the limit of the right hand side exists in an appropriate sense. Such an operator is called the Hilbert transform of a function f. Note that it exists for any function f that is locally Hölder and belongs to L (R). Indeed, note that for any ε > we can write dy x y =. y x ε