Parabolic L p L q estimates

Transcription

1 Parabolic L p L q estimates Dietmar A. Salamon ETH Zürich 7 July 27 Abstract The purpose of this expository paper is to give a self-contained proof of maximal L p /L q regularity for the heat equation on R n, and to explain the role of the Besov space Bq 2 2/q,p for the initial conditions. Contents Introduction 2 2 Proof of Theorem Riesz Thorin and Stein interpolation 6 4 Marcinkiewicz interpolation 22 5 The Calderón Zygmund inequality 27 6 The Mikhlin multiplier theorem 34 7 The Khinchin inequality 4 8 The Littlewood Paley inequality 46 9 Maximal regularity for semigroups 52 Coifman Weiss Transference 6

2 Proof of Theorem. 7 2 Besov spaces 9 3 Besov, Littlewood Paley, Peetre, Triebel 99 4 Besov spaces and heat kernels 2 5 Proof of Theorem.2 3 Introduction This is an expository paper. Its purpose is to give self-contained proofs of the following three theorems. For n N let (t, x) = (t, x,..., x n ) be the coordinates on R n+ and denote the Laplace operator on R n by := n i= 2 / x 2 i. Abbreviate t := / t and i := / x i for i =,..., n. The gradient of a smooth function f : R n C is the function f := ( f,..., n f) : R n C n. Theorem.. For every positive integer n and every pair of real numbers p, q > there exists a constant c = c(n, p, q) > such that every compactly supported smooth function u : R n+ C satisfies the estimate ( ( ) /q t u q L p (R n ) dt c Proof. See page 9. ) /q t u u q L p (R n ) dt. (.) Theorem. leads to the question under which assumption on the initial condition u the solution u : [, ) R n C of the heat equation t u = u, u(, ) = u, (.2) belongs to the space W,q ([, ), L p (R n, C)) L q ([, ), W 2,p (R n, C)). The answer involves the Besov space Bq s,p (R n, C) for < s < 2 and p, q >. This space is the completion of C (R n, C) with respect to the norm f B s,p q := f L p + f b s,p q, (.3) for f C (R n, C), where ( f q sup h r f(x + h) 2f(x) + f(x h) p dx ) q/p R := n b s,p q 2 r sq dr r. (.4)

3 Theorem.2. For every positive integer n and every pair of real numbers p, q > there exists a constant c = c(n, p, q) > with the following significance. Let u C (R n, C) and suppose that u : [, ) R n C is the unique solution of the heat equation (.2) such that u t := u(t, ) is square integrable for all t. Define the number < s < 2 by Then s := 2 2/q. ( ) /q c u b s,p q t u q L p (R n ) dt c u B s,p. (.5) q If < s < (or, equivalently, < q < 2) then the norm u B s,p on the right q can be replaced by the norm u b s,p of the homogeneous Besov space. q Proof. See page 3. Theorem.3. For every positive integer n and every real number p 2 there exists a constant c = c(n, p) > such that every compactly supported smooth function u : R n+ C satisfies the estimate for all T R. Proof. See page. u(t, ) L p (R n ) c ( T t u u 2 L p (R n ) dt ) /2 (.6) Theorem. is called maximal regularity and was proved in the sixties by desimon [] for p = 2 and Ladyshenskaya Solonnikov Uralćeva [2] for p = q. In [2] Benedek Calderón Panzone proved that the assertion is independent of q for general analytic semigroups (Theorem 9.3). In our proof for p = q we follow the approach of Lamberton [2]. A proof of Theorem. for all p and q (which applies to general analytic semigroups and extends the result of Lamberton) can be found in Hieber Prüss [5]. The Besov spaces Bq s,p (R n, C) were introduced in 959 by Besov [3]. Theorem.2 is due to Peetre [3] and Triebel [38, 39] for < q < 2, and to Grigor yan Liu [3, Thm.5 & Rmk.8] for q 2 (see also [7, Thm 6.7] and [23, Thm 5.8]). The present exposition follows the argument in [3]. We will use without proof the theory of strongly continuous semigroups and the basic properties of the Fourier transform. 3

4 Before entering into the proofs we formulate some consequences of these results. The inhomogeneous heat equation on R n with a compactly supported smooth function u : R n C as initial condition and a smooth compactly supported inhomogeneous term f : (, ) R n C has the form t u = u + f, u(, ) = u. (.7) This equation has a unique solution u : [, ) R n C such that u(t, ) is square integrable for all t. This solution can be expressed in the form t u(t, x) = K t (x y)u (y) dy + K t s (x y)f(s, y) dy ds (.8) R n R n for t, where K : R n+ R denotes the fundamental solution of the heat equation. It is given by K(t, x) := K t (x) := (4πt) n/2 e x 2 /4t for t > and by K(t, x) := K t (x) := for t. (.9) Remark.4. (i) A simple computation shows that, for every n N, there exists a constant c = c(n) > such that K t L =, K t L c t, t K t L c t, 2 t K t L c t 2 for all t >. (ii) Let u : R n+ C be a smooth function with compact support contained in [, ) R n and define f t (x) := f(t, x) := t u(t, x) ( u)(t, x) for t R and x R n. Then u satisfies equation (.7) with u = and hence is given by equation (.8). Thus the gradien of u t is given by u t = t K t s f s ds for t. Hence, by (i) and Young s inequality, u t L p t t K t s f s L p ds K t s L f s L p ds t c t s f s L p By Hölder s inequality this implies the estimate (.6) with the exponent 2 on the right replaced by any number q > 2 (and a constant depending on T ). To prove the estimate for q = 2 requires different arguments that will be spelled out in Section 2. 4 ds.

5 The next corollary asserts that the heat equation defines a strongly continuous semigroup on the Besov spaces. The condition s = 2 2/q is not needed for this result but it suffices for our purposes. Corollary.5. Fix an integer n N and real numbers p, q >. Define s := 2 2/q. Then the solutions of the heat equation (.2) on R n define a strongly continuous semigroup S(t) on the Besov space Bq s,p (R n, C) given by { Kt f, for t >, S(t)f := f Bq s,p (R n, C). f, for t =, It is a contraction semigroup with respect to the norm ( /q f p,q := f L p + (K t f) q L dt). (.) p Proof. By Theorem.2 the norm p,q in (.) is equivalent to the norm in (.3). Hence the completion of C (R n, C) with respect to the norm (.) is the Besov space Bq s,p (R n, C). Moreover, by definition, ( /q S(t)f p,q = K t f L p + (K τ f) q L dτ) f p p,q for all t > and all f C (R n, C). This implies that the function t [, ) B s,p q (R n, C) : t S(t)f (.) is continuous for every f Bq s,p (R n, C). (Choose a sequence f k C (R n, C) that converges to f with respect to the norm (.); then the function [, ) Bq s,p (R n, C) : t S(t)f k is continuous for each k and converges uniformly to (.); so the latter is continuous.) This proves Corollary.5. The next corollary implies that the solution to the inhomogeneous heat equation with an inhomogeneous term in L q ([, T ], L p (R n, C)) is a continuous function with values in the appropriate Besov space. Corollary.6. Fix an integer n N, real numbers p, q >, and a compact interval I = [, T ]. Define s := 2 2/q. Then there exists a constant c > such that every smooth function I R n C : (t, x) u(t, x) = u t (x) with compact support satisfies the inequality sup u t B s,p c q tt ( T ( u t ql p + t u t ql p + u t ql p ) dt) /q. (.2) 5

6 Proof. Throughout denote by c = c (n, p, q) the constant of Theorem. and by c 2 = c 2 (n, p, q) the constant of Theorem.2. Fix a smooth function u : [, T ] R n C with compact support and abbreviate u t (x) := u(t, x), f t (x) := t u t (x) u t (x) for t T and x R n. Then for < t T and hence u t K t u L p u t = K t u + t f t L p t K t τ f τ dτ dt t /q ( t f t q L p dt) /q (.3) for t [, T ] by Hölder s inequality. Fix a constant < t T and define { ut K v t := t u, for t t, K t t (u t K t u ), for t > t, { ft, for t t g t :=,, for t > t. Then v t = t K t τ g τ dτ, for all t > and hence, by Theorem., t v t v t = g t ( ) /q ( t v t q L dt c p g t q L p ) /q ( t /q dt = c f t q L dt). p Since v t = K t t (u t K t u ) for t > t, it follows from Theorem.2 that u t K t u b s,p q Combine this with (.3) to obtain ( ) /q ( t /q c 2 t v t q L dt c p c 2 f t q L dt). p t ( T /q sup u t K t u B s,p c q 3 t u t u t q L dt). (.4) p tt where c 3 := c 3 (n, p, q, T ) := c c 2 + T /q. 6

7 It remains to estimate u. First, since we have Second, define u = T T (u t + (t T ) t u t ) dt, u L p T /q u L q (I,L p ) + T /q t u L q (I,L p ). (.5) h t := t u t u t, w t := for t T. Then, by Theorem., t K t τ h τ dτ t w L q (I,L p ) c h L q (I,L p ) = c t u u L q (I,L p ). (.6) Third, K t u = u t w t and so ( T /q (K t u ) q L dt) p t u L q (I,L p ) + tw t L q (I,L p ). (.7) Fourth, since K t L C/t for some constant C = C(n) >, we have ( /q ( ) T (K t u ) q q /q L dt) C u p q Lp. (.8) T By (.5), (.6), (.7), (.8), there is a constant c > such that every compactly supported smooth function I R n C : (t, x) u(t, x) = u t (x) satisfies the estimate ) u p,q c ( u L q (I,L p ) + tu L q (I,L p ) + u L q (I,L p ). (.9) By Corollary.5 the function K t u satisfies the same estimate with the same constant. Moreover, K t u B s,p q c 2 K t u p,q by Theorem.2. This implies ) K t u B s,p c q c 2 ( u L q (I,L p ) + tu L q (I,L p ) + u L q (I,L p ). (.2) for all t [, T ]. The estimate (.2) follows directly from (.4) and (.2). This proves Corollary.6. 7

8 Corollary.7. Let n, p, q, s and I = [, T ] be as in Corollary.6. Then the identity on the space of complex valued smooth functions on I R n with compact support extends to a bounded linear operator L q (I, W 2,p (R n, C)) W,q (I, L p (R n, C)) C(I, Bq s,p (R n, C)). Proof. This follows directly from the estimate (.2) in Corollary.6. The next corollary shows that the result of Corollary.7 is sharp. Corollary.8. Let n, p, q, s and I = [, T ] be as in Corollary.6 and let f L p (R n, C). Then the following are equivalent. (i) f Bq s,p (R n, C) (ii) There exists a function such that u(, ) = f. u L q (I, W 2,p (R n, C)) W,q (I, L p (R n, C)) Proof. If f Bq s,p (R n, C) then the function u t := K t f satisfies the requirements of part (ii) with t u = u, by Theorem.2 and the Calderón Zygmund inequality in Corollary 6.2. That (ii) implies (i) follows immediately from Corollary.7. This proves Corollary.8. Corollary.9. Let n, p, q, s and I = [, T ] be as in Corollary.6 and consider the Banach spaces W q,p := L q (I, W 2,p (R n, C)) W,q (I, L p (R n, C)), F q,p := Bq s,p (R n, C) L q (I, L p (R n, C)). Define the operators D : W q,p F q,p and T : F q,p W q,p by Du := (u(, ), t u u), t (T (f, g))(t, x) := (K t f)(x) + K t s (x y)g(s, y) dy ds R n Then D and T are bijective bounded linear operators and T = D (.2) (.22) Proof. That D is a bounded linear operator follows from Corollary.6 and that T is a bounded linear operator follows from Theorems. and.2. Moreover, it follows from the basic properties of the heat kernel K t that DT (f, g) = (f, g) for every pair of smooth functions f : R n C and g : I R n C with compact support. Hence D T = id and so D is surjective. That D is injective follows from a standard uniqueness result for solutions of the heat equation. Thus D is a bijective bounded linear operator and D = T. This proves Corollary.9. 8

9 2 Proof of Theorem.3 For p = q = 2 the estimates of Theorem. and Theorem.3 follow from a straight forward integration by parts argument. An extension of this argument leads to a proof of Theorem.3. For a compactly supported smooth function f : R n C with mean value zero and a real number p > define f W,p (R n ) := sup φ Re(φf) R n, q := p φ L q (R n ) p. Here the supremum is understood over all nonvanishing smooth functions φ : R n C with compact support. Theorem 2.. Let T > and let u : [, T ] R n supported smooth function. Write C be a compactly and suppose that u t (x) := u(t, x) R n u t (x) dx = for all t [, T ]. Then, for every p 2, T ( ) u T 2 L p (R n ) + u t p 2 u t p 2 u t 2 dt L p (R n ) R n (2.) T u 2 L p (R n ) + (p )2 t u t u t 2 W,p (R n ) dt. The integrand on the left is taken to be zero for each t with u t. Proof. It suffices to prove the assertion under the assumption that u t for all t [, T ]. Define f t (x) := f(t, x) := t u(t, x) ( u)(t, x) for t [, T ] and x R n. Then f t : R n C is a smooth function with compact support and mean value zero for all t [, T ]. Moreover, t u p = p u p 2 Re(u t u), ( u p 2 u) = p 2 u p 2 u + p 2 2 u p 4 u 2 u. 9

10 Hence, by Hölder s inequality, ( u p 2 u) p L = q (R n ) = = ( ( ( R n ( u p 2 u ) p p ) p p R n ( u p 2 u 2) p 2p 2 u p p 2 2p 2 R n u p 2 u 2 ( u p 2 L p (R n ) ) p p ) ( ) p 2 2 u p 2p Rn ) 2. R n u p 2 u 2 This implies d u p = u p 2 Re(u t u) = u p 2 Re(u(f + u)) dt p R n R n R n = u p 2 uf p u p 2 u 2 p 2 u p 4 Re(u 2 ( i u) 2 ) R 2 n R 2 n R n i ( u p 2 u) f L q (R n ) W,p (R n ) u p 2 u 2 R ( ) n (p ) u p 2 L p (R n ) u p 2 u 2 2 f W,p (R n ) u p 2 u 2. R n R n It follows that d d dt 2 u 2 L p (R n ) = u p 2 u p dt p L p (R n ) R n ( ) 2 (p ) u p 2 u p 2 u 2 f W,p (R n ) L p (R n ) R n u p 2 u p 2 u 2 L p (R n ) R n (p )2 2 f 2 W,p (R n ) 2 u p 2 u p 2 u 2. L p (R n ) R n The assertion of Theorem 2. follows by integrating this inequality over the interval t T.

11 The next corollary shows that Theorem.3 holds (for p 2) with the constant c = n(p ). Corollary 2.2. Let p 2 and T. Then every compactly supported smooth function u : R n+ C satisfies the estimates and T u T 2 L p (R n ) n u 2 L p (R n ) + n(p )2 t u u 2 L p (R n ) dt T u T L p (R n ) u L p (R n ) + t u u L p (R n ) dt. Proof. Let u : R n+ C be a compactly supported smooth function. Then the function i u t : R n C has mean value zero for every t. Moreover, it follows directly from the definition and the Hölder inequality that i f W.p (R n ) f L p (R n ) for every compactly supported smooth function f : R n C and every index i =,..., n. Hence it follows from Theorem 2. that T i u T 2 L p (R n ) iu 2 L p (R n ) + (p )2 i ( t u u) 2 W,p (R n ) dt T u 2 L p (R n ) + (p )2 t u u 2 L p (R n ) dt for all i. Since f 2 L = p i if 2 L p/2 i if 2 L p/2 = i if 2 L for p all f C (R n, C), the first inequality follows by taking the sum over all i. The second inequality follows from (.8) with f t (x) := t u(t, x) ( u)(t, x). Namely, T u T L p (R n ) = K T u + K T t f t dt L p (R n ) T K T u L p (R n ) + K T t f t L p (R n ) dt T K T L (R n ) u L p (R n ) + K T t L (R n ) f t L p (R n ) dt T u L p (R n ) + f t L p (R n ) dt. Here the third step follows from Young s inequality and the last step follows from part (i) of Remark.4. This proves Corollary 2.2.

12 Corollary 2.3. For every p 2 and every T > the identity on the space of compactly supported smooth functions on [, T ] R n extends to a continuous inclusion operator W,2 ([, T ], W 2,p (R n, C)) L 2 ([, T ], L p (R n, C)) C([, T ], W,p (R n, C)). Proof. For f C (R n, C) define ( ) /p f W,p (R n ) := f p + f p R n and Then, in particular, f W 2,p (R n ) := ( R n f p + f p + ) /p n i j f p. i,j= f L p (R n ) n/p f W 2,p (R n ). By Corollary 2.2, every smooth function function u : [, T ] R n C with compact support satisfies the inequality u t 2 W,p (R n ) u t 2 L p (R n ) + u t 2 L p (R n ) ( t u s 2 L p (R n ) + t u u L p (R n ) s ) 2 t + n u s 2 L p (R n ) + n(p )2 t u u 2 L p (R n ) (n + ) u s 2 W,p (R n ) + (n(p )2 + T ) (n + ) u s 2 W 2,p (R n ) + 2n 2/p (n(p ) 2 + T ) T s T t u u 2 L p (R n ) ( ) t u 2 L p (R n ) + u 2 W 2,p (R n ) for s t T. Replacing u(t, x) with u(t t, x), we obtain the same inequality for t s T. Integrate the resulting inequality over the interval s T to obtain T sup u t 2 W,p (R n ) c tt ( ) t u 2 L p (R n ) + u 2 W 2,p (R n ), where c := n+ T + 2n 2/p (n(p ) 2 + T ). This proves Corollary

13 Corollary 2.4. Theorems. and.3 hold for p = q = 2. Proof. For p = 2 Theorem 2. asserts that the inequality u T 2 L 2 (R n ) + T u t 2 L 2 (R n ) dt T t u t u t 2 W,2 (R n ) dt holds for every T R and every compactly supported smooth function u : R n+ C such that R n u t = for all t. For T sufficiently large it follows under the same assumption on u that u t 2 L 2 (R n ) dt t u t u t 2 W,2 (R n ) dt. Replace u by i u, use the inequality i f W,2 (R n ) f L 2 (R n ) for every smooth function f C (R n, C), and take the sum over all i to obtain t u t 2 L 2 (R n ) dt (n + ) (n + ) t u u 2 L 2 (R n ) dt + (n + ) i i u 2 L 2 (R n ) dt t u u 2 L 2 (R n ) dt + (n + ) n (n + ) 2 t u t u t 2 L 2 (R n ) dt. i= i u t 2 L 2 (R n ) dt This proves the assertion of Theorem. for p = q = 2 with c = n +. Moreover, it follows from Corollary 2.2 that the assertion of Theorem.3 holds for p = 2 with c = n. This proves Corollary 2.4. Corollary 2.5. Assume < p 2, let u : R n C be a compactly supported smooth function, and define u t := K t u : R n C for t. Thus u(t, x) := u t (x) is the unique solution of (.7) with f = such that u t is square integrable for all t. Then ( u t 2 L p (R n ) dt ) /2 p u L p (R n ). (2.2) 3

14 Proof. Define q := p/(p ) 2 and let g : [, ) R n C be any smooth function with compact support such that g t := g(t, ) has mean value zero for all t. Define v t (x) = v(t, x) by ( ) v t (x) := (K s t g s )(x) ds = K s t (x y) g s (y) dy ds t t R n for t and x R n. Then t v + v = g, v t is square integrable for all t, and Theorem 2. implies that v 2 L q (R n ) (q )2 g t 2 W,q (R n ) dt. Moreover, d Re(v t u t ) = Re(( t v t )u t ) + Re(v t ( t u t )) dt R n R n R n = Re(( t v t )u t ) + Re(v t ( u t )) R n R n = Re(( t v t + v t )u t ) R n = Re(g t u t ). R n Integrate this equation over the interval t < to obtain ( ) Re(g t u t ) dt = Re(v u ) u L p (R n ) v L q (R n ) R n R n ( ) /2 (q ) u L p (R n ) g t 2 W,q (R n ) dt. Since ( u t 2 W,p (R n ) dt ) /2 = sup g ( Re(g R n t u t ) ) dt ( ) /2, g t 2 W,q (R n ) dt where the supremum on the right is over all nonvanishing compactly supported smooth functions g : [, ) R n C such that g t := g(t, ) has mean value zero for all t, it follows that ( ) /2 u t 2 W,p (R n ) dt (q ) u L p (R n ). Since (q ) = (p ), this proves Corollary

15 Corollary 2.6. Assume < p 2, let u : R n C be a compactly supported smooth function, and define u t := K t u : R n C for t as in Corollary 2.5. Then ( ) /2 t u t 2 L p (R n ) dt + n ( ) /2 i u t 2 L p (R n ) dt i= 2n p u L p (R n ). (2.3) Proof. Apply the estimate (2.2) in Corollary 2.5 to the function i u and take the sum over all i to obtain n ( ) /2 n i u t 2 L p (R n ) dt p iu L p (R n ) Since i= it follows that ( ) /2 t u t 2 L p (R n ) dt This proves Corollary 2.6. t u = u = i= n i i u i= n p u L p (R n ). n ( ) /2 i u t 2 L p (R n ) dt i= n p u L p (R n ). Corollary 2.6 is a kind of converse of Corollary 2.3. While Corollary 2.3 asserts (for p 2) that every function in the space W p := W,2 ([, T ], L p (R n, C)) L 2 ([, T ], W 2,p (R n, C)) is a continuous function on the intervall [, T ] with values in W,p (R n, C), Corollary 2.6 asserts (for p 2) that every element u W,p (R n, C) extends to a function in W p that agrees with u at t =. 5

16 3 Riesz Thorin and Stein interpolation Assume throughout that (X, A, µ) and (Y, B, ν) are measure spaces. For p denote by L p (X, µ) and L p (Y, ν) the complex L p -spaces. Also denote by X the set of all equivalence classes of A-measurable step functions f : X C with support of finite measure and by Y the set of all equivalence classes of B-measurable step functions g : Y C with support of finite measure. The equivalence relation in both cases is equality almost everywhere. Whenever convenient we abuse notation and denote by f either an equivalence class of measurable functions on X (respectively Y ) or a representative of the corresponding equivalence class. We begin our exposition with the Riesz Thorin Interpolation Theorem [32, 37]. Theorem 3. (Riesz Thorin). Let p, p, q, q, let T : L p (X, µ) L p (X, µ) L q (Y, ν) L q (Y, ν) be a linear operator, and suppose that there exist positive real numbers c, c such that, for all f L p (X, µ) L p (X, µ), T f L q c f L p, T f L q c f L p (3.) Fix a real number < λ < and define the numbers p λ, q λ, c λ by := λ + λ, p λ p p If q λ = assume that (Y, B, ν) is semi-finite. Then for all f L p (X, µ) L p (X, µ) L p λ (X, µ). Proof. See page 7. := λ + λ, c λ := c λ c λ q λ q q. (3.2) T f L q λ c λ f L p λ (3.3) The proof requires Hadamard s Three Lines Theorem. Define S := { z C Re(z) }. (3.4) Theorem 3.2 (Hadamard Three Lines Theorem). Let Φ : S C be a bounded continuous function that is holomorphic in int(s). Then ( ) λ ( for all λ [, ]. sup Φ(z) Re(z)=λ sup Φ(z) Re(z)= 6 sup Φ(z) Re(z)= ) λ

17 Proof. Define c := sup Re(z)= Φ(z) and c := sup Re(z)= Φ(z). Then the function Ψ n : S C defined by Ψ n (z) := Φ(z) c z e z2 c z n is continuous, is holomorphic in int(s), and converges to zero as z tends to infinity. Hence it attains its maximum on the boundary of S. Since Ψ n (z) e Re(z2 ) n = e Re(z)2 Im(z) 2 n for Re(z) =,, it follows that Ψ n (z) for all z C with R(z). Take the limit n to obtain the inequality Φ(z) c z c z = c Re(z) c Re(z) for all z C with R(z). This proves Theorem 3.2. Proof of Theorem 3.. The proof follows the exposition in [8]. Step. The assertion holds when p = p =: p. In this case p λ = p and it follows from Hölder s inequality and equations (3.) and (3.2) that T f L q λ for all f L p (X, µ). This proves Step. T f λ L q T f λ L q c λ c λ f L p Step 2. Let h L q (Y, ν) L q (Y, ν). Thus h L q λ (Y, ν) and so gh is integrable for all g Y. Define r λ [, ] by let c >, and assume that for all g Y. Then h L q λ c. Y q λ + r λ =, (3.5) gh dν c g L r λ (3.6) Assume first that < q λ < and so < r λ <. Hence Y is dense in L r λ (Y, ν) by [33, Lemma 4.2]. Thus the inequality (3.6) continues to hold for all g L r λ (Y, ν). Define g : Y C by g(y) := h(y) q λ 2 h(y) whenever h(y) and g(y) := otherwise. Then g L r λ (Y, ν) and g L r λ = h q λ L q λ. Hence h q λ L q λ = gh dν Y c g L r λ = c h q λ L q λ and so h L q λ c. 7

18 Next assume q λ = and so r λ =. Then Y is dense in L (Y, ν) by [33, Lemma 4.2] and so (3.6) continues to hold for all g L (Y, ν). Assume, by contradiction, that h L > c. Then there is a δ > such that the set B := {y Y h(y) > c + δ} has positive measure. Since (Y, B, ν) is semifinite, there exists a measurable set E B such that < ν(e) <. Define g by g(y) := h(y) h(y) for y E and by g(y) := for y Y \ E. Then g L (Y, ν) and gh dν = h dν (c + δ)ν(e) > cν(e) = c g L Y E in contradiction to (3.6). Next assume q λ = and so r λ =. Suppose, by contradiction, that h L > c. Since Y is dense in L (Y, ν) there is a k Y such that h k L < h L c. 2 Define g : Y C by g(y) := k(y) k(y) whenever k(y) and g(y) := otherwise. Then g Y, g L =, and gk dν = k Y L. Hence gh dν k L h k L h L 2 h k L > c = c g L Y in contradiction to (3.6). This proves Step 2. Step 3. Let r λ be as in Step 2. Then the inequality (T f)g dν c λ f L p λ g L r λ (3.7) holds for all f X and all g Y. Y This is the heart of the proof of Theorem 3.. Write k l f = a i χ Ai, g = b j χ Bj, (3.8) i= where a,..., a k and b,..., b l are nonzero complex numbers, the A i are pairwise disjoint measurable subsets of X with finite measure, and the B j are pairwise disjoint measurable subsets of Y with finite measure. Here χ A denotes the characteristic function of a set A X and χ B denotes the characteristic function of a set B Y. Choose φ i, ψ j R such that a i = a i e iφ i, for i =,..., k and j =,..., l. 8 j= b j = b j e iψ j

19 For z S define f z : X C and g z : Y C by k f z (x) := a i ( z)p +zp ( λ)p +λp e iφ i χ Ai, g z (y) := i= (3.9) l b j q q ( z)q zq q q ( λ)q λq e iψ j χ Bi j= for x X and y Y. Then f λ = f and g λ = g. Moreover, the function Φ : S C defined by Φ(z) := (T f z )g z dν Y = (3.) a i ( z)p +zp ( λ)p +λp b j q q ( z)q zq q q ( λ)q λq e i(φ i+ψ j ) (T χ Ai ) dν i,j B j for z S is holomorphic. Let z C with Re(z) =. Then for x A i and f z (x) p = a i p p ( λ)p +λp = a i p λ = f(x) p λ r (q )q g z (y) r q = b i q ( λ)q λq = b i for y B j. Hence q q q q ( λ)q λq = b j q λ q λ = g(x) r λ Φ(z) T f z L q g z L r c f z L p g z L r = c f p λ/p L p λ g r λ/r L r λ for all z C with Re(z) =. A similar argument shows that Φ(z) T f z L q g z L r c f z L p g z L r = c f p λ/p L p λ g r λ/r L r λ for all z C with Re(z) =. Hence it follows from Hadamard s Three Lines Theorem 3.2 that ( ) λ ( Φ(λ) sup Φ(z) Re(z)= ( c f p λ/p L p λ g r λ/r L r λ = c λ f L p λ g L r λ. sup Φ(z) Re(z)= ) λ ) λ ( c f p λ/p L p λ g r λ/r L r λ The last equation uses the identities c λ = c λ c λ and p λ as r λ = q λ = λ q λ q = λ r + λ r. This proves Step 3. 9 ) λ = λ p + λ p as well

20 Step 4. T f L q λ c λ f L p λ for all f X. Let f X. Then Step 3 shows that h := T f satisfies the hypotheses of Step 2 with c := c λ f L p λ. Hence the assertion follows from Step 2. Step 5. We prove the theorem. For p = p the assertion holds by Step. Hence assume p p and, without loss of generality, that p <. Then p λ <. Fix a function f L p (X, µ) L p (X, µ). We prove that there exists a sequence f n X such that lim f f n n L p λ =, lim (T f n )(y) = (T f)(y) (3.) n for almost every y Y. To see this assume first that f. Then there exists a monotone sequence of measurable step functions f n : X [, ) such that s (x) s 2 (x) and lim n f n (x) = f(x) for all x X (see [33, Theorem.26]). Since p, p λ < and f L p λ (X, µ) L p (X, µ), the functions f p λ and f p are integrable. Since f n (x) f(x) p λ f(x) p λ and f n (x) f(x) p f(x) p for all x X, it follows from the Lebesgue Dominated Convergence Theorem (see [33, Theorem.45]) that lim f n f n L p λ =, lim f n f n L p =. Since T f n T f L q c f n f L p, we have lim n T f n T f L q =. Hence there exists a subsequence, still denoted by f n, such that T f n converges almost everywhere to T f (see [33, Corollary 4.]). This is the required sequence in the case f. To obtain the result in general apply this argument to the positive and negative parts of the real and imaginary parts of an arbitrary function f L p (X, µ) L p (X, µ). This proves the existence of a sequence f n X that satisfies (3.). Since T f n T f m L q λ c f n f m L p λ for all n, m N, by Step 4, it follows from (3.) that T f n is a Cauchy sequence in L q λ (Y, ν) and hence converges in L q λ (Y, ν). Since T fn converges to T f almost everywhere, its limit in L q λ (Y, ν) agrees with T f. Hence T f L q λ(y, ν) and By (3.), (3.2), and Step 4 we have T f L q λ lim n T f n L q λ = T f L q λ. (3.2) = lim n T f n L q λ This proves Step 5 and Theorem 3.. c λ lim n f n L p λ = c λ f L p λ. 2

21 The Stein Interpolation Theorem in [36] is an extension of the Riesz Thorin Interpolation Theorem, where the operator T is replaced by a holomorphic operator family {T z } z S, parametrized by the elements of the strip S = {z C Re(z) } in (3.4). Denote by L loc (Y, ν) the space of all equivalence classes of measurable functions g : Y C such that the restriction of g to every measurable subset of Y with finite measure is integrable. Recall that X denotes the set of all equivalence classes of A-measurable step function f : X C with support of finite measure. Theorem 3.3 (Stein). Let p, p, q, q and let T z : X L loc(y, ν), z S, be a family of linear operators satisfying the following two conditions. (a) For all f X and all g Y the function S C : z g(t z f) dν is continuous and is holomorphic in int(s). (b) There exist positive real numbers c, c such that Y T it f L q c f L p, T +it f L q c f L p (3.3) for all f X and all t R. Let < λ < and define p λ, q λ, c λ by := λ + λ, p λ p p If q λ = assume that (Y, B, ν) is semi-finite. Then for all f X. := λ + λ, c λ := c λ c λ q λ q q. (3.4) T λ f L q λ c λ f L p λ (3.5) Proof. The proof is a straight forward extension of the proof of Theorem 3.. Namely, let f X and g Y, and define the function Φ : S C by Φ(z) := g z (T z f z ) dν Y for z S, where f z : X C and g z : Y C are given by (3.9) (with f and g given by (3.8)). Then it follows as in Step 3 in the proof of Theorem 3. that Y g(t λf) dν = Φ(λ) c λ f L p λ g L q λ. By Step 2 in the proof of Theorem 3. this implies the assertion of Theorem

22 4 Marcinkiewicz interpolation The Marcinkiewicz interpolation theorem provides a criterion for a linear operator on L 2 (µ) to induce a linear operator on L p for < p < 2. Let (X, A, µ) be a measure space. For a measurable function f : X R define the (Borel measurable) function κ f : [, ) [, ] by κ f (t) := µ(a(t, f)), A(t, f) := { x X f(x) > t }, for t. (4.) Lemma 4.. Let p < and let f, g : X R be measurable functions. Then, for all t >, κ f+g (t) κ f (t/2) + κ g (t/2), (4.2) t p κ f (t) f p dµ = p s p κ f (s) ds. (4.3) X Proof. The inequality (4.2) holds because A(t, f + g) A(t/2, f) A(t/2, g). We prove (4.3) in four steps. Step. t p κ f (t) X f p dµ for all t. Since t p χ A(t,f) f p it follows that t p κ f (t) = X tp χ (A(t,f) dµ X f p dµ for all t. This proves Step. Step 2. If κ f (t) = for some t > then X f p dµ = = t p κ f (t) dt. By Step, we have X f p dµ =. Moreover, t p κ f (t) = for t > sufficiently small and hence t p κ f (t) dt =. This proves Step 2. Step 3. Assume (X, A, µ) is σ-finite and κ f (t) < for all t >. Then equation (4.3) holds. Let B 2 [, ) be the Borel σ-algebra and denote by m : B [, ] the restriction of the Lebesgue measure to B. Let (X [, ), A B, µ m) be the product measure space in [33, Def 7.]. We prove that Q(f) := { (x, t) X [, ) t < f(x) } A B. To see this, assume first that f is an A-measurable step-function. Then there exist finitely many pairwise disjoint measurable sets A,..., A l A and positive real numbers α,..., α l such that f = l i= α iχ Ai. In this case Q(f) = l i= A i [, α i ) A B. Now consider the general case. Then there is a sequence of A-measurable step-functions f i : X [, ) such that f f 2 and f i converges pointwise to f (see [33, Thm.26]). Since Q(f i ) A B for all i, we have Q(f) = i= Q(f i) A B. 22

23 Now define h : X [, ) [, ) by h(x, t) := pt p. This function is A B-measurable and so is hχ Q(f). Hence, by Fubini s Theorem, ( ) f(x) f p dµ = pt p dt dµ(x) X This proves Step 3. = = = X X ( ) (hχ Q(f) )(x, t) dm(t) dµ(x) ( ) (hχ Q(f) )(x, t) dµ(x) dm(t) X pt p µ(a(t, f)) dt. Step 4. Assume κ f (t) < for all t >. Then (4.3) holds. Define X := {x X f(x) }, A := { A A } A X, and µ := µ A. Then the measure space (X, A, µ ) is σ-finite because X n := A(/n, f) is a sequence of A n -measurable sets such that µ (X n ) = κ f (/n) < for all n and X = n= X n. Moreover, f := f X : X R is A -measurable and κ f = κ f. Hence it follows from Step 3 that f p dµ = f p dµ = t p κ f (t) dt = t p κ f (t) dt. X X This proves Step 4 and Lemma 4.. Fix real numbers p q. Then Hölder s inequality implies q p q(p ) p(q ) p(q ) f p f f q (4.4) for every measurable function f : X R and hence L (µ) L q (µ) L p (µ). Since the intersection L (µ) L q (µ) contains (the equivalences classes of) all characteristic functions of measurable sets with finite measure, it is dense in L p (µ) (see [33, Lem 4.2]). The following theorem was proved in 939 by Józef Marcinkiewicz (a PhD student of Antoni Zygmund). To formulate the result it will be convenient to slightly abuse notation and use the same letter f to denote an element of L p (µ) and its equivalence class in L p (µ). 23

24 For a measurable function f : X C define f, := sup tκ f (t) f L. (4.5) t> We emphasize that the map f f, satisfies the weak triangle inequality is not a norm because it only f + g /2, f /2, + g /2,. However the formula d, (f, g) := f g /2, defines a metric on L (R n, C) and the completion of L (R n, C) with respect to this metric is the topological vector space L, (R n, C) of weakly integrable functions (see [33, Section 6.]). Theorem 4.2 (Marcinkiewicz). Let q > and let T : L q (µ) L q (µ) be a linear operator. Suppose there are constants c > and c q > such that T f, c f, T f q c q f q (4.6) for all f L (µ) L q (µ). Fix a constant < p < q. Then ( p(q ) T f p c p f p, c p := 2 (q p)(p ) ) /p c q p q(p ) p(q ) p(q ) c q, (4.7) for all f L (µ) L q (µ). Thus the restriction of T to L (µ) L q (µ) extends (uniquely) to a bounded linear operator from L p (µ) to itself for < p < q. Proof. Let c > and let f L (µ) L q (µ). For t define f t (x) := { f(x), if f(x) > ct,, if f(x) ct, g t (x) := {, if f(x) > ct, f(x), if f(x) ct. Then A(s, f t ) = κ ft (s) = { A(s, f), if s > ct, A(ct, f), if s ct, { κf (s), if s > ct, κ f (ct), if s ct, A(s, g t ) = κ gt (s) = {, if s ct, A(s, f) \ A(ct, f), if s < ct, {, if s ct, κ f (s) κ f (ct), if s < ct. 24

25 By Lemma 4. and Fubini s Theorem, this implies ( ) t p 2 f t dt = t p 2 κ ft (s) ds dt ) = t (ctκ p 2 f (ct) + κ f (s) ds dt and = c p t p κ f (t) dt + = c p t p κ f (t) dt + = c p p p = c p p X t p κ f (t) dt f p dµ ( t p q g t q q dt = t p q = = q = q = cq p p q p = cq p q p ct s/c ) qs q κ gt (s) ds dt t p 2 dt κ f (s) ds (s/c) p p κ f(s) ds ( ct ) t p q qs q (κ f (s) κ f (ct)) ds dt s/c t p q dt s q κ f (s) ds c q t p κ f (ct) dt s p c q p q p κ f(s) ds c q p X t p κ f (t) dt f p dµ. t p κ f (t) dt Moreover, f = f t + g t for all t. Hence, by Lemma 4. and (4.6), κ T f (t) κ T ft (t/2) + κ T gt (t/2) 2 t T f t, + 2q t T g t q q q 2c t f t + (2c q) q 25 t q g t q q.

26 Hence, by Lemma 4. and the identities on page 25, T f p dµ = p t p κ T f (t) dt X p2c t p 2 f t dt + p(2c q ) q t p q g t q q dt ( p2c c p = + p(2c ) q) q c q p f p dµ p q p X = p(q )2p c (q p)/(q ) c (qp q)/(q ) q f p dµ (q p)(p ) Here the last equation follows with the choice of c := (2c ) /(q ) /(2c q ) q/(q ). This proves Theorem 4.2. Theorem 4.2 extends to Banach space valued functions. Here is an example for such an extension that is used in Section 9. Consider the positive real axis equipped with the Lebesgue measure. For a strongly Lebesgue measurable function f : [, ) X with values in a Banach space X define the function κ f : (, ) [, ] by κ f (r) := µ ({ t }) f(t) > r for r >. This function depends only on the equivalence class of f up to equality almost everywhere. Corollary 4.3 (Marcinkiewicz). Fix a real number < q <. Let X be a Banach space and let T : L q ([, ), X) L q ([, ), X) be a linear operator. Suppose there exist positive constants c, c q such that T f L q c q f L q, sup rκ T f (r) c f L r> for all f L q ([, ), X) L ([, ), X). Then ( ) /p p(q ) T f L p c p f L p, c p := 2 c (q p)(p ) X q p p(q ) c q(p ) p(q ) q for < p < q and f L q ([, ), X) L ([, ), X) L p ([, ), X). Proof. The proofs of Lemma 4. and Theorem 4.2 carry over verbatim to Banach space valued functions. 26

27 5 The Calderón Zygmund inequality The next definition is taken from the exposition in Parissis [29]. Definition 5.. Let n N and let n := {(x, y) R n R n x = y} be the diagonal in R n R n. Fix two constants C > and < σ. A Calderón Zygmund pair on R n with constants C and σ is a pair (T, K), consisting of a bounded linear operator T : L 2 (R n, C) L 2 (R n, C) and a continuous function K : (R n R n ) \ n C, satisfying the following axioms. (CZ) T f L 2 C f L 2 for all f L 2 (R n, C). (CZ2) If f : R n C is a continuous function with compact support then the restriction of T f to the open set R n \ supp(f) is continuous and (T f)(x) = K(x, y)f(y) dy for all x R n \ supp(f). (5.) R n (CZ3) Let x, y R n such that x y. Then K(x, y) C x y n. (5.2) (CZ4) Let x, x, y, y R n such that x y, x y, and x y. Then y y < 2 x y = K(x, y) K(x, y ) C y y σ x y n+σ, x x < 2 x y = K(x, y) K(x, y) C x x σ x y n+σ. (5.3) We remark that if (T, K) is a Calderón Zygmund pair then so is (T, K), where the operator T : L 2 (R n, C) L 2 (R n, C) is given by T f = T f +bf for all f L 2 (R n, C) and some bounded measurable function b : R n C. Thus the operator T is not uniquely determined by K. However, it is easy to see that the function K is uniquely determined by the operator T. (Exercise!) Theorem 5.2 (Calderón Zygmund). Fix an integer n N, a real number < p <, and two constants C > and < σ. Then there exists a constant c = c(n, p, σ, C) > such that every Calderón Zygmund pair (T, K) on R n with constants C and σ satisfies the inequality for all f L 2 (R n, C) L p (R n, C). T f L p c f L p (5.4) 27

28 Proof. The proof has four steps. Denote by µ the Lebesgue measure on R n. Step. There is a constant c = c(n, σ, C) with the following significance. Let (T, K) be a Calderón Zygmund pair on R n with the constants C and σ, let B R n be a countable union of closed cubes Q i R n with pairwise disjoint interiors, and let h L 2 (R n, C) L (R n, C) such that h R n \B, h(x) dx = for all i N. (5.5) Q i Then ( κ T h (t) c µ(b) + ) t h for all t >. (5.6) For i N define h i : R n R by h i (x) := h(x) for x Q i and by h i (x) := for x R\Q i. Denote by q i Q i the center of the cube Q i and by 2r i > its side length. Then x q i nr i for all x Q i. Fix an element x R n \ Q i. Then it follows from (5.) that (T h i )(x) = K(x, y)h i (y) dy Q i ( = K(x, y) K(x, qi ) ) (5.7) h i (y) dy. Q i The function h i need not be continuous. Since x / Q i one can approximate h i in L 2 (R n, C) by a sequence of compactly supported continuous functions that vanish near x. For the approximating sequence the first equation in (5.7) holds by (5.); now take the limit. The second equation follows from (5.5). Now choose x R n such that x q i > 3 nr i. Then d(x, Q i ) := inf y Q i x y > 2 nr i and so y q i nr i < x y for all y Q 2 i. Hence, by (5.3) and (5.7), (T h i )(x) K(x, y) K(x, q i ) h i (y) dy Q i sup y Q i K(x, y) K(x, q i ) h i sup y Q i c y q i σ x y h i n+σ c r σ i d(x, Q i ) n+σ h i. Here C is the constant in (5.3) and c := Cn σ/2. 28

29 Define P i := { x R n x qi 3 nr i }. Then d(x, Q i ) x q i nr i for all x R n \ P i. Hence R n \P i (T h i )(x) dx c r σ i = c r σ i = c r σ i = c ω n r σ i ( x q i nr i ) n+σ dx h i R n \P i y >3 nr i ( y nr i ) n+ dy h i ω n r n dr 3 nr i (r nr i ) n+σ h i c 2 n ω n r σ i = c 2 h i. 2 nr i (s + nri ) n ds s n+σ h i 2 nr i ds s +σ h i Here c 2 := c 2 n σ n σ/2 σ ω n and ω n := Vol n (S n ). The third step in this computation follows from Fubini s Theorem in polar coordinates. Thus we have proved that R n \P i (T h i )(x) dx c 2 h i for all i N. (5.8) Recall that T h and T h i are only equivalence classes in L 2 (R n ). Choose square integrable functions on R n representing these equivalence classes and denote them by the same letters T h, T h i. We prove that there exists a Lebesgue null set E R n such that (T h)(x) (T h i )(x) for all x R n \ E. (5.9) i= To see this, note that the sequence l i= h i converges to h in L 2 (R n ) as l tends to infinity and so the sequence l i= T h i converges to T h in L 2 (R n ). By [33, Cor 4.] a subsequence converges almost everywhere. Hence there exists a Lebesgue null set E R n and a sequence of integers < l < l 2 < l 3 < such that the sequence l ν i= (T h i)(x) coverges to (T h)(x) as ν tends to infinity for all x R n \E. Since l ν i= (T h i)(x) i= (T h i)(x) for all x R n, this proves (5.9). 29

30 Now define A := P i. Then it follows from (5.8) and (5.9) that (T h)(x) dx (T h i )(x) dx R n \A = i= R n \A i= i= i= R n \A c 2 h i i= = c 2 h. (T h i )(x) dx R n \P i (T h i )(x) dx Moreover, where Hence µ(a) µ(p i ) = c 3 i= i= µ(q i ) = c 3 µ(b), c 3 = c 3 (n) := µ(b 3 n) µ([, ] n ) = µ(b 3 n/2) = ω n3 n n n/2. 2 n n tκ T h (t) tµ(a) + tµ ({ x R n \ A }) (T h)(x) > t tµ(a) + (T h)(x) dx R n \A c 3 tµ(b) + c 2 h c 4 ( tµ(b) + h ) for all t >, where This proves Step. c 4 := max{c 2, c 3 }. 3

31 Step 2 (Calderón Zygmund Decomposition). Let f L 2 (R n, C) L (R n, C) and t >. Then there exists a countable collection of closed cubes Q i R n with pairwise disjoint interiors such that µ(q i ) < f(x) dx 2 n µ(q i ) for all i N (5.) t Q i and where B := i= Q i. For ξ Z n and l Z define f(x) t for almost all x R n \ B, (5.) Q(ξ, l) := { x R n 2 l ξ i x i 2 l (ξ i + ) }. Let Q := { Q(ξ, l) ξ Z n, l Z } and define the subset Q Q by { Q := Q Q tµ(q) < f(x) dx and, for all Q Q Q, Q Q = f(x) dx tµ(q ) Q }. Then every decreasing sequence of cubes in Q contains at most one element of Q. Hence every element of Q satisfies (5.) and any two cubes in Q have disjoint interiors. Define B := Q Q Q. We prove that x R n \ B, x Q Q = µ(q) Q f(x) dx t. (5.2) Suppose, by contradiction, that there exists an element x R n \ B and a cube Q Q such that x Q and tµ(q) < f(x) dx. Then, since Q f <, there exists a maximal cube Q Q such that x Q and tµ(q) < f(y) dy. Such a maximal cube would be an element of Q Q and hence x B, a contradiction. This proves (5.2). Now the Lebesgue Differentiation Theorem [33, Thm 6.4] asserts that there exists a Lebesgue null set E R n \B such that every element of R n \(B E) is a Lebesgue point of f. By (5.2), every point x R n \ (B E) is the intersection point of a decreasing sequence of cubes over which f has mean value at most t. Hence it follows from [33, Thm 6.6] (a corollary of the Lebesgue Differentiation Theorem) that f(x) t for all x R n \ (B E). This proves Step 2. 3

32 Step 3. Let c = c(n, σ, C) be the constant in Step and let (T, K) be a Calderón Zygmund pair on R n with the constants C and σ. Then for all f L 2 (R n, C) L (R n, C). T f, ( 2 n+ + 6c ) f (5.3) Fix a function f L 2 (R n ) L (R n, C) and a constant t >. Let the Q i be as in Step 2 and define B := Q i. i Then µ(q i ) < t Q i f(x) dx for all i by Step 2 and hence µ(b) = µ(q i ) f(x) dx t i Q i t f. Define g, h : R n R by Then g := fχ R n \B + i i Q i f(x) dx χ Qi, h := f g. µ(q i ) g f, h 2 f. Moreover, h vanishes on R n \B and Q i h(x) dx = for all i. Hence it follows from Step that ( κ T h (t) c µ(b) + ) t h 3c t f. (5.4) Moreover, it follows from Step 2 that g(x) t for almost every x R n \ B and g(x) 2 n t for every x int(q i ). Thus g 2 n t almost everywhere. Hence it follows from [33, Lemma 7.36] that κ T g (t) Rn g(x) 2 dx 2n g(x) dx 2n t 2 R t t f. (5.5) n Now combine (5.4) and (5.5) with the inequality (4.2) to obtain κ T f (2t) κ T g (t) + κ T h (t) 2n+ + 6c f 2t. Here the splitting f = g + h depends on t but the constant c does not. Multiply the inequality by 2t and take the supremum over all t to obtain (5.3). This proves Step 3. 32

33 Step 4. Fix a real number < p < as well as n N and C > and < σ. Then there exists a constant c = c(n, p, σ, C) > such that T f L p c f L p for all f L 2 (R n, C) L p (R n, C) and every Calderón Zygmund pair (T, K) on R n with constants C and σ. For p = 2 this holds by assumption, and for < p < 2 it follows from Step 3 and the Marcinkiewicz Interpolation Theorem 4.2 with q = 2. Now assume 2 < p < and choose < q < 2 such that /p + /q =. Define the function K : (R n R n ) \ n R by K (x, y) := K(y, x) and let T : L 2 (R n, C) L 2 (R n, C) be the adjoint operator of T. Then (T, K ) is again a Calderón Zygmund pair on R n with constants C and σ. To see this, note that T g, f L 2 = g, T f L 2 = g(x)k(x, y)f(y) dy dx R n R n = K (y, x)g(x)f(y) dx dy R n R n for any two continuous functions f, g : R n C with disjoint compact supports. This implies (T g)(y) = K (y, x)g(x) dx for every continuous R n function g : R n C with compact support and every y R n \ supp(g). Moreover, the function K evidently satisfies (5.2) and (5.3) and T has the same operator norm as T. Now define c := c(n, q, σ, C). Then, by what we have already proved, T g L q c g L q for all g C (R n, C). Hence T f L p = g, T f sup L 2 g C (Rn,C) g L q = T g, f sup L 2 g C (Rn,C) g L q sup T g L q f L p g L q g C c q f L p (Rn,C) for all f L 2 (R n, C) L p (R n, C). Here the first equality follows from [33, Thm 4.33] and the fact that C (R n, C) is a dense subspace of L q (R n, C). This proves Theorem

34 6 The Mikhlin multiplier theorem The Fourier transform on R n is the unique bounded linear operator F : L 2 (R n, C) L 2 (R n, C) given by (F (u))(ξ) := û(ξ) := e i ξ,x u(x) dx (6.) R n for ξ R n and u L 2 (R n, C) L (R n, C). Its inverse is (F (û))(x) = u(x) = e i ξ,x û(ξ) dξ (6.2) (2π) n R n for x R n and û L 2 (R n, C) L (R n, C). If m : R n C is a bounded measurable function, it determines a bounded linear operator given by T m : L 2 (R n, C) L 2 (R n, C) T m u := F (mf (u)) for u L 2 (R n, C). The Mikhlin Multiplier Theorem gives conditions on m under which this operator extends to a (unique) bounded linear operator from L p (R n, C), still denoted by T m, which agrees with the original operator on the intersection L 2 (R n, C) L p (R n, C). We state and prove this result in a slightly weaker form than in Mikhlin [25] and Hörmander [6]. This version suffices for the purposes of the present exposition. Theorem 6. (Mikhlin). For every integer n N, every constant C >, and every real number < p < there exists a constant c = c(n, p, C) > with the following significance. Let m : R n \ {} C be a C n+2 function that satisfies the inequality α m(ξ) C ξ α (6.3) for every ξ R n \ {} and every multi-index α = (α,..., α n ) N n with α = α + + α n n + 2. Then for all f L 2 (R n, C) L p (R n, C). T m f L p c f L p. (6.4) 34

35 Proof. The proof is based on the generalized Calderón Zygmund inequality in Theorem 5.2 and follows the argument in Parissis [29]. The main idea is to show that there exists a function K m : R n \ {} C such that T m and the function R n R n \ m C : (x, y) K m (x y) form a Calderón Zygmund pair as in Definition 5.. One would like to choose K m such that m is the Fourier transform of K m. The difficulty is that, in the interesting cases, m is not the Fourier transform of any integrable function. To overcome this problem one can use the Littlewood Paley decomposition (Section 8). More precisely, let φ : R n R be a Schwartz function such that φ(x) = φ( x) for all x R n so that its Fourier transform φ := F (φ) is a smooth real valued function that satisfies φ(ξ) = φ( ξ) for all ξ R n. Assume in addition that φ satisfies the conditions φ(ξ) > for / 2 ξ 2, φ(ξ) for /2 ξ 2, φ(ξ/2) + φ(ξ) = for ξ 2, (6.5) φ(ξ) = for ξ / [/2, 2]. In Definition 8. below a function φ with these properties is called a Littlewood Paley function and that it exists is shown in Example 8.2. For j Z define the function φ j : R n R by φ j (x) := 2 nj φ(2 j x), φj (ξ) := φ(2 j ξ) (6.6) for x, ξ R n. Then it follows from (6.5) that j= φ k (ξ) = and hence j= φ j (ξ)m(ξ) = m(ξ) for all ξ R n \ {}. (6.7) For j Z the function φ j m : R n C is of class C n+2 and has compact support and we denote its inverse Fourier transform by K j := F ( φ j m). We prove in three steps that the series K m := j= K j defines a continuous function on R n \ {} and that the pair (T m, (x, y) K m (x y)) satisfies the requirements of Definition

36 Step. There exists a constant c = c (n, C) > with the following significance. Let m : R n C be a C n+2 function that satisfies (6.3) for α n+2. Then, for all x R n \ {}, the limit K m (x) := K j (x) j= = lim N N j= N e i ξ,x φ(2 j ξ)m(ξ) dξ (2π) n 2 j ξ 2 j+ exists, the resulting function K m : R n \ {} C is C, and (6.8) K m (x) + x K m (x)) c x n for all x R n \ {}. (6.9) Each function K j has a C n+2 Fourier transform K j = φ j m with support in the compact set {ξ R n 2 j ξ 2 j+ }. Hence K j is smooth and, for every α N n, every integer k n + 2, and every x R n, we have α K j (x) = (2π) n = (2π) n = ( )k (2π) n = ( )k (2π) n (iξ) α φ(2 j ξ)m(ξ)e i ξ,x dξ R n ( n ) k (iξ) α φ(2 j x i ξ)m(ξ) e i ξ,x dξ R i x n 2 ξ i= i ( n ) k e i ξ,x x i (iξ) α φ(2 j ξ)m(ξ) dξ R i x n 2 ξ i= i e ( ) β i ξ,x k! x ( β R β! i x n 2 ξ (iξ) α φ(2 j ξ)m(ξ) ) dξ β =k The integrand is supported in the domain {ξ R n 2 j ξ 2 j+ }. Hence there exists a constant c 2 >, depending only on n, φ, α, and the constant C in (6.3), such that β ξ ((iξ)α φ(2 j ξ)m(ξ)) c 2 2 j( α k) for all ξ R n, all j Z, and all β N n with β = k n + 2. This implies α K j (x) c 2 (2π) n β =k k! 2 j( α k) ξ 2j+ 2 j(n+ α k) dξ = c β! x k 3 x k (6.) for all α N n, all j Z, all k {,,..., n + 2}, and all x R n \ {}. Here c 3 := c 2 π n n k ω n and ω n := Vol n (S n ), so ω n /n is the volume of the unit ball in R n, and we have used the identity β =k k!/β! = nk. 36

37 Now fix a nonzero vector x R n and let j Z be the largest integer such that 2 j x. Then 2 j x and 2 (j +) < x. Hence it follows from (6.) with k = that j j= α K j (x) c 3 j = j= c 3 2 j(n+ α ) 2 n α 2j (n+ α ) 2c 3 x, n+ α c because 3 2c 2 n α 3 and 2 j(n+ α ) x (n+ α ). This holds for all α N n. Now assume α {, } and use (6.) with k = n + 2 > n + α to obtain α 2 j( α 2) K j (x) c 3 x n+2 c 3 j=j + = < j=j + c 3 2 (j+)( α 2) 2 α 2 x n+2 2c 3 x, n+ α because 2c 2 α 2 3 and 2 (j+)( α 2) < x 2 α. This proves Step with the constant c = 4(n + )c 3. Step 2. Let c, m, K m be as in Step. Then the function R n R n \ { n ) C : (x, y) K m (x y) (6.) satisfies conditions (5.2) and (5.3) in Definition 5. with σ = and with C replaced by 2 n+ c. The estimate (5.2) follows directly from (6.9). To prove (5.3), fix a vector x R n \ {} and let y R n such that y x /2. Then K m (x ty) c x ty 2n+ c n+ x n+ for t. Hence it follows from the mean value inequality that K m (x) K m (x y) 2n+ c y x n+. Hence the function (6.) satisfies (5.3) with σ = and this proves Step 2. 37

38 Step 3. Let c, m, K m be as in Step. Then the function (6.) satisfies condition (5.) in Definition 5. with T = T m. Let f, g : R n C be continuous functions with compact support and assume supp(f) supp(g) =. Define the function h : R n C by h(x) := g(y)f(x + y) dy for x R n R n Then h vanishes near the origin and ĥ = ĝ f. Hence g, T m f L 2 = ĝ, (2π) T n m f L 2 = = (2π) n (2π) n = lim N = lim N = = ĝ, m f R (2π) n L 2 m(ξ)ĥ(ξ) dξ n N N R n j= N R n j= N R n K m (x)h( x) dx R n supp(g) = g, K m f L 2 φ j (ξ)m(ξ)ĥ(ξ) dξ K j (x)h( x) dx K m (x)g(y)f(y x) dy dx Here the first equality follows from Plancherel s Theorem, the second equality follows from the definition of the operator T m, the third equality uses the formula ĥ = ĝ f, the fourth equality uses Lebesgue dominated convergence, the fifth equality follows again from Plancherel s Theorem, the sixth equality follows from Lebesgue dominated convergence and the fact that h has compact support and vanishes near the origin, the seventh equality follows from the definition of h, and the last equality follows from Fubini s theorem. It follows that (T m f)(x) = R n K(x)f(x y) dx for all x R n \ supp(f). This proves Step 3. By Step 2 and Step 3 the operator T m and the function (6.) form a Calderón Zygmund pair on R n with constants 2 n+ c and σ =. Hence the assertion follows from Theorem 5.2. This proves Theorem

39 Corollary 6.2 (Calderón Zygmund). For every integer n N and every real number < p < there exists a constant c = c(n, p) > such that n i j u L p c u L p (6.2) for all u C (R n, C). i,j= Proof. For i, j {,..., n} define T ij f := i (K j f), K j (x) := x j ω n x n, (6.3) for f C (R n ). Then Poisson s identity asserts that T ij u = i j u, (K j f) = j f (6.4) for all u, f C (R n, C) (e.g. [33, Thm 7.4]). The second equation in (6.4) implies that K j f(ξ) = iξ j ξ 2 f(ξ) for all ξ R n \ {}, and hence T ij f = m ij f, mij (ξ) = ξ iξ j ξ 2. (6.5) This implies that T ij extends to a bounded linear operator from L 2 (R n, C) to itself (see also [33, Lem 7.44]). Since the function m ij satisfies the requirements of Theorem 6., there is a constant c > such that T ij f L p c f L p for all f C (R n ). Take f := u and use the first equation in (6.4) to obtain the estimate (6.2). This proves Corollary 6.2. Corollary 6.3. For every real number < p < and every C > there exists a constant c = c(p, C) > with the following significance. Let m : R \ {} C be a C 3 function such that m (k) (ξ) C ξ k for k =,, 2, 3 and ξ R \ {}. If (X, A, µ) is a σ-finite measure space, L p (X, µ) is the complex L p -space, f L p (R, L p (X, µ)) L 2 (R, L 2 (X, µ)), and the function T m f L 2 (R, L 2 (X, µ)) is defined by T m f(ξ) := m(ξ) f(ξ) for ξ R, then T m f L p (R, L p (X, µ)) and T m f L p (R,L p (X,µ)) c f L p (R,L p (X,µ)). (6.6) Proof. For functions f : R C, where X is a singleton, the assertion follows from Theorem 6.. The general case then follows from Fubini s theorem. 39