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Interpolation of Operators: Riesz-Thorin, Marcinkiewicz

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Interpolation of Operators: Riesz-Thorin, Marcinkiewicz INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE If we notice that, for 1 < p < r < q < 1, Lp \ Lq ⊆ Lr ⊆ Lp + Lq then, if we have a linear operator T which is bounded on Lp and on Lq, it’s natural to wonder whether or not T must be bounded on Lr. The theorems of Riesz-Thorin and Marcinkiewicz will provide us with an answer to this question. In this project, we will discuss the interpolation theorems of Riesz-Thorin and Marcinkiewicz along with their proofs and several applications. Our first goal is to prove the Riesz-Thorin Theorem but we will need to prove several lemmas beforehand. The Riesz-Thorin Theorem pertains to complex valued functions so this is assumed for all lemmas along with the theorem. Lemma 1.1 (Hadamard’s Three-Line Theorem). Suppose f(z) is a bounded and continuous function on 0 ≤ <z ≤ 1 and analytic in the interior. Denote Mθ = sup jf(θ + iy)j y2R 1−θ θ Then Mθ ≤ M0 M1 for 0 ≤ θ ≤ 1. Proof. First, we may assume that M0;M1 > 0 else f = 0 and the proof is trivial. Case 1: M0 = M1 = 1. Since f(z) is bounded on 0 ≤ <z ≤ 1, there exists k > 0 so that jf(z)j ≤ k on 0 ≤ <z ≤ 1. Let > 0 and define f(z) F = : 1 + z −k k Further, let Ω = fz : 0 < <z < 1g and R = fz = x+iy : x 2 [0; 1]; y 2 [ ; ]g. Let z = x+iy 2 δΩ. Then either x = 0 or x = 1. Well, if x = 0, then jf(z)j jf(z)j jf(z)j jF(z)j = = = ≤ jf(z)j ≤ M0 = 1: j1 + zj j1 + iyj p1 + (y)2 Now, suppose x = 1. Then, jf(z)j jf(z)j jf(z)j jF (z)j = = ≤ ≤ jf(z)j ≤ M = 1: j1 + zj j1 + + iyj 1 + 1 Also, jf(z)j jf(z)j jf(z)j jf(z)j k jF (z)j = = ≤ = ≤ : j1 + zj j1 + x + iyj j=(1 + x + iy)j jyj jyj So, for z 2 δR − Ω, we have k jf(x ± i k )j jf(x ± i k )j k jF (x ± i )j ≤ = ≤ = 1: k k k j j Date: December 2011. 1 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE So, for a fixed > 0, jF(z)j ≤ 1 on δR. So, by the maximum modulus principle, maxz2R jF(z)j ≤ 1. k Also for z = x + iy 2 Ω − R, we have jyj > , so, by (1), k k jF (z)j ≤ ≤ ≤ 1: jyj k j j Hence jF(z)j ≤ 1 for all z 2 Ω. Then we have that jf(z)j ≤ j1 + zj ≤ 1 + jzj. Since this is true 1−θ θ for all > 0, we have that jf(z)j ≤ 1. So, for 0 ≤ θ ≤ 1, we have Mθ ≤ 1 = M0 M1 . Case 2: M0;M1 > 0. Let f(z) G(z) = 1−<z <z M0 M1 1−<z <z Since M0 and M1 are both entire functions, G(z) is continuous on 0 ≤ <z ≤ 1 and analytic in the interior. Further, for 0 ≤ <z ≤ 1, jf(z)j k jG(z)j = ≤ 1−<z <z min (1;M ) min (1;M ) M0 M1 0 1 So G(z) is bounded on 0 ≤ <z ≤ 1. Also, jf(iy)j M sup jG(iy)j = sup = 0 = 1 y2R y2R M0 M0 and jf(1 + iy)j M sup jG(1 + iy)j = sup = 1 = 1 y2R y2R M1 M1 Hence, by case 1, for 0 ≤ θ ≤ 1 we have sup jG(θ + iyj ≤ 1 y2R i.e., jf(θ + iy)j sup ≤ 1 1−θ θ y2R M0 M1 1−θ θ Therefore Mθ = supy2R jf(θ + iy)j ≤ M0 M1 . Lemma 1.2. If f 2 Lp and 1 ≤ p < 1 then kfkp = sup jhf; gij khkp0 =1 R 1 1 where hf; gi = U fgdµ, p + p0 = 1 and h is a simple function with compact support. Proof. First we will establish the validity of the statement when the supremum is taken over all g 2 Lp0 where kgkp0 = 1. If we let g be such then Z jhf; gij ≤ jfgjdµ U 1=p 1=p0 Z Z 0 ≤ jfjpdµ jgjp dµ Holder U U = kfkpkgkp0 = kfkp 2 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ So sup khf; gij ≤ kfk . Let kgkp0 =1 p e−i arg f(x)jf(x)jp−1 h(x) = p−1 kfkp Well, p0 −i arg f(x) p−1 0 Z e jf(x)j khkp = dµ p0 p−1 U kfkp 1 Z 0 = jf(x)jp (p−1)dµ p0(p−1) kfkp U Z 1 p 0 = p jf(x)j dµ since p = p (p − 1) kfkp U = 1 Also, Z e−i arg f(x)jf(x)jp−1 hf; hi = f(x) p−1 dµ U kfkp Z jf(x)jjf(x)jp−1 = p−1 kfkp Z 1 p = p−1 jf(x)j dµ kfkp U = kfkp So we have that kfk = sup jhf; gij. Now we want to show p kgkp0 =1 sup jhf; gij = sup jhf; gij khkp0 =1 kgkp0 =1 where g 2 Lp0 and h is simple with compact support. Clearly sup jhf; gij ≤ sup jhf; gij khkp0 =1 kgkp0 =1 since h 2 Lp0 for all h. For the other direction, let > 0. Then there exists g0 2 Lp0 so that kg0k = 1 and sup jhf; gij ≤ jhf; g ij + 0 2 kgkp0 =1 3 INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ MATT ZIEMKE Since g 2 L 0 there exists a simple function h with compact support so that kg −h k 0 < . 0 p 0 0 0 p 2kfkp+1 Then sup jhf; gij ≤ jhf; g ij + 0 2 kgkp0 =1 = jhf; h i + hf; g − h ij + 0 0 0 2 ≤ jhf; h ij + jhf; g − h ij + 0 0 0 2 Z = jhf; h0ij + f(g0 − h0)dµ + U 2 ≤ sup jhf; hij + kfk kg − h k 0 + Holder p 0 0 p 2 khkp0 =1 < sup jhf; hij + khkp0 =1 We are now ready to prove the Riesz-Thorin Interpolation Theorem. Theorem 1.3 (Riesz-Thorin). Let T be a linear operator and let 1 ≤ p0; p1; q0; q1 ≤ 1 where p0 6= p1 and q0 6= q1. Suppose T : Lp0 (µ) −! Lq0 (ν) is bounded with norm M0 and T : Lp1 (µ) −! L (ν) is bounded with norm M . Let 1 = 1−θ + θ and 1 = 1−θ + θ where 0 < θ < 1. Then q1 1 p p0 p1 q q0 q1 1−θ θ T : Lp(µ) −! Lq(ν) is bounded with norm M ≤ M0 M1 . Proof. Case 1: 1 ≤ p0; p1; q0; q1 < 1 1−θ θ Let f 2 Lp where kfkp = 1. We want to prove that kT (f)kq ≤ M0 M1 . By lemma 1.2, it suffices R 1−θ θ to show, for any simple function g with finite support, j U T (f)gdµj ≤ M0 M1 . Case 1: f is simple with compact support. Pn Pm Let f = j=1 aj1Aj and let g = k=1 bk1Bk where aj; bj 2 C and (An) and (Bn) are both disjoint collections of subsets of U(not necessarily disjoint from each other) where µ(Aj) < 1 and ν(B ) < 1 for all j; k. Let α(z) = 1−z + z and let β(z) = 1−z + z . For any z = x + iy where k p0 p1 q0 q1 0 ≤ x ≤ 1, define α(z) i arg f(t) fz(t) = jf(t)j α(θ) e and 1−β(z) i arg g(t) gz(t) = jg(t)j 1−β(θ) e Also, let Z F (z) = T (fz)gzdν V 4 MATT ZIEMKE INTERPOLATION OF OPERATORS: RIESZ-THORIN, MARCINKIEWICZ . Claim: F (z) is analytic on 0 < <z < 1. Z F (z) = T (fz)gzdν V Z α(z) 1−β(z) = T jf(t)j α(θ) ei arg f(t) jg(t)j 1−β(θ) ei arg g(t)dν V 0 n 1 m ! Z α(z) 1−β(z) X α(θ) i arg aj X 1−β(θ) i arg bk = T @ jajj e 1Aj A jbkj e 1Bk dν V j=1 k=1 0 n 1 m ! Z α(z) 1−β(z) X α(θ) i arg aj X 1−β(θ) i arg bk = @ jajj e T (1Aj )A jbkj e 1Bk dν V j=1 k=1 n m α(z) 1−β(z) Z X X α(θ) 1−β(θ) i(arg aj +arg bk) = jajj jbkj e T (1Aj )1Bk dν j=1 k=1 V If we recall how α and β are defined then we see that the above is just a linear combination of exponential functions so we actually have that F (z) is entire. From this, we also see that F (z) is bounded on 0 ≤ <z ≤ 1. Claim: jF (iy)j ≤ M0 and jF (1 + iy)j ≤ M1 for all y 2 R. Well, Z jF (iy)j = T (fiy)giydν V 0 1=q0 1=q Z Z 0 0 q0 q ≤ jT (fiy)j jgiyj 0 Holder V V = kT (f )k kg k 0 iy q0 iy q0 ≤ kT k kf k kg k 0 p0!q0 iy p0 iy q0 = M kf k kg k 0 0 iy p0 iy q0 since T is bounded from Lp0 to Lq0 with norm M0.
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