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INTERPOLATION THEOREMS and APPLICATIONS Contents 1 INTERPOLATION THEOREMS AND APPLICATIONS CALISTA BERNARD Abstract. We discuss and prove two major theorems, the Riesz-Thorin In- terpolation Theorem and the Marcinkiewicz Interpolation Theorem. We then address several important applications of these theorems, including the bound- edness of the Hardy-Littlewood maximal operator, the boundedness of the Fourier transform, Young's inequality for convolution, and the boundedness of the Hilbert transform. Contents 1. Introduction 1 2. The Riesz-Thorin Interpolation Theorem 2 3. The Marcinkiewicz Interpolation Theorem 5 4. The Hardy-Littlewood Maximal Operator 11 5. The Fourier Transform and Convolution 12 Acknowledgments 14 References 14 1. Introduction In this paper we present two main classical results of interpolation of operators: the Riesz-Thorin Interpolation Theorem and the Marcinkiewicz Theorem. The former allows us to show that a linear operator that is bounded on two Lp spaces is bounded on every Lp space in between the two. The latter allows us to show that a sublinear operator that satisfies two weak-type estimates is bounded on any Lp space in between the two weak Lp spaces. Thus, we may simplify the proof of the boundedness of an operator by proving the statement in two \simpler" cases (such as L1 and L1 or L1 and L2) and then interpolating to prove the statement for every Lp space in between. Following our discussion of the two interpolation theorems, we consider several applications: the boundedness of the Hardy-Littlewood maximal operator (section 4), the boundedness of the Fourier transform (section 5), and a proof of Young's inequality (section 5). We assume the reader is familiar with Lp spaces, the Lebesgue integral, and basic measure theory. See Chapter one of [2] for background on Lp spaces, Chapter 2 of [1] for the Lebesgue integral, and Chapter 1 of [1] for measure theory. Date: 30 August 2013. 1 2 CALISTA BERNARD 2. The Riesz-Thorin Interpolation Theorem We begin by proving a few useful lemmas. Lemma 2.1. Let 1 ≤ p; q ≤ 1 be conjugate exponents. If f is integrable over all sets of finite measure (and the measure µ is semifinite if q = 1) and Z sup fg = M < 1 kgkp≤1;g simple q then f 2 L and kfkq = M. Proof. First we consider the case where p < 1 and q < 1. Note that by Holder's inequality, M ≤ kfkqkgkp ≤ kfkq since kgkp ≤ 1. Now we show the reverse inequality. We can find a sequence of q simple functions ffngn2N in L that converges to f pointwise from below. Now define q−1 jfn(x)j · sgnf gn(x) = q−1 : kfnkq It follows that q 1 Z kfnk kg kp = jf (x)jp(q−1) = q = 1; n p p(q−1) n q kfnkq kfnkq where the middle equality follows from the fact that p and q are conjugate expo- nents. Furthermore, since for large n, sgnfn = sgnf, we see that Z R q q jfnj kfnkq (2.2) fngn = q−1 = q−1 = kfnkq kfnkq kfnkq R for sufficiently large n. Thus, it follows from our assumption that fngn = kfnkq ≤ R q q M. Then 2:2 implies that jfnj ≤ M . By Fatou's lemma, we have that Z Z q q q jfj ≤ lim inf jfnj ≤ M : It follows that kfkq ≤ M. Now we consider the case where p = 1 and q = 1. Fix > 0 and let E = fx j jf(x)j ≥ M + g. Since µ is semifinite, if µ(E) were positive, then there would exist −1 F ⊂ E such that 0 < µ(F ) < 1. Let g = µ(F ) χF sgn f. Then kgk1 = 1 and Z 1 Z M ≥ fg = jfj ≥ M + . µ(F ) F Since this is clearly impossible, we have that µ(E) = 0. Then f 2 L1 and M ≥ kfk1. The opposite inequality is, once again, given by Holder's inequality. Lemma 2.3. (Three Lines Lemma) Suppose Φ: C ! C is holomorphic on the inside of the strip 0 ≤ Re z ≤ 1 and continuous and bounded on the closure of the strip. Furthermore, suppose jΦ(z)j ≤ M0 on the boundary Re z = 0 and jΦ(z)j ≤ M1 on the boundary Re z = 1. Then for all y 2 R, x 2 (0; 1), 1−x x jΦ(x + iy)j ≤ M0 M1 : INTERPOLATION THEOREMS AND APPLICATIONS 3 z−1 −z z(z−1) Proof. Let > 0. Let Φ(z) = Φ(z)M0 M1 e . Then if Re z = 0, we have 2 iy −iy iy(iy−1) iy log (M0=M1) −y −iy jΦ(iy)j ≤ jM0 M1 e j = je e e j ≤ 1; where the last inequality follows from the fact that eiy log M0=M1 and e−iy are both on the unit circle. Similarly, 2 1+iy−1 −iy (1+iy)(1+iy−1) iy log (M0=M1) −y −iy jΦ(1 + iy)j ≤ jM0 M1 e j = je e e j ≤ 1: Since Φ is clearly holomorphic inside the strip, we see that it satisfies the assump- tions from the statement (with 1 as M0 and M1). Now for any x 2 (0; 1), x+iy−1 −x−iy (x+iy)(x+iy−1) jΦ(x + iy)j = jΦ(x + iy)M0 M1 e j 2 2 = jΦ(x + iy)jje(x+iy−1) log M0 e(−x−iy) log M1 e(x −x−y )eiy(2x−1)j x−1 −x (x2−x) −y2 = jΦ(x + iy)jjM0 M1 e je 2 ≤ Ce−y 2 for some positive constant C. This follows from the fact that Φ and ex −x are −y2 bounded on the strip. Since e ! 0 as jyj ! 1, it follows that jΦ(z)j ! 0 as jIm(z)j ! 1. Thus, we can pick A large enough that jΦ(z)j ≤ 1 on the boundary of the rectangle 0 ≤ Re z ≤ 1 and −A ≤ Im z ≤ A. Then the maximum modulus principle implies that jΦ(z)j ≤ 1 on the interior of the rectangle as well. Since this holds for any arbitrarily large A, we have that jΦ(z)j ≤ 1 on the entire strip. Now let ! 0. From above, we now have x−1 −x lim jΦ(z)j = jΦ(z)jM0 M1 ≤ 1 !0 for x = Re z. The result follows immediately. We are now ready to prove the Riesz-Thorin Interpolation Theorem, which allows us to establish boundedness of a linear operator on certain Lp spaces. Theorem 2.4. (Riesz-Thorin Interpolation Theorem) Consider a linear func- tion T , which maps the measure space (X; µ) to the measure space (Y; ν). Suppose p0; q0; p1; q1 2 [1; 1] and 1 1 − t t 1 1 − t t = + = + p p0 p1 q q0 q1 for t 2 (0; 1). If q0 = q1 = 1, we further suppose that ν is semifinite. Suppose p0 p1 q0 q1 T maps L (µ) + L (µ) into L (ν) + L (ν) and we have kT fkq0 ≤ M0kfkp0 for p0 p1 f 2 L and kT fkq1 ≤ M1kfkp1 for f 2 L , for constants M0;M1 > 0. Then T is p 1−t t p bounded on L and furthermore, kT fkq ≤ M0 M1kfkp for all f 2 L . Our general strategy will be to construct a function Φ that satisfies the assump- tions of the Three Lines Lemma, then use Φ to bound T f in Lq(ν) for f simple. We then extend the result from f simple to all Lp. Proof. We begin with the case p0 < p1. Let f be a simple function on X. Recall that simple functions are dense in Lp for any p 2 [1; 1]. Note that we can scale f because of the linearity of T , so it suffices to show the statement for kfkp = 1. Pn Pn iθj iθj Write f = j=1 ajχEj = j=1 jajje χEj , where jajje is the polar form of aj and the Ej are disjoint. Now let g be a simple function on Y . We further take g to 4 CALISTA BERNARD have q0-norm equal to 1 (where q0 is the conjugate exponent of q). Again, we write Pm Pm iθk g = k=1 bkχFk = k=1 jbkje χFk . We now define two complex functions α and β as follows: −1 −1 −1 −1 α(z) = (1 − z)p0 + zp1 β(z) = (1 − z)q0 + zq1 : Note that if t 2 (0; 1), then α(t) = p−1 and β(t) = q−1. Fixing t 2 (0; 1), we let n X α(z)/α(t) iθj fz = jajj e χEj j=1 (note that α(t) > 0) and m X (1−β(z))=(1−β(t)) iθk gz = jbkj e χFk k=1 if β(t) 6= 1 and gz = g if β(t) = 1. Finally, we define Z Φ(z) = (T fz)gz dν: Then if β(t) 6= 1, Z X α(z)/α(t) (1−β(z))=(1−β(t)) i(θj +θk) Φ(z) = jajj jbkj e (T χEj )χFk j;k P α(z)/α(t) i(θj +θk) R and if β(t) = 1, Φ(z) = j;k jajj jbkje (T χEj )χFk : Then Φ is bounded and holomorphic in the strip 0 ≤ Re z ≤ 1. We claim that jΦ(z)j ≤ M0 for Re z = 0 and jΦ(z)j ≤ M1 for Re z = 1. −1 −1 −1 −1 If z = iy for y 2 R, then α(z) = p0 + iy(p1 − p0 ) and 1 − β(z) = 1 − q0 − −1 −1 iy(q1 − q0 ). Now note that since the Ej are disjoint, for any x 2 X at most one term of the sum equal to f(x) or fz(x) may be nonzero.
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