MATH6081A Homework 8. In addition, when 1 < p 2 the above inequality can be refined using Lorentz spaces: f

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1 MATH68A Homework 8. Prove the Hausdorff-Young inequality, namely f f L L p p for all f L p (R n and all p 2. In addition, when < p 2 the above inequality can be refined using Lorentz spaces: f L p,p f L p for all f L p (R n and all < p (a In Homework we gave a direct proof of Young s convolution inequality, namely f g L r f L p g L q if p, q, r [, ] with + r = p + q. Can you now give an alternative proof using interpolation? (b We now give a refinement of Young s convolution inequality in the scale of Lorentz spaces. (i Suppose < p <. Show that if f L p, and g L p,, then f g L f L p, g L p,. (Hint: Use Question 6 of Homework 3. (ii Hence show that f g L r,q f L p, g L q and f g L r f L p, g L q,r if p, q, r (, with + = +. In particular, for the same range of r p q p, q, r, we have f g L r f L p, g L q. This proves the last theorem in Lecture In Question 6 of Homework 6 we proved a Schur s lemma. We mentioned that there is an interpolation proof. Can you carry that out now? More generally, suppose T is an integral operator defined by T f(x = K(x, yf(ydy, where sup x K(x, y L (dy A and sup K(x, y L (dx B. y Show that if p [, ], then T is bounded on L p with norm A /p B /p. 4. Let p (,, r (, ], and f be a measurable function. Show that f L p,r 2 k W /p k l r (Z where W k := {2 k < f 2 k }. (Hint: Since { f > 2 k } l N W k+l, when r <, the essence is in showing that k Z 2 kr ( l N W k+l r/p

2 27-8 First Term MATH68A 2 when k Z 2kr W r/p k this by =. If r/p, use the triangle inequality for l r/p, to bound l N ( k Z 2 kr W r/p k+l on the other hand, if r/p, just observe that The case r = is easier. p/r ( r/p W k+l l N l N r/p ; W r/p k+l. 5. (a Show that L p,r L p,r 2 if p (, and < r < r 2. (b Show that L p, L p, L p,r if < p < p < p and r (,. 6. Let (X, µ be a measure space. In Question 6 of Homework 3, we effectively showed that if p (,, then { } f L p, sup fgdµ: g L p,, X and the right hand side is a norm on L p, that is comparable to L p,. In this question, we will adapt the approach taken there, and show that if p (,, r [,, then { } f L p,r sup fg dµ: g L p,r. X In particular, the right hand side defines a norm on L p,r that is comparable to the quasi-norm L p,r on L p,r. (With a bit more work one can also show that the dual space of L p,r is L p,r, when p (,, r [,. See also Stein and Weiss Introduction to Fourier Analysis, Chapter V.3, for another approach of defining a norm on L p,r, which proceeds via Hardy s inequality. (a Show that if p (, and r (,, then ( f L p,r = [ t /p f (t ] /r r dt t where f is the decreasing rearrangement of f, namely the unique decreasing non-negative right-continuous function on [, so that µ{ f > α} = { f > α} for all α >. (Hint: Approximate by simple functions. (b Show that X fgdµ f (tg (tdt.

3 27-8 First Term MATH68A 3 Hence show that if p [, ] and r [, ], then fgdµ f L g p,r L p.,r X (Hint: For the first part, it suffices to consider non-negative f and g s. Use f(x = χ f(x u du and similarly for g, to evaluate the integral on the left hand side; then manipulate using Fubini s theorem, and note that dµ{x: f(x u and g(x v} min {dµ{x: f(x u}, dµ{x: g(x v}} = min { {t: f (t u}, {t: g (t v} } = {t: f (t u and g (t v} the last line following because either the set {t: f (t u} is contained in {t: g (t v}, or vice versa. Reverse the manipulation done earlier, but to f and g instead of to f and g, to finish the proof of the first inequality. The second part now follows from the first part by Hölder s inequality. (c Suppose now p (, and r [,. Let f be a non-negative function in L p,r with f L p,r =. For j Z, let E j = {x: 2 j < f(x 2 j+ }. Let also g(x = j Z ( 2 j µ(e j /p r µ(ej /p χ Ej (x. Show that g L p,r, and that fgdµ. Together with the previous X part, we prove the desired comparison of f L p,r with a genuine norm. (Hint: Note that g(x 2 k χ Fk where F k = E j. k Z 2 k <(2 j µ(e j /p r µ(e j /p 2 k We want to show that g L p,r. Suppose now r =. It suffices to show µ(f k 2 kp for all k Z. But since f L p,r =, we have µ(e j 2 jp for all j Z. Hence for any k Z, we have F k = E j E j, µ(e j 2 kp 2 jp 2 kp which shows that µ(f k µ(e j 2 jp 2 kp, 2 jp 2 kp 2 jp 2 kp as desired. Next suppose r (,. If r = p then g(x = j Z 2jp χ Ej f(x p /p, so

4 27-8 First Term MATH68A 4 g L p,r = g L f p L p = f L =. Hence we may assume r p. We p,r want to show that k Z 2kr µ(f k r /p. But when r p, F k = E j, so µ(f k µ(e j (2 k 2 j(r p r p µ(e j (2 k 2 j(r p r p µ(e j. The right hand side is a sum of a subset of an essentially geometric series, so µ(f k r p µ(e j r p. µ(e j (2 k 2 j(r p r p (We used p, r (, here to ensure that the exponent r /p is finite; indeed the implicit constant here blows up like /(p as p +. Since µ(e j r p r p = µ(e j p r p r = 2 kr 2 jr when µ(e j (2 k 2 j(r p r p, the right hand side above is bounded by 2 kr 2 jr µ(e j r p. It follows that as desired. k Z µ(e j (2 k 2 j(r p r p 2 kr µ(f k r p j Z 2 jr µ(e j r p, 7. Show that there is no norm on L, that is comparable to L,. (Hint: We have N N N log N, while x n x n N. L, n= L, Obtain a contradiction as N. 8. (a Prove the following Phragmén-Lindelöf principle: Suppose f is a holomorphic function on the right half-space H := {Re z > } that extends continuously to the closure H of H. Assume f(z on the boundary of H, and that there exist α <, and constants C, c, such that n= f(z Ce c z α for all z H. Then f(z on the half-space H. (Hint: It is crucial that α < here, for this allows one to rule out the enemy e z. We turn this enemy to our advantage: for ε >, consider g ε (z := f(ze εzβ where β (α, and z β is defined using the principal branch of logarithm. Then g ε (z tends to zero as z tends to infinity within the closed half-plane. Thus the maximum modulus principle shows that g ε (z for all z in the closed half-plane. It remains to let ε + to obtain the desired conclusion.

5 27-8 First Term MATH68A 5 (b Prove the following extension of the result in the previous part: Suppose f is a holomorphic function on the half-space H := {Re z > } that extends continuously to H \ {}. Assume f(z on the boundary of H except possibly at, and that there exist α <, and constants C, c, such that f(z Ce c( z α + z α for all z H. Then f(z on the half-space H. (Hint: Consider g ε (z := f(ze ε(zβ +z β instead; note that g ε (z as z within H. (c Prove the following three-lines lemma: Suppose F is a holomorphic function on the strip S := { < Re z < } that extends continuously to the closure of the strip. Assume F (z A when Re z =, and F (z A when Re z =. If that there exist α <, and constants C, c, such that F (z Ce ceπα z for all z S, then F (z A Re z A Re z on the strip S. (Hint: By considering A ( z A z F (z in place of F (z, we may assume A = A =. The conformal map z w defined by w = ie πiz maps S conformally onto the right half-space H, with inverse z = Log(iw. It remains to apply the πi result of the previous part to the holomorphic function f(w = F ( Log(iw πi defined for w H. Another way of presenting the same proof is to directly consider F (z exp ( ε cosh(πi(z β for z S Let (X, µ, (Y, ν be measure spaces. Let < p, p, q, q, and T : (L p + L p (X (L q + L q (Y be a linear operator. Suppose there exist constants A, A such that T f L q A f L p and T f L q A f L p for all f L p L p. Show that if θ (, and p = θ + θ, p p q = θ + θ, q q then T f L q A θ A θ f L p for all f L p. This extends the Riesz-Thorin complex interpolation theorem to Lebesgue exponents below. (Hint: Pick q (, such that q < min{q, q }. Let r (, be such that r = q q. For all f Lp, we write T f L q = T f q /q L q/q = sup g L r simple g L r = T f q g q /q

6 27-8 First Term MATH68A 6 so that A ( θ T f L q = sup g L r simple g L r = A ( θ T f(xg(x q dx /q. Now given simple functions f and g, let {f z }, {g z } be holomorphic families of functions on { Re z } such that f θ = f, g θ = g, where In other words, f z (x = f(x f(x f z L p f L p and g z L r g L r when Re z =, f z L p f L p and g z L r g L r when Re z =, f(x p ( z p =, r q q p =. r q q ( f p z f Lp, g z (x = p p L p g(x g(x Then for each ε >, the following function of z A ( z A z e εz2 T f z (xg z (x q dx ( g(x r z r g r ( z r L r r r g Lr. is subharmonic in the strip { < Re z < }, continuous up to the boundary of the strip and decays to as z tends to infinity along the strip. Now invoke the maximum modulus principle for subharmonic functions; letting ε + and then taking supremum over all simple functions f and g, we see that A ( θ T f L q f L p for all f L p, if p ; if p = then an easy adaptation of the above argument works since then p = p = p.. Let < p, p, q, q, and S be the strip { < Re z < }. Suppose {T z } z S is a family of linear operators mapping compactly supported simple functions on R n to locally integrable functions on R n. Assume that for every compactly supported simple function f, the following conditions are satisfied: (a for almost every x R n, the map z T z f(x is holomorphic on S and continuous up to S; (b there exists some locally integrable function ϕ(x on R n such that for every z S, we have T z f(x ϕ(x for almost every x R n. Suppose further that there exist constants A, A such that for all compactly supported simple functions f on R n, we have T z f L q j A j f L p j whenever Re z = j, for j =,. Show that for any θ (,, and any compactly supported simple functions f on R n, we have T θ f L q A θ A θ f L p,

7 27-8 First Term MATH68A 7 where p = θ + θ, p p q = θ + θ. q q In particular, T θ extends to a continuous linear map from L p (R n to L q (R n. (Hint: As before, pick q (, such that q < min{q, q }. Let r (, be such that = r q. For a compactly supported simple function f, we have q A ( θ T θ f L q = sup g simple, compactly supported g L r = A ( θ T θ f(xg(x q dx Now given compactly supported simple functions f and g, let {f z }, {g z } be holomorphic families of functions on { Re z } such that f θ = f, g θ = g, where f z L p f L p and g z L r g L r when Re z =, f z L p f L p and g z L r g L r when Re z =, =, r q q =. r q q Then for each ε >, the following function of z A ( z A z e εz2 T z f z (xg z (x q dx is subharmonic in the strip { < Re z < }, continuous up to the boundary of the strip, and decays to as z tends to infinity along the strip. Now apply the maximum modulus principle for subharmonic functions; letting ε + and then taking supremum over all compactly supported simple functions g, we see that A ( θ T θ f L q f L p, as desired.. Let (X, dx, (Y, dy be measure spaces, and S be the strip { < Re z < }. Let /q. f(x, y = j,k c j,k χ Aj (xχ Bk (y be a simple function on X Y. Suppose < p, p, q, q, θ (, and p = θ p + θ p, q = θ + θ. q q Show that there exists a family of simple functions {f z (x, y} z S, holomorphic for z S, continuous and bounded as z varies on S, such that f θ (x, y = f(x, y, and f z L p x L q y f L p x L q y when Re z =, Here f z L p x L q y f L p x L q y when Re z =. f L p x L q := f(x, y y L q (dy L p (dx.

8 27-8 First Term MATH68A 8 This allows one to formulate a vector-valued complex interpolation theorem involving the anisotropic Lebesgue spaces L p xl q y. (Hint: Just take f z (x, y = ( f(x, y q z q f(x, y f(x, y f(x, y q ( z L q y q q q ( f(x, y p z p p f(x, y p L q y ( z p p L p xl q y f(x, y L p x L q y ; c.f. Question The goal of this question is to prove a bilinear complex interpolation theorem. Let (X, µ, (X 2, µ 2, (Y, ν be measure spaces. Let < p, p, q, q, r, r, and T : (L p + L p (X (L q + L q (X 2 (L r + L r (Y be a bilinear operator. Suppose there exist constants A, A such that T (f, f 2 L r A f L p f 2 L q and T (f, f 2 L r A f L p f 2 L q for all f L p L p (X and f 2 L q L q (X 2. Show that if θ (, and p = θ p + θ p, q = θ + θ, q q r = θ + θ, r r then T (f, f 2 L r A θ A θ f L p f 2 L q for all f L p, f 2 L q. Also generalize this to a multilinear complex interpolation theorem. (Hint: Mimic Question In the last question, we saw an easy extension of the complex interpolation theorem to the multilinear setting. In this question, we will see that the real method of interpolation does not extend to the multilinear setting the way one might naively expect. In particular, we show that there exists a bilinear operator T, such that T is continuous from (say L 2 L 2 into L 2,, continuous from L L into L, but does not map L p L p continuously into L p, even if p p. So two endpoints are not enough, if we only begin with weak-type hypothesis and want a strong-type conclusion! (The following example is due to Strichartz. We will consider the measure space X = (, with Lebesgue measure dx. The bilinear operator involved is given by (a Show that T (f, g(x = f(xyg(ydy. T (f, g L p, f L p g L p for all p. Indeed, T (f, g(x x /p f L p g L p. In particular, T is continuous from (say L 2 L 2 into L 2,, continuous from L L into L.

9 27-8 First Term MATH68A 9 (b Let p [,. (,, we have Show that for any non-negative measurable function g on sup T (f, g L p = f L p y /p g(ydy. This shows that sup T (f, g L p =. f L p, g L p Hence T is not continuous from L p L p to L p. (Hint: By scaling, we may assume that y /p g(ydy =. The triangle inequality shows that sup T (f, g L p. f L p On the other hand, let α (, be sufficiently close to, and let ε (, be sufficiently close to. Let f ε (x = x /p χ [ε,ε ](x. Then and For x [ε α, ε α ], we have Thus which gives f ε L p = [2 log(ε ] /p, ε x T (f ε, g(x = x /p y /p g(ydy. εx εα T (f ε, g(x x /p y /p g(ydy. ε α T (f ε, g L p x /p L p [ε α,ε α ] T (f ε, g L p α /p f ε L p εα ε α εα ε α y /p g(ydy, y /p g(ydy. By taking α (, sufficiently close to, and ε (, sufficiently close to, we can make the right hand side > ( δ f ε L p for any δ >. Thus as well, as desired. sup T (f, g L p f L p 4. Suppose < p <, and h L p (R n for some < p p. Show that if M h L p (R n, then Mh L p (R n, and Mh L p (R n n,p M h L p (R n. Here M h is the Hardy-Littlewood maximal function of h. In particular, we have h L p (R n, and h L p (R n n,p M h L p (R n.

10 27-8 First Term MATH68A (Hint: The key is a relative distribution inequality. For any n, we claim that there exists b n (,, such that for any b, c > with b b n, we have {x R n : Mh(x > α, M h(x cα} n c {x R n : Mh(x > bα} If this is true, then by taking c sufficiently small, we can use Question 8 of Homework 3 to conclude the proof. To prove the above relative distributional inequality, let b (, first. Let h L p (R n. Decompose the open set {x R n : Mh(x > bα} into an essentially disjoint union of Whitney cubes {Q}, so that the distance of each Q from the complement of this set is bounded by 4 times the diameter of Q. Now since {x R n : Mh(x > α, M h(x cα} is a subset of {x R n : Mh(x > bα}, we just need to show that for each Whitney cube Q as above, we have {x Q: Mh(x > α, M h(x cα} n c Q. This inequality would be trivial if the set on the left hand side were empty. So let s assume there exists a point x Q such that M h(x cα. Now let Q be any cube that intersects Q and that has diameter at least that of Q. Then 2 Q will contain a point y where Mh(y bα. Hence ffl Q h 2 n bα for all such cubes Q. If x Q and Mh(x > α, then by taking b < 2 n, we see that M(hχ 3Q (x > α. We also have ffl 3Q h 2n bα. Thus {x Q: Mh(x > α, M h(x cα} whose measure is bounded by C n hχ ( 2 n 3Q (y h 3Q dy bα Q { ( x Q: M hχ 3Q } h (x > ( 2 n bα, 3Q C n ( 2 n bα 3Q M h(x 3n C n ( 2 n b c Q where C n is the constant arising in the weak-type (, bound of M : L (R n L, (R n. This proves the desired relative distributional inequality. 5. The goal of this question is to prove a complex interpolation theorem involving the Hardy space H on R n (see Homework 7 for its definition. Let p [, ]. Suppose {T z } z S is a family of continuous linear operators from L p (R n to L loc (Rn, analytic in the sense that for every f and g that are bounded measurable with compact support, the map z R n T z f gdx is holomorphic on S, continuous up to S and bounded on S. Let q, q [, ]. Assume T z f L q A f H whenever Re z = and f H L p, T z f L q A f L p whenever Re z = and f L p. Then for any θ (,, we have where T θ f L q A θ A θ f L p for all f L p L p, p = ( θ + θ p, q = θ q + θ q.

11 27-8 First Term MATH68A In particular, T θ extends to a bounded linear map from L p (R n to L q (R n, with norm A θ A θ. (Hint: For each z S and every bounded measurable g with compact support, let Tz g be the L p function on R n, such that f Tz gdx = T z f gdx R n R n for all f L p. Now let θ (,, and let g be a compactly supported simple function on R n. Let {g z } be a holomorphic family with g θ = g and g z L q g L q when Re z =, g z L q g L q when Re z =. Then Tz g z is an analytic family of L loc function, and satisfies the hypothesis of the proposition we used in the lecture notes to prove the complex interpolation theorem for BMO; note in particular that Tz g z L p for all z S, Thus T θ g Lp, with Tz g z BMO A g L q when Re z =, Tz g z L p A g L q when Re z =. T θ g L p As a result, if f L p L p, then T θ f gdx R n A θ A θ g. L q A θ A θ f L p g L q. By density of compactly supported simple functions in L q, this shows that for all f L p L p, as desired. T θ f L p A θ A θ f L p