SOLUTIONS TO HOMEWORK ASSIGNMENT 4

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1 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 Exercise. A criterion for the image under the Hilbert transform to belong to L Let φ S be given. Show that Hφ L if and only if φx dx = 0. Solution: Suppose first that φ S is such that Hφ L. We know that the function Hφ is then continuous. Now, Hφ ξ = i signξ φξ. We know that φ is a Schwartz function. In particular, it is continuous. Hence, if i signξ φξ is continuous, then φ0 = φx dx = 0. Conversely, suppose that φx dx = φ0 = 0. We then use the Fourier Inversion Formula to compute: Hφx = Hφ ξ e 2πix ξ dξ = i signξ φξ e 2πix ξ dξ + = i φξ e 2πix ξ dξ + i φξ e 2πix ξ dξ. 0 This first equality in this step is made rigorous by recalling the result of Exercise 3 from Homework Assignment. Namely, we note that, for φ S, we have Hφ L 2. Consequently: Hφx = Hφ ξ e 2πix ξ dξ lim M ξ M where the limit is taken in L 2. Furthermore, the pointwise limit: Hφ ξ e 2πix ξ dξ = i signξ φξ dξ equals lim M ξ M lim M 0 ξ M i signξ φξ dξ for almost very x by using φ S and the Dominated Convergence Theorem. This limit has to coincide with the L 2 limit, almost everywhere. Hence, we can really apply the Fourier Inversion Formula as above. Since φ S, it follows from that Hφ L. We now consider each of the two integrals in separately. { + i φξ e 2πix ξ u = dξ = φξ, du = φ ξ dξ 0 dv = e 2πix ξ dξ, v = e2πix ξ 2πix = + 0 2πx e2πix ξ φ ξ dξ Here, we integrated by parts and used the assumption that φ0 = 0. Integrating by parts again in the same way, we see that this expression equals: + 2πx φ 0 + i 4π 2 x 2 φ ξ e 2πix ξ dξ. 0 An analogous argument shows that the second integral equals: + 2πx φ 0 i 4π 2 x 2 0 φ ξ e 2πix ξ dξ.

2 2 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 Consequently Hφ is a bounded function which is also O x 2. Therefore, Hφ L. Exercise 2. L p -boundedness of restriction of the Fourier transform in one dimension and its applications In order to avoid working with tempered distributions, we work with p 2 throughout this exercise. Namely, then the Fourier transform is then a bounded linear map from L p to L p by the Hausdorff-Young inequality. For technical reasons, we also assume that p >. Therefore, we fix p, 2] throughout this exercise. a Let a, b with a < b be given. Let T be an operator defined as: T f ξ := χ [a,b] ξ fξ. Here, χ [a,b] denotes the characteristic function of the interval [a, b]. In other words, T = mχ [a,b]. Show that T is strong p, p with operator norm independent of a and b. b Show that the set of all g L p with ĝ compactly supported is dense in L p. c In light of part a, we can define for > 0 the operator S on L p by Show that for all f L p, we have S f ξ := χ [,] ξ fξ. 2 lim S f f L p 0. Solution: a In what follows, we will work with functions f in the Schwartz class. The general result for f L p will follow from density. Given c, we let M c denote the modulation operator: M c φ := e 2πix c φ. Then M c is an isometry on L p. We recall that: for φ S. In particular, for all f S: M c φ ξ = φξ c M a HM a f ξ = HM a ξ a = i signξ a M a f ξ a We note that, for almost all ξ : = i signξ a fξ. χ [a,b] ξ = 2 signξ a signξ b. Here, we ignore the ambiguity at the endpoints. It hence follows that: T f ξ = χ [a,b] ξ = i 2 M ahm a M b HM b f ξ for almost every ξ. Thus: T = i 2 M ahm a M b HM b as operators on S. Since the Hilbert transform is strong p, p, it follows that T extends uniquely to an operator on L p which is strong p, p with operator norm independent of a and b. b Let us fix φ S with φx dx = such that φ is compactly supported. To see that such a φ exists, we can take the inverse Fourier transform of a nonzero function ψ Cc and

3 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 3 then multiply it by an appropriate constant in order to ensure φx dx =. Furthermore, we consider the rescaling φ ɛ := ɛ φ ɛ for ɛ > 0. Then by Theorem 4. in Chapter, we know that for all f L p 3 lim φ ɛ f f L p = 0. Moreover 4 φ ɛ f ξ = φɛξ fξ. This identity holds for all f S. Note that, when viewed as operators applied to f, both sides are bounded linear maps from L p to L p n. For the left-hand side, this follows from Young s inequality and from the Hausdorff-Young inequality. For the right-hand side, we use only the Hausdorff-Young inequality. Therefore, we deduce that 4 holds for all f L p. In particular, φ ɛ f is compactly supported and the density claim follows from 3. c From part a, it follows that there exists C > 0, depending only on p, such that for all > 0 and for all h L p we have 5 S h L p C h L p. Let f L p be given. For given ɛ > 0, by part b, we can find g L p such that ĝ L p is compactly supported and such that 6 f g L p < ɛ C +, where C > 0 is the constant from 5. Note that, for sufficiently large > 0, we have supp ĝ [, ]. Therefore, for such we have S g = g and hence S f f L p = S f g f g L p S f g L p + f g L p C + f g L p < ɛ. Here, we used 5 and 6. The claim 2 now follows. emark : The proof of part a also shows that T, and hence S are weak,. emark : The analysis in this exercise extends to p 2,, but then we need to work with tempered distributions recall that the Fourier transform of an L p function in this case is no longer a function, but a tempered distribution. emark 3: Analogous results do not hold in higher dimensions unless p = 2. This is a highly non-trivial result which was proved by Charles Fefferman in 972 and it goes under the name of the Ball Multiplier Problem. Exercise 3. Pointwise characterization of the Hilbert transform on L p for < p < In this exercise, we make sense of the limit fy Hfx = lim π x y dy x y >ɛ for f L p with < p <. ecall that in class, we defined the Hilbert transform on L p by using the Fourier transform when p = 2 and additional interpolation and duality properties to define it for the other values of p in this range.

4 4 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 a Given ɛ > 0 define the function h ɛ on by: { h ɛ x = x if x > ɛ 0 if x ɛ. Consider the truncated Hilbert transform defined by: 7 H ɛ f := π h ɛ f. Show that H ɛ is well defined linear operator on L p for all < p <. Moreover, show that H ɛ : L p L is continuous. b For the family of linear operators H ɛ ɛ defined in 7 we consider the associated maximal operator 8 H f := sup H ɛ f. ɛ>0 Note that, for f L p with < p < and for almost every x we have 9 H fx = sup ɛ Q + H ɛ fx. Deduce that the function H f is measurable. c Show that there exists a constant C > 0 such that for all f S we have: 0 H f MHf + CMf, where M denotes the Hardy-Littlewood maximal operator. The inequality 0 is understood to hold almost everywhere. This result is known as Cotlar s inequality. In order to prove 0, it is useful to make several reductions. Note that it suffices to show that H ɛ f MHf + CMf for a fixed ɛ and for a constant C > 0 which is independent of ɛ. In order to prove, a possible way to argue is to take φ C even, nonnegative, decreasing on [0, + with supp φ [ /2, /2] and φx dx =. Moreover, we define φ ɛ x := x ɛ φ. ɛ Write 2 π h ɛ c Show that one has 3 = π φ ɛ p.v. + h ɛ φ ɛ p.v. x π x π h ɛ f MHf. =: π h ɛ + π h2 ɛ At this step, one can use the following identity which we did not prove in class 4 ψ p.v. ψ 2 = ψ p.v. x x ψ 2, for all ψ, ψ 2 S. c2 Show that 5 h 2 ɛ = ɛ Φ, ɛ where 6 Φ := h φ p.v. x.

5 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 5 Moreover, show that Φ defined in 6 satisfies 7 Φx C + x 2 for some constant C > 0. Deduce from 5 and 7 that 8 π h2 ɛ f CMf. d Explain why 0 holds for all f L p with < p <. e Use part d and the proof of Proposition 3.3. from class to deduce that for all f L p with < p < we have 9 Hf = lim H ɛ f almost everywhere. emark: The result 9 still holds for f L, but the analysis is a bit more involved and it relies on weak, bounds. This is discussed in Theorem 3.4. of the textbook by Duoandikoetxea. Solution: a Let us fix ɛ > 0 and < p <. We note that h ɛ belongs to L q for all < q. Note that the L q norm tends to infinity as ɛ tends to zero. In particular h ɛ belongs to L p where p denotes the Hölder conjugate of p, i.e. p + p =. By Young s inequality, it follows that H ɛ = π h ɛ f is an L function. Therefore, H ɛ is a well-defined operator on L p. It is linear by construction. Furthermore, H ɛ : L p n L is continuous; this follows from Young s inequality. Namely, for all f, f 2 L p, we have H ɛ f H ɛ f 2 L π h ɛ L p f f 2 L p. b We show that ɛ H ɛ fx is continuous. This implies 9 which, in turn implies that H f is measurable since it is a countable supremum of measurable functions. Consider a sequence ɛ n such that ɛ n as n. Note that, by construction Hence, by Young s inequality lim n h ɛ n = h ɛ in L p. H ɛn f H ɛ f L = π h ɛ n h ɛ f L π h ɛ n h ɛ L p f Lp 0 as n. The wanted continuity claim now follows. c We note that 0 indeed follows from by taking suprema in ɛ. We now prove. Let us recall the decomposition 2. c From 4, we deduce that [ π h ɛ f = π φ ɛ p.v. x ] f = φ ɛ π p.v. x f = φ ɛ Hf. By Proposition 3.6. from Chapter 3, the above quantity is Mf, thus proving 3. c2 We compute: 20 ɛ h ɛ ɛy = ɛ ɛy χ { ɛy >ɛ}y = y χ { y >}y = h y.

6 6 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 Above, χ A denotes the characteristic function of a set A. Moreover, 2 ɛ p.v. x φ φ ɛ ɛy w ɛɛy = lim ɛ δ 0 w >δ w φ ɛ ɛy ɛw = lim ɛ δ 0 w >δ ɛw ɛ dw = lim δ 0 w >δ φ ɛ ɛy ɛw dw = lim ɛ δ 0 w >δ/ɛ ɛw ɛ dw φy w ɛw ɛ dw = lim δ 0 w >δ φy w w dw = φ p.v. x. Here, we used the change of variables w = w/ɛ and we recalled the definition of φ ɛ. The identity 5 for Φ given by 6 now follows from 20 and 2. We now show 7. Case : x >. Using φy dy = and supp φ [ /2, /2] we hence write 22 Φx = x φw x w dw = φw x w dw x w /2 w /2 = w /2 φw w xx w dw. In the above calculation, we also used that, by the triangle inequality, x w 2 x > 2 for all w 2. This allowed us to write φ p.v. φw y = x y w dw w /2 w /2 for all y >. Using the identity 22, as well as x w > 2 x for w 2 it follows that 23 Φx φw x 2 dw = x 2 φw dw C x 2, for some constant C > 0. w /2 Case 2: x. Using supp φ [ /2, /2] and the triangle inequality as w x w + x x w +, we compute 24 Φx φx w = lim dw δ 0 w = lim φx w φx dw δ 0 w 3/2 w >δ w 3/2 3/2 w >δ φx w φx dw 4 sup φ <. w }{{} sup φ The estimate 7 now follows from 23 and 24. Finally, we note that 7 implies 8 by using the remark on Page 73 of the lecture notes since +x is an nonnegative even integrable function 2 that is decreasing on [0, +. d We find a sequence f n in S such that 25 lim n f n = f in L p.

7 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 7 Since M and H are strong p, p we deduce that lim MHf n = MHf, n lim Mf n = Mf in L p. n In particular, up to a subsequence, which we also denote by f n, it follows that 26 lim n MHf n = MHf, lim Mf n = Mf pointwise almost everywhere. n For fixed ɛ > 0, it follows from part c that Hɛ f n H f n MHf n + CMf n. In particular, using 25 combined with the continuity result in part a on the left-hand side and 26 on the right-hand side, it follows that Hɛ f MHf + CMf. Taking suprema over ɛ we obtain the claim. e Let us first note that H is strong p, p. This follows immediately from part d since M and H are strong p, p. We now consider the set { } B := f L p : Hf = lim H ɛ f almost everywhere, which we can write as B B 2, where { } B := f L p : Hf = lim sup H ɛ f almost everywhere and { } B 2 := f L p : Hf = lim inf ɛf almost everywhere. We now argue as in the proof of Proposition 3.3. from Chapter 3 to prove that B = B 2 = L p and therefore B = L p. By replacing f with f, it suffices to show that B = L p. Note that, by construction of the Hilbert transform, we know that S B. Since S is dense in L p, it suffices to show that B is closed in L p. To this end, let g n be a sequence in B such that g n g in L p. We want to show that g B. We observe that, for λ > 0 and n N, we have { µ x : Hgx lim sup H ɛ gx > λ} { } = µ x : Hg g n x lim sup H ɛ g g n x > λ {x µ : Hg gn x > λ/2 } { } + µ x : lim sup H ɛ g g n x > λ/2 {x µ : Hg gn x > λ/2 } { } + µ x : H g g n x > λ/2 which, by Markov s inequality and the strong p, p boundedness of H is p C g gn L p, λ for some constant C > 0. This quantity tends to zero as n tends to infinity. In particular, for all λ > 0, we have { µ x : Hgx lim sup H ɛ gx > λ} = 0. Therefore { µ x : Hgx lim sup H ɛ gx > 0} { µ x : Hgx lim sup H ɛ gx > /k} = 0. k= This implies that B = L p and the claim follows.

8 8 SOLUTIONS TO HOMEWOK ASSIGNMENT 4 Exercise 4. iesz transforms and Elliptic regularity Given n N, we define the iesz transform j for j =, 2,..., n as j = m j D, where m j ξ := iξj ξ. This is an n-dimensional analogue of the Hilbert transform. a Show that j is strong p, p for all < p <. b Show that, for all j, k =, 2,..., n, we have, for all f S n : 2 f j k f =. x j x k c Prove that, for all f S n, and for all < p <, we have: 2 f L p n n,p f L p n. We recall here that 2 = n k= 2. The result in part c is called a Elliptic regularity estimate x 2 k or a Maximal regularity estimate. Heuristically, it tells us that we can control the sum of all second partial derivatives by the Laplacian. Solution: a The fact that j is strong p, p for < p < follows from the Hörmander-Mikhlin multiplier theorem. Namely, we know that m j ξ = iξj ξ is in L n \ {0} and that any mixed partial derivative of order m is ξ m in absolute value for ξ 0. b It suffices to check that: j k f = 2 f. x j x k We observe that: j k f ξ = m j ξ m k ξ 4π 2 ξ 2 fξ = 4π 2 ξ j ξ k fξ. 2 f ξ = 2πiξ j 2πiξ k x j x fξ = 4π 2 ξ j ξ k fξ. k The identity now follows. c From part b, it follows that: Hence, for all < p < : By part a, this expression is: 2 f = n j,k= 2 f Lp n 2 f x j x k = n j,k= n j,k= n,p f Lp n. j k f. j k f Lp n.

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