Math 565: Introduction to Harmonic Analysis - Spring 2008 Homework # 2, Daewon Chung

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1 Math 565: Introduction to Harmonic Analysis - Spring 8 Homework #, Daewon Chung. Show that, and that Hχ [a, b] : lim H χ [a, b] a b. H χ [a, b] : sup > H χ [a, b] a b. [Solution] Let us pick < min a, b and we consider three cases < a, > b, and < a < < b. By definition, H χ [a, b] K χ [a, b] χ [a, b] y y χ { y >}ydy χ [ a, b] y dy y y dy y > y > b<y< a If < a, then b < y < a < <, and H χ [a, b] a b y dy a b. If > b, then < b < y < a, and we have the same result as above. If a < < b, then b < y < or < y < a, thus H χ [a, b] a b y dy + y dy b a + a b,. Thus, for every case, we have Hχ [a, b] lim H χ [a, b] a b. Consider now H χ [ a, b]. If we consider > ma a, b, then the integral region becomes an empty set. We consider the case when is between a and b. If < a, then b < y < < a <, and Since < b and H χ [a, b] b <, b H χ [a, b] y dy If > b, then b < < y < a, and we have H χ [a, b] a y dy b > a b. b < a b. a < a b.

2 If a < < b, then there are two possibilities. One is b < < a, i.e. H χ [a, b] a y dy a < a b. Another is a < < b, i.e. H χ [a, b] b y dy b b < a b. Hence, by first part, we can conclude that H χ [a, b] sup H χ [a, b] a b,. Show that { : Hχ[ a, b] > λ} 4 b a e λ e λ. More generally show that for any measurable subset E of of finite measure E, { : HχE > λ} 4 E E e λ e λ λ. [Solution] By first problem, we know Hχ [ a, b] b. First, we need to observe the function. This function has a horizontal asymptote y, vertical asymptote b and has a function value a b a at a + b/. Therefore we can figure out the graph of a b. a a b λ E E λ b Figure. Graph of a b. Let E { : Hχ [ a, b] > λ} E E, we can find E E + E as follows. Let E [α, β ]. Then α a be λ e λ

3 can be found with a b λ because of α < a < b, and β be λ + a + e λ can be found with a b e λ, a < β < b. Also, if we set E [α, β ] then we can find similarly Then α beλ + a + e λ, β beλ a e λ. E β α + β α be λ + a a be λ + e λ e λ + beλ a e λ beλ + a + e λ b + aeλ + e λ aeλ b e λ + beλ a e λ beλ + a b a + a beλ + eλ + e λ + b aeλ + b a e λ e λ b a + e λ + eλ + e λ 4e λ b a e λ 4b a e λ e λ. Now, to see the general case let us assume E is the union of finitely many disjoint intervals, each of finite length. We may epress E in the form E n a j, b j, where the a j and b j, j,,..., n, satisfy a < b < a < b < < a n < b n. It follows from the linearity of the Hilbert transform that Hχ E a b + a b + + a n b n n a j b j. Fi λ > and set F { : Hχ E > λ} Then F can be decomposed into the disjoint union where g is the rational function defined by F { g > e λ } { g < e λ } F F, g n a j b j. Here, we claim that if µ, then the equation g µ has n distinct root r, r,..., r n which satisfy Furthermore, if µ >, then b j r j + µ b j a j. µ {g > µ} µ + {g < µ} b j a j. 3

4 µ a b r a b r b n r n Figure. Graph of g n a j b j. Since g has a simple pole at each b j, j,,..., n, and g approach to as goes to infinity, there are eactly n distinct solutions, say r, r,..., r n, to the equation g µ, µ Figure.. Since the numbers r, r,..., r n are the roots of g µ, those are also n roots of the n-th degree polynomial equation p, where n n p p j j a j µ b j. j The sum r j of the roots is equal to p n /p n so, equating coefficients of n and of n in p, we can have p n µ and p n a j + µ b j, thus we obtain j r j which is equivalent to b j µ r j µ µ r j µ r j r j + µ µ j a j + µ b j, a j n µ + r j + µ a j µ µ µ r j µ µ a j + µ b j a j a j µ µ r j + µ b j + µ b j a j. Thus we prove the our first claim. If µ >, then { g > µ} n b j, r j Figure., so the identity; µ {g > µ} n b j a j follows from directly from the previous result. The other one can be established in similar fashion. Now we go back to our case, then we obtain from the claim, F {g > e λ } + {g < e λ } E e λ + 4 E e λ + E e λ e λ. a j

5 By considering the rational function /g instead of g, and applying the anaous version of claim we obtain a similar estimate F E /e λ e λ. Since F F + F, we have { : HχE > λ} 4 E e λ e λ, where E is the union of finitely many disjoint intervals. For given general measurable subset E of with finite measure, we can find E n such that each E n is a finite union of intervals and E n \ E as n. Then, as n, χ En χ E L χ En χ E d d E n \ E. Also, by L boundedness of Hilbert transform, Hχ En converge to Hχ E in L -sense, also it converges in weak L. Thus, given > find an N such that for n > N, we have E n \E Hχ En Hχ E L, sup α { : Hχ En Hχ E > α} < +. α> Taking α, we have Hχ En converge to Hχ E in measure. Then some subsequence of Hχ En converges to Hχ E almost everywhere. Hence, by Lebesgue s Dominated Convergence Theorem with dominate function χ { HχE >λ}. lim { : Hχ E nk > λ} lim d k k { : Hχ Enk >λ} lim χ { HχE k n k >λ}d lim χ { Hχ k E n k >λ}d χ { HχE >λ}d { : Hχ E > λ}. This and the fact: e e give us desired result. t 3. Let P t be the Poisson kernel, and Q +t t be the Conjugate Poisson kernel defined +t for all t >. Check that {P t } t> is an approimation of the identity as t, but {Q t } t> is not. Verify that P t ξ e t ξ, Qt ξ isgnξe t ξ. [Solution] Let s try to see {P t } t> is an approimation of the identity as t. After substitute y /t for fied t, we can see P t d t + t dt y + dy tan y. Because P t is positive for all, P t L <. And we need to show, for all δ >, P t d. lim t >δ 5

6 For fied δ, >δ P t d P t d >δ / tan δ t sec θ tan θ + dθ δ t + t d tan δ t. Since, δ/t as t, lim P t d lim t >δ t δ tan. t On the other hand, + t d + t d + + t. + t d + t d Thus {Q t } t> is not an approimation of the identity. Now, we ll see Fourier transform of each kernel. If we show that e ξ t e iξ t dξ P t + t, then, by the Fourier inversion theorem in the case of moderate decrease function, we can have P t e iξ d e ξ t. This mean e ξ t is a Fourier transform of P t. Now, we try to see the previous equality by calculate the integral separately. e ξt e iξ dξ e i+itξ dξ ei+itξ i + it i + it, and e ξt e iξ dξ e i itξ dξ ei itξ i it By adding these two integration, we get desired result, e ξ t e iξ dξ i + it + i it i it. t + t. To avoid the sign confuse, we ll calculate the Q t ξ for the two cases. If ξ, then Q t ξ + t e iξ d cosξ sinξ + t d i + t d e + t eiξ d i Im + t eiξ d. 6

7 Let fz zeiξz z + t. Consider the integral of fz over a closed semicircular contour C [, ] Γ with radius in the upper half plane. Then ze C iξz z z dz iesfz, it i + t z + it eiξz ie ξt. zit On the other hand, as, because of Γ ze C iξz z + t dz + t eiξ d + Γ z z + t eiξz dz e iθ e iθ + t e ξ sin θ dθ z z + t eiξz dz / which tends to as approaches. Thus, + t eiξ d ie ξt, + t eiξ d e 4ξθ dθ e ξ, ξ that implies, for ξ, Q t ξ ie ξt. If ξ <, denote ξ η η >, then Q t ξ + t eiη d cosη + t d + i + t eiη d ie ηt ie tξ. sinη + t d by previous contour integral. Hence, we can see Fourier transform of Conjugate Poisson kernel is Q t ξ isgnξe t ξ. 4. Eercise 4.. a in Grafakos characterization of the Hilbert transform via invariances Prove that if T is a bounded operator on L that commutes with translations and dilations and anticommutes with the reflection f f f, then T is a constant multiple of the Hilbert transform. [Solution] Since T commutes with translations, then T is a convolution type operator.grafakos, Theorem.5.. Thus, there eist unique tempered distribution v such that After taken Fourier transform, we can get T f f v. T fξ uξ fξ, 7

8 where uξ vξ. Since T commutes with dilations, for a >, we have that However, T anticommutes with reflection, we have T δ a f δ a T f. T δ a f sgna δ a T f, for all values of a. Let gξ fξ, and observe dilation of uξgξ with Time-Frequency Dictionary. a uξ/agξ/a δ auξgξ δ a T ǧ ξ a sgna T δ a a ǧ ξ sgna a This gives us, for a, δ a T ǧ ξ a sgna T δ a ǧ ξ uξ δ a ǧ ξ sgna a uξaδ agξ sgna uξ a gξ/a. uξ/a sgna uξ. Thus, uξ must be a constant multiple of sgnξ, and if we denote this constant with C then where D Ci is a constant. T f idsgnξ fξ D isgnξ fξ DHf, 8

9 [emark.] In the class, we have a little problem to prove the weak type, of Hilbert transform. The question was for L function, two definitions of Hilbert transform are coincide. Actually we ve solved in the class. However, it is still interesting question for more general function which is in L p. First, we may check this for characteristic function with Hf isgnξ fξ. We already found one in the problem. Now, we ll use the followings to see the other. For any number c, d >, d d I e y e y d ddy y dy c dy y d. By Fubini s theorem, c I d c c e y dyd We will use the last equality for following calculation. c e d e c d c d. Hχ [ a, b] isgnξ χ [ a, b] ξ isgn ξ e iξa e iξb iξ sgn ξ e iξ a e iξ b dξ ξ e iξ a e iξ b e iξ b e iξ a dξ + dξ ξ< ξ ξ> ξ e iξ b e iξ a + e iξb e iξa dξ ξ> ξ e iξ b e iξ a ξ i a ia + i b ib dξ + e iξb e iξa ξ a b. We may show, for any f L, that H f converge to Hf isgnξ fξ in L sense, as. This give us strong type of,. Since the step functions are dense in L, for f L, there is c j and A j [a j, b j ] for j,,..., so that f c j χ [aj, b j ] as n. L Thus, for given >, we can choose N such that f c j χ [aj, b j ] L. Also, we have seen, from the previous observationproblem, lim H χ [a,b] and Hχ [a,b] are coincide. By linearity of H and H, we can have lim H c j χ [aj, b j ] H 9 c j χ [aj, b j ]. dξ

10 Then we can have lim H f Hf L lim H f lim H N c j χ [aj, b j ] + H N c j χ [aj, b j ] Hf lim H f lim H N c j χ [aj, b j ] Hf L + N H c j χ [aj, b j ] L lim H f f c j χ [aj, b j ] H L + N f c j χ [aj, b j ] L c j χ [aj, b j ] L. Then, we can finished the proof of weak type, of H as we did in the class. Since the step functions are also dense in L, for f L, there is c j and A j [a j, b j ] for j,,... and we can choose some positive integer N so that f c j χ [aj, b j ], L for any given. Then, we claim that, for all δ >, { : lim H f Hf > δ}. Choose N so that f N c jχ [aj, b j ] L δ /4, then we have { : lim H f Hf > δ} { : lim H f lim H c j χ [aj, b j ] + H c j χ [aj, b j ] Hf > δ} { : lim H f 4C δ f N c j χ [aj, b j ] > δ/} + { : H f c j χ [aj, b j ] > δ/} c j χ [aj, b j ] C. L Last inequality due to the weak type, of Hilbert transform. Therefore we can conclude, for any f L p, p, fy lim y > y dy isgnξ fξ almost everywhere. Then, by duality argument and interpolation, we can conclude that fy lim y dy isgnξ fξ a.e., for given f L p, p <, y > L