Harmonic Analysis Homework 5

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1 Harmonic Analysis Homework 5 Bruno Poggi Department of Mathematics, University of Minnesota November 4, 6 Notation Throughout, B, r is the ball of radius r with center in the understood metric space usually R. m is the Lebesgue measure. χ X is the characteristic function of the set X. The distributional pairing is denoted by,. δ is the Dirac-delta distribution. For a >, δ a is the dilation operator. f is the reflection of f. Γ is the gamma function. Problem a Prove that there is some absolute constant C > C = 4 would work such that for any < a < b < b d C. b Prove that the value of the Dirichlet integral is π/, i.e. a d = π.. Show that this integral is defined in the sense of improper Riemann integral, but is not defined in the sense of Lebesgue integration on,, in particular, it is not absolutely integrable on,. Deduce that for b R, b sinb d = π sgn b.

2 Solution. a. First assume b. Then we have b a d d. Now assume a. Then we note by integration by parts that b d = cos b b cos a d a cos b b + cos a + a a b a cos d + d = 3. Finally, in the general case that a < < b, we combine the previous two estimates, and the result is obtained immediately. b. To show that the integral is not defined in the sense of Lebesgue integration on,, it is enough to show that But this is clear: d = + then d = k= Now for a > define πk+ πk d Ia = lim Ia = a k= π k + e a d, d. πk+ πk d = +.

3 To see this, we note that b e a d e a d + b e a d d + b d. Per part a, we know that the last term can be made arbitrarily small as we increase b, and therefore, so too can we make the second term arbitrarily small as b increases. So, for fied b so that the last two terms are small, we look at the first term: b e a d e ab b d e ab + ln b, and so, for fied b, we can choose a as small as necessary to make the right-hand side of the above estimate small too. This proves the continuity of I at. For a >, the integrand and its derivative are uniformly integrable, and so as a consequence of the Lebesgue Dominated Convergence Theorem, we can write I a = e a d = a + where the last equality is obtained by a straightforward integration by parts twice. By the Fundamental Theorem of Calculus, Ia = arctan a + C but from the definition of I, it is clear that as a, Ia, and thus from this we can obtain C = π. Thus this yields the calculation I = π/, which is.. since result. Finally, for b R, b, perform the change of variable = bu, then we have sin b π = d = sgn b sin bu u du = sgn b sin bu u is an even function. This calculation immediately yields the last desired du 3

4 Problem Use the Fourier multiplier representation of the Hilbert transform to define Hf as an element of S R for bounded functions f whose distributional Fourier transform vanishes in the neighborhood of the origin. Show that with this definition Solution. Hcos =. First we note that in the distributional sense, as is seen from the following calculation: e πi = δ e πi, φ = e πi, ˆφ = φ = δ, φ, φ S, and the second equality follows by the definition of the inverse Fourier Transform. Since e i = δ π e πi, it follows that e i = πδ π δ = δ π in the distributional sense. Therefore, we note that He i, φ = Ĥei, φ = i sgn ξe i, φ = isgn ξ δ π, φ for each φ S. We have sgn ξ δ π = δ π in the distributional sense. Hence we can write He i, φ = iδ π, φ = ie i, φ for each φ S. From the above and using the properties of reflection, it follows that He i = ie i. Net, using Euler s Identity, we observe Hcos, φ = H ei + e i, φ = = as desired. He i + He i, φ = [ ] e ie i, φ + ie i i e i, φ =, φ =, φ, φ S, i 4

5 3 Problem 3 Let E R be a measurable set with finite measure. Show that the distribution function of H E satisfies 4 E d HE λ = e πλ e πλ 3. for all λ >. Conclude that for all f of the form f = E the Hilbert transform satisfies a weak, inequality, i.e. Hf, C f. Remark: this is called restricted weak type and it suffices for real interpolation See Stein or Grafakos. Solution. The solution to this problem is rather long, so it is broken up into steps. The first step is to prove 3. when E is a finite union of disjoint intervals on the line. Having proved the first step, we use the boundedness in L of the Hilbert transform to etend 3. to arbitrary measurable E of finite measure. The last step is to use the identity to prove the restricted weak type estimate. Step. Let E be a finite union of disjoint intervals on R. Thus for some a j, b j we can write E = N a j, b j, with a j < b j < a j+, j =,..., N set a N+ =. We observe d HχE λ = m { HχE > λ } = m {HχE > λ } + m {HχE < λ } 3. Since Hχ a,b = π log a b, then by linearity of the Hilbert transform it is easy to calculate Hχ E = π log N a j b j. 3.3 From 3.3 we make the following remarks: Hχ E has singularities at each of the points a j, b j, j =,..., N. In fact, Hχ E blows up to + from either side of b j, and blows down to from either side of a j. Also, Hχ E goes to as ±. Moreover, Hχ E is clearly differentiable and monotone this can be shown easily by a computation of the derivative on each of the intervals, a, a j, b j, b N, +. It follows by the Intermediate Value Theorem and the monotonicity that there is eactly one point in each of the intervals a j, b j, b N, + for which Hχ E intersects 5

6 the graph of y = λ for λ > likewise, there is eactly one point in each of the intervals, a, a j, b j such that Hχ E = λ. Let r j denote the point such that Hχ E r j = λ and b j < r j < a j+, while let ρ j denote the point such that Hχ E ρ j = λ and ρ j a j, b j, for each j =,..., N. It is then easy to see that {Hχ E > λ} = N ρ j, r j and the latter intervals are disjoint, from which we assert m {HχE > λ } = r j ρ j = r j ρ j. 3.4 To write the right-hand side of 3.4 in terms of a j, b j, we note that, per 3.3, the r j s and ρ j s solve the equation N log a j b j = πλ. Since r j s lie outside all the intervals a j, b j, then the epression in the absolute value is positive for r j s. Hence the r j s are roots of the polynomial N N a j e πλ b j, which has degree N. Since there are N r j s, it follows the above polynomial can be factorized as follows: N N N a j e πλ b j = C r j, where C is the coefficient of the highest-degree term. Thus we note that C = e πλ and, matching the coefficients of the term of degree N, we have C r j = a j + e πλ N b j so we conclude r j = e πλ a j e πλ b j. 6

7 Similarly, by virtue of the ρ j s falling on one of the intervals a j, b j, it follows the epression in the absolute value in 3.3 is negative. Hence the ρ j s are roots of the polynomial N N a j + e πλ b j, and by an argument as above, we calculate ρ j = + e πλ a j + e πλ b j, so that, per 3.4, a quick algebraic calculation yields {HχE m > λ } = eπλ e πλ b j a j = The second term in the right-hand side of 3. is seen to be {HχE m < λ } = eπλ e πλ b j a j = E e πλ e πλ. E e πλ e πλ in a perfectly analogous way. Thus we have established 3. for the case when E is a finite union of disjoint intervals. Remark. In the case when E is a finite interval, it is possible to eplicitly compute r, ρ there is only one of each in this case in terms of a, b, and the identity is then proved easily. Due to this, and due to the linearity of the Hilbert Transform, one can see that for finite unions of disjoint intervals the distribution function of the Hilbert Transform of the characteristic functions of these sets satisfy a linearity property. This might lead one to think, in hindsight that there might be an easier way to prove Step. Step. Let E be an arbitrary measurable set of finite measure of the real line. Since E is measurable, there eists a sequence {E n } of sets, which are finite unions of disjoint open intervals, such that see, for instance, []. Equivalently, me n E as n, χ En χ E in L R as n 7

8 and, since χ E, χ En L R with a uniform bound in n, it follows χ En χ E in L R. Since H is an isometry in L R, we thus have Hχ En Hχ E in L R as n. In particular, this implies convergence of Hχ En to Hχ E in measure. Thus, using Lemma 7. from the Appendi, it follows that d HχEn λ d HχE λ as n, for almost every λ >. On the other hand, it is clear that d HχEn λ 4 E e πλ e πλ, whence 3. is shown on {λ > }, modulo a set of measure. However, since the distribution function is right-continuous, it follows 3. is true for all λ >. Step 3. Fi λ > and E an arbitrary measurable subset of the real line with finite measure. We observe λ λd HχE λ = 4 e πλ e πλ χ E. Let fλ = λ e πλ e πλ. It is clear that f is continuous on,, lim λ fλ = and lim fλ = lim λ λ πe πλ + πe πλ = π so that f is uniformly bounded on,. So there eists M > such that sup λ> fλ M. Thus, Hχ E, = sup λd HχE λ 4M χ E λ> which is the desired result. 4 Problem 4 Prove the distributional i.e. in the sense of S R n identity j n+ = n p.v. j. 4. n+ 8

9 Use it to give a proof of the Fourier multiplier representation of the Riesz transform. Solution. For each φ S, we observe j n+, φ = n+, j φ = R n n+ φ j d since the last integral is finite this is easily seen by switching to spherical coordinates and using that φ S, the calculation below shows eactly what to do with the singularity at. For each ε >, we can write = R n >ε n+ φ j d = j n φ d + n+ = >ε j >ε =ε n φ d n+ n+ φ d j n+ φν j ds ε ε ε n+ φ d j n+ φ d j ε n+ + n+ φ j d 4. where ν j is the outward unit normal vector at Sε n, the sphere of radius ε; in the second and third equalities we used integration by parts. Now, by switching to spherical coordinates, we note ε n+ + n+ φ d C ε n+ + r n+ r n drds = Oε j ε B,ε so that due to 4. and the definition of the principal value, 4. follows. Recall that the Riesz Transform R j is given by R j f = f W j where W j is the tempered distribution defined by W j, φ = Γ n+ j p.v. n+, φ, φ S, whence Ŵ j = Γ n+ nπ n+ π n+ n p.v. j n+ 9 = Γ n+ nπ n+ j n+ 4.3

10 in the distributional sense, using 4.. We continue: j n+ = πiξ j n+ Γ = πiξ j u π n+ = πiξ j u = πi π n Γ n ξ j ξ, 4.4 the second identity is valid since n + > n, the third is valid since Γ/ = π see Appendi A of [] and due to Theorem.4.6 in [], and the fourth is valid since > n we might as well consider only dimensions bigger than, since on dimension the Riesz Transform is just the Hilbert Transform. Using 4.4 on 4.3, we have Ŵ j = Γ n+ nπ n+ πi π n Γ n ξ j ξ = i ξ j ξ so that as desired. R j f = f W j = i ξ j ξ ˆf, 5 Problem 5 Prove that if T is a linear operator, which is bounded on L R, commutes with translations and dilations, and anticommutes with reflections i.e. T f = T f, where f = f, then T is a constant multiple of the Hilbert transform. Solution. Due to Theorem.5. in [], the fact that T is bounded on L R and that T commutes with translations implies that there is a unique tempered distribution w such that T f = f w a.e. for allf S. Actually, ŵ L R by the boundedness of T in L R, so ŵ is a function. Net, in the distributional sense, T δ a f = ŵδ a f = ŵa n δ a ˆf, δ a T f = a n δ a T f = a n δ a ŵ ˆf

11 and due to commutativity with translations, we must have a n ŵ δ a ˆf = ŵa n δ a ˆf = δ a ŵ = ŵ, a > which implies that ŵξ cannot depend on ξ, but only its sign. Hence at this point we know that ŵ θsgnξ, where θ is a comple-valued function defined on the set {,, }. Such a function assigns numbers θ, θ, θ to each of,, respectively. Since the point is a set of measure in R, we might as well assume that θ = in any case, w is determined only a.e.. Now, ŵ ˆ f = T f = T f = ˆf ŵ = ˆ f ŵ = ŵ = ŵ which shows that ŵ must be an odd function of ξ. Thus θ = θ. It follows that ŵ satisfies the representation ŵ = c sgn ξ for some constant c C. So ŵ is a constant multiple of the Fourier Transform of the Hilbert Transform, which implies the desired result immediately because T is bounded on L R. 6 Problem 6 Assume that a linear operator T maps L p R n to itself for some p > and assume that it satisfies the following property: if f is supported in a cube Q, then T f is supported in a fied multiple of Q, i.e. a cube Q = cq with the same center as Q and c-times the side length for some fied c >. Prove that T is of weak type,, i.e. T maps L into L,. Solution. Fi λ > and f L R n. Write f = g + b, where g, b are the good and bad functions respectively given from the Calderon-Zygmund decomposition of f at level λ. We note m { T f > λ} m { T g > λ/} + m { T b > λ/}. 6. The first term on the right-hand side of 6. is estimated as follows: Let C p be a constant such that T f p C p f p

12 and recall that from the CZ decomposition, the following estimate holds true: g p n p λ p p f where p is the Holder conjugate of p. As such, p m { T g > λ/} T g p d λ R n λ p C p p np p λ p p f C p f λ 6. where C p depends only on p and n. To estimate the second term on the right-hand side of 6., let Ω = Q j, where the Q js are the dyadic cubes in the definition of b for the CZ decomposition, and define Ω c = cq j that is, Ω c is the union of the dilated cubes cq j. Then write m { T b > λ/} mω c + m { / Ω c : T b > λ/}. 6.3 We readily observe mω c cmω c j Q j c f λ while the second term of 6.3 is identically. To see this, first note that T b j T b j, 6.4 which is true by the following argument: If the summation is over finitely many j s, this is trivial. Otherwise, note that due to the boundedness of T on L p R n, we know that N j T b j T b in L p R N, and so there is a subsequence such that this convergence occurs pointwise a.e.. Now, since T b j j T b j uniformly over N, by the a.e. convergence of a subsequence mentioned above, it must be the case that 6.4 holds. Therefore, m { / Ω c : T b > λ/} m { λ/} / Ω c : T b j >, but if / Ω c, then T b j = for all j owing to the fact that each b j is supported only on Q j and so T b j is supported on cq j. Therefore, the right-hand side of the above inequality is identically. So, from 6., 6., and the estimates on the right-hand side of 6.3, the weak-type, estimate follows. j j

13 7 Appendi The following result is analogous to a basic fact in probability theory. It was actually given in an assignment in Professor Safonov s Real Analysis course last year. Lemma 7.. Let f be a function on the real line, and {f n } be a sequence of functions on the real line such that f n f in measure. Then for almost every λ >. d fn λ d f λ as n Proof. The distribution function is obviously monotone non-increasing, and as such, it can have at most countably many discontinuities. So almost every λ > is a point of continuity for the function d f. Now fi ε > and let λ > be a point of continuity for d f. For any η > we can write m{f n > λ} = m {f n > λ} { f n f < η} + m {f n > λ} { f n f η} and we can choose η so small that m{f > λ η} + m{ f n f η} 7. m{f > λ η} < m{f > λ} + ε/. Having chosen η, the second term on the right-hand side of 7. goes to as n since f n f in measure, so it becomes small. Thus we have shown that d fn λ < d f λ + ε. The estimate in the other direction is shown similarly: We observe m{f > λ + η} = m {f > λ + η} { f n f < η} + m {f > λ + η} { f n f η} which implies the estimate m{f n > λ} + m{ f n f η} m{f n > λ} > m{f > λ} ε as desired. 3

14 Acknowledgements For all of the problems, there are hints in [] which are very useful. Problem 3 was done with some collaboration with Ryan Matzke and Brendan Cook. Problem 4 was done with some collaboration with Jonathan Hill and Brendan Cook. References [] G. Folland, Real Analysis: Modern Techniques and their applications Wiley 984. [] L. Grafakos, Classical Fourier Analysis 3rd.Ed., Springer 4. 4