Lewko's blog

L^{p} convergence of Fourier transforms

Posted in expository, Fourier Analysis, math.CA by Mark Lewko on August 3, 2009

Let {\chi_{B}(x)} denote the characteristic function of the unit ball {B} in {d} dimensions. For a smooth function of rapid decay, say {f}, we can define the linear operator {S_{1}} by the relation

 \displaystyle \widehat{S_{1}f}(\xi) = \chi_{B}(\xi)\hat{f}(\xi)

where {\hat{f}(\xi)} denotes the Fourier transform of {f}, as usual. This operator naturally arises in problems regarding the convergence of Fourier transforms (which we discuss below). A fundamental problem regarding this operator is to determine for which values of {p} and {d} we can extended {S_{1}} to a bounded linear operator on {L^{p}(\mathbb{R}^d)}. The {1}-dimensional case of this problem was settled around 1928 by M. Riesz, however the higher dimensional cases proved to be much more subtle. In 1954 Herz showed that 2d/(d+1) <p< 2d/(d-1) was a necessary condition for the boundedness of S_{1}, and sufficient in the special case of radial functions. It was widely conjectured that these conditions were also sufficient in general (this was known as the disc conjecture). However,  in 1971 Charles Fefferman proved, for {d\geq2}, that {S_{1}} does not extend to a bounded operator on any {L^p} space apart from the trivial case when {p=2} (which follows from Parseval’s identity). Recently, I needed to look at Fefferman’s proof and decided to spend some time trying to figure out what is really going on. I will attempt to give a motivated account of Fefferman’s result, in a two post presentation. In this (the first) post I will describe the motivation for the problem, as well as develop some tools needed in the proof. The problems discussed here were first considered in the context of Fourier series (i.e. functions on the {d}-dimensional torus {\mathbb{T}^d}). It turns out, however, that these problems are slightly easier to address on Euclidean space, and are equivalent thanks to a result of de Leeuw. In light of this, we will work exclusively on {\mathbb{R}^d}.

As usual, we will denote the Schwartz class of smooth rapidly decreasing functions as {\mathcal{S}(\mathbb{R}^d)}. Let {f(x) \in \mathcal{S}(\mathbb{R}^d)}, we define the Fourier transform of {f(x)} to be

 \displaystyle \hat{f}(\xi) = \int_{\mathbb{R}^d} f(x) e^{-2 \pi i \xi \cdot x}.

It is well known that {\hat{f}(\xi) \in \mathcal{S}(\mathbb{R}^d)} if {f(x) \in \mathcal{S}(\mathbb{R}^d)}, and that the domain of the Fourier transform can be extended to {L^{p}(\mathbb{R}^d)} for {1\leq p < \infty} (although we have to allow the Fourier transform map to distributions when {p>2}). A central problem in Euclidean Fourier analysis is to understand how well we can recover a function from its Fourier transform. This problem comes in many flavors, here we will be interested in the following situation. Define the partial summation operator {S_{R}} by

 \displaystyle S_{R}f(x) = \int_{|x|\leq R} \hat{f}(\xi)e^{2\pi i \xi \cdot x} d \xi.

Is it the case that these partial sums (this terminology is borrowed from the analogous situation on the torus, where {S_{R}} would be a partial sum of the Fourier expansion of {f}) converge to {f} in the {L^{p}}-norm? More specifically, if {f \in L^{p}(\mathbb{R}^{d})} does the following hold

\displaystyle \lim_{R\rightarrow \infty} ||S_{R}f - f||_{L^{p}}=0? \ \ \ \ \ (1)

Of course it is possible that the answer may depend on the choice of {p}. In fact this must be the case. Consider the case when {p=2}, by assumption we have {f \in L^{2}(\mathbb{R})}. Since {\left(S_{R}f - f\right)\hat{}(\xi)=\chi_{|x|>R}(\xi)\hat{f}(\xi)}, Parseval’s identity gives us that 

 \displaystyle \lim_{R\rightarrow \infty} ||S_{R}f - f||_{L^{2}} =\lim_{R\rightarrow \infty} ||\chi_{|x|>R}(\xi)\hat{f}(\xi)||_{L^{2}} = 0,

where the last equality is a simple consequence of {\hat{f}(\xi)} being a {L^2} function. Conversely, it isn’t hard to show that (1) does not hold when {p=1} in any dimension, although we won’t prove this here.

If {f \in \mathcal{C}_{0}^{\infty}(\mathbb{R}^d)} then (1) holds for all {p\geq 1}. To see this notice that the infinite differentiability of {f(x)} implies that {\hat{f}(\xi)} has rapid decay. Thus we can think of {\hat{f}(\xi)} as essentially having compact support (of course this is never quite true by the uncertainty principle). If {\text{supp}( \hat{f}) \in B(0,R')} was true then we’d have that {S_{R}f(x)=f(x)} for {R\geq R'}, which immediately implies (1). A little care can make this argument rigerious.

For {f \in L^{p}(\mathbb{R}^d)} and {\epsilon>0}, we can find {\phi(x)\in \mathcal{C}_{0}^{\infty}(\mathbb{R}^d)} such that {||f-\phi||_{p}<\epsilon}. Using the fact that {\lim_{R\rightarrow\infty}||S_{R}\phi-\phi||_{p}=0} we have that 

\displaystyle \limsup_{R\rightarrow\infty}||S_{R}(f) - f ||_{p}= \limsup_{R\rightarrow\infty}||S_{R}(f-\phi) - f -\phi ||_{p}

\displaystyle \leq \limsup_{R\rightarrow\infty}||S_{R}(f-\phi)||_{p} + ||f -\phi ||_{p} \leq \sup_{R}||S_{R}||_{L^p\rightarrow L^p}\times \epsilon + \epsilon.

Since {\epsilon>0} can be taken arbitrarily close to {0}, it suffices to show that the operators {S_{R}} are uniformly bounded (in {R}) as operators on {L^{p}(\mathbb{R}^d)}. However, scale invariance shows that each {S_{R}} has the same norm. Conversely, if {S_{1}} is not bounded on {L^{p}(\mathbb{R}^{d})} then we have that there exists, by the uniform boundedness principle, a function {f \in L^{p}({\mathbb R}^{d})} such that {||S_{R}f||_{p}=\infty} for every {R}. The triangle inequality would then imply that

 \displaystyle ||S_{R}f -f||_{p} \geq ||S_{R}f||_{p} - ||f||_{p} = \infty,

thus (1) could not hold. Thus the problem of spherical {L^{p}} convergence of Fourier integrals is equivalent to determining if {S_{1}} is a bounded operator on {L^{p}(\mathbb{R}^d)}. In summary, we have shown that

 Lemma 1  For {p \geq 1} the relation {\lim_{R\rightarrow \infty} ||S_{R}f - f||_{L^{p}}=0} holds for all {f \in L^{p}(\mathbb{R}^d)} if and only if there exists a constant {C_{d,p}} such that for every {\phi \in \mathcal{S}(\mathbb{R}^{d})} we have that {||S_{1}\phi||_{p} \leq C_{d,p}||\phi||_{p}}.  

{L^P} Convergence of Fourier Transforms and the Hilbert Transform

The function {S_{1}} is called a Fourier multiplier since it acts by multiplying the Fourier transform of {f} by some function, which is referred to as the symbol of the multiplier. More specifically, we have that 

\displaystyle \widehat{S_{1}f}(\xi) = \chi_{|x|\leq 1}(\xi) \hat{f}(\xi).

In the {1}-dimensional case the operator {S_{1}} is just a variant of the Hilbert transform. To see this recall that the Hilbert transform {\mathcal{H}} is defined for Schwartz functions (and then extended by density) by

\displaystyle \mathcal{H}f(x) = \text{p.v.} \frac{1}{\pi}\int_{-\infty}^{\infty} \frac{f(x-y)}{y}. \ \ \ \ \ (2)

One can verify that {\mathcal{H}} is a Fourier multiplier with symbol {i \text{sgn}(\xi)}. In other words {\widehat{\mathcal{H}f}(\xi)= \hat{f}(\xi)i \text{sgn}(\xi)}. Furthermore, it isn’t hard to derive from this that { \chi_{(-\infty,0)}(\xi) = \frac{1+i\times i \text{sgn}(\xi)}{2}} and { \chi_{(0,\infty)}(\xi) = \frac{1-i\times i\text{sgn}(\xi)}{2}} . One can further deduce that 

\displaystyle \chi_{(-1,1)}(\xi)=\chi_{(-\infty,0)}(\xi-1) \chi_{(0,\infty)}(1-\xi).

So we can write {S_{1}=\mathcal{H}_{1}\mathcal{H}_2} where {\mathcal{H}_{1}} and {\mathcal{H}_2} are the Fourier multiplier operators with symbols {\chi_{(-\infty,0)}(\xi-1)} and {\chi_{(0,\infty)}(1-\xi)}, respectively. In fact, one can compute that 

\displaystyle \mathcal{H}_{1} = e^{-2\pi i x}\frac{1}{2} \left(I -i\mathcal{H}\right)e^{2\pi i x} .

where {I} denotes the identity operator. The operator {\mathcal{H}_{2}} can be expanded similarly. Now 

\displaystyle ||S_{1}||_{L^{p}\rightarrow L^{p}} \leq ||\mathcal{H}_{1}||_{L^{p}\rightarrow L^{p}} ||\mathcal{H}_{2}||_{L^{p}\rightarrow L^{p}} \leq \frac{1}{4}\left(1 + ||\mathcal{H}||_{L^{p}\rightarrow L^{p}} \right)^2 .

Of course, a fundamental result regarding the Hilbert transform, which can be found in many harmonic analysis textbooks, is the following 

Theorem 2 (M. Riesz, 1928) For {1<p<\infty} there exists a constant {C_{p}} such that

\displaystyle ||\mathcal{H}f||_{p} \leq C_{p} ||f||_{p}.

In other words, the Hilbert transform is a bounded operator on {L^{p}(\mathbb{R})}

Riesz’s theorem solves the problem of {L^{p}} convergence in the {1}-dimensional case. In higher dimensions, however, the operator {S_{1}} is a multiplier with symbol {\chi_{|\xi|<1}(\xi)}, that is the characteristic function of the unit ball. Understanding this operator is a much thornier issue, and the question of its boundedness remained open until Fefferman’s 1971 paper. We can, however, obtain some higher dimensional results from the Hilbert transform. To do this, let us rename \mathcal{H}_{1} as $\mathcal{P}$ (this will help us avoid someconfusion later on), again we have

\displaystyle \mathcal{P}= e^{-2\pi i x}\frac{1}{2} \left(I -i\mathcal{H}\right)e^{2\pi i x}.

Now let us consider the following variant of {S_{1}} defined by 

\displaystyle \tilde{S}_{1} = \int_{|x_1|,|x_2|,\ldots,|x_d|<1}\hat{f}(\xi) e^{2 \pi i x \cdot \xi} d\xi.

By the same considerations made above, the {L^p}-boundedness of this operator is equivalent to the {L^p} convergence of the rectangular partial sums of {f}. This is to say that if we can show that {\tilde{S}_{1}} is a bounded operator on {L^{p}} then we’d establish that

\displaystyle \lim_{R\rightarrow \infty} ||\int_{|x_1|,|x_2|,\ldots,|x_d|<R} \hat{f}(\xi) e^{2\pi i x \cdot \xi} d\xi - f(x)||_{p}=0.

It turns out the question of the boundedness of {\tilde{S}_{1}} is far less subtle than the boundedness of {S_{1}}. Again, this follows from the boundedness of the Hilbert transform. To see this notice that the operator {\mathcal{P}} defined above, has symbol {\chi_{(1,\infty)}(\xi)}. Using the fact that unit cube in {\mathbb{R}^d} can be written as the intersection of a finite number of half-planes, one can see that the Fourier multiplier with symbol {\chi_{|x_1|,\ldots,|x_d|<R}(\xi)} (that is the characteristic function of a cube with side-length {R}) can be written as the composition of a finite number of rotations of the operator {\mathcal{P}}. Since rotations do not effect {L^{p}} norms, the {L^p} boundedness (for {1<p<\infty}) of {\tilde{S}_{1}} again follows from that of the Hilbert transform. We omit the details. Before we leave the operator {\mathcal{P}}, let us record a lemma that we will need later. 

Lemma 3 Define the intervals {t=[0,b]} and {\tilde{t}=[10b, 10b+b]}. Then there exists a positive constant {c} such that 

\displaystyle |\mathcal{P}\left(e^{-2\pi i x}\chi_{t}\right)(x)| \geq c 

for all {x \in \tilde{t}}, where {c} is independent of the choice of {b}.

Proof: Recall that {\mathcal{P}=e^{-2\pi i x}\frac{1}{2} \left(I-i\mathcal{H}\right)e^{2\pi i x}}. Thus for {x \notin t}, such as {x \in \tilde{t}}, we have that

\displaystyle |\mathcal{P}\left(e^{-2\pi i x}\chi_{t}\right)(x)| = |\mathcal{H}\chi_{t}(x)|.

The Hilbert transform of the characteristic function of the interval {t=[0,b]} can be computed directly from the definition (2) as 

\displaystyle |\mathcal{H}\chi_{t}(x)|= \frac{1}{\pi} \left| \ln\left|\frac{x+b}{x} \right| \right|.

Now if {x \in \tilde{t}=[10b, 10b+b]} we have that {\frac{x+b}{x} \geq \frac{12}{11}} thus

\displaystyle \frac{1}{\pi} \left| \ln\left|\frac{x+b}{x-a} \right| \right| \geq \frac{\ln(12)-\ln(11))}{\pi}>0

which establishes the lemma. \Box

This presentation is continued in the post: Fefferman’s Ball Multiplier Counterexample.

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4 Responses

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  1. [...] Posted in Fourier Analysis, expository, math.CA by Mark Lewko on August 4, 2009 In the previous post we saw the connection between the ball multiplier and spherical convergence of Fourier [...]

  2. Marcelo Fernandes said, on May 30, 2012 at 10:31 pm

    Reblogged this on Marcelo Fernandes de Almeida.

  3. Being simple said, on September 2, 2012 at 11:42 pm

    [...] as Marcinkiewicz, Minhklin, Hörmander, Lizorkin and more recentily Fefferman’s work about ball multiplier conjecture. In this post will be initiated the saga related to spaces so far. Before, we recall some [...]

  4. Hormander-Minlin | Being simple said, on February 2, 2013 at 11:05 pm

    [...] operator does not extend to a bounded operator on for any and . This result answer the famous disk conjecture which states that is bounded on for . In this post we will work with symbols more regularities [...]


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