# Lewko's blog

## Fefferman’s ball multiplier counterexample

Posted in expository, Fourier Analysis, math.CA by Mark Lewko on August 4, 2009

In the previous post we saw the connection between the ball multiplier ${S_{1}}$ and spherical ${L^{p}}$ convergence of Fourier transforms. Recall that the operator ${S_{1}}$ is defined in ${d}$ dimensions by the relation

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

where ${B}$ denotes the ${d}$-dimensional unit ball.  The focus of this post will be to prove the following result

Theorem 1 (Fefferman, 1971) The operator ${S_{1}}$ is not bounded on ${L^{p}(\mathbb{R}^d)}$ if ${d>1}$ and ${p\neq2}$.

We will restrict our attention to the ${2}$-dimensional case. The general case can be obtained by a straightforward modification of these arguments, or by appealing to a theorem of de Leeuw which states that the boundedness of ${S_{1}}$ on ${L^{p}(\mathbb{R}^d)}$ implies the boundedness of ${S_{1}}$ on ${L^{p}(\mathbb{R}^{d-1})}$ for ${d\geq3}$. Before turning our attention to the operator ${S_{1}}$ we must make one more detour. We will need the following geometric result of Besicovitch.

Theorem 2For ${\delta>0}$ there exists a collection of rectangles ${T=\{t_{i}\}_{i=1}^{K}}$ with the following properties: (1) Each ${t_{i}}$ is a rectangle of dimensions ${[0,R^2]\times[0,aR]}$ oriented such that the long axis of the rectangle makes an angle of ${0\leq \theta_{i} \leq \pi}$ with the ${x}$-axis. (2) The collection of rectangles ${\tilde{T} = \{\tilde{t}_{i}\}_{i=1}^{K}}$ obtained by translating each ${t_{i}}$ ${10R^2}$ units in the direction ${\theta_{i}}$ are disjoint, and (3) ${\sum_{i=1}^{K} |t_{i}|=\left| \bigcup_{i=1}^{K} \tilde{t}_{i} \right| \leq \delta \left| \bigcup_{i=1}^{K} t_{i} \right|}$.

This theorem is a key tool in Besicovtich’s construction of a compact subset of ${\mathbb{R}^2}$ of measure ${0}$, which contains a unit line segment in every direction. Such sets are known as Besicovitch (or Kakeya) sets. Theorem 2 can be thought of as a discrete analog of a Besicovitch set. The proof of this theorem, which is based on elementary geometry, can be found in the references listed below.

We now turn our attention to the study of the operator ${S_{1}}$. Let us first try to understand how the operator ${S_{1}}$ behaves when applied to specific functions. Arguably the simplest ${2}$-dimensional function is the characteristic function of a rectangle. The Fourier transform, however, interacts more cleanly with Schwartz functions than discontinuous functions, so we will smooth out the characteristic function of an interval.

Let ${\psi_{0}(x) \in \mathcal{S}(\mathbb{R})}$ be a non-negative bump function supported on ${[0,1]}$ and equal to ${1}$ on the sub-interval ${[\frac{1}{3},\frac{2}{3}]}$. Moreover, we can take ${\hat{\psi_{0}}(\xi)}$ to be nearly supported on the interval ${[-\frac{1}{2},\frac{1}{2}]}$. Of course, ${\hat{\psi}(\xi)}$ can’t be made to be completely supported on ${[-\frac{1}{2},\frac{1}{2}]}$ (by the uncertainty principle) but we can construct it to be rapidly decreasing away from this interval. For ${x=(x_1,x_2)\in\mathbb{R}^2}$ we can define ${\psi(x)=\psi_{0}(x_1)\psi_{0}(x_2)}$ which is a non-negative function supported on the rectangle ${[0,1] \times [0,1]}$ and equal to ${1}$ on the sub-square ${[\frac{1}{3},\frac{2}{3}]\times[\frac{1}{3},\frac{2}{3}]}$. The function ${\psi(x)}$ should be thought of as being localized on the rectangle ${[0,1]\times[0,1]}$. We will use the phrase localized on informally to mean supported or nearly supported on. We then have that ${\hat{\psi}(\xi)}$ is localized on the interval ${[-\frac{1}{2},\frac{1}{2}]\times[-\frac{1}{2},\frac{1}{2}]}$. The guiding principle here is that if ${f(x)}$ is localized on some rectangle ${t}$ then one can assume that ${|f(x)| \approx |\chi_{t}(x)|}$ in most estimates. We now define a function, ${f(x)}$, localized to the rectangle ${[0,R^2] \times [0,aR]}$ (this choice for the rectangle is connected to the statement of Theorem 2, which we will eventually apply) by

$\displaystyle f(x_1,x_2)= \psi( \frac{x_1}{R^2} , \frac{x}{aR}).$

Moreover, we see that

$\displaystyle \hat{f}(\xi_1, \xi_2) =a R^{3} \hat{\psi}(R^2 \xi_1, a R \xi_2).$

and hence ${\hat{f}}$ is localized on the rectangle ${[-\frac{1}{2R^2} ,\frac{1}{2R^2}] \times [-\frac{1}{2aR},\frac{1}{2aR}]}$. So what does the disc multiplier ${S_{1}}$ do to the function ${f}$? For large values of ${R}$ and ${a}$ we see that ${\hat{f}}$ is localized on a small rectangle near ${0}$, which of course is contained inside the unit ball. Thus we expect that ${S_{1}f \approx f}$. This doesn’t seem too helpful in showing that ${S_{1}f }$ is an unbounded operator. We’ll need to modify ${f}$ a bit. Consider the modulation of ${f}$ (at frequency ${c}$) defined by

$\displaystyle g(x) = e^{-2 \pi i c \cdot x}f(x).$

Since ${|e^{-2 \pi i c \cdot x}|=1}$ we see that ${|g(x)|=f(x)}$ (moreover, ${||g||_{L^{p}}=||f||_{L^{p}}}$). The Fourier transform of a modulation is a translation, so ${g(x)}$ is localized to the interval ${I_{c}:=[c_1-\frac{1}{2R^2},c_1+\frac{1}{2R^2}]\times[c_2-\frac{1}{2aR},c_2+\frac{1}{2aR}]}$. Thus if either ${c_1}$ or ${c_2}$ is large then the interval ${I_{c}}$ is far from the unit ball and we expect that ${S_{1} f \approx 0}$. Again, this doesn’t seem very helpful in showing that ${S_{1}}$ is unbounded.

In order to see some of the non-trivial behavior of ${S_{1}}$ let us localize the Fourier transform of ${g}$ to a rectangle that lies near the edge of the unit ball. To do this take ${c=(1,0)}$. Now ${\hat{g}(\xi)}$ is localized on the rectangle ${[1-\frac{1}{2R^2},1+\frac{1}{2R^2}]\times[-\frac{1}{2aR},\frac{1}{2aR}]}$. For large ${a}$ and ${R}$ we see that ${I_c}$ is a very small rectangle centered at ${(1,0)}$. More specifically, we see that this rectangle has length ${1/R}$, and a (even smaller) height of ${1/aR^2}$. Since ${g}$ is localized on ${I_c}$ we have that

$\displaystyle |\widehat{S_{1} g}(\xi)| = |\chi_{B}(\xi) \hat{g}(\xi)| \approx |\chi_{B}(\xi) \chi_{I_c}(\xi)|.$

Recall that ${I_{c}:=[1 - \frac{1}{2R^2},1+\frac{1}{2R^2}]\times[-\frac{1}{aR},\frac{1}{aR}]}$. Thus ${\chi_{B}(\xi)\chi_{I_c}(\xi) \approx \chi_{[1,1+\frac{1}{2R^2}]\times [-\frac{1}{2aR},\frac{1}{2aR}]}(\xi)}$. In fact, as we increase the size of ${a}$ the approximation ${\chi_{B}(\xi)\chi_{I_c}(\xi) \approx \chi_{[1,1+\frac{1}{2R^2}]\times [0,\frac{1}{aR}]}(\xi)}$ tends to the truth (pointwise). In other words, if we define ${H=\{ x_1 \leq 1 : (x_1,x_2) \in \mathbb{R}^2\}}$ we are saying that ${\chi_{B}(\xi)\chi_{I_c}(\xi) \approx \chi_{H}(\xi)\chi_{I_c}(\xi)}$ for large ${a}$ (and ${R \gg 1}$). This can be made precise without too much difficulty, but we’ll skip the calculation for the sake of clarity. (Heuristically, all we are saying is that, to a very small rectangle, the edge of the unit ball looks like a half-plane. One might compare this to the fact that the surface of the Earth looks like a plane at small scales.) In other words, for large ${a}$, we can show that

$\displaystyle |S_{1}g (x_1, x_2)| \approx |\left( \mathcal{P} \chi_{[0,R^2]} (\cdot) e^{-2\pi i \cdot} \right)(x_1)\chi_{[0,aR]}(x_2) | \ \ \ \ \ (1)$

where ${\mathcal{P}_{x_1}}$ is the variant of the (${1}$-dimensional) Hilbert transform that we introduced in the previous post, applied in the variable ${x_1}$. Specifically, we have that ${\mathcal{P}}$ is defined (on ${\mathbb{R}^{1}}$) by the relation ${\widehat{\mathcal{P}f}(\xi) = \chi_{(-\infty,1)}(\xi) \hat{f}(\xi)}$. The operator ${\mathcal{P}_{x_1}}$ is just the operator ${\mathcal{P}}$ applied to a function on ${\mathbb{R}^2}$ in the first co-ordinate. One then has that ${|\widehat{\mathcal{P}_{x_1}f}(\xi_1, \xi_2)| =|\chi_{(-\infty,1)}(\xi_1)\hat{f}(\xi_1, \xi_2)| =|\chi_{H}(\xi_1,\xi_2)\hat{f}(\xi_1, \xi_2)|}$. This gives (1).

Rewriting Lemma 3 from the previous post with the interval ${t}$ (which we rename as ${y}$ to avoid confusion with the rectangle ${t}$) taken to be ${[0,R^2]}$, gives

Lemma 3Consider the intervals ${y=[0,R^2]}$ and ${\tilde{y}=[10R, 10R+R]}$. Then there exists a positive constant ${c}$ such that

$\displaystyle |\mathcal{P}\left(e^{-2\pi i \cdot}\chi_{y}(\cdot)\right)(x)|$

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

Putting this lemma together with (1) we see that ${|S_{1} g(x)| \geq c \chi_{\tilde{t}}(x)}$, where ${g}$ is localized to the rectangle ${t=[0,R^2]\times[0,aR]}$ and ${\tilde{t}}$ is defined to be the rectangle ${[10R, 10R+R]\times[0,aR]}$. Moreover, using the fact that rotations commute with the Fourier transform, we can obtain a function ${g}$ with the properties described above localized to any rotation of the interval ${t=[0,R^2]\times[0,aR]}$. More specifically, we claim the following theorem

Theorem 4Let ${t}$ be the rectangle ${[0,R^2]\times[0,aR]}$ rotated so that the longest axis is oriented in the direction ${\theta}$, and let ${\tilde{t}}$ be the rectangle obtained by translating ${t}$ ${10R^2}$ units in the direction ${\theta}$. Then, for large ${a}$ and ${R}$, there exists a function ${g}$ supported on ${t}$ such that ${|g(x)|\leq 1}$ but ${|S_{1} g(x)| \geq c}$ for all ${x \in \tilde{t}}$.

Let ${t=[0,R^2]\times[0,aR]}$ be a rectangle as described in the above theorem, and let ${g_{t}}$ be the bump function localized to this tile. The theorem then states that ${|S_{1}g_{t}(x)| \geq c \chi_{\tilde{t}}(x)}$. In other words we can, very roughly, think of the operator ${S_{1}}$ as an operator that translates the rectangle ${t}$ to ${\tilde{t}}$. We are now in a position to apply Theorem 2, which roughly states that there exists a family of disjoint rectangles ${\{t_{i}\}_{i=1}^{K}}$ with the property that their translates ${\{\tilde{t}_{i}\}_{i=1}^{K}}$ have a large amount of overlap. We now want to exploit the following heuristic, the quantity ${||\sum_{i=1}^{K} g_{t_i} ||_{p} \approx || \sum_{i=1}^{k} {\chi_{t_i}}||_{p}= \sum_{i=1}^{k} ||\chi_{t_i}||_{p} }$ should be very small compared to the quantity ${||\sum_{i=1}^{K} S_{1} g_{t_i} ||_{p} \approx || \sum_{i=1}^{K} {\chi_{\tilde{t}_i}}||_{p}}$ since the function ${\sum_{i=1}^{K} \chi_{t_i}(x)}$ takes small values (either ${0}$ or ${1}$) on a set of large measure, while ${\sum_{i=1}^{K} {\chi_{\tilde{t}_i}}(x)}$ takes large values on a set of small measure (due to the overlap of the rectangles ${\tilde{t}_{i}}$). It turns out that this line of reasoning isn’t in itself enough to create a counterexample. However, we can amplify this approach by introducing some randomness via Khinchin’s inequality. We recall Khinchin’s inequality.

Lemma 5(Khinchin) Let ${g_1(x),\ldots,g_K(x)}$ be a collection of functions and ${\epsilon_{1},\ldots,\epsilon_{K}}$ a collection of random ${\pm}$-signs. Then, for any ${1, we have that

$\displaystyle c^{-1}_{p} || (\sum_{i=1}^{K} | g_i|^2 )^{1/2} ||_{p}^{p} \leq \mathbb{E}\left(|| \sum_{i=1}^{K} \epsilon_{i} g_i ||_{p}^{p}\right) \leq c_{p} || (\sum_{i=1}^{K} | g_i|^2 )^{1/2} ||_{p}^{p} .$

Instead of considering the function ${\sum_{i=1}^{K} g_{t_{i}}}$ used in the heuristic reasoning above, we will consider ${\sum_{i=1}^{K} \epsilon_{i} g_{t_{i}}}$. First we have that

$\displaystyle ||\sum_{i=1}^{k} \epsilon_{i} g_{t_{i}}||_{p} = \sum_{i=1}^{k} || g_{t_{i}}||_{p} \leq \sum_{i=1}^{K} || \chi_{t_{i}}||_{p} \leq (\sum_{i=1}^{K}|t_{i}|)^{1/p} = |\bigcup_{i}^{K} t_{i}|.\ \ \ \ \ (2)$

Here the choice of signs ${\epsilon_{i}}$ was inconsequential. Next, using Khinchin’s inequality we see that

$\displaystyle \mathbb{E}\left(||\sum_{i=1}^{K} \epsilon_{i} S_{1}g_{t_{i}}||_{p}^{p} \right) \geq c^{-1}_{p} || (\sum_{i=1}^{K} | g_i|^2 )^{1/2} ||_{p}^{p} \geq c || (\sum_{i=1}^{K} | \chi_{\tilde{t}_{i}}|^2 )^{1/2} ||_{p}^{p} \geq c || \sum_{i=1}^{K} \chi_{\tilde{t}_{i}} ||_{p/2}^{p/2}.$

Thus, for some choice of signs ${\{\epsilon_{i}\}_{i=1}^{k}}$ we have that ${||\sum_{i=1}^{k} \epsilon_{i} S_{1}g_{t_{i}}||_{p}^{p} \geq c || \sum_{i=1}^{K} \chi_{\tilde{t}_{i}} ||_{p/2}^{p/2}}$. Now let us estimate the right-hand side of this expression. Here we’ll use property (3) from Theorem 2 which states that we can take ${|\bigcup_{i} \tilde{t}_{i} | = \delta |\bigcup_{i} t_{i}| }$ (for any ${\delta >0}$). This gives us that the function ${\sum_{i=1}^{K} \chi_{\tilde{t}_{i}} (x)}$ is supported on a set of measure at most ${\delta |\bigcup_{i} t_{i}|}$. One now can easily see that ${|| \sum_{i=1}^{K} \chi_{\tilde{t}_{i}} ||_{p/2}^{p/2}}$ is as small as possible when the support is as large as possible (that is ${\delta |\bigcup_{i} t_{i}| }$) and ${\sum_{i=1}^{K} \chi_{\tilde{t}_{i}}(x) = \delta^{-1}}$ for all ${x}$ in the support of this function. Thus we have that

$\displaystyle ||\sum_{i=1}^{K} \epsilon_{i} S_{1}g_{t_{i}}||_{p}^{p} \geq c || \sum_{i=1}^{K} \chi_{\tilde{t}_{i}} ||_{p/2}^{p/2} \geq c \delta^{1-p/2} |\bigcup_{i=1}^{K} t_{i}|.$

Putting this together with (2) we have that

$\displaystyle \frac{||\sum_{i=1}^{K} \epsilon_{i} S_{1}g_{t_{i}}||_{p}^{p}}{||\sum_{i=1}^{K} \epsilon_{i} g_{t_{i}}||_{p}^{p}} \geq \delta^{1-p/2}.$

If ${p>2}$, taking ${\delta}$ arbitrarily close to zero shows that ${S_{1}}$ is not a bounded operator on ${L^{p}(\mathbb{R}^2)}$. Notice that the argument breaks down when ${p=2}$. This should be reassuring since ${S_1}$ is bounded in this case. We can, however, extend our counterexample to ${p}$ such that ${1 by duality. Let ${p'}$ be the conjugate exponent to ${p}$ defined ${1/p+1/p'=1}$. If ${1 then ${p'>2}$ and we have that

$\displaystyle ||S_{1}||_{L^p \rightarrow L^{p}} =\sup_{||f||_{p}=1} \sup_{||g||_{p'}=1} \left = \sup_{||g||_{p'}=1}\sup_{||f||_{p}=1} \left = ||S_{1}||_{L^{p'} \rightarrow L^{p'}}$

which shows that ${S_{1}}$ is unbounded on ${L^{p'}(\mathbb{R}^2)}$ as well.

In preparing these notes, I relied on the following references: Davis and Changs’s Lectures on Bochner-Riesz Means, Fefferman’s paper The Multiplier Problem for the Ball, Stein’s Harmonic Analysis, and Tao’s Math 254B notes.

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### 5 Responses

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1. […] Fourier Multipliers, Marcel Riesz leave a comment « Lattice Points in l^{1} Balls Fefferman’s Ball Multiplier Counterexample […]

2. Thomas said, on October 20, 2010 at 1:36 pm

I really like this post, I’m doing a paper on Kakeya sets and their applications and this (and its prequel) helped me a lot. I hope you have time to answer a (probably simple) question about the very last line of the post: why is

sup sup [S_1 f, g]

equal to

sup sup [f, S_1 g]

It looks like some kind of self-adjointness, do we know that the operator is its own adjoint?

Thanks,
Thomas

• Mark Lewko said, on October 20, 2010 at 6:33 pm

Thomas,

Thanks for reading! You’re right that this follows from self-adjointness. By Parseval’s identity we have

$(S_{1}f,g) =\int_{\mathbb{R}^2} \widehat{S_{1} f}(\xi)\overline{\widehat{g}(\xi)}d\xi=\int_{B}\widehat{f}(\xi) \overline{\widehat{g}(\xi)}d\xi$.

The last equality follows from the definition of $S_{1}$ where $B$ denotes the unit ball. Now the last expression is also equal to (by essentially the same argument just given) $\int_{\mathbb{R}^2} \widehat{f}(\xi) \widehat{S_{1}g}(\xi)d\xi=(f,S_{1}g)$.

• Thomas said, on October 22, 2010 at 5:35 am

Of course, that makes a lot of sense. Thank you very much.

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

[…] for be a Fourier multiplier on (see e.g., [1]). However, Fefferman (see [2],[3]) gave an intricate proof which show us that this condition is not sufficient, that is, he showed that the operator does […]