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Suppose that $f \in L^p(\mathbb R^n)$ such that $1\leq p < \infty$. Let $\hat f$ be the Fourier transform of $f$. Clearly, if $p=1$ or $p=2$ then the support of $\hat f$ has positive Lebesgue measure, provided that $f \neq 0$. Using Hausdorff-Young this also holds if $p \in [1,2]$. On the other hand, if $p=\infty$ then this statement is not true as the Delta distribution shows.

What about $p \in (2,\infty)$?

That is, let $f \in L^p(\mathbb R^n)$ such that $f \neq 0$ and $p \in (2,\infty)$. Does the support of $\hat f$ (in the sense of distributions) have positive Lebesgue measure? Or does it at least contain a limit point?

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2 Answers 2

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No. In dimension $d\ge 2$, the surface measure of the sphere gives a counterexample: its Fourier transform has power decay.

For $d=1$, there are also singular measures whose Fourier transform has power decay. One can consider a closed set of positive Fourier dimension.

As for the last part of your question, the support of $\widehat{f}$ cannot be a bounded discrete (= finite) set since then $\widehat{f}$ is a combination of $\delta$'s and their derivatives, so $f$ is a linear combination of functions of the type $x^n e^{iax}$, which cannot be in $L^p$.

In the case of an infinite discrete set it's clear that a measure supported by such a set will no longer work as a counterexample. In fact, if $\mu$ is any measure with a point part, then, by Wiener's theorem, $f=\widehat{\mu}$ satisfies $$ T\lesssim \int_{-T}^T |f|^2\, dt \le \left( \int_{-T}^T|f|^p\, dt\right)^{2/p} (2T)^{1-2/p} , $$ so $f\notin L^p$.

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    $\begingroup$ I would add that the example is obtained by taking the inverse Fourier transform and I would perhaps show the explicit formula for the Fourier transform when $n=3$ so it would be clear that it works with any $p>3$ when $n=3$. $\endgroup$ Oct 28, 2022 at 17:13
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    $\begingroup$ In any dimension, one can generate counterexamples by working with Salem sets of suitable dimension in place of the sphere. One recent explicit construction of such sets is in arxiv.org/abs/1909.04581 $\endgroup$
    – Terry Tao
    Oct 29, 2022 at 16:29
  • $\begingroup$ @TerryTao: One does need to be a bit careful with such examples of sets of positive Fourier dimension since we need closed sets here or the support will get larger. (It's not clear to me from a very quick look at the paper you cited if the examples in the paper are of this type, or if the authors are even interested in this.) $\endgroup$ Oct 29, 2022 at 16:36
  • $\begingroup$ Fair enough; but many constructions of Salem sets are closed. For instance, the constructions in Section 6 of arxiv.org/abs/0712.3882 are Cantor-type sets and thus compact (they are in one dimension, but as in your answer, this can then be boosted to arbitrary dimension by taking Cartesian products). $\endgroup$
    – Terry Tao
    Oct 29, 2022 at 16:50
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I can prove that when $p\in (2,\infty)$, then the Fourier transform need not be represented by a locally integrable function. The proof presented below is not constructive. For an explicit example of a function whose Fourier transform has support of Lebesgue measure zero, see the example of Christian Remling.

Theorem. If $p\in (2,\infty)$, then there is $f\in L^p(\mathbb{R}^n)$ such that the distributional Fourier transform $\hat{f}$ is not given by a locally integrable function.

In the proof we will need the following result.

Lemma. Fix $z\in\mathbb{C}$ with ${\rm re}\, z>0$ and choose $z^{1/2}$ to have ${\rm re}\, z^{1/2}>0$. Then the Fourier transform of $$ f(x)=z^{-1/2} e^{-\pi x^2/z}, \quad x\in\mathbb{R} $$ equals $$ \hat{f}(\xi) = e^{-\pi z\xi^2}, \quad \xi\in\mathbb{R}\, . $$ In particular for $\delta>0$ we have \begin{equation} \label{a} \left(\frac{e^{-\pi x^2/(1+i\delta)}}{(1+i\delta)^{1/2}}\right)^\wedge(\xi)= e^{-\pi(1+i\delta)\xi^2}\, . \end{equation}

Proof. If $f(x)=e^{-4\pi^2tx^2}$, $t>0$, then $\hat{f}(\xi)=(4\pi t)^{-1/2} e^{-\xi^2/(4t)}$. Hence also $$ \left( (4\pi t)^{-1/2} e^{-x^2/(4t)}\right)^\wedge(\xi)= e^{-4\pi^2 t\xi^2}\, . $$ Taking $4\pi t=a$ we have \begin{equation} \left(\frac{e^{-\pi x^2/a}}{a^{1/2}}\right)^\wedge(\xi)= e^{-\pi\xi^2 a} \quad \mbox{for $a>0$.} \tag1 \end{equation} Let $$ f_z(x)= z^{-1/2} e^{-\pi x^2/z}, \quad {\rm re}\, z>0\, . $$ Fix $\xi\in\mathbb{R}$. It is easy to see that the function $$ z\mapsto \hat{f}_z(\xi), \quad {\rm re}\, z>0 $$ is holomorphic. This easily follows from the integral formula that defines $\hat{f}_z(\xi)$ and the fact that we can differentiate this formula with respect to $z$ under the sign of the integral.

The function $$ z\mapsto g_z(\xi)=e^{-\pi z\xi^2}, \quad {\rm re}\, z>0 $$ is holomorphic (on $\mathbb{C}$). Since $\hat{f}_z(\xi)=g_z(\xi)$ on the half line ${\rm re}\, z>0$, ${\rm im}\, z=0$ by (1), we conclude that $\hat{f}_z(\xi)=g_z(\xi)$ for all ${\rm re}\, z>0$. $\Box$

Proof of the theorem. Let $p>2$. Suppose that for every $f\in L^p(\mathbb{R}^n)$, $\hat{f}$ is represented by a locally integrable function. let $\eta\in C_0^\infty(\mathbb{R}^n)$, $\eta(x)=1$ for $|x|\leq 1$. Then the operator $$ L^p(\mathbb{R}^n)\ni f\stackrel{T}{\mapsto} \eta\hat{f}\in L^1(\mathbb{R}^n) $$ is linear. We will prove that $T$ is bounded. Recall that according to the closed graph theorem in order to prove that a linear operator between Banach spaces $T:X\to Y$ is bounded it suffices to show the implication $$ x_k\to x, \quad Tx_k\to y \qquad \Rightarrow \qquad Tx=y $$ which is easily equivalent to the implication $$ x_k\to 0, \quad Tx_k\to y \qquad \Rightarrow \qquad y=0. $$ Suppose that $f_n\to 0$ in $L^p$ and $Tf_n\to g$ in $L^1$. Then for every $\psi\in\mathscr{S}_n$ we have $$ Tf_n[\psi]\to \int_{\mathbb{R}^n} g(x)\psi(x)\, dx\, . $$ On the other hand $$ Tf_n[\psi] = \eta\hat{f}_n[\psi] = f_n[(\eta\psi)\,\hat{}\,]=\int_{\mathbb{R}^n} f_n(x)(\eta\psi)\,\hat{}\,(x)\, dx\to 0\, , $$ so $$ \int_{\mathbb{R}^n} g(x)\psi(x)\, dx =0 $$ for all $\psi\in \mathscr{S}_n$ and hence $g=0$ a.e. This proves boundedness of $T$. In particular $$ \int_{B(0,1)} |\hat{f}| \leq\int_{\mathbb{R}^n}|\eta\hat{f}|\leq \Vert f\Vert_p\, . $$ Let now $$ f_\delta(x)=\frac{e^{-\pi x^2/(1+i\delta)}}{(1+i\delta)^{1/2}}\, . $$ Then according to the lemma $$ \hat{f}_\delta(\xi) = e^{-\pi(1+i\delta)\xi^2} $$ and according to our estimate $$ \int_{-1}^1 |\hat{f}_\delta|\leq C\Vert f_\delta\Vert_p\, . $$ We have $$ \int_{-1}^1 |\hat{f}_\delta| = \int_{-1}^1 e^{-\pi\xi^2}\, d\xi = A >0\, , $$ where $A$ is a constant independent of $\delta$. Hence $$ A^p\leq C^p\int_\mathbb{R} |f_\delta|^p = C\int_\mathbb{R} \frac{e^{-\pi x^2p/(1+\delta^2)}}{(1+\delta^2)^{p/2}}\, dx \to 0 \quad \mbox{as $\delta\to 0$} $$ which is a contradiction. $\Box$

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