Does the (distributional) support of the Fourier transform of an $L^p$-function with $p<\infty$ have positive measure?

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?

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 $$f\notin L^p$$.

• 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 $n=3$. Commented Oct 28, 2022 at 17:13
• 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 Commented Oct 29, 2022 at 16:29
• @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.) Commented Oct 29, 2022 at 16:36
• 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). Commented Oct 29, 2022 at 16:50

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 $$$$\label{a} \left(\frac{e^{-\pi x^2/(1+i\delta)}}{(1+i\delta)^{1/2}}\right)^\wedge(\xi)= e^{-\pi(1+i\delta)\xi^2}\, .$$$$

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 $$$$\left(\frac{e^{-\pi x^2/a}}{a^{1/2}}\right)^\wedge(\xi)= e^{-\pi\xi^2 a} \quad \mbox{for a>0.} \tag1$$$$ Let $$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. $$\Box$$