# Fourier transform of a locally integrable tempered distribution

The following statement is an exercise in a set of lecture notes, and I cannot prove nor find a proof of it.

Let $u \in \mathcal{S}'(\mathbb{R}^n)\cap L^1_{\mathrm{loc}}(\mathbb{R}^n)$ be non-negative, and assume that the Fourier transform $\hat{u}$ of $u$ is in $L^{\infty}(\mathbb{R}^n)$. Then in fact $u \in L^1(\mathbb{R})$, and we have the equality $(2\pi)^{n/2}\|\hat{u}\|_{\infty} = \|u\|_{1}$.

The exercise is supplemented by the hint: Consider the sequence $u_k(x) = u(x)\varphi(x/k)$ for some $\varphi \in C_0^{\infty}(\mathbb{R}^n)$ is a standard bump function with support around zero and $\varphi(0) = 1$.

If we drop the assumption that $u \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ then the statement has easy counterexamples, so this must be key. I do not see an immediate counterexample if we drop the assumption that $u$ is non-negative.

I have tried to use the function given in the hint to prove the statement. We can see that $u_k$ converges almost everywhere to $u$ as $k \to \infty$. However, without the dominated convergence theorem at my disposal (as we have no bounds by integrable functions), I am unable to use this to relate the norm of $u$ to that of $\hat{u}$. I may require another approach.

• It is indeed true that there is no uniform bound for $u_k$, as $\varphi(x)\geqslant 1/2$ on a ball $B(0,\delta)$ hence $\sup_k\left\lvert u_k(x)\right\rvert\geqslant \left\lvert u(x)\right\rvert /2$. Maybe a uniform integrability argument can help. – Davide Giraudo Aug 19 '17 at 11:25
• Such an argument could be that $\phi(0) = \tanh(x/k)$, this way $(2\pi)^{-n/2} u_{n+1} = u_{n}$ will have a uniform integral. – McTaffy Aug 19 '17 at 12:41

## 1 Answer

Let $u_k(x) = u(x) e^{-\pi |x|^2/k^2}$, $\widehat{u_k} = \widehat{u} \ast k^n e^{-\pi |x|^2 k^2}$. Then since $u_k(x) \ge 0$ and is increasing in $k$ $$\|u\|_{L^1}=\lim_{k \to \infty}\|u_k\|_{L^1} = \lim_{k \to \infty} \widehat{u_k}(0) \le \|\widehat{u}\|_{L^\infty}\lim_{k \to \infty} \|1 \ast k^n e^{-\pi |x|^2 k^2}\|_\infty =\|\widehat{u}\|_{L^\infty}$$