## Elegant Proof of Fourier Uniqueness

I stumbled upon the 2012 AMS release Complex Proofs of Real Theorems by Peter Lax and Lawrence Zalcman while browsing online a few days ago, and I finally got around to checking out the library’s copy today. It’s a short publication–part of the AMS University Lecture Series–of around 90 pages focused on (particularly elegant) applications of complex analysis to proving statements about functions of a real variable. I have only looked at a small fraction of the text, which is not saying much, but one result concerning the uniqueness of the Fourier transform on ${ L^{1}(\mathbb{R})}$ stood out for me.

If ${f \in L^{1}(\mathbb{R})}$ and

$\displaystyle \widehat{f}(\xi) = \int_{\mathbb{R}}f(x)e^{-2\pi i x \xi}dx = 0, \indent \forall \xi \in \mathbb{R},$

then ${f \equiv 0}$.

The argument in the following proof of Fourier uniqueness was originally given D.J. Newman, Fourier uniqueness via complex variables, Amer. Math. Monthly, 81 (1974), 379-380. Proof:Suppose ${\widehat{f}(\xi) = 0}$ for all ${\xi \in \mathbb{R}}$. Fix ${a \in \mathbb{R}}$ and define ${F_{a}: \mathbb{R} \rightarrow \mathbb{C}}$ by

$\displaystyle F_{a}(\xi) = \int_{-\infty}^{a}f(x)e^{-2\pi i(x-a)\xi}dx = -\int_{a}^{\infty}f(x)e^{-2\pi i(x-a) \xi}dx$

We now extend the domain of ${F_{a}}$ to the complex plane. For ${\xi \in \mathbb{H}^{+} = \left\{z \in \mathbb{C} : \text{Im}(z) > 0\right\}}$, define ${F_{a}(\xi)}$ by

$\displaystyle F_{a}(\xi) = \int_{-\infty}^{a}f(x)e^{-2\pi i (x-a)\xi}dx$

and if ${\xi \in \mathbb{H}^{-} = \left\{z \in \mathbb{C} : \text{Im}(z) < 0\right\}}$, define

$\displaystyle F_{a}(\xi) = -\int_{a}^{\infty}f(x)e^{-2\pi i(x-a)\xi}dx$

It is clear that this defines a continuation of ${F_{a}}$, also denoted by ${F_{a}}$, which is bounded on ${\mathbb{C}}$. Moreover, ${F_{a}}$ is continuous on ${\mathbb{C}}$ as a consequence of the Lebesgue dominated convergence theorem. I claim that ${F}$ is analytic in ${\mathbb{C}}$. By Morera’s theorem, it suffices to show that ${\int_{\partial R}F_{a}(\xi)d\xi = 0}$ for any rectangle ${R}$ with positively oriented boundary. Without loss of generality, we can assume that ${R \subset \overline{H}^{+}}$ (the argument for ${R \subset \overline{H}^{-}}$ is completely analogous). Since ${\partial R}$ is compact, ${\int_{-\infty}^{a}\left|f(t)e^{-2\pi i(x-a)\xi}\right|dx < \infty}$, and therefore

$\displaystyle \int_{\partial R}\int_{-\infty}^{a}\left|f(t)e^{-2\pi i(x-a)\xi}\right|dxd\xi < \infty,$

we can apply Fubini’s theorem to obtain

$\displaystyle \int_{\partial R}F_{a}(\xi)d\xi = \int_{-\infty}^{a}f(x)\left(\int_{\partial R}e^{-2\pi i(x-a)\xi}d\xi\right)dx = \int_{-\infty}^{a}f(x)(0)dx = 0$

, since ${e^{-2\pi i(x-a)\xi}}$ is an analytic function of ${\xi}$ for fixed ${x}$. Since ${F_{a}}$ is a bounded entire function, ${F_{a}}$ must be constant by Liouville’s theorem. I claim that ${F_{a} \equiv 0}$. Indeed, by the Lebesgue dominated convergence theorem,

$\displaystyle \lim_{t \uparrow \infty} F_{a}(it) = \lim_{t \uparrow \infty} \int_{-\infty}^{a}f(x)e^{-2\pi i(x-a)it}dx = \lim_{t \uparrow \infty} \int_{-\infty}^{a}f(x)e^{2\pi(x-a)t}dx = 0$

Since ${a \in \mathbb{R}}$ was arbtrary, we obtain

$\displaystyle 0 = F_{a}(a) = \int_{-\infty}^{a}f(x)dx, \indent \forall a \in \mathbb{R}$

It follows from the Lebesgue differentiation theorem that ${f(a) = 0}$ a.e. $\Box$

It is worth remarking that we invoked non-trivial results of the Lebesgue theory of integration through the proof in the forms of the dominated convergence theorem, Fubini’s theorem, and the Lebesgue differentiation theorem.