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.

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