I am attempting to extend this question, which says that for a nonzero continuous $f$ with compact support on the real line, the Fourier transform cannot decay exponentially. Suppose $f \in C_0(\mathbb{R}^n)$ and let $\alpha$ be a multi-index with $|\alpha|=m$. Extending robojohn's answer to that question, suppose as a contradiction that $|\hat f(\xi)|\leq Ce^{-k|\xi|}$ for some positive constants $C$ and $k$. Then by Fourier inversion,
$$
|D^{\alpha}f(x)| \leq \int_{\mathbb{R}^n} Ce^{-k|\xi|}(2\pi|\xi|)^m \; d\xi = C\frac{(2\pi)^m(m+n-1)!}{k^{m+n}}.
$$
This matches the previous estimate if we take $n=1$. In this case, you can use the Hadamard formula to claim that $f$ must be real analytic at every point on the real line, thus $f$ must be identically zero.
Can the above estimate be used to show that $f\equiv0$ in the general case $n>1$? I believe my knowledge of power series in several variables is lacking. If not, how to show that for $f \in C_0(\mathbb{R}^n)$, the Fourier transform cannot decay exponentially?
Best Answer
Let $g(t):=f(x_1,\dots,x_{n-1},t)$ for fixed $x_1,\dots,x_{n-1}$. By the previous estimate for $\alpha:=(0,\dots,0,p)$, that for all $t$ $$|g^{(p)}(t)|\leqslant C\frac{(2\pi)^p(p+n-1)!}{k^{p+n}},$$ so the Taylor series of $g$ has a radius of convergence $\leqslant \frac{2\pi}k$ everywhere. This implies, as $g$ has compact support, that $g(t)=0$ for all $t\in\Bbb R$. We conclude as the $x_j$ are arbitrary.