On weighted Fourier transforms

Question

Suppose that $f\in L^{\infty}((0,1))$ and that there exists $c_1,c_2>0$ such that $$ \left|\int_0^1 e^{i \xi x} e^{-|\xi|^{-1}x}f(x)\,dx \right| \leq c_1 e^{-c_2|\xi|} \quad \forall\, |\xi|>1.$$ Does it follow that $f$ must vanish almost everywhere?

Answer 1

The answer is yes. I will prove it using the assumption only for $\xi > 0$ (almost the same proof works if we only assume it for $\xi < 0$).

For $z\in U = \{z: Im (z) \ge 0, |z| \ge 1\}$, consider the function

$$F(z) = \int_0^1 e^{izx - \frac{x}{z}}f(x)dx.$$

We have $|F(z)| \le \int_0^1 e^x |f(x)|dx \le e ||f||_{L^\infty} \le e ||f||_{L^2} < \infty$, so the function $F$ is bounded in $U$. By differentiating under the integral sign we can also see that the function $F$ is analytic. Then, unless $F\equiv 0$, the logaritmic integral over the boundary must converge, in particular

$$\int_1^\infty \frac{|\log |F(x)||}{1+x^2} dx < \infty.$$

(here there's a slight issue in that I am writing the logarithmic integral for the full upper half-plane, and in general we have to integrate against the harmonic measure, but it is not hard to see that for the domain $U$ it is proportional to the half-plane).

On the other hand, we know that $|\log |F(x)|| \ge c_2 x - \log c_1$ from our assumption, so the integral actually diverges. Thus, $F\equiv 0$.

Now, we are gonna show that $\hat{f} \equiv 0$ from which it would follow that $f\equiv 0$. We have from the same differentiation under the integral sign that $\hat{f}$ is an entire function (that is, analytic in the whole plane). On the other hand, $F(z) = \hat{f}(z + \frac{i}{z})$. We know that $F$ is identically zero on the whole of $U$, so $\hat{f}$ is identically zero on some open part of $\mathbb{C}$, hence it is zero everywhere by the identity theorem, in particular it is zero on $\mathbb{R}$, so $\hat{f} \equiv 0$, thus $f\equiv 0$, as required.

As can be seen from the argument, it is enough to assume that $$\int_1^\infty \frac{|\log |F(x)||}{1+x^2} dx = \infty,$$

say the bound $c_1 e^{-c_2 |\xi|/\log(|\xi|)}$ would also have been enough (and also it is enough to assume that $f\in L^1((0, 1))$, no need for $L^\infty$).