Fourier transform of compact supported function is entire

Question

Suppose $f\in L^1(\mathbb{R})$ has compact support, say $\operatorname{supp}f\subset[-r,r]$. I want to show that its Fourier transform $$ \hat{f}(z) = \int_{-a}^{a} f(t) e^{-2\pi itz}dt$$ for $z\in\mathbb{C}$ is entire, so $\hat{f}\in H(\mathbb{C})$. I think this follows from Morera's theorem, but I am not very familiar with contour integrals. Could someone help me out with the first steps?

Answer 1

"No, but I actually would like to learn how one could apply Morera's theorem here"

Well, for that we have to prove two things:

(a) $\hat f$ is continuous.

(b) The integral of $\hat f$ over each triangle is zero.

So, let's start with (a). Since for $z$ close to $z_0$, $$ |e^{2\pi itz}-e^{2\pi itz_0}|\le 4\sqrt 2 \pi ae^{2\pi a(|1+Im(z_0)|)}|z-z_0| $$ (that's at least what I could achieve), we have $$ |\hat f(z)-\hat f(z_0)| = \left|\int_{-a}^af(t)\left(e^{2\pi itz} - e^{2\pi itz_0}\right)\,dt\right|\,\le\,4\sqrt 2 \pi ae^{2\pi a(|1+Im(z_0)|)}\|f\|_{L^1}|z-z_0|. $$ Hence, $\hat f$ is locally Lipschitz. Now, let $\Delta$ be any triangle. Then with Fubini we get $$ \int_\Delta\hat f(z)\,dz = \int_\Delta \int_{-a}^af(t)e^{2\pi itz}\,dt\,dz = \int_{-a}^af(t)\left(\int_\Delta e^{2\pi itz}\,dz\right)\,dt = 0. $$

Answer 2

Here is a proof based on Moreras theorem that $\hat{f}(z) = \int_{-a}^a f(t) e^{-2 \pi i tz} dt$ is holomorphic.

Let $\gamma(t)$ be a simply closed curve in the complex plane.

Now, $\int_\gamma \hat{f}(z) dz=\int_\gamma \int_{-a}^a f(t) e^{-2 \pi i tz} dt dz = \int_{-a}^a \int_\gamma f(t) e^{-2 \pi i tz} dzdt$ by Fubinis theorem.

But the function $g(z)=f(t)e^{-2 \pi i tz}$ is holomorphic so $\int_\gamma g(z) dz = 0 $ by Cauchys theorem.

Hence, we have shown that for any simply closed curve $\gamma$, $\int_\gamma \hat{f}(z) dz =0$. Moreras theorem, then, implies that $\hat{f}$ is holomorphic.

Answer 3

Otherwise it is immediate that $$\hat{f}(z)=\int_{-a}^a \sum_{k\ge 0}z^k\frac{(-2i\pi t)^k}{k!} f(t)dt= \sum_{k\ge 0}z^k\int_{-a}^a \frac{(-2i\pi t)^k}{k!} f(t)dt$$

where inverting $\sum,\int$ is allowed by convergence of $\int_{-a}^a \sum_{k\ge 0}\frac{|-2i\pi z t|^k}{k!} |f(t)|dt$