How to show that Fourier transform $\hat{f}(\gamma)$ analytic in whole complex plane if $f\in L^2(\mathbb{R}^1)$ is compact?
Attempt: Since $f \in L^2(\mathbb{R}^1)$ we have that
$\hat{f}(\gamma)=\int_{-\infty}^\infty f(y)e^{-2\pi i y \gamma}dy$
is well defined. Because $f$ is bounded we have $f(y)\leq M$ for all $y$ and some constant $M$. Thus by the triangle inequality.
$\hat{f}(\gamma) \leq M\int_{-\infty}^\infty |e^{-2\pi i y \gamma}|dy \leq M\int_{-\infty}^\infty dy$.
But since the integral is diverging that did not seem to help as we want to show, for $\hat{f}(\gamma)$ to be analytic, that $\hat{f}(\gamma)$ can be represented as a e.g. convergent power series or that:
$ \hat{f}'(\gamma)=-2\pi i y\int_{-\infty}^\infty f(y)e^{-2\pi i y \gamma}dy $
exists.
It should not be "$f$ is compact" which has no meaning to my knowledge but "$f$ is compactly supported". Then the theorem is known as the Paley-Wiener theorem.
Let $K$ be the compact support of $f$. In this case indeed, $\hat f$ is bounded since the integral defining the Fourier transform is reduced to a bounded domain and by the Cauchy-Schwarz inequality $$ \hat f(y) = \int_K f(x) \,e^{-2i\pi x\cdot y}\, \mathrm d x \leq \|f\|_{L^2} \,|K|^{1/2} $$ More generally, taking $n$ derivatives and defining $R$ such that $K$ is included in the ball of size $R$, $$ |(\hat f)^{(n)}(y)| = \left|\int_K (2\pi x)^n f(x) \,e^{-2i\pi x\cdot y}\, \mathrm d x\right| \\ \leq (2\pi)^n \|f\|_{L^2} \left(\int_{-R}^R |x|^{2n} \,\mathrm d x\right)^{1/2} \\ \leq \sqrt{\tfrac{2R}{2n+1}} (2\pi R)^n \|f\|_{L^2} $$ from which you can easily prove the convergence of the Taylor series, and the convergence to $0$ of the remainder.