If I have some function $f(x)$ that is periodic with a period going from $-L$ to $L$, its Fourier coefficients in exponential form can be written as
$$c_{k}=\frac{1}{2 L} \int_{-L}^{L} f(x) e^{\pi i k x / L} d x $$
Suppose I generalize these coefficients to a complex function of $z$ as
$$C(z)=\frac{1}{2 L} \int_{-L}^{L} f(x) e^{\pi i z x / L} d x $$
Is there anything I can know a priori about the continuity of $C(z)$?
For any complex $h$ observe that: $$\left|\frac{e^{ih}-1}{h}\right|^2=|h|^{-2}(e^{-2\Im h}\cos^2(\Re h)+1-2e^{-\Im h}\cos(\Re h)+e^{-2\Im h}\sin^2(\Re h))$$Which may be bounded from above by: $$|h|^{-2}(e^{-2\Im h}+1-2e^{-\Im h}(1-(\Re h)^2/2))\le e^{-\Im h}\left(1+2\frac{\cosh(\Im h)-1}{(\Im h)^2}\right)$$This last quantity tends to $2$ as $h\to0$ in the complex plane, hence is bounded for sufficiently small $h$. That means the dominated convergence theorem applies, when $f$ is integrable, and shows $C$ is even a holomorphic function by a simple bit of rearranging the expression for the difference quotients.
