Fourier tranform of decaying periodic function $f(t+k)=e^{-k^2/2}e^{-kt}f(t)$

Question

I have a function which I know is almost periodic, except that it is decaying

$$f(t+k)=e^{-k^2/2}e^{-kt}f(t)=e^{-k(k/2+t)}f(t)$$

for integers $k\in\mathbb Z$. For example if I choose $k=1$ I get

$$f(t+1)=e^{-1/2}e^{-t}f(t)$$ or if I choose $k=-1$ I get

$$f(t-1)=e^{-1/2}e^{t}f(t)=e^{-k(k/2+t)}f(t)$$

This means that if we choose some point $t_0=0$ we could define $f_0=f(t_0=0)$ and write

$$f(t=k)=e^{-k^2/2}f_0$$

I would like to know whether it is possible to use this to extract information about the Fourier transform of $f$. I suspect that using this information it will be possible to identify the Fourier transform, at least in some region.

If this is not possible in general, then what information can we identify about $\hat f(x)$?

I have tried taking the Fourier transform of both sides of the initial expression but I don't know how to evaluate the Fourier transform of the right hand side (which reduces to something like a Laplace transformation)

$$e^{-k^2/2}\int f(t) e^{-kt}e^{ixt}\text{d} t=e^{-k^2/2}\int f(t) e^{(ix-k)t}\text{d} t$$

Answer 1

So if you define $g(t) = e^{t^2/2} f(t)$, you find $$ g(t+k) = e^{k^2/2+tk+t^2/2} f(t+k) = e^{t^2/2} f(t) = g(t) $$ hence you condition is equivalent to say that $e^{t^2/2} f(t)$ is a $1$-periodic function. Equivalently, it means you can write $f(t) = g(t)\,e^{-t^2/2}$ for some $1$-periodic function $g$. Taking the Fourier transform yields $$ \widehat{f}(x) = \widehat{g} * \widehat{e^{-t^2/2}} = \widehat{g} * e^{-2(\pi x)^2} $$ But since $g$ is periodic, its Fourier transform is related to its Fourier series since $$ \widehat{g}(x) = \sum_{k\in\Bbb Z} c_k(g) \,\widehat{e^{2i\pi k x}} = \sum_{k\in\Bbb Z} c_k(g) \,\delta_k $$ where $\delta_k$ is the Dirac delta distribution centered at $x=k$ and the $c_k(g)$ are the Fourier coefficients of $g$. Now doing the convolution gives $$ \widehat{f}(x) = \int\sum_{k\in\Bbb Z} c_k(g)\, e^{-2(\pi (x-y))^2} \,\delta_k(\mathrm d y) = \sum_{k\in\Bbb Z} c_k(g)\, e^{-2(\pi (x-k))^2}. $$ You cannot really simplify it further since $g$ can be any $1$-periodic function in general.