For example, is there a function $g:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ such that the condition $\int\_\mathbb{R}\int\_\mathbb{R}|f(u,v)g(u,v)|dudv<\infty$ implies $\sum\limits\_{k\in\mathbb{Z}}\int\_\mathbb{R}\int\_\mathbb{R}e^{ik(u+v)}f(u,v)dudv<\infty$?



For which $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$, $\sum\limits_{k\in\mathbb{Z}}\int_\mathbb{R}\int_\mathbb{R}e^{ik(u+v)}f(u,v)dudv<\infty$?
Writing $g(t) = \int\_{-\infty}^\infty f(t+s,-s) ds$, we reduce the question to the following one: when $$\sum\_{k = -\infty}^\infty \int\_{-\infty}^\infty e^{i k t} g(t) dt$$ is finite? Setting $h(t) = \sum\_{k = -\infty}^\infty g(t + 2 \pi k)$, we see that the summands in the above expression are simply the coefficients of the Fourier series of $h$. Thus, the question is: when the Fourier coefficients of a periodic function $h$ form a summable sequence. (I assume your question asks for absolute summability.)


As far as I know, there is no simple "if and only if" condition. It is clearly sufficient to assume that $h$ is sufficiently smooth, say: twice continuously differentiable (or differentiable, with $f'$ Hölder continuous). There is no sufficient condition of the form $\int\_0^{2 \pi} |h(t) \phi(t)| dt < \infty$, unless $\phi$ is infinite almost everywhere (indeed: take $h$ to be any discontinuous function supported on the set $\{t : |\phi(t)| < M \}$ for some $M$; if the Fourier coefficients of $h$ were summable, $f$ would be a continuous function).


So the answer to you question is **no**: there is no such function $g(u,v)$.


