Here is a basic theorem from locally compact topological groups. Suppose $G$ is a locally compact abelian topological group with closed subgroup $H$, and Haar measures $dg$ and $dh$. Suppose that $f: G \rightarrow \mathbb C$ is integrable. Then for almost all $g \in G$, the integral $\int\limits_H f(gh)dh$ converges absolutely and becomes an integrable function of $gH$ on the quotient group $G/H$ with respect to its Haar measure $d\bar{g}$. Furthermore, we have

$$\int\limits_G f(g)dg = \int\limits_{G/H} \int\limits_H f(gh)dh d\bar{g}$$

This is proved first when $f$ is continuous and compactly supported, and then extended to arbitrary $f \in L^1(G)$ by density.

I was wondering whether there is anything more about this that can be said when $f$ is not only integrable, but also continuous on $G$. If it makes the results nicer, let's assume $G$ is $\sigma$-compact. If it makes it even nicer, let's assume $G = \mathbb R$ and $H = \mathbb Z$.

Assuming $f: G \rightarrow \mathbb C$ is continuous and integrable...

  • Is $\int\limits_H f(gh)dh$ absolutely convergent for all $g \in G$?

  • Is $g \mapsto \int\limits_H f(gh)dh$ continuous?

  • If $\int\limits_H f(gh)dh$ is absolutely convergent for all $g \in G$, is $g \mapsto \int\limits_H f(gh)dh$ continuous?

  • $\begingroup$ no try with $f(x)=\sum_n n\phi (n^2(x−n))$ where $\phi \in C^\infty_c([-1/2,1/2]),\phi(0)=1$. If $f \in L^1$ and $f' \in L^\infty$ then $f= o(1)$, if $ (1+|x|^{1+\epsilon}) f' \in L^\infty$ then $f = o(1/x^{1+\epsilon})$ and $\sum_n f(x+n)$ converges absolutely and locally uniformly. $\endgroup$ – reuns Jul 6 '19 at 21:00

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