# The absolute convergence of (say) $\sum\limits_{n \in \mathbb Z} f(x+n)dx$ when $f$ is continuous and in $L^1(\mathbb R)$

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?

• 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. – reuns Jul 6 '19 at 21:00