given a continuous function $f \in L^1(\mathbb{R})$, I can't seem to find out if $\forall a, b \in \mathbb{R_+^*}$, the sequence $(u_n := f(a + bn))_{n \in \mathbb{Z}}$ is absolutely summable for sure or not. I don't have a lead for this, but I can't find a counter-example either.
Thanks by advance !
EDIT : I had the idea to use the fact that for any non-zero natural integer $N$, $b = \frac{a + Nb - a}{N}$, so the sum can be understood as some sort of Riemann sum over $[a, a + Nb]$ maybe possibly. Then, using the fact that $f \in L_1(\mathbb R)$, say that there exists some $N_l$ so that $f$ is "negligeable" on $[N_l, +\infty)$ in order to replace the right bound of the summation's interval with $N_l$ in order to get rid of the dependency to $N$. Not sure where to go with this, but that's what I have.
Just consider a $L^1$ function with spikes of height $1$ on the natural numbers and take $a,b=1$.
Explicitly (although not of height $1$ on the naturals per se), let $f:\mathbb{R}_{ \geq 0} \to \mathbb{R}$ be given by $f(x)=(x-(n+1))^{2n^2}$ if $n \leq x \leq n+2$, where $n$ is even. By the glueing lemma, this is continuous. The integral is given by $\sum\limits_{n \text{ even}}\frac{2}{2n^2+1}$. Mirroring $f$, we get a function $g$ defined on the real numbers which is in $L^1$ (since its integral is $\sum\limits_{n \text{ even}} \frac{4}{2n^2+1}$). However, $f(2+2n)=1$ for every $n$, and the series $\sum\limits_{n \in \mathbb{N}} 1$ diverges.