# Non-summable sampling of integrable function?

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.

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.

• Regarding the quantifiers: you want to find if "for every $a,b$, the sequence is summable" or if "there exist some $a,b$ for which it is sommable"? – Clement C. Mar 30 '17 at 22:52
• @ClementC. Didn't he specify the former? – zhw. Mar 30 '17 at 22:53
• "any" sounds a bit ambiguous to me, but that's how I'd interpret it my default. @zhw. – Clement C. Mar 30 '17 at 22:56
• I'm trying to determine if the sequence is summable for every value of $a$ and $b$, yes. I'll edit to clarify – Matrefeytontias Mar 30 '17 at 22:59
• This has been asked a billion times here at MSE. Since I am never able to search well here: Visualize triangles centered over $1,2, \dots$ of heights $1$ having bases equal to $1/2^1, 1/2^2, \dots$. Define the function to be $0$ everywhere else. What do you think? – zhw. Mar 30 '17 at 23:03

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.
• This poses the same question as with the triangles example in the comments of my question. Does this remain continuous "towards infinity" ? I didn't try this time, but seeing that your "$n$th elementary function" tends towards something that is 1 for $x = 2n$ and 0 everywhere else, I think that the same kind of proof works to show that it in fact isn't continuous on $\mathbb R$. – Matrefeytontias Mar 30 '17 at 23:44
• It is continuous on $\mathbb{R}$. It does not have a limit as $x \to \infty$, which are completely different things. – Aloizio Macedo Mar 30 '17 at 23:46
• Or do you think that $\sin(x)$ is not continuous on $\mathbb{R}$, say? – Aloizio Macedo Mar 30 '17 at 23:47