2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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"? $\endgroup$ – Clement C. Mar 30 '17 at 22:52
  • $\begingroup$ @ClementC. Didn't he specify the former? $\endgroup$ – zhw. Mar 30 '17 at 22:53
  • $\begingroup$ "any" sounds a bit ambiguous to me, but that's how I'd interpret it my default. @zhw. $\endgroup$ – Clement C. Mar 30 '17 at 22:56
  • $\begingroup$ I'm trying to determine if the sequence is summable for every value of $a$ and $b$, yes. I'll edit to clarify $\endgroup$ – Matrefeytontias Mar 30 '17 at 22:59
  • 1
    $\begingroup$ 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? $\endgroup$ – zhw. Mar 30 '17 at 23:03
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\mathbb R$. $\endgroup$ – Matrefeytontias Mar 30 '17 at 23:44
  • $\begingroup$ It is continuous on $\mathbb{R}$. It does not have a limit as $x \to \infty$, which are completely different things. $\endgroup$ – Aloizio Macedo Mar 30 '17 at 23:46
  • $\begingroup$ Or do you think that $\sin(x)$ is not continuous on $\mathbb{R}$, say? $\endgroup$ – Aloizio Macedo Mar 30 '17 at 23:47
  • 1
    $\begingroup$ @Matrefeytontias What your argument (although a little sketchy) shows is that the function is not uniformly continuous. It indeed isn't. But it is continuous, and this is a mere consequence of the glueing lemma, as I said. The case of zhw is the same. $\endgroup$ – Aloizio Macedo Mar 30 '17 at 23:58
  • 1
    $\begingroup$ Oh wow. Turns out I actually can't read (or remember), and mixed both definitions. That's entirely my bad, this is valid then. $\endgroup$ – Matrefeytontias Mar 31 '17 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.