3

I am asking about the sort-of converse to this question: under what additional conditions on $f:\mathbb{R}_+\rightarrow\mathbb{R}$ does the following hold? $$ \int_{\mathbb{R}_+}f<\infty\implies \lim_{x\rightarrow\infty}f(x)=0 $$

Where the limit above is made in the topological sense: for every increasing diverging sequence $x_n$, $f(x_n)\rightarrow 0$.

Surely, by the linked question, the set of such functions includes uniformly continuous ones, but can we expand the set and completely characterize it? Is it the set of BV functions? Absolutely continuous?

Community
  • 1
VF1
  • 1,993
  • In what sense do you consider your $\lim$? Note that $f$ is a class of functions that are equal almost everywhere. So depending on the version of $f$ the limit can vary – Glinka Jan 22 '17 at 22:36
  • Yes, I'm asking if the set of all such vanishing versions $\left{\tilde{f}\right}$ can be characterized in some nontrivial way. – VF1 Jan 22 '17 at 22:40

2 Answers2

1

Other than a condition that is essentially a restatement of $\lim_{x \to \infty} f(x) = 0$, I don't see how you can expect to completely characterize it. It's easy to construct some quite nasty functions (discontinuous and not BV) which have limit $0$ at $\infty$ and finite integral.

Robert Israel
  • 448,999
  • Hmm, you're right, it can get pretty arbitrary. Mind if I change the question slightly to require continuity, then? – VF1 Jan 22 '17 at 22:52
  • It wouldn't help much. The function can be about as nasty as a continuous function can be... – Robert Israel Jan 22 '17 at 22:53
  • OK, you're right. But I can't think of any counterexamples that BV - perhaps what I was trying to get at with the question is that the uniformly continuous condition could be a bit weakened. Does that merit asking in another thread? – VF1 Jan 22 '17 at 23:04
  • @VF1, BV seems enough. If $f(x)$ does not converge as $x \to \infty$ so that $\limsup_{x\to\infty} f(x) > \liminf_{x\to\infty} f(x)$. then $f$ cannot be of BV. Knowing now that $f$ converges as $x \to \infty$, the integrability condition forces the limit to be zero. – Sangchul Lee Jan 22 '17 at 23:26
1

I think that I have made some sort of generalization of your statement. $$\int f= \sum_{i=0}^{\infty} \int_{[i,i+1]} f$$ Therefore $\lim_{i\to \infty} \int_{[i,i+1]} f=0$ or else the sum would diverge. This implies (from this fact here) that $\lim_{i\to \infty} f|_{[i,i+1]}=0$ almost everywhere. This means that $f$ restricted to the set $[i,i+1]$ must be $0$ in the limit $i \to \infty$ except on a subset of measure zero.

Alternatively, $$\int f = \lim_{a\to \infty} \int_{[0,a]} f = \lim_{a\to \infty} (\int f - \int_{[a,\infty]} f) = \int f - \lim_{a\to \infty} \int_{[a,\infty]} f$$

So $\lim_{a\to \infty} \int_{[a,\infty]} f=0$ if $\int f < \infty$

This like before implies that the $\lim_{a\to \infty} f|_{[a,\infty]} =0$ almost everwhere.

Community
  • 1
edenstar
  • 944
  • 6
  • 16