$f \in L^1 \implies \lim_{ |x| \to \infty } f(x) = 0$

Question

Suppose $f\in L^{1}(\mathbb R) $

My Question is:

Can we show, $\lim_{ |x| \to \infty } f(x) = 0$?

Thanks,

(this question is obviously related to $f, f'\in L^{1}(\mathbb R) \implies \lim_{x\to \infty} f(x)=0 ?$ a thing I've already proven.)

You may have noticed that the link you provide also states that the claim is false if $f$ is not differentiable everywhere. This automatically applies to your question since the conditions on the other question are stronger. — Clayton, May 18 '19 at 11:51

score 0 · Accepted Answer · answered May 18 '19 at 11:50

0

No you can't : for a simple counterexample, look at $f$ defined by $f(x)=0$ if $x\le 0$, and for $n\in\mathbb N$, $f(n)=n$, $f(n-\frac{1}{n2^{n+1}})=f(n+\frac{1}{n2^{n+1}})=0$ and $f$ affine on any interval between such points.

The graph of $f$ is an union of triangles, the total area is (if I remember correctly) $1$, and $f$ has no limit at $+\infty$.

answered May 18 '19 at 11:50

Nicolas FRANCOIS

4,960

thanks François. Do you know a sufficient condition that is weaker than the derivative, like uniformly continuous or someting like this? – Marine Galantin May 18 '19 at 12:19
Uniform continuity, if I remember correctly... – Nicolas FRANCOIS May 18 '19 at 13:43

$f \in L^1 \implies \lim_{ |x| \to \infty } f(x) = 0$

1 Answers1