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This question already has an answer here:

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

My Question is:

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

Thanks,

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

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marked as duplicate by Martin R, Clayton, Community May 18 at 12:17

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – Clayton May 18 at 11:51
  • $\begingroup$ oh sorry you're right $\endgroup$ – Marine Galantin May 18 at 12:16
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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 $+\infty$.

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

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