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

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.)

## 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.

• 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 at 11:51
• oh sorry you're right – Marine Galantin May 18 at 12:16

## 1 Answer

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 $$+\infty$$.

• 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 at 12:19
• Uniform continuity, if I remember correctly... – Nicolas FRANCOIS May 18 at 13:43