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

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

• 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
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 $$+\infty$$.