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Possible Duplicate:
If $\int_0^\infty fdx$ exists, does $\lim_{x\to\infty}f(x)=0$?

Let $ \int_{-\infty}^\infty |f| < \infty$. Then $ \lim_{x \to \infty} f(x) =0 \;?$ If this is true, then how can I prove this?

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    If you let $f$ be uniformly continuous,then $\lim_{x\to\infty}f(x)=0$2012-10-13

1 Answers 1

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Let $f(x)=\begin{cases} n,&\text{if }n\le x\le n+\frac1{n^3}\text{ for some }n\in\Bbb Z^+\\ 0,&\text{otherwise}\;; \end{cases}$

then $\int_{-\infty}^\infty f(x)\,dx=\sum_{n\ge 1}\frac{n}{n^3}=\frac{\pi^2}6\;,$

but $\limsup\limits_{x\to\infty}f(x)=\infty$. You can replace the steps with ‘tents’ to make $f$ continuous without qualitatively affecting the example; you can even round off the corners to make it arbitrarily differentiable.

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    Noting that $\liminf\limits_{x\to\infty}f(x)=0$ shows that the limit does not exist. However, if the limit exists...2012-10-13