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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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1Basically the same question as this one: If $\int_0^\infty fdx$ exists, does $\lim_{x\to\infty}f(x)=0$? – Martin Sleziak Oct 13 '12 at 06:32
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2Consider $$f(x) = \begin{cases}1 & \text{if }x \in \mathbb{Q} \0 & \text{otherwise} \end{cases}$$ for a counterexample. – Oct 13 '12 at 06:32
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@Marvis Thank you for your comment. But how about the case of $f \in L^1(\Bbb R)$ ? – Ann Oct 13 '12 at 06:35
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1@Ann $f(x)=\sum_{n=0}^\infty 2^{n} \chi_{[n,n+4^{-n}]}$. – JSchlather Oct 13 '12 at 06:40
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3Marvis's function is in $L^1$! – Mariano Suárez-Álvarez Oct 13 '12 at 06:48
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If you let $f$ be uniformly continuous,then $\lim_{x\to\infty}f(x)=0$ – Tomas Oct 13 '12 at 09:17
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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.
Brian M. Scott
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An easy way to make it $C^\infty$ is to use the same idea, but scale Gaussian functions instead. – Potato Oct 13 '12 at 07:00
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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... – robjohn Oct 13 '12 at 07:37