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I want to find a function, such that the function shall be $p$-integrable in the sense of $\int_{\mathbb{R}}|f(x)|^pdx<\infty$ for any $p\in[1,\infty)$ and that this does NOT imply $\lim_{x\to\infty}f(x)=0$.

Now it's similar to the question here Continuous unbounded but integrable functions, but I'm not looking for a function which is unbounded in any point, just one which limit in infinity does not converge to $0$ or better say does not exist.

Does anyone know of any examples?

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  • $\begingroup$ Do you care if $f$ is continuous? $\endgroup$ – πr8 May 7 '17 at 10:37
  • $\begingroup$ nope, it doesn't matter to me $\endgroup$ – N. Maks May 7 '17 at 10:39
  • $\begingroup$ and do you mean Lebesgue or Riemann integable? $\endgroup$ – πr8 May 7 '17 at 10:39
  • $\begingroup$ Let's take Lebesgue $\endgroup$ – N. Maks May 7 '17 at 10:41
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    $\begingroup$ Sure, then take $f(x) = \exp(-|x|) + \mathbb{I}[x\in\mathbb{Q}]$. (Or, as one of the answers states, even just $f(x) = \mathbb{I}[x\in\mathbb{Q}]$ ). $\endgroup$ – πr8 May 7 '17 at 10:43
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There are, of course, a lot of examples of $p$ integrable functions such that the limit at infinity does not exist. For example, one can take the characteristic function of $\mathbb{Q}$.

It should be noted, however, that there is no function such that $\lim_{x \rightarrow \infty} f(x)$ exists and is $ \ne 0$.

Indeed, assume $\lim_{x \rightarrow \infty} f(x) = \alpha > 0$. Then, for every $\epsilon >0$, there exists $M>0$ such that $$ f(x) > \alpha - \epsilon \qquad \text{for every } x > M. $$ If you take $\epsilon $ such that $ \alpha - \epsilon >0$ than you have that $$ \int_{\mathbb{R}} |f(x)|^p \, dx \ge \int_{\{x > M\}} (\alpha - \epsilon)^p \, dx= \infty. $$

A similar argument works for $\alpha < 0$.

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Take $f$ to be a piecewise affine function. At each natural number $n$ the graph of $f$ is a triangle of height $1$ (so $f(n)=1$) and basis $\frac{1}{n^{2/p}}$ and $f$ is zero elsewhere. Then $\limsup_{x\to\infty}f(x)=1$ while $$\int_0^\infty |f(x)|^p\,dx=\sum_n \frac1{n^2}<\infty.$$

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  • $\begingroup$ when I replied $p\ge 1$ was not specified, so it could have been $p<1$. $\endgroup$ – Gio67 May 7 '17 at 17:12
  • $\begingroup$ Sorry, I deleted my previous comment. There is a bigger problem: The base doesn't get raised to the $p$th power, $f$ does. $\endgroup$ – zhw. May 7 '17 at 17:15

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