PDF with a finite second moment converges to zero? Suppose for a probability density function $p(x)$, $x\in \mathbb{R}$, we have $\int_{-\infty}^{+\infty} x^2p(x)dx < \infty$. Can we claim $\lim_{x\rightarrow\infty} p(x) = 0$? 
I think it boils down to under what conditions $p(x)$ has a limit as $x\rightarrow\infty$. If it has a limit, the limit must be zero. Are there any conditions that guarantee the above convergence? 
I also have a hard time coming up counter examples. 
Edit: Initially i thought absolute continuity of $p(x)$ would help. now i am not sure.
 A: Absolute continuity might do the trick, but given only that $p(x)$ is a PDF and the convergence of $\int_{-\infty}^\infty x^2 p(x)dx$, you cannot conclude that $p(x)\to 0$ as $x\to\infty$.
For a counterexample, consider the PDF $p_c(x)$ which is equal to $1$ in the intervals $[1,1+1/2^1]$, $[2,2+1/2^2]$, $[3,3+1/2^3]$, and in general all intervals of the form $[n,n+1/2^n]$ for $n\in\mathbb Z_+$, but equal to $0$ everywhere else. 
The fact that $p_c$ is a PDF follows from the fact that 
$$\sum_{n=1}^\infty \frac{1}{2^n}=1$$
And, as specified in your question, the second moment converges:
$$\begin{align}
\int_{-\infty}^\infty x^2 p_c(x)dx &=
\sum_{n=1}^\infty \int_n^{n+1/2^n}x^2 dx\\ &=\sum_{n=1}^\infty \frac{n^2}{2^n}+\frac{n}{4^n}+\frac{1}{3\cdot 8^n}\\
&= \frac{409}{63}
\end{align}$$
Thus, our peculiar function $p_c(x)$ serves as a counterexample. However, it is not continuous, let alone absolutely continuous, so there’s still hope that adding continuity conditions could entail the desired result.
