Basic probability limit problem Let $g(x)$ be a smooth probability density function, whose mean value is $0$.
The variance of $g(x)$ is finite, and $\forall x\in R\ \ g(x)>0$.
Under the above conditions, does the following formula holds?
$$\lim_{x\to\infty}xg(x) = 0.$$
I assumed this formula is true, and tried to apply Chebyshev's inequality but
this approach didn't succeed. Please give me some hint.
 A: Outline: We define $f(x)$ to be $0$ except near non-zero integers. We will describe $f(x)$ when $x$ is near the positive integer $n$. For the negative integers, reflect across the $y$-axis.
Let $a_n=n-\frac{1}{100\cdot 2^{n^2}}$ and let $b_n=n+\frac{1}{100\cdot 2^{n^2}}$. Over the interval $a_n\le x\le n$, define $f(x)$ by using the line that joins $(a_n,0)$ to $(n,\frac{1}{n})$.  Over the interval $n\le x\le b_n$, define $f(x)$ analogously, using the line that falls from $(n,\frac{1}{n})$ to $(b_n,0)$.
The areas under the spikes decrease very rapidly, and the sum of these areas is small. Let $g(x)=ke^{-x^2}+f(x)$, where $k$ is chosen so that $\int_{-\infty}^\infty f(x)\,dx=1$. This also takes care of the positivity requirement on $g(x)$. 
Because far out the spikes have very small area, the mean and variance of a random variable with density function $g(x)$ exist. Since the mean exists, we can conclude by symmetry that it is $0$. Note that $xg(x)=1$ whenever $x$ is a positive integer. 
The function $g(x)$ is continuous but not smooth: the derivative does not exist at $a_n$, $n$, and $b_n$.
However, the spikes can be smoothed out. A small modification will get us differentiability. Indeed we can make $g(x)$ infinitely differentiable everywhere by replacing $f(x)$ by an infinite sum of bump functions that vanish outside $(a_n,b_n)$ and reach height $\frac{1}{n}$ at $n$.  
