Upper bound or maximization for $xf(x)$

Question

Suppose a random variable $x$, with density function $f$ (continuous and differentiable). $f$ is unimodal and symmetric around zero. Under what conditions can I show that $xf(x)$ does not exceed $1$ for $x>0$? I tried the normal distribution and it was clearly satisfied. But I wonder whether there is a more general condition for $f$.

Thanks.

Answer 1

For $b>0$, $1/2>bf(b)-\int_{0}^b{xf'(x)dx}$. If the distribution $f(x)$ has only one 'peak', and is symmetric around zero, then this peak occurs at zero. So then we can assume that $f'$ is negative, so the integral part of the inequality is negative and so \begin{equation}1/2>bf(b)+K,\end{equation} for some positive $K$.