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Let $f$ be a nonnegative (probably not needed) function on $\mathbb{R}$ such that for all $t$, $xf(x)f(t-x)$ and $f(x)f(t-x)$ are both integrable in $x$.

Is it true that $$ \int\nolimits_{-\infty}^\infty x f(x)f(t-x) dx =\frac{t}{2} \int_{-\infty}^\infty f(x)f(t-x) dx $$ for all $t$?

I found that this is true if $f(x)=e^{-|x|} $ or $\frac{1}{1+x^2}$.

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up vote 9 down vote accepted

It is true. Notice that $$\int_{-\infty}^\infty xf(x)f(t-x)dx = \int_{-\infty}^\infty(t-x)f(t-x) f(x)dx.$$ This follows from the substitution $x=t-u$, or the fact that convolution is symmetric.

From here, use linearity of the integral to split up $(t-x)$, and you will find your equation.

Hope that helps

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