How can we use characteristic functions to show that if $E[Xf(X)] = E[f'(X)]$ for all differentiable functions $f$, then $X$ is standard normal? I am trying to use the characteristic function, defined as $\phi(t) = E[e^{itX}] = E[\cos(tX)+i\sin(tX)]$, to show that if $E[Xf(X)] = E[f'(X)]$ for all differentiable functions $f: \mathbb{R} \to \mathbb{R}$, with $X$ a real valued random variable, then $X$ is standard normal? My approach was that since the equation holds for all $f$, then I set $f(X) = \phi_X(t) = E[e^{itX}]$. However, that didn't seem to work. Is there a way to do this without resorting to integration? Thanks.
 A: Presumably the hypothesis should be that $E[X f(X)] = E[f'(X)]$ for some 
set of functions $f$ including $\sin(st)$ and $\cos(st)$ for real $s$ (and therefore, by linearity, their linear combinations, including complex ones).
In particular, taking 
$f(t) = 1$ we see that $E[X]$ must exist.  Now taking $f(t) = \exp(ist)$ where $s$ is real, we have
$E[X \exp(isX)] = is E[\exp(isX)]$.  Let $\phi(s) = E[\exp(isX)]$.
Now
$$ \dfrac{d}{ds} \exp(isX) = \lim_{r \to 0} \dfrac{\exp(i(s+r)X)- \exp(isX)}{r} $$
with $$\left| \dfrac{\exp(i(r+s)X)-\exp(isX)}{r}\right| \le |X| \ \text{for real $r$}
$$
so that by the Lebesgue dominated convergence theorem and the finiteness of $E[|X|]$ we have
$$ E \left[ \dfrac{d}{ds} \exp(isX)\right] = \lim_{r \to 0} E \left[ 
\dfrac{\exp(i(s+r)X)- \exp(isX)}{r} \right]  = \dfrac{d}{ds} \phi(s)$$
But
$$ E \left[ \dfrac{d}{ds} \exp(isX)\right] = E \left[ iX f(X)\right]
= i E[f'(X)] = - s \phi(s)$$
Solve this differential equation with $\phi(0)=1$, and you get the characteristic function of the standard normal distribution.
The standard normal random variable does satisfy $E[X f(X)] = E[f'(X)]$ for trigonometric polynomials $f$.  I'm not sure exactly what class of functions $f$ this can be generalized to, but as noted in my comments it does not include all differentiable functions.
