The converse of Stein's lemma We know from Stein's lemma that if $ X\sim N(0,1)$, then (suppose the expectations exist)
$$E(Xf(X)) = E(f'(X))$$ 
How to prove that if for all the smooth functions  $f$ with compact support, the property above holds, then the random variable $ X\sim N(0,1)$?
 I can not come up with a strict proof, since I do not know how to handle $dF(x)$. (The distribution function may not be continuous)
 A: First assume that the distribution of $X$ has a $C^1$ density $h$. This case follows readily via integration by parts, as noted by C. Lowen [1].
Suppose  $f$ is $C^1$ with compact support.  Then the hypothesis means that
$$\int_{-\infty}^{\infty} f'(u)h(u) \, du = \int_{-\infty}^{\infty}uf(u)h(u)\, du $$
Integration by parts   yields
$$\int_{-\infty}^{\infty} -f(u)h'(u) \, du = \int_{-\infty}^{\infty}uf(u)h(u)\, du $$
Since $f$ is an arbitrary $C^1$ function with compact support,   the last equation holds iff
$h'(u) + uh(u)$ = 0 identically, or equivalently,
$$\frac{d}{du} (e^{u^2/2} h(u))=0 \,.$$
Thus $e^{u^2/2} h(u)$ is a constant, and that constant is determined by the condition that $\int_{-\infty}^{\infty} h(u) \, du=1$.
Now for the general case. Let $W$ be a standard normal variable independent of $X$, and write $Y=aX+bW$ where $a,b>0$ satisfy $a^2+b^2=1$. If for every such $a,b$ the resulting $Y$ is standard normal, then by letting $b$ tend to zero (keeping $a^2+b^2=1$ throughout) we can deduce that $X$ is standard normal as well. Now $Y$ itself has a smooth density since convolution is a smoothing operation (Indeed, if $g$ is the normal density of $bW$, then the density of $Y$ is $y \mapsto E[g(y-aX)]$ which we can differentiate inside the expectation as many times as desired.)
Thus our goal will be to verify that $Y$ also satisfies the Stein identity, whence it will be standard normal by the argument  for the smooth density case. Fix a real number $w$. Applying the hypothesis to the function $f_w(x)=f(ax+bw)$, we see that
$$E[af’(aX+bw)]=E[f_w'(X)] = E[Xf_w(X)]=E[Xf(aX+bw)] \,.$$
Replacing $w$ by the random variable $W$ and taking another expectation (this time with respect to $W$) we get via Fubini that
$$ E[af’(aX+bW)]= E[Xf(aX+bW)]\,,\quad  (*)$$
where here the expectation is in the product space (with respect to both variables $X,W$.) Now $W$, being standard normal, also satisfies Stein’s identity, so the same argument (switching the roles of Z and W) gives
$$E[bf’(aX+bW)]= E[Wf(aX+bW)] \,.  \quad (**) $$
Finally, if we add $(*)$ multiplied by $a$ to $(**)$ multiplied by $b$, and recall that $a^2+b^2=1$ and $Y=aX+bW$, we conclude that
$$E[f’(Y)]=E[Yf(Y)] \,, \quad (***)$$
that is, Stein’s identity holds for $Y$.
For a shorter, but more "magical" argument, see Lemma 2.1 in [2].
[1] https://www.facebook.com/groups/1923323131245618
[2] Ross, Nathan. "Fundamentals of Stein’s method." Probability Surveys 8 (2011): 210-293.
https://projecteuclid.org/journals/probability-surveys/volume-8/issue-none/Fundamentals-of-Steins-method/10.1214/11-PS182.pdf
