TITLE: Stein's lemma condition
QUESTION [14 upvotes]: (Apologies if I break some conventions, this is my first time posting!)
I am working on proving Stein's characterization of the Normal distribution: for Z $\sim N(0,1)$ and some differentiable function $f$ with $E[|f'(Z)|] < \infty$, $$E[Zf(Z)] = E[f'(Z)]$$
Writing the LHS expression in integral form and integrating by parts, I eventually obtain:
$$E[Zf(Z)] = \frac{1}{\sqrt{2\pi}} \left[ -f(z) \cdot \exp \left\{ \frac{-z^2}{2} \right\} \right] \Bigg|_{-\infty}^{\infty} + E[f'(Z)]$$
Now I need to show that the first expression on the right hand size is zero. Intuitively, this seems clear because of the exponential term, but I am having trouble explicitly applying the condition on $f'$ to prove this rigorously. Any ideas?

REPLY [5 votes]: This might be too late in the game but here is an answer to Gogurt's original question, that shows that the assumptions on $f$ imply that
$f(z)\phi(z)$ vanishes at infinity.
Student45 establishes that $f$ and $zf(z)$ are integrable with respect to the normal distribution using Fubini's theorem. This implies that $f'(z)\phi(z)+ f(z)z\phi(z)=(\phi(z)f(z))'$ is Lebesgue integrable in $\mathbb{R}$. In particular,
$$ f(z)\phi(z) = f(0)\phi(0) + \int^z_0 (\phi\cdot f)'(t)dt $$
for all $z$. This is due to the Fundamental Theorem of Calculus that says that if $f$ is differentiable in an interval $[a,b]$ and if $f'$ is Lebesgue integrable in $[a,b]$, then
$$f(x)-f(a)=\int^x_af'(t)\,dt,\quad a\leq x\leq b.$$
Therefore, the limits $\lim_{z\rightarrow\pm\infty}f(z)\phi(z)=A_{\pm}$ exist. As $f(z)\phi(z)$ is Lebesgue integrable in $\mathbb{R}$, then one must have that $A_{\pm}=0$ as Student45 mentioned in his comment above.