Proving a Poincare inequality using Stein's characterization Question: 
Let X be a standard Gaussian r.v. 
Use Stein's characterization $Ef'(X) = E(Xf(X))$
to prove the Poincare inequality $E|f(X)-Ef(X)|^2 \leq E|f'(X)|^2$.
This looks like Markov's inequality could be used to prove the inequality. 
However, I'm not quite sure how to piece it together. 
 A: You want to prove that $E|f(X)-Ef(X)|^2 \leq E|f'(X)|^2$. This inequality doesn't change if you add a constant to $f(x)$, so without loss of generality you can assume that $Ef(X)= 0$. So, all we need is that $E|f(X)|^2 \leq E|f'(X)|^2$.
To show that we will use Stein's method and define a function $g$ such that $g'(x) - x g(x) = f(x)$. This implies that $f'(x) = g''(x) - x g'(x) - g(x)$.
We will use the following two equalities derived by Stein's Lemma:
$$E[ f'(X)g(X) ] + E[ f(X)g'(X) ] = E(f(X)g(X))' = E [ X f(X) g(X)] \,\,\,\,\,\, \textrm{(1)}$$
also
$$E[ g'(X) g'(X) ] + E[ g''(X) g(X) ] = E(g'(X)g(X))' = E [ X g'(X) g(X)]\,\,\,\,\,\, \textrm{(2)}$$
So, we have that $E[ f(X)^2 ] = E[ f(X) f(X) ] = E[ (g'(X) - X g(X)) f(X) ] = -E[ f'(X)g(X) ]$, where the last equality follows by $(1)$.
Moreover, we also have that: 
$$E[ f(X)^2 ] = -E[ f'(X)g(X) ] = -E[ (g ''(X) - X g'(X) - g(X) ) g(X) ] = E[ g'(X)^2 + g(X)^2 ]$$
where we applied $(2)$ for the last equality.
This implies that $E[ f(X)^2 ] \le E[ g(X)^2 ]$.
Finally, by the Cauchy-Swartz we have that:
$$E[ f(X)^2 ] = -E[ f'(X)g(X) ] \le \sqrt{ E[ f'(X)^2 ] E[ g(X)^2 ] } \le \sqrt{ E[ f'(X)^2 ] E[ f(X)^2 ] }$$
This completes the proof of the Poincare Inequality.
