Inequality for the derivative of a density of a random variable convolved with a normal r.v. I have a question about the following proof.
The statement is: Let $X$ be a random variable and $Z_\tau \sim N(0,\tau)$ be an independent random variable. Then $Y_\tau := X + Z_\tau$ has a differentiable density $g_\tau$ which satisfies
\begin{align*}
(g'_\tau(y))^2 \le c_\tau g_{2\tau}(y)  g_\tau(y), \quad \forall y \in \mathbb R
\end{align*}
where $c_\tau = \frac{4\sqrt 2}{e \tau}$.
This statement comes from a paper by Barron published in 1986. The idea of the proof is to use the Cauchy-Schwarz inequality. Then one obtains
\begin{align*}
(g'_\tau(y))^2 \le \frac{g_\tau(y)}{\tau^2} E((y-X)^2 \phi_\tau(y-X))
\end{align*}
where $\phi_\tau$ denotes the density of $Z_\tau$. Then he concludes with
\begin{align*}
\frac{g_\tau(y)}{\tau^2} E((y-X)^2 \phi_\tau(y-X))
\le c_\tau g_\tau(y) E(\phi_{2\tau}(y-X))
= c_\tau g_{2\tau}(y) g_\tau(y).
\end{align*}
I understand everything except the last inequality. How does he obtain that? I tried to use the inequality $\exp(x) \ge 1 + x$ but that did not help me either. Any ideas?
Thanks in advance.
 A: Let we assume that $u(y)$ is the density of $X$ and 
$$ f_\tau(y) = \frac{1}{\tau\sqrt{2\pi}}e^{-\frac{y^2}{2\tau^2}} $$
is the density of $Z_\tau$. Then $g_\tau = u * f_\tau$, and since $f_\tau$ is differentiable, $g_\tau$ is differentiable and:
$$ g_\tau' = u * f_t'.$$
So we have:
$$ g_\tau(y) = \int_{\mathbb{R}} u(y-x)\frac{1}{\tau\sqrt{2\pi}}e^{-\frac{x^2}{2\tau^2}}\,dx $$
$$ g_{2\tau}(y) = \int_{\mathbb{R}} u(y-x)\frac{1}{2\tau\sqrt{2\pi}}e^{-\frac{x^2}{8\tau^2}}\,dx $$
and:
$$ g_\tau'(y) = -\int_{\mathbb{R}} u(y-x)\frac{x}{\tau^3\sqrt{2\pi}}e^{-\frac{x^2}{2\tau^2}}\,dx.$$
By the Cauchy-Schwarz inequality we have:
$$\begin{eqnarray*} g_\tau'(y)^2 &\leq& g_\tau(y)\cdot\int_{\mathbb{R}}u(y-x)\frac{x^2}{\tau^5\sqrt{2\pi}}e^{-\frac{x^2}{2\tau^2}}\,dx\\&=&g_\tau(y)\cdot\int_{\mathbb{R}}u(y-x)\frac{1}{2\tau\sqrt{2\pi}}e^{-\frac{x^2}{8\tau^2}}\left(\frac{2x^2}{\tau^4}e^{-\frac{3x^2}{8\tau^2}}\right)\,dx,\end{eqnarray*}$$
but since the function $h(x)=2x^2 e^{-\frac{3x^2}{8}}$ is non-negative and attains its maximum value at $x=\pm 2\sqrt{\frac{2}{3}}$, we have $h(x)\leq\frac{16}{3e}$ and the inequality:

$$ g_\tau'(y)^2 \leq \frac{16}{3e\tau^2}g_\tau(y)g_{2\tau}(y) $$

holds. It is stronger than Barron's inequality for any $\tau\geq 1$.

Here I assumed that $\tau$ is the square root of the variance of $Z_\tau$. If you are assuming that $\tau$ is the variance of $Z_\tau$, you have to suitably modify the above lines, but the steps are just the same.
