= Solution
For $t>0$, differentiate
$$
u(t)=\mathbb E[f(\sqrt tZ)].
$$
The growth assumptions justify differentiation under the expectation, and <Gaussian integration by parts> gives
$$
\begin{aligned}
u'(t)
&=\frac1{2\sqrt t}\mathbb E[Zf'(\sqrt tZ)]\\
&=\frac1{2\sqrt t}\mathbb E\left[
\frac d{dZ}f'(\sqrt tZ)\right]\\
&=\frac12\mathbb E[f''(\sqrt tZ)].
\end{aligned}
$$
Since $u(0)=f(0)$, integration from $0$ to $t$ yields the <Brownian transition semigroup> identity
$$
\boxed{
\frac12\int_0^t\mathbb E[f''(\sqrt sZ)]\,ds
=\mathbb E[f(\sqrt tZ)]-f(0)}.
$$
Continuity at zero covers $t=0$.
Back to article page
