Brownian motion semigroup is $C^{1,2}$

118 Views Asked by At

This is a problem from Rene Schilling's Brownian Motion.

Let $u \in B_b(\mathbb{R}^d)$, i.e. bounded Borel measurable map. And set $u(t,z):= P_t u(z)=p_t *u(z)=(2\pi t)^{-d/2} \int_{\mathbb{R}^d} u(y)e^{-|z-y|^2 /2t} dy.$

Show that $u(t,\cdot) \in C^\infty$ and $u \in C^{1,2}$.

Solution is given below. In the proof to showing that $x \mapsto \partial_t u(t,x)$ is in $C^\infty$ for $t>0$, we need to find a uniform bound for $\frac{1}{2}(\frac{|z-y|^2}{t^2} - \frac{d}{t}) p_t(z-y)$ to use the domination argument. I am struggling to come up with a bound for this. I would appreciate some help here.

enter image description here