 Killed geometric Brownian heat kernel

The differential operator $L=\frac{1}{2}{x}^2{\partial }_{xx}+\frac{1}{2}x{\partial }_x-\frac{1}{2}$ has solution operator $P_{t}g(x)=e^{-t/2}\mathbb{E}[g(x{e}^{B_t})]$. For $x\ne 0$ its integral kernel with respect to Lebesgue measure is $p(t,x,y)=e^{-t/2}(|y|\sqrt{2\pi t}{)}^{-1}\exp [-{\log }^2(|y/x|)/(2t)]$ when $xy>0$, and zero otherwise. For $x=0$ the kernel is the measure $e^{-t/2}{\delta }_0$, not a density. The Itô formula applied to $e^{-s/2}u(t-s,X_s)$ proves this Feynman-Kac formula.