If $u>0$ is $C^0(\Omega)\cap H^1(\Omega)$ and vanishes on the boundary of $\Omega$, then up to a zero extension, is the new function also $H^1$?

I'm trying to show that if $u$ is positive, continuous and $\Delta u\geq 0$ in $\Omega$, and if $u=0$ on $\partial \Omega$, then let $\tilde u$ be the zero extension of $u$, I want to say that $\tilde u$ is also subharmonic. This conclusion holds when $\tilde u$ is $H^1(\mathbb R^n)\cap C^0(\mathbb R^n)$, however, since I do not have any regularity of $\partial \Omega$, I'm not sure if $\tilde u\in H^1$.