# A bounded subharmonic function $u \to 0$ on a specific subset of $\partial D$, then $u \leq 0$ on all of $D$

Let $D \subset \Bbb C$ be a bounded domain. Let $E$ be a subset of $\partial D$ such that there exists a subharmonic function $v<0$ on $D$ that tends to $-\infty$ when $z$ approaches $E$. Finally let $u$ be a subharmonic function on $D$, $u \leq M$, that tends to $0$ when $z$ approaches $\partial D \setminus E$. Show that $u \leq 0$ on $D$.

I don't see any data on $E$ that can be deduced from the existence of $v$.

This is a variant of the usual maximum principle, which says that if $f$ is subharmonic on $D$ and all of its partial limits at $\partial D$ are $\leq 0$, then $u \leq 0$ on $D$. If $E$ is empty then our statement follows immediately. the proof of the principle starts with extending $u$ continuously to the compact set $D \cup \partial D$, but here $D \cup \partial D$ is not necessarily compact. To apply this principle, I try to devise a subharmonic function with $\limsup \leq 0$ on all of the boundary. I know that subharmonicity is preserved under addition and under $\max$, but none of these two seems to produce such a function, or use the boundedness of $u$.

For $\varepsilon > 0$, consider
$$u_{\varepsilon}(z) = u(z) + \varepsilon\cdot v(z).$$
Then $u_{\varepsilon}$ is a subharmonic function on $D$, and
$$\limsup_{z \to \zeta} u_{\varepsilon}(z) \leqslant 0$$
for all $\zeta \in \partial D$. Hence $u_{\varepsilon}(z) \leqslant 0$ for all $z\in D$.
Now let $\varepsilon \to 0$ to conclude $u\leqslant 0$.