Subharmonic on $U$ iff subharmonic on each $U_{\alpha}$, where $(U_{\alpha})$ is an open cover of $U$

Question

I am working with this definition of subharmonicity:

Definition Let $U$ be an open subset of $\mathbb{C}$. A function $u: \, U \to [-\infty,\infty)$ is called subharmonic if it is upper semicontinuous and satisfies the local submean inequality, i.e. given $w \in U$, there exists $\rho >0$ such that \begin{eqnarray} u(w) \leq \frac{1}{2 \pi} \int_0 ^{2 \pi} u(w+re^{it})dt \quad (0 \leq r < \rho). \end{eqnarray}

I would like to show that if $(U_{\alpha})$ is an open cover of $U$, then $u$ is subharmonic on $U$ if and only if it is subharmonic on each $U_{\alpha}$. This makes sense intuitively to me, but I am having trouble showing it formally.

Can somebody help?

Answer 1

• $u$ subharmonic in $U$ $\implies$ $u$ subharmonic in each $U_\alpha$:

Let $w \in U_\alpha$. $u$ is subharmonic in $U$, therefore $u$ satisfies the local submean inequality in some $B_\rho(w) \subset U $. Now choose $\rho' \le \rho$ such that $B_{\rho'}(w) \subset U_\alpha$, then $u$ satisfies the local submean inequality in $B_{\rho'}(w) \subset U_\alpha$.

• $u$ subharmonic in each $U_\alpha$ $\implies$ $u$ subharmonic in $U$:

Let $w \in U$. Then $w \in U_\alpha$ for some $\alpha$. $u$ is subharmonic in $U_\alpha$, therefore $u$ satisfies the local submean inequality in some $B_\rho(w) \subset U_\alpha \subset U $.

($B_\rho(w)$ is the open disk with center $w$ and radius $\rho$.)