# Subharmonicity is a local property.

Let $$D\subseteq \mathbb{C}$$ be a domain.

Definition. A function $$u:D\to\mathbb{R}\cup\{-\infty\}$$ is called subharmonic if $$u$$ is upper semicontinuous and if for every compact set $$K\subseteq D$$ and every function $$h:K\to\mathbb{R}$$ continuous which is harmonic on the interior of $$K$$ and satisfies $$u\leq h$$ on $$\partial K$$ it follows that $$u\leq h$$ on $$K$$.

Does anyone have any ideas to prove the following?

As above

Proposition. $$u$$ is subharmonic if, and only if, each $$x\in D$$ has a neighborhood $$U_{x}\subseteq D$$ such that $$u|_{U_{x}}$$ is subharmonic on $$U_{x}$$.

Thanks

This easily follows form the definition of subharmonic functions as upper semicontinuous functions with the local sub-mean property, which is obviously local and the equivalence of the two definitions is a standard part of theory.

(so for all $$w \in D$$ and all $$r$$ small enough depending on $$w$$ we have $$f(w)\le \frac{1}{2\pi}\int_0^{2\pi}f(w+re^{it}) dt$$)

Here is a proof using the maximum principle for usc functions, namely that on any compact set $$K$$ there is $$w_K$$ st $$f(w_K) =\max_{z\in K}f(z)$$ which is a standard property of usc functions and follows from the fact that usc is equivalent to the preimages of $$(-\infty, a)$$ being open for all $$a$$ real as well as the fact that usc functions can be decreasingly approximated by continuous functions on any compact (and more generally on any set) which is a bit more involved but part of the theory.

So assume $$u$$ locally subharmonic as above and pick $$K, h$$ as in the OP, so $$h$$ harmonic on the interior on $$K$$ and $$u \le h$$ on $$\partial K$$ and we need to prove $$u \le h$$ on $$K$$.

Then $$g=u-h$$ is usc since $$h$$ is continuous so $$-h$$ is continuous hence usc and the sum of two usc functions is usc, so $$g$$ attains its maximum on $$K$$. If that is at most $$0$$ we are done, but if not there is $$w$$ in the interior of $$K$$ st $$g(w)>0$$ and $$g(w) \ge g(z)$$ for all $$z \in K$$ as $$g\le 0$$ on the boundary.

Take a small disc $$D_w$$ centered at $$w$$ and included in the interior of $$K$$ on which $$u$$ (hence $$g$$) is subharmonic and note that if $$C_w$$ is the boundary circle, we have $$g(w) \ge g(z), z\in C_w$$. Since $$g|_{C_w}$$ is usc there is a sequence of continuous functions $$g_n$$ converging decreasingly to $$g$$ on $$C_w$$ hence if $$g_n$$ is also their harmonic extension to $$D_w$$ we have $$g \le g_n$$ on $$C_w$$ hence $$g \le g_n$$ on $$D_w$$.

But by the mean value theorem for harmonic functions we have $$g_n(w)=\frac{1}{2\pi}\int_0^{2\pi}g_n(w+re^{it}) dt$$ where $$r$$ is the radius of $$D_w$$ and by the monotone convergence theorem $$\int_0^{2\pi}g_n(w+re^{it}) dt \to \int_0^{2\pi}g(w+re^{it}) dt$$ hence putting it together we get $$\frac{1}{2\pi}\int_0^{2\pi}g(w+re^{it})dt \le g(w) \le \frac{1}{2\pi}\int_0^{2\pi}g(w+re^{it})dt$$ so we must have equality and $$g(w)=g(z), z \in C_w$$. But clearly, we can take any smaller radius $$r_1 and repeat so $$g$$ is constant on $$D_w$$.

From here a standard argument shows that $$g$$ must be constant positive at least on the component $$K_1$$ of the interior of $$K$$ containing $$w$$ and that is impossible since if $$z_n \to z \in \partial K_1$$ we have $$0 \ge g(z) \ge \lim g(z_n)=g(w)>0$$ etc

(take $$K_1$$ the component of the interior of $$K$$ containing $$w$$ and $$A$$ the set of $$\zeta$$ where $$g(\zeta)=\max_{z \in K_1}g(z)$$ and $$\zeta$$ has a small neighborhood on which $$g$$ is constant and immediately it follows that $$A$$ is open and closed as $$z_n \to z$$ implies $$\limsup g(z_n) \le g(z)$$ by usc, while the above shows it is nonempty, so $$A=K_1$$ hence $$g$$ is constant on $$K_1$$)

• In the following case: but if not there is $w$ in the interior of $K$ st $g(w)>0$ and $g(w) \ge g(z)$ for all $z \in K$ as $g\le 0$ on the boundary. Why there is $w$ in the interior of $K$ st $g(w)>0$ and $g(w) \ge g(z)$ for all $z \in K$ as $g\le 0$ on the boundary? Nov 21 at 0:00
• by assumption, we have $g=u-h \le 0$ on the boundary of $K$ and we need to prove that $g \le 0$ on $K$; that is the OP definition of subharmonic which we need to show fulfilled; because $g$ is usc it attains its maximum on any compact so if $w$ is such then either $g(w) \le 0$ so $g \le 0$ done, or $g(w)>0$ and we continue as above Nov 21 at 0:03