Superharmonic functions, Barriers - Complex Analysis (Conway) Let $G$ be an open connected subset of the plane $\mathbb{C}$, and let $a \in \partial_{\infty}G$. Suppose $\psi_{r}(z)$ is a function from $G \cap B(a,r)$ to $\mathbb{R}$ with the following properties:


*

*$\psi_{r}$ is superharmonic on $G \cap B(a,r)$ with $\psi_{r} \geq 0$;

*$\lim_{z \rightarrow a} \psi_{r}(z) = 0$;

*$\lim_{z \rightarrow w} \psi_{r}(z) = 1$ for $w \in G \cap \{w : |w-a| = r\}$.


Define $\hat{\psi_{r}}$ by letting $\hat{\psi_{r}}=\psi_{r}$ on $G \cap B(a,r)$ and $\hat{\psi_{r}}=1$ for $z$ on $G-B(a,r)$.
Show that $\hat{\psi_{r}}$ is superharmonic. (Conway Chapter X Section 4. The Dirichlet Problem.)
Thanks. I just have a presentation to do on Harmonic Functions in two days, and I'd prefer to have all the details under my belt. Just reading Conway and this is the section about barriers. The book says (Verify!), but I've been trying for a while, and even though I managed to prove something similar earlier, I can't seem to do this one.
 A: I'm Dongryul, preparing my presentation on Dirichlet Problem for Complex Function Theory Course, based on the Conway's book, too.
I concerned the same question as you did, and I think that I can answer as follows.
To show that $\hat{\psi}$ is superharmonic, it suffices to prove that $\hat{\psi} - u$ satisfies the minimum principle for any harmonic function $u$ on $G_1 \subseteq G$.
Denote $G^1 = \{z \in G : |z - a| > r\}$ and $G^2 = \{ z \in G : |z-a| < r\}$, and suppose to the contrary that $\hat{\psi} - u$ is non-constant and attains minimum at $z_0 \in G_1$.
If $z_0 \in G^1$, then $$1 - u(z_0) \le \hat{\psi}(z) - u(z) \le 1 - u(z)$$ so $u$ is constant.
If $z_0 \in G^2$, then $\hat{\psi} - u$ is constant on a connected component of $G_1 \cap G^2$ containing $z_0$ since it is superharmonic there. Since $\hat{\psi} - u$ is non-constant, $G^1 \cap G_1 \neq \emptyset$ or there is another connected component, but connectedness of $G_1$ implies $G^1 \cap G_1 \neq \emptyset$ even in this case. Then by continuity we may assume that $|z_0 - a| = r$. Now $$1 - u(z_0) \le \hat{\psi}(z) - u(z) \le 1 - u(z)$$ so $u$ is constant.
Finally if $|z_0 - a| = r$, $$\hat{\psi}(z) - u(z) \ge \hat{\psi}(z_0) - u(z_0) = 1 - u(z_0)$$ and thus $$u(z_0) \ge 1 - \hat{\psi}(z) + u(z) \ge u(z).$$ Since $u$ is harmonic, it means that $u$ is constant.
In either case, $u$ is constant, and the minimum of $\hat{\psi} - u$ is $1 - u(z_0)$. However, $\hat{\psi} - u$ is assumed to be non-constant so $\hat{\psi}(w) < 1$ for some $w \in G_1$ and it contradicts to the minimality.
Therefore, we conclude that $\hat{\psi}$ is superharmonic.
I think that I'm late, but I hope that your presentation was successful.
