Let $u : D \rightarrow \mathbb R$ be a bounded harmonic function on a unit disk $D$. Suppose that $\limsup_{z \rightarrow a} u(z) \leq 0$ for all $a \in D\setminus \{1\}$. I am trying to show that $u \leq 0$ on $D$.
To apply the Maximum modulus principle, I need to show that $\limsup_{z \rightarrow 1} u(z) \leq 0$.
One direction I am pursuing is to reduce the problem to holomorphic functions.
In particular, since $D$ is simply connected, there is a harmonic conjugate, $v : D \rightarrow \mathbb R$.
But beyond this, I am not sure how to proceed.
I would appreciate any hint or reference.
Best Answer
Here is a direct proof using harmonic functions only (one can definitely do a complex analytic proof along the lines in the comments and the following arguments adapted appropriately)
Let $u(z) \le M, |z| < 1$ and we have $M < \infty$ by hypothesis (this is crucial as otherwise result is not true).
For $\epsilon >0$ consider the harmonic function (defined in the open unit disc) $$g_{\epsilon}(z)=u(z)+\epsilon \log (|z-1|/2)$$
Note that since $|z-1|/2 \le 1$ for $|z| \le 1$ we have that $g_{\epsilon}(z) \le u(z) \le M$ in the unit disc, while since $ \log (|z-1|/2) \to -\infty, z \to 1$, for fixed $\epsilon>0$ there is a small neighborhood $U_{\epsilon}$ of $1$ st $g_{\epsilon}(z) < 0, z \in D \cap U_{\epsilon}$ which together with the fact that $g_{\epsilon}(z) \le u(z)$ and the hypothesis on $u$ at the boundary, ensures that $\limsup_{|z| \to 1}g_{\epsilon}(z) \le 0$ so by the original maximum modulus theorem we get that $g_{\epsilon}(z) \le 0$ in the open unit disc.
But if we fix $z, |z|<1$ and let $\epsilon \to 0$ we get $u(z) \le 0$ since $\log |z-1|/2$ is a fixed finite number for $z$ fixed and we are done!