# Maximum modulus principle for harmonic functions on a disk

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.

• Let $f$ be a holomorphic function so that ${\rm Re}\,f=u$ then $g=e^f$ is holomorphic and the condition $u\le 0$ is equivalent to $|g|\le 1.$ Aug 31, 2022 at 17:00
• @RyszardSzwarc Thank you for the comment. If I could show that $\limsup_{z \rightarrow 1} |g(z)| < \infty$, then I can apply the Phragmen-Lindelof thereom to conclude that $|g| \leq 1$, and done. Was this the argument you had in mind? I am now stuck with proving that $\limsup_{z \rightarrow 1} |g(z)| < \infty$. Would you be able to share a little bit more insight?
– Luke
Sep 2, 2022 at 4:19
• I do not know how to solve your problem. I have just translated it to a question concerning holomorphic functions, as you suggested. Clearly $\limsup_{z\to 1}|g(z)|<\infty$ is equivalent to $\limsup_{z\to 1}u(z)<\infty$ Sep 2, 2022 at 9:53

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!

Here is a complex analytic proof.

As Ryszard Szwarc pointed out in the comment, we have a holomorphic function $$f$$ on $$D$$ such that $$\mathrm{Re}\,f = u$$ (this works as $$D$$ is simply connected), and $$u \leq 0$$ on $$D$$ is equivalent to $$|e^f| \leq 1$$ on $$D$$.

Recall the Phragmen-Lindelof Theorem, which says

Let $$G\subseteq \mathbb C$$ be simply connected and $$g$$ be holomorphic on $$G$$. Suppose there is a holomorphic $$\varphi : G \rightarrow \mathbb C$$ which never vanishes and bounded on $$G$$. If $$M$$ is a constant and $$\partial_\infty G = A\cup B$$ such that:
(a) $$\limsup_{z \rightarrow a} |g(z)| \leq M$$ for all $$a \in A$$
(b) $$\limsup_{z \rightarrow b} |g(z)| |\varphi(z)|^{\eta} \leq M$$ for all $$b \in B$$, $$\eta > 0$$
then $$|g(z)| \leq M$$ for all $$z \in G$$.

(Proof of this particular version of the Phragmen-Lindelof Theorem can be found in Functions of one complex variable by Conway)

By the assumption, $$\limsup_{z \rightarrow a} |e^{f(z)}| \leq 1$$ for $$a \in \mathbb D \setminus \{1\}$$.
For $$b = 1$$, let $$\varphi(z) := z-1$$ which never vanishes on $$D$$. Since $$u$$ is bounded, $$e^{f(z)}$$ is bounded on $$D$$, and so $$\limsup_{z \rightarrow 1} |e^{f(z)}||\varphi(z)|^\eta = 0 \leq 1$$.
Therefore, by the Phragmen-Lindelof Theorem, $$|e^f| \leq 1$$ on $$D$$, and so $$u \leq 0$$ on $$D$$, as desired.