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.

  • 2
    $\begingroup$ 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 $|g|\le 1.$ $\endgroup$ Aug 31, 2022 at 17:00
  • $\begingroup$ @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? $\endgroup$
    – Luke
    Sep 2, 2022 at 4:19
  • $\begingroup$ 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 $ $\endgroup$ Sep 2, 2022 at 9:53

2 Answers 2


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .