Let $u$ be a nonconstant harmonic function on $\mathbb C$. Show that for any $c\in\mathbb R, u^{-1}(c)$ is unbounded. Hint: $\{|z|>R\}$ is connected for any $R>0$.

It seems like this proof might require some sort of a "trick", because I'm not sure how to attack this one directly. I know what I want to show is that for any potential upper bound $M \in \mathbb{R}$, there exists $z \in \mathbb{C}$ such that $|f(z)|>M$.

However, I don't know much about $u^{-1}$ except that $u$ is harmonic. (Aside: is there a mistake in the question? That is, isn't $u^{-1}(c)$ just a complex number? It seems like it should ask to show the function $u^{-1}$ is unbounded, not $u^{-1}(c)$, but I could be wrong.)

  • 1
    $\begingroup$ I think that $u^{-1}(c)$ here means the preimage: $u^{-1}(c)=\{z\in \mathbb{C}\, |\, u(z)=c\}$ $\endgroup$ – Daniele A Mar 18 '15 at 11:38

Suppose that $u^{-1}(c)=\{z\in \mathbb{C}\, |\, u(z)=c\}$ is bounded (that is, $u^{-1}(c)$ is contained in $\{|z|<R\}$). Then, by the continuity of $u$ and the fact that $\{|z|>R\}$ is connected, (i) $u(z)>c$ for all $z\in \{|z|>R\} $ or (ii) $u(z)<c$ for all $z\in \{|z|>R\}$. In case of (i) consider $f(z)$ holomorphic in $\mathbb{C}$ with $\operatorname{Re}f(z)=-u(z)$, and in case of (ii) consider $f(z)$ with $\operatorname{Re}f(z)=u(z)$. We conclude that $f(z)$ is constant in either case since $g(z)=e^{f(z)}$ is holomorphic in $\mathbb{C}$ and bounded.

  • $\begingroup$ Thank you very much, ts375_zk26! $\endgroup$ – Mathemanic Mar 18 '15 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.