Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $u$ be a real valued harmonic function on the complex plane $\mathbb{C}$, such that \begin{equation*} u(z)\le a\big|\ln|z|\big|+b, \end{equation*} for all $z$, where $a,b$ are positive constants. Prove that $u$ is constant.

My idea: let an entire function $f=u+iv$. Then $f(z)=\sum_{n=0}^{\infty}a_n z^n$. The order of a polynomial is higher-infinity when $z\to\infty$ compared with $\ln$. But this is not a valid argument. How to prove?

share|cite|improve this question
up vote 0 down vote accepted

Let $\tilde u=\frac{u-b}{a}$, then $\tilde u(z)\le \big|\log |z|\big|$. Let $f$ an entire function, such that, $\mathrm{Re}\,f(z)=\tilde u(z)$. Then $$ \left|\mathrm{e}^{\,f(z)}\right|=\mathrm{e}^{\tilde{u}(z)}\le \mathrm{e}^{|\log |z||} =\max\big\{|z|,|z|^{-1}\big\}. $$ So if we set $g(z)=\mathrm{e}^{\,f(z)}-\mathrm{e}^{\,f(0)},$ then $g(0)=0$, and for $|z|\ge 1$, we have $$ |g(z)|=\big|\mathrm{e}^{\,f(z)}-\mathrm{e}^{\,f(0)}\big|\le \big|\mathrm{e}^{\,f(z)}\big|+ \big|\mathrm{e}^{\,f(0)}\big|\le |z|+c\le (c+1)|z|, $$ and hence $$ \left|\frac{g(z)}{z}\right|\le c+1. $$ But $g(z)/z$ is also entire, since $g(0)=0$, and as it is bounded, it has to be constant. Thus there is an $a\in\mathbb C$, such that $$ g(z)=az \,\,\Longrightarrow\,\, \mathrm{e}^{\,f(z)}-\mathrm{e}^{\,f(0)}=az \,\,\Longrightarrow\,\, \mathrm{e}^{\,f(z)}=\mathrm{e}^{\,f(0)}+az, $$ and as $\mathrm{e}^{\,f(z)}$ does not vanish, this implies that $a=0$, and hence $f$ is constant, and so is $u$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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