

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

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


