2
$\begingroup$

I have a function $f$ that is harmonic on $\mathbb{C}$, s.t.

$|f(z)|\leq\sqrt{1+|z|}$.

I would like to show that it is constant. I think that for a bounded harmonic function, there is a holomorphic function of which it is the real part, and which is then also bounded? But then why specify the bounds? I would appreciate some help, thanks!

$\endgroup$
  • $\begingroup$ Probably better/simpler just to use the Poisson formula? $\endgroup$ – paul garrett Aug 27 '16 at 23:22
  • $\begingroup$ how can I use Poisson's formula here? $\endgroup$ – mj_indefinite Aug 28 '16 at 0:59
2
$\begingroup$

Hint: Suppose $u$ is harmonic on $\mathbb C$ and $|u(z)|\le (1+|z|)^{1/2}.$ WLOG $u$ is real. Then $u = \text { Re } g,$ where $g$ is entire. Write $g(z) = \sum_{n=0}^{\infty}a_nz^n.$ Then $u = (g + \bar g)/2.$ Compute

$$\int_0^{2\pi} |u(re^{it})|^2\, dt$$

using Parseval and the orthogonality of the exponentials $e^{int}.$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Can I ask you why calculating this integral is useful? sorry I saw your answer, I couldn't complete it. $\endgroup$ – Parisina May 31 '18 at 23:12
  • 1
    $\begingroup$ @Parisina The integral is bounded above by $2\pi [(1+r)^{1/2}]^2=1+r.$ This implies most of the $a_n$ equal $0.$ $\endgroup$ – zhw. Jun 1 '18 at 13:46

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.