# harmonic function bounded is constant

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!

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

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 $e^{int}.$
• @Parisina The integral is bounded above by $2\pi [(1+r)^{1/2}]^2=1+r.$ This implies most of the $a_n$ equal $0.$ – zhw. Jun 1 '18 at 13:46