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!
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 $e^{int}.$