Let $u:\mathbb{R}^n \to \mathbb{R}$ be harmonic function and suppose that exist $C>0$ such that $|u(x)| \leq C(1+\sqrt{\|x\|})$. I want show that $u$ is constant.
My first idea: Show that $C(1+\sqrt{\|x\|})$ is harmonic and use the property of volumetric mean, but $C(1+\sqrt{\|x\|})$ is not harmonic.
I also can not limit for $C(1+\sqrt{\|x\|})$.
I have no more ideas.
For harmonic functions, the following is valid:
$$|D_iu(y)| \leq \frac n R \sup_{\partial B_R(y)} |u|.$$
(Look Estimates of derivatives of harmonic function for a simple proof.)
Thus, we have for any $y\in\mathbb{R}^n$ and a fixed $R$:
$$|D_iu(y)|\leq\frac{nC}{R}(1+\sqrt{R})=\frac{nC}{R}+\frac{nC}{\sqrt{R}}.$$
The RHS of the previous inequality goes to zero when $R\to\infty$. Thus, $D_iu(y)\equiv0,\forall y\in\mathbb{R}^n$ and $1\leq i\leq n$. Therefore, $$Du\equiv0\Rightarrow u=constant.$$