Prove: For a harmonic function $u: \mathbb{R}^3 \rightarrow \mathbb{R}$, if for all $(x,y,z) \in \mathbb{R}^3$ $|z|^{1-\epsilon}|u(x,y,z)| \leq (x^2+y^2)^{0.5-\epsilon}$ then $u = 0$ for $0 < \epsilon < 0.5$.
I'm pretty sure that this can be proved with the mean value theorem for harmonic functions over spheres starting with the point $(0,0,0)$ and then showing that the proof can be done for any point.
I tried using spherical coordinates to show that the integral over a sphere has to be $o(R^3)$ for a sphere with a radius of $R$, but I'm having a bit of trouble showing that the integral converges and I would appreciate any help.
Probably, my answer is not that strict; but I hope it'll give you some useful idea.
Let us rewrite our condition in the form:
$$ |u(x,y,z)|\leq\frac{(x^2+y^2)^{0.5-\epsilon}}{|z|^{1-\epsilon}}.\tag 1 $$
Let us consider some ray starting at $(0,0,0)$. We will evaluate right-hand side (RHS) of $(1)$ along this ray as $r\to\infty$. $x=\alpha r$, $y=\beta r$, $z=\gamma r$. $\alpha,\beta,\gamma=\mathrm{const}$, $r\to\infty$.
$$ x\propto r,~~y\propto r;~~x^2\propto r^2;~~y^2\propto r^2;~~ x^2+y^2\propto r^2 \\ (x^2+y^2)^{0.5-\epsilon}\propto r^{1-2\epsilon} \\ z\propto r;~~ |z|\propto r^{1-\epsilon} \\ \mathrm{RHS}\propto r^{-\epsilon}\propto \frac{1}{r^\epsilon} $$ For any $0<\epsilon<0.5$ RHS decays at $r\to\infty$ faster, than just a constant $\propto r^0$. Our $u(x,y,z)$ should as well decay along this ray not slower than $r^0$.
The general expression for the harmonic function is $$ u(r,\theta,\varphi)=\sum_{l=0}^\infty\sum_{m=-l}^{l}(A_lr^l+B_lr^{-l-1})Y_l^m(\theta,\varphi).\tag 2 $$ (see Eq. (20) here http://mathworld.wolfram.com/LaplacesEquationSphericalCoordinates.html)
I suppose, that we consider only $u(x,y,z)$, that are finite at $(0,0,0)$. So we may disregard $r^{-l-1}$ terms in $(2)$. Our $u(x,y,z)$ should decay at $r\to\infty$ faster that $r^0$, so all $r^l$ terms in $(2)$ also vanish. The only opportunity for $u(x,y,z)$ is to be zero. Spherical functions $Y_l^m(\theta,\varphi)$ should not cause any trouble here.