Let $u \in C^{\infty}(\mathbb R^3)$ be harmonic. Suppose that $u$ has no critical points outside the unit ball but that it has at least one critical point inside the unit ball.


Does it follow that $u$ is a polynomial?



On critical points of harmonic functions
This is not true in dimension 2.
Function $f(z)=ze^z$ is entire, and $f'(z)=0$ at one point,
$z=-1$. It follows that the function
$$u(x,y)=\mathrm{Re}f(x+iy)=e^x(x\cos y-y\sin y)$$
is harmonic, has one critical point $(x,y)=(-1,0)$, and evidently not a polynomial.


Now one can construct a similar example in even dimension, for example:
$$v(x\_1,x\_2,x\_3,x\_4)=u(x\_1,x\_2)+u(x\_3,x\_4).$$
But for $n=3$ I could not figure out.


