Show that a given harmonic function has a specific form Suppose $u$ is a harmonic function in $\mathbb{R}^n$ and satisfies 
$$|u(x)| \leq C |x|~\forall x \in \mathbb{R}^n.$$
Show that $u(x)=q \cdot x$ where $q$ is a constant vector.
I'm struggling with this problem. I tried defining a new function, $v(x)=u(x)/{x}$ for $x \neq 0$ and using the given bound $C,$ that leads to a constant vector.  Being dividing by a vector, that couldn't be correct. My gut feeling says that this could be done using Maximum principle. Any help is much appreciated. Thank you.

Hint: Because $u$ is harmonic, it is real analytic, and so it can be uniformly approximated in some large open ball by a high-degree polynomial. Deduce that most of the polynomial's coefficients are zero.
