Corollary to mean value property for harmonic functions? For $\Omega \subset\mathbb{R}^n$ open, and $u_i:\Omega \to \mathbb{R}$ a sequence of harmonic functions which are uniformly bounded. Prove that for any multi-index $\alpha$ and for any $K \subset \Omega$ there exists $u_{i_{k}}$ such that $D^\alpha u_{i_{k}}$ converges uniformly in $K$.
The mark scheme apparently uses a corollary to the mean value property which says that $$|u(x) - u(y)| \leq \sup_\xi |\nabla f(\xi)| |x-y|$$ but I can't see where this corollary is coming from? Could someone prove this?
 A: The inequality $$|u(x) - u(y)| \leq \sup_\xi |\nabla f(\xi)| |x-y|\tag1$$ in which the supremum is taken over $\xi$ on the line segment from $x$ to $y$, has nothing to do with harmonic functions, or their mean value property. It is a corollary of the Mean Value Theorem. 
As for the original question, the key point is the interior regularity of harmonic functions. Namely, for any multiindex $\alpha$ and every locally uniformly bounded family of harmonic functions $u_j$, the derivatives $u_j^\alpha$ also form a locally uniformly bounded family of harmonic functions. By induction, it suffices to prove this for first order derivative. The quickest way is probably to use the Harnack inequality, which gives an estimate of the form $$|\nabla u(x)|\le \frac{C}{r}\sup_{|y-x|= r} |u(y)-u(x)| \tag2$$ with a universal constant $C$. Note that (2) goes in the direction opposite to (1). 
A: By Ascoli-Arzelá Thorem, there exists a subsequence that converges uniformly to a function $u$ over compacts subsets of $\Omega$. This will define the candidate $u$. 
After that, see that a uniform limit of harmonic functions is harmonic. So $u$ is analitic. In particular, you can apply the Mean Value Theorem and obtain the desired inequality.
