Prove family $A_j = \{\frac{\partial u}{\partial x_j}: u\in A\}$ has property $\sup_K |\frac{\partial u}{\partial x_j}|\le C$

Question

Let $\Omega$ be an open subset of $\mathbb{R}^N$ and $A\subset C^{\infty}(\Omega)$ a family of harmonic functions satisfying the following property: for all compact $k\subset \Omega$ there exists $C>0$ such that $\sup_K |u|\le C, \forall u\in A$. Show that the same property is valid for the families $$A_j = \{\frac{\partial u}{\partial x_j}: u\in A\}, j=1,\cdots,N$$ Finish, using the Arzelà-Ascoli theorem that, for all compact $K\subset\Omega$, the set $\{u_K: u\in A\}\subset C(K)$ is relatively compact in $C(K)$. Here, as usual, we're providing $C(K)$ with the distance $d(f,g0 = \sup_K |f-g|$

I know how to prove that $\sup \frac{\partial u}{\partial x_j}\le C^{|\alpha|+1}\alpha!$ from here $u$ harmonic then $|D^{\alpha} u(x_0)|\le \frac{C_k}{r^{n+k}}||u||$ or $\sup_{x\in B}|\partial^{\alpha} f(x)|\le C^{|\alpha|+1}\alpha!$ however I don't know if the sup bound was supposed to depend on $\alpha$ and this proof requires $u$ harmonic.

It isn't clear if $A_j$ is like $A$: a family of harmonic functions. If it is, maybe I can use something like the volumetric mean property, just like in the proof for the question I linked. Anyone has some hints?

Answer 1

Let us first observe that $A_j $ has the same property as $A$. Indeed, $u$ is smooth we can interchange the order of differentiation, i.e. $\partial_i \partial_j u = \partial_j \partial_i u$ for any $1\leq i, j \leq N$. Hence, for any $1\leq j \leq N$ we get $\Delta( \partial_j u ) = \partial_j ( \Delta u) = 0 $ for any $u \in A$. It follows that functions in $A_j$ are $C^\infty(\Omega)$ and harmonic. To prove the upper bound on compact subsets of $\Omega$, observe that for each $v \in A_j$ there is $u\in A$ such that $v = \partial_j u$. Invoking the estimate on derivatives of harmonic functions quoted in your post, we get $$ \sup\limits_{K}|v| \leq C \sup\limits_{K} |u| \leq C_1. $$

For the final part on Arzelà-Ascoli, observe that the equi-boundedness of the family $A$ is given, and for equi-continuity, we observe that for any $x,y \in K$ and any $u \in A$ one has $$ |u(x) - u(y)| \leq C (\sup_K|\nabla u|) |x-y| \leq C_1 |x-y|, $$ where we used the mean value theorem and a uniform bound on the gradient of $u$ by its supremum norm (which is bounded itself) due to harmonicity of $u$. Now Arzelà-Ascoli can be applied to conclude the claim on relative compactness.