Let $\Omega\subset\mathbb{R}^N$ open and $\{u_j\}$ be a sequence in $C^2(\Omega)$. If each $u_j$ is harmonic in $\Omega$ and the sequences $\{\partial^{\alpha}, |\alpha|\le 2\}$ converge uniformly over compact subspaces of $\Omega$, then show that $u = \lim_j u_j$ is harmonic in $\Omega$
$u_j$ harmonic implies
$$\sum_{i}^n\frac{\partial^2 u_j}{\partial x_i^2} = 0$$
We also have by hypotesis that
$$\frac{\partial^2 u_j}{\partial x_i\partial x_j}$$
converge uniformly over compacts of $\Omega$.
I have to prove that $u = \lim u_j$ is harmonic, that is
$$\sum_{i}^n\frac{\partial^2 u}{\partial x_i^2} = 0$$
If I remember correctly, there's a theorem that guarantees I can take the derivative of the limit and pass the derivative to the limit elements, when the limit converges uniformly. If this is possible then proving the exercise would be just a matter of doing:
$$\sum_{i}^n\frac{\partial^2 u}{\partial x_i^2} = \lim \sum_{i}^n \frac{\partial^2 u_j}{\partial x_i^2}$$
and proving it's equal to $0$. I have some questions:
What does it mean to say $\{\partial^{\alpha}, |\alpha|\le 2\}$ converges uniformly? These are operators.
Under which conditions exactly I can interchange the limits like I did? (a partial derivative is a limit)
What's the role of compacity?
It is given that partial derivatives up to order 2 converge uniformly, but it is not given that the limits are equal to partial derivatives of $u$. So we are required to identify the limits in terms of $u =\lim_n u_n$. This can be done by repeated application of the following elementary result from one variable calculus: Suppose $f_n \to f$ uniformly , $f_n$'s are differentiable and $f_n' \to g$ uniformly. Then $f$ is differentiable and $f'=g$.