# Prove partial derivatives of uniformly convergent harmonic functions converge to the partial derivative of the limit of the sequence.

I think the title says it all.

If you have a sequence of harmonic functions from a bounded complex domain to the real numbers, show that on a subset at a positive distance from the boundary of the domain, e.g. a compact subset of the domain, the derivatives of the harmonic functions converge uniformly to the derivative of the limit of the sequence of the harmonic functions.

Thank you!

My attempt: I tried to apply the mean value property (like Gamelin does for analytic functions) to find a bound (unsuccessfully). I know the question has been asked before, but I did not understand the solution. I also tried to come up with something similar to Cauchy estimates for harmonic functions, but I wound up more confused than when I started.

EDIT: Keep in mind by derivative I meant partial derivatives. My attempt was to find an analogue of the Cauchy integral formula (for derivatives) but I seem to get the same formula for both partial derivatives, which does not make sense to me because it seems like the partial derivatives can be different.

Your approach is correct, I would use the mean value property to try to derive an analogue of the Cauchy estimates. Let $\Omega$ be the domain, and let $u$ be harmonic. See if you can show that if $B(z,r) \subset \Omega$, then $$|\partial_i u(z)| \le Cr^{-1} \|u \|_{L^{\infty}(\partial B(z,r))}$$ You will be using both the mean value property (since $\partial_i u$ is itself harmonic) and the divergence theorem. If you have further questions I can elaborate more.
Elaboration: The $L^{\infty}$, in the context of continuous functions (such as the harmonic functions in this problem), is just the norm corresponding to uniform convergence. So the sequence of harmonic functions $u_n$ converging uniformly on a compact set $K$ can be rewritten as $\|u - u_n\|_{L^{\infty}(K)} \rightarrow 0$ as $n \rightarrow \infty$.
To obtain the estimate in question, first use the mean value property to obtain: $$\partial_i u(z) = \frac{1}{\pi r^2} \int_{B(z,r)} \partial_i u(x,y) \, dx \, dy,$$ which follows from $\partial_i u$ itself being harmonic. By the divergence theorem, then this is equal to $$\frac{1}{\pi r^2} \int_{\partial B(z,r)} u \nu^i \, dS,$$ where $\nu^i$ is the $i$-th component of the unit vector normal to the surface $\partial B(z,r)$. Thus $$|\partial_i u(z)| \le \frac{1}{\pi r^2} \int_{\partial B(z,r)} |u| \, dS = \frac{2}{r} \|u\|_{L^{\infty}(\partial B(z,r))}$$ This allows you to control the pointwise convergence of the partial derivatives of $u_n$ by the uniform convergence of $u_n$. However, this in turn allows you to control the uniform convergence of the derivatives on a compact set, since then you can just use larger balls.