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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.

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1 Answer 1

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.

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Hello. I'm unfamiliar with L norms. I'm learning about harmonic functions in a one variable complex analysis course. I was wondering if you could elaborate on how you derived the given inequality. I'm still unsure of where divergence theorem comes in. I think I was able to derive the Cauchy integral formula for harmonic functions but the formula for derivative I'm getting seems to be the same for all the partial derivatives... –  MR1992 Feb 7 '13 at 0:35

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