# Derivatives of Harmonic Functions converge normally?

Let $h_{n}$ be a sequence of harmonic functions that converge normally to $h$. Prove that $h_{n}'\rightarrow h'$ normally.

Certainly, this theorem holds when we have analytic functions (using Cauchy estimates). But how to generalize from here, using that every harmonic function is locally the real part of an analytic function, or Poisson integral...?

• – anon Aug 23 '11 at 9:29
• Harmonic functions satisfy Cauchy estimates, almost by definition. – Henri Aug 23 '11 at 9:46

Harmonic functions also satisfy Cauchy estimates. There are various ways of proving this. One way is to use the mean value property: Let $\phi(x,y)$ be a radial function with integral $1$, and let $\phi_R(x,y) = R^2\phi(Rx,Ry)$. Then $$u \ast \phi_R(x,y) = R^2\int u(x - X,y - Y)\phi(Rx,Ry)\,dX\,dY$$ Converting this integral into polar coordinates the using the mean value property for a fixed radius will give that $$u \ast \phi_R(x,y) = u(x,y)$$ Then differentiating under the integral sign gives $$(u \ast \partial^{\alpha} \phi_R)(x,y) = \partial^{\alpha}u(x,y)$$ Bounding $\partial^{\alpha} \phi_R$ and integrating gives the Cauchy estimates. (I'm not writing every step due to time constraints). One thing worth pointing out here is that this argument works for higher derivatives too, as well as harmonic functions in $n$ dimensions for $n > 2$.