I have the following exercise in my complex analysis course:
Let $u(x,y)$ be harmonic on the closed disk: $$D=\{(x,y)\text{ } |\text{ } (x-x_0)^2+(y-y_0)^2\leq r^2\}$$ Prove the following equality (the mean value property): $$u(x_0,y_0)=\frac{1}{2 \pi}\int_{-\pi} ^{\pi}u(x_0+r\cos t,y_0+r\sin t)dt$$
I understand how to prove it using complex analysis by using the fact that there exists a conjugate harmonic function $v(x,y)$, which allows me to define a holomorphic function $f=u+iv$ and finally use the mean value property for complex holomorphic functions and equate the real parts of this equation to get the result I need.
The problem is that when I look at the conditions for the following theorems, I need to state that $D$ is a simply connected domain, but $D$ here is not a domain since it's not an open set in $\mathbb{C}$.
I have two questions:
- What does it mean here that $u(x,y)$ is harmonic on a closed set?
- How can I use the reasoning I outlined even though $D$ is closed?
Thank you
I wouldn't bother defining harmonicity on closed sets. Instead I would just interpret the question as $u$ being a harmonic function on the open disk, and being a continuous function on the closed disk.
I hope you know that for all $0< \rho<r$, the mean-value property for $u$ holds on the circle of radius $\rho$: \begin{align} u(x_0,y_0)=\frac{1}{2\pi}\int_0^{2\pi}u((x_0,y_0)+\rho (\cos t,\sin t))\,dt. \end{align} Now, the task is to show that the same thing holds with $\rho=r$. This is a standard exercise in analysis; here things are easy enough because $u$ is continuous on the closed disk, hence uniformly continuous there. If you're still struggling, then take a look at Cauchy's integral formula at the border for some inspiration with the details.
You can actually generalize this result to arbitrary dimensions, because harmonic functions always satisfy the mean-value property on open sets (and proving it for closed balls is a simple analysis exercise with uniform continuity and integrals).