I'm looking at the following problem.
Prove that if $h$ is harmonic on an open neighborhood of the disc $B(w,\rho)$, then for $0 \leq r < \rho, 0 \leq t < 2\pi$, $$h \left(w+r\mathrm{e}^{it} \right) = \frac{1}{2\pi} \int_0^{2\pi} \frac{\rho^2 - r^2}{\rho^2 - 2\rho r \cos(\theta - t) + r^2} h\left(w + \rho\mathrm{e}^{i\theta} \right) \, \mathrm{d}\theta.$$
We know that the Poisson integral of a continuous function $\phi: \partial B \to \mathbb{R}$ where $D=U(w,\rho)$, $w \in \mathbb{C}, \rho > 0$ has the exact form $$P_D \phi \left( w + r \mathrm{e}^{it} \right) = \frac{1}{2\pi} \int_0^{2\pi} \frac{\rho^2 - r^2}{\rho^2 - 2\rho r \cos(\theta - t) + r^2} \phi \left(w + \rho\mathrm{e}^{i\theta} \right) \, \mathrm{d}\theta$$ for $0 \leq r < \rho, 0 \leq t < 2\pi$. To solve the problem, is it as simple as saying that the function $h$ is itself continuous on the boundary and so $h$ will act as $\phi$ in the Poisson integral?
Hope you can help!