I am working out an example from Pinchover's PDE book (p. 198).
Suppose we look for solutions to $\Delta u = 0$ on the circular sector $\{ (r, \theta) \mid 0 < r < a, 0 < \theta < \gamma \}$ with initial conditions $u(a, \theta) = g(\theta), \, u(r, 0) = u(r, \gamma) = 0$.
Using separation of variables, I showed that the solution is of the form
$$ u(r, \theta) = \sum_{n=1}^\infty \frac{2a^{-\frac{n\pi}{\gamma}}}{\gamma} \left(\int_0^\gamma g(\phi)\sin\left(\frac{n\pi \phi}{\gamma}\right) \, d\phi\right) \, r^{\frac{n\pi}{\gamma}}\sin\left(\frac{n\pi\theta}{\gamma}\right)$$
Now I'm interested in figuring out what conditions on $g$ will guarantee that $u \in C^2$. For the same problem on the disk, I knew something about the rate of decay of the coefficients, but I'm having trouble translating that proof into this case.
My guess would be $g \in C^2$ is sufficient. Is it possible to show $u \in C^2$ for $g \in C^1$?
I presume that you mean $g$ instead of $h$. Continuity of $g$ (or even integrability) is enough for $u$ to be $C^\infty$ on the open circular sector. Under those conditions the Fourier coefficients $\int_0^\gamma g(\phi)\sin\left(\frac{n\pi \phi}{\gamma}\right) \, d\phi$ are bounded, and the term $\left(\frac{r}{a}\right)^{n\pi/\gamma}$ guarantees that the series can be differentiated term by term on sectors $0<r<b<a$.
Regularity up to the boundary is another matter. If you want a continuous solution on the closed sector, $g$ must be continuous and satisfy $g(0)=g(\gamma)=0$.