Why does $v \in C^2(D) \cap C(\overline{D})$ and $\Delta v = - \lambda v$ in $D$ implies $v \in C^2(\overline{D})$

Question

The actual problem is to show that the pde, where $D = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 <1\}$ and $v = 0$ in $\partial D$ plus the above hypothesis, has non-trivial solution only if $\lambda \geq 0$. Which is quite trivial if you use one of the Green's identity. But for that is necessary that $v \in C^2(\overline{D})$, and I don't see why is that true.

Answer 1

If $v\in C^2(D)\cap C(\bar{D})$, and $\Delta v=-\lambda v$, then $\Delta v\in C^2(D)\cap C(\bar{D})$, which implies that $v\in C^2(\bar{D})$