# If $f$ is a eigenfunction of $-\Delta$ in $L^2[0,1]$, is it necessarily $C^\infty$?

I am a little bit confused about the properties of the Laplacian $$-\Delta$$ on $$L^2[0,1]$$ with the periodic boundary conditions.

At least I know that $$-\Delta$$ is an unbounded self-adjoint operator on $$L^2[0,1]$$ and its eigenvalues are all nonnegative. Moreover, each eigenvalue has a finite multiplicity.

Now, my confusions are as follows:

1. If $$f \in L^2[0,1]$$ is an eigenfunction of $$-\Delta$$, then is $$f$$ necessarily $$C^\infty$$? I vaguely remember some regularity theorems from PDE context, but I cannot find an exactly relevant reference.
2. If the first item is correct, then each eigenspace of $$-\Delta$$ must be a finite dimensional subspace of $$L^2[0,1]$$, consisting of smooth functions. Is this also true?
3. Lastly, let $$g$$ be a smooth periodic function on $$[0,1]$$ such that $$-\Delta g$$ is an eigenfunction of $$-\Delta$$ with the eigenvalue $$\lambda (\geq 0)$$. Then, I suspect that $$g$$ itself is an eigenfunction with the eigenvalue $$\lambda^2$$. But I cannot really prove this rigorously.

All these issues seem to be related with the regularity of the eigenfunctions for the Laplacian and a bit subtle to me. Could anyone please clarify?

1. It's a bit of ping-pong between $$f$$ and $$f''$$: $$f''$$ being $$L^2$$ means $$f'$$ is $$H^1$$ hence $$\mathcal{C}^0$$, thus $$f$$ is $$\mathcal{C}^1$$, but since $$\lambda f = -f''$$ this implies that $$f''$$ is $$\mathcal{C}^1$$, which means that $$f$$ is $$\mathcal{C}^3$$, and so on and so forth... at least for $$\lambda > 0$$.
As for the eigenvalue $$0$$ (when it is an eigenvalue, but for the periodic conditions you chose it is an eigenvalue), you need to use the fact that $$f'' = 0$$ has only affine solutions even if $$f''$$ is only seen as $$L^2$$ due to $$f'$$ being then a constant function thanks to the uniqueness of weak derivatives, and thus these eigenfunctions are smooth too.
3. If $$\lambda g'' = -g^{(4)}$$ then by integrating twice you'll find that there exists $$a,b$$ two scalars such that $$(\lambda g + g'')(x) = ax + b$$, thus if $$g$$ is an eigenfunction it would be for $$\lambda$$ and not $$\lambda^2$$. I think you're confused with it then being an eigenvalue of $$(-\Delta) \circ (-\Delta)$$ in which case it would be a $$\lambda^2$$-eigenfunction?