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 Answer 1

  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.

  2. Well, that's what "each eigenvalue has a finite multiplicity" tells you, so yes that would be true 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?

  • $\begingroup$ Oh, I was confused about the 3rd issue. Yes.. Thank you so much! $\endgroup$
    – Keith
    Commented May 28, 2023 at 14:38

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