# Why does the Laplace operator extend to $L^2(X)$?

Suppose $X$ is a Riemannian manifold. Then we get a Laplace operator on $C^\infty(X)$. In most texts I see the Laplace operator extended to $L^2(X)$, but I don't see how, since it does not seem to be continuous for the $L^2$ norm (There are small functions with big second derivatives, e.g. $f(x) = \epsilon\sin\frac{x}{\epsilon}$ on $[0, 2\pi]$).

1. Is my example wrong? If not, how do we extend the Laplace operator to $L^2(X)$ in a canonical way, and what good is an extension if it's not continuous?

2. If I just care about the eigenfunctions of the Laplace operator, do I need to extend it to $L^2$, or do they already occur in $C^\infty$?

3. There is a Laplace operator on $\Omega^k(X)$. Is there a Hilbert space analogous to $L^2$ to which this Laplace operator extends?

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You are entirely correct that the Laplacian cannot really extend to the whole space in any reasonable way. (Sure, we could use the axiom of choice to define it quasi-randomly but consistently on too-not-differentiable functions, but this extension would not be useful.) Instead, what is usually meant is to extend the Laplacian to an "unbounded" $L^2$-valued operator defined on a dense subspace... arranged so that the domain of its adjoint is the domain of the operator. But these domains are not the whole space, indeed. – paul garrett Nov 13 '12 at 16:25
Regarding item 2: my understanding is that going to $L^2$ is important because $L^2$ is complete (and Hilbert), while $C^{\infty}$ is not. You cannot make $C^{\infty}$ into a Banach space, although you could use stronger topologies to make it a complete Frechet space. But in this setting, it's much more complicated to use fundamental tools like the inverse function theorem. That's why people tend to avoid $C^{\infty}$, and the $L^p$'s (or the Sobolev spaces, if you need derivatives) come into play. – student Nov 13 '12 at 17:39