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If $M$ is a complete Riemannian manifold and $L$ is the Friedrichs extension of the Laplacian $-\Delta$, then it is known (first proven by Gaffney in the '50) that $C_0 ^\infty (M)$ is a core for $L$. In particular, this is true for compact manifolds without boundary.

If $M$ is now compact, but with boundary, does the above still hold?

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    $\begingroup$ If $C_0^\infty(M)$ stands for smooth functions with support far away from the booundary, then they are not a core for example when $M$ is a bounded open subset of $R^n$. In fact you can have two (actually many) maximal acceretive extensions corresponding to Dirichlet and Neumann boundary conditions. $\endgroup$
    – Giorgio Metafune
    Commented Oct 28, 2021 at 7:23
  • $\begingroup$ @GiorgioMetafune: Funny, I just talked to somebody who told me the exact opposite of what you say. In particular, I was told that what I ask is true on the interval $[0,1]$, while you claim that it is false... I imagined that this couldn't be possible in mathematics. $\endgroup$
    – Alex M.
    Commented Oct 28, 2021 at 8:49
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    $\begingroup$ I have only one doubt about the definition. For "core" I mean a dense subset of the domain of the operator with respect to the graph norm (which is a Sobolev case in the smooth case). Sometimes one means "form core", that is a dense subset for the domain of the associated quadratic form (which is usually $H^1$ for $p=2$). In the first case the assertion is clearly false in the interval: smooth functions with compact support are not dense in the domain of the operator, which is $H^2\cap H^1_0$ (that is only $u(0)=0$). In the closure one has also $u'(0)=0$.) $\endgroup$
    – Giorgio Metafune
    Commented Oct 28, 2021 at 11:08
  • $\begingroup$ @GiorgioMetafune: You guessed it correctly, I meant "core", not "form core". Thank you. $\endgroup$
    – Alex M.
    Commented Oct 28, 2021 at 11:14
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    $\begingroup$ In that case it is true, provided $\partial M$ is smoooth. The usual way is to prove first for the half space and then use local coordinates to flatten the boundary. Basically one needs a $C^k$, $ k \geq 2$, boundary to approximate with $C^k$ functions vanishing at the boundary. $\endgroup$
    – Giorgio Metafune
    Commented Oct 28, 2021 at 17:04

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