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Alexey Ustinov
Alexey Ustinov
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If the boundary is Lipschitz, one can prove only that the operator is bounded, it is not necessarily even compact. If the surface is smooth the operator is in fact a pseudodifferential operator of order -1$-1$, and Weyl asymptotics gives the answer. The latter result holds if the boundary is just a tiny little bit better than Lipschitz, but a hard analysis and perturbation theory is neded.

If the boundary is Lipschitz, one can prove only that the operator is bounded, it is not necessarily even compact. If the surface is smooth the operator is in fact a pseudodifferential operator of order -1, and Weyl asymptotics gives the answer. The latter result holds if the boundary is just a tiny little bit better than Lipschitz, but a hard analysis and perturbation theory is neded.

If the boundary is Lipschitz, one can prove only that the operator is bounded, it is not necessarily even compact. If the surface is smooth the operator is in fact a pseudodifferential operator of order $-1$, and Weyl asymptotics gives the answer. The latter result holds if the boundary is just a tiny little bit better than Lipschitz, but a hard analysis and perturbation theory is neded.

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Grigori Rozenblum
Grigori Rozenblum
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If the boundary is Lipschitz, one can prove only that the operator is bounded, it is not necessarily even compact. If the surface is smooth the operator is in fact a pseudodifferential operator of order -1, and Weyl asymptotics gives the answer. The latter result holds if the boundary is just a tiny little bit better than Lipschitz, but a hard analysis and perturbation theory is neded.