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Let $X$ be a compact Riemannian manifold and let $\mathcal{M}(X)$ be the space of regular finite Borel measures with the total variation as norm.

The Laplace-Belrami-Operator $\Delta$ on $X$ with domain $C^2(M)$ is not densely defined, as $C^2$ is not dense in $\mathcal{M}(X)$.

Question: Can the Laplacian be realized as a densely defined, closed operator on $\mathcal{M}(X)$ that generates an operator semigroup $e^{t\Delta}$?

More specifically, does there exist a subset $$\mathrm{dom}(\Delta) \subseteq \{ \mu \in \mathcal{M}(X) \mid \Delta \mu \in \mathcal{M}(X)\}$$ (where $\Delta \mu$ is the distributional derivative) such that $\mathrm{dom}(\Delta)$ is dense in $\mathcal{M}(X)$ such that $\Delta$ is closed on $\mathrm{dom}(\Delta)$?

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  • $\begingroup$ Not an answer, but a comment which might be useful. There are many indications that the norm topology on the space of measures on a compact set is not the most suitable for many situations--- it is simply too strong. A suitable substitute is the topology of compact convergence on the space of contnuous functions. This is not normed but it is complete and has many other nice properties. It might be the right setting for your needs. $\endgroup$
    – couperin
    Commented May 31, 2014 at 10:54
  • $\begingroup$ Have you looked at the case of an interval on the real line? If the second derivative of a function is a measure, this makes the function continuous, and in particular integrable. Such functions are not dense in the space of measures. $\endgroup$
    – Michael Renardy
    Commented May 31, 2014 at 11:14
  • $\begingroup$ Is this a fact? How come? If this is so, this gives a negative answer already. $\endgroup$
    – Matthias Ludewig
    Commented May 31, 2014 at 11:19

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The semigroup exists --- it is the adjoint of the heat semigroup on $C(X)$. However, it is not strongly continuous at $t=0$ and thus does not have a densely defined generator. The semigroup for $t>0$ is given by $$e^{t\Delta}\mu(E) \ = \ \int_{E} H_t(x,y) d V(x) d\mu(y) , $$ where $H_t(x,y)$ is the heat kernel on $X$ and $V$ is the volume measure on $X$. Note that $e^{t\Delta}\mu$ is absolutely continuous with respect to $V$ for $t>0$. It follows that $e^{t\Delta}\mu$ is discontinuous at $t=0$ whenever $\mu$ is not absolutely continuous with respect to $V$.

Note that $e^{t\Delta}M(X) \subset L^1(d V)$ as soon as $t>0$ and on $L^1$ the semigroup is strongly continuous. Thus $e^{t\Delta}$ is strongly continuous for $t>0$. In some sense the semigroup has a non densely defined generator, which is $\Delta$ on $L^1(dV)$.

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