I am currently trying to prove an elliptic regularity type result for the Laplace Beltrami operator $\Delta\_g$ on a Riemannian manifold $(M^n,g)$. As a matter of convention, I will assume $\Delta\_g$ to be a nonpositive (self-adjoint) unbounded operator on $L^2(M)$, with domain $$
\mathcal{D}(\Delta\_g):=\{f\in L^2(M)\ :\ \Delta\_g f\in L^2(M)\}
$$ where we define $\Delta\_g f\in\mathcal{D}'(M)$ by using symmetry of $\Delta\_g$ and taking the formal adjoint. Then by functional calculus, we can define a semigroup $$
[0,\infty)\to\mathscr{L}(L^2(M)),\quad t\mapsto e^{t\Delta\_g}
$$ which is strongly continuous, and continuous in norm for $t>0$. Now, I have shown that $e^{t\Delta\_g}$ is smooth as a function of $t$, with $$
\frac{\mathrm d^k}{\mathrm d t^k}e^{t\Delta\_g} = \Delta\_g^k e^{t\Delta\_g}\in\mathscr{L}(L^2(M))
$$ as a derivative, however, it is less clear to me how to prove smoothness in space. My issue is that I can only see how to define $e^{t\Delta\_g}$ via the spectral theorem. This is done by fixing a unitary map $U:L^2(M)\to L^2(X,\mu)$ for some measure space $(X,\mu)$, and fixing some measurable $g:X\to(-\infty,0]$ such that the multiplication operator $M\_g$ corresponds to $\Delta\_g$, and by setting $e^{t\Delta\_g}$ to correspond with $M\_{e^{tg}}\in\mathscr{L}(L^2(X,\mu))$.


Can smoothness of $u:=e^{t\Delta\_g}f$ be proved from the fact that $\Delta\_g^k u\in L^2(M)$ for all $k\ge0$, or is more needed?



Elliptic regularity of Laplace-Beltrami operator on a manifold
What about M? I will give an answer supposing it is compact without boundary.


You want to prove that for fixed $t\in \mathbb{R}$, $e^{t\Delta}f\in C^\infty(M)$. This comes from an eigenvalue estimate of the Laplacian + elliptic regularity. For $-\Delta$ (positive Laplacian), due to the compactness of the manifold and Sobolev Embedding theorem you have that the Laplacian is Discrete operator with $\lambda\_j \geq Cj^c, c>0$ (For your Laplacian you get the opposite). (A proof of this you can find on the book "Invariance theory, Heat equation and the Atiyah-Singer Index Theorem" Ch. 1.6)


Now on M you have Sobolev Embedding that states $H^{k+m}\hookrightarrow C^k$ and using that $e^{t\Delta}f(x) = \sum\_j e^{t\lambda\_j}\psi\_j(x)<f,\psi\_j>$ you can easily show that $\Delta^k e^{t\Delta}f = e^{t\Delta} \Delta^kf$ (For this you need the eigenvalue estimate, because it shows that $\sum\_j \lambda\_j^ke^{t\lambda\_j}$ converges absolutely for every k.


But now you have that $\Delta^k (e^{t\Delta}f)\in L^2(M)\Rightarrow (e^{t\Delta}f)\in H^{2k}(M) \ \forall\ k$ and by Sobolev Embedding you get smoothness.


