Do Degree Zero Pseudo-Differential Operators on a Manifold Send Smooth Functions to Smooth Functions? I'm not an analyst, so forgive me if what I'm asking is not suitable for Mathoverflow.
For convenience, let $X$ be a compact complex manifold, and $E$ a holomorphic vector bundle on $X$. Let $H$ be the Hilbert space of $L^2$-integrable sections of $E$ on $X$, and let $F$ be a degree zero pseudo-differential operator in $\mathfrak{B}(H)$. Then my question is whether $F$ maps smooth(or complex/real analytic) sections to the same class of functions? What if $F$ is a degree $-1$ pseudo-differential operator?
More specifically, I'm thinking about the case when $E$ is $\oplus _i \wedge ^{0,i} T^*X $ and $F$ is either $\frac{1}{\sqrt{1+D^2}}$ or $\frac{D}{\sqrt{1+D^2}}$, where $D= \bar{\partial}^* + \bar{\partial}$, where $\bar{\partial}$ is the Dolbeault operator.
Any helpful comments and references to learn about this subject is greatly appreciated.
 A: As Deane Yang pointed out, in general, an elliptic operator of order $s$ maps $H^{k}\rightarrow H^{k-s}$ for functions defined on $\mathbb{R}^{n}$. The pseudo-differential operator on a compact manifold is defined by patching over local coordinates. A smooth function over the whole compact manifold is of course smooth in the local coordinates as well. So a standard partition of unity argument is suffice. Since a smooth functions' derivatives are bounded on $M$, it is trivially inside every Sobolev space. Together this gives you the desired argument. 
The argument is usually expressed in terms of wave front sets. A standard theorem in distribution theory asserts that if $P$ is an operator with Schwartz kernel $K\in D'(\mathbb{R}^{m}\times \mathbb{R}^{n})$, and $K$'s wavefront set is contained in
$$
\{\xi\not=0, \eta\not=0\}
$$
Then $P$ defines maps 
$$
P:C^{\infty}_{c}(\mathbb{R}^{n})\rightarrow C^{\infty}(\mathbb{R}^{n})
$$
(see Friedlander and Joshi, for example)
In your case the Schwartz kernel of both operators can be explicitly written down and then the above theorem can be applied. But I am not sure if this is the best way to solve the problem at here. In general, the Fourier transform of a compactly supported function is a Schwartz function, and that of a compactly supported distribution is an analytic function. So I think analyticity should follow as well. But I do not have a proof over the top of my head. 
