# 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.

• If the symbol of the pseudodifferential operator is smooth away from the zero section of the cotangent bundle, then it always maps smooth function to smooth functions. The order tells you how it maps between $L^2$ Sobolev spaces. An operator of order $K$ is a bounded map from $W^{2,j+k}$ to $W^{2,j}$ for any $j$. – Deane Yang Mar 12 '18 at 0:36
• Thanks for your comment. Do you have a reference for why smoothness of symbol results in the p.d.operator to send smooth functions to themselves? Also, if the symbol is say, analytic, can I say the same thing? So it seems in case of the Dolbeault operator, if I understand correctly, the symbol at $(x,\zeta)$ is $\frac{1}{ i \Vert \zeta \Vert }$, which is smooth when $\zeta \neq 0$. – Kashayar Mar 12 '18 at 3:14
• One way to see it is that smooth functions is within the intersection of Sobolev spaces. So Deane Yang's comment just follow. – YYY Mar 12 '18 at 3:29
• @Kashayar: This is not trivial. See math.mit.edu/~rbm/18.157-F05-Chapter5.ps, or math.mit.edu/~rbm/iml90c2.ps, for example. Your manifold is compact, so the estimates are not difficult. – YYY Mar 12 '18 at 4:12
• For the specific operators you mentioned, they do map smooth sections to smooth sections. – Liviu Nicolaescu Mar 12 '18 at 16:11

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)