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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.

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    $\begingroup$ 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 $j$. $\endgroup$
    – Deane Yang
    Mar 12, 2018 at 0:36
  • $\begingroup$ 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 $\zeta \neq 0$. $\endgroup$
    – Kashayar
    Mar 12, 2018 at 3:14
  • $\begingroup$ One way to see it is that smooth functions is within the intersection of Sobolev spaces. So Deane Yang's comment just follow. $\endgroup$
    – YYY
    Mar 12, 2018 at 3:29
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    $\begingroup$ @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. $\endgroup$
    – YYY
    Mar 12, 2018 at 4:12
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    $\begingroup$ For the specific operators you mentioned, they do map smooth sections to smooth sections. $\endgroup$
    – Liviu Nicolaescu
    Mar 12, 2018 at 16:11

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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.

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  • $\begingroup$ Thanks for the explanation. In the analytic case, can we still work locally? The partition of unity argument no longer works there. $\endgroup$
    – Kashayar
    Mar 13, 2018 at 17:19
  • $\begingroup$ @Kashayar: I honestly do not know. Thanks for asking. $\endgroup$
    – Bombyx mori
    Mar 13, 2018 at 17:51
  • $\begingroup$ @Kashayar: Not sure if this is relevant: mathoverflow.net/questions/110410/…. I will spend some time look into it. $\endgroup$
    – Bombyx mori
    Mar 13, 2018 at 17:53
  • $\begingroup$ For the real and complex analytic questions, I think you need hyperfunctions as developed by Sato and Kashiwara. $\endgroup$
    – Deane Yang
    Mar 19, 2018 at 5:19
  • $\begingroup$ @DeaneYang: I guess you meant something like this: math.harvard.edu/~siu/math212b/division_of_distribution.pdf. I never really had time to read it carefully in the past. I will take a look at it carefully in future, I am not sure if it is related. $\endgroup$
    – Bombyx mori
    Mar 19, 2018 at 5:26

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