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Let $\mathbb{D}$ be an self-adjoint elliptic operator of a compact manifold and $G(x,y)$ the Green's function of $\mathbb{D}$. By definition $G(x,y)$ is the right inverse of $\mathbb{D}$ in the sense that $$ \mathbb{D} \int G(x,y)f(y)dy=f(x) $$ for any function $f$. What can we say about the left inverse of the Green's function? The naive guess is that $$ \int G(x,y)\mathbb{D}f(y)dy=f(x), $$ but I think this is not true in general as $f$ can be in the kernel of $\mathbb{D}$. My question is, what can we say about the left inverse of the Green's function?

I asked a simpler question in this post with no answer so far. In this case, it seems to me that the integration-by-part argument still works assuming that $c$ is not in the eigenvalue of the operator. Does this generalize to arbitrary elliptic operators?

Lastly, I would appreciate it if someone could suggest a good textbook to learn this kind of materials on the elliptic operators and Green's functions on compact manifolds.

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    $\begingroup$ You have to be careful. What if $f\notin Domain(\mathbb{D})$? What if $\mathbb{D}$ is not self adjoint (i.e. how do you know that the second integral you wrote down holds true if $\mathbb{D}$ is not self adjoint)? Also, as you pointed out the Fredholm alternative says that you can't solve $\mathbb{D} u = f$ if $f\in ker(\mathbb{D})$. $\endgroup$
    – k3thomps
    Commented May 16, 2014 at 17:26
  • $\begingroup$ I need self-adjointness. Thank you for pointing out that. I am not sure what to do with the case where $f \not \in \mathrm{Dom}(\mathbb{D})$. This subtlety is part of my question and so I want to understand how to formulate the left inverse. $\endgroup$
    – Shiu
    Commented May 16, 2014 at 18:15
  • $\begingroup$ Can I ask what you need the left inverse for? It isn't clear to me how it would be useful. $\endgroup$
    – k3thomps
    Commented May 16, 2014 at 18:21
  • $\begingroup$ This question is motivated by my post on MathStackExchange (please follow the link above). $\endgroup$
    – Shiu
    Commented May 16, 2014 at 19:50

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First quick comment. If you want to integrate on manifolds, you need some kind of volume form. You didn't say anything specific about your $dy$, but you did say that the operator $\mathbb{D}$ is self-adjoint. So I guess we can assume that it is self-adjoint with respect to the specific volume form $dy$.

As was discussed in the comments above, if $\mathbb{D}$ has a kernel, the operator $\mathbb{D}$ is not invertible, unless one restricts its domain of definition. So a Green function $G(x,y)$ such that your first equation holds exists only if $\mathbb{D}$ has a trivial kernel. In that case, your second equation holds as well, following a simple integration by parts argument that you've pointed out.

In general, the best you can do is have an identity of the form $G\mathbb{D}f = Pf$, where $P$ is some projection operator that annihilates the kernel of $\mathbb{D}$ and depends on the precise way the Green function was defined.

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