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Let $(M,g)$ be a smooth Riemannian manifold. The celabrated kernel theorem of Schwartz shows that for any linear and continuous operator $A:C_{c}^{\infty}(M)\to C^{\infty}(M)$, there exists a distribution $K_{A}\in\mathcal{D}^{\prime}(M\times M)$ such that $$\langle A(f),g\rangle_{M}=\langle K_{A},g\otimes f\rangle_{M\times M}$$

where $\langle u,f\rangle_{M}=u(f)$ denotes the distributional pairing for $u\in\mathcal{D}^{\prime}(M)$ and $f\in C^{\infty}_{c}(M)$ and where $A(f)\in\mathcal{D}^{\prime}(M)$ is the distribution $\langle A(f),g\rangle_{M}:=\int_{M}A(f)g\,\mathrm{vol}_{g}$.

Let $\mathcal{S}\subset C^{\infty}(M)$ be a submodule and $\mathcal{S}_{c}:=\mathcal{S}\cap C^{\infty}_{c}(M)$. Under which assumptions on $\mathcal{S}$ is there an analogues theorem as Schwartz kernel theorem for linear and continuous operators $A:\mathcal{S}_{c}\to\mathcal{S}$ and suitable kernels living in some subspace of $\mathcal{D}^{\prime}(M\times M)$?

I am actually interested in the bundle-valued case, but lets not complify things.

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    $\begingroup$ Since the question is a bit open ended, here are just some possibly relevant comments. If the operator $A$ can be extended from $\mathcal{S}_c$ to $C_c^\infty$, then it has a kernel representation. Perhaps you want a condition on $A$ rather than on $\mathcal{S}$? If you find a kernel for $A$, then it automatically extends to all of $C_c^\infty$. The extension might not be unique tough. The difference between two extensions will annihilate $\mathcal{S}$. $\endgroup$
    – Igor Khavkine
    Commented Jul 13, 2023 at 15:34

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