Does every smoothing operator have a unique kernel? Consider smooth manifolds $M, N$ and the spaces of smooth functions, $C^\infty(M)$, $C^\infty(N)$ and $C^\infty(M \times N)$. A smoothing operator $T_k$ with kernel $k$ is defined as
$$
(T_k f)(x) = \int_N k(x, y) f(y) dy,
$$
where $f \in C^\infty(N)$ and $k \in C^\infty(M \times N)$ and $T_k f \in C^\infty(M)$.
In this way, every smooth kernel $k \in C^\infty(M \times N)$ induces a linear operator $T_k : C^\infty(N) \rightarrow C^\infty(M)$. My first question is whether the kernel $k$ of a smoothing operator $T_k$ is unique, or whether it is possible that $T_k = T_{k'}$ for $k \neq k'$. My second question is whether any (bounded?) linear operator $T : C^\infty(N) \rightarrow C^\infty(M)$ can be written as an integral transform with a smooth kernel $k \in C^\infty(M \times N)$. Finally, I'm interested in the same questions for the case of smooth vector valued functions and operator-valued / matrix-valued kernels (i.e. $f : N \rightarrow V$, $k : M \times N \rightarrow \operatorname{Hom}(V, W)$ and $T_k f : M \rightarrow W$).
I think that the Schwartz kernel theorem is relevant here, but I am not too familiar with the theory of distributions, and for my application I need to stay within the realm of smooth functions and kernels.
In addition to an explanation or proof, references for the theorems would be much appreciated.
 A: There is a follow-up theorem to the Schwartz-Kernel Theorem in Hörmanders Linear Partial Differential Equations 1, Thm., 5.2.6, saying that every continiuous map from $\mathcal{C}^\infty_0(N)\rightarrow \mathcal{C}^\infty(M)$ is defined by a smooth integral kernel (with no requirements to the support of $k$) and vice versa.
Jumping into distribution theory, the extention from $\mathcal{C}^\infty_0(N)\rightarrow \mathcal{C}^\infty(M)$ to $\mathcal{E}'(N)\rightarrow \mathcal{C}^\infty(M)$ gives you the uniqueness of $k$, since $\delta_q\in\mathcal{E}'(N)$  defined by $\delta_q g:=g(q)$ yields that
\begin{align*} \int_N k(x,y)\delta_q(y)dy=k(x,q) \end{align*}
and hence for any continiuous operator $T:\mathcal{C}^\infty_0(N)\rightarrow \mathcal{C}^\infty(M)$ we obtain a function $f_q\in\mathcal{C}^\infty(M)$ by $f_q:=T\delta_q$ for each $q\in N$. Define $k(\cdot,q):=f_q$ which gives you the integral kernel.
Note that the "integral" is to be understood as "applying $\delta_q$ to $k$", which turns out to be usefull in the theory of distributions. You can probably recover the same result by assuming $T$ is given by two different integral kernels $k$ and $k'$ and then evaluate these at a sequence $\tau_n$ of compactly supported simple functions with the condition that $\{q\}=\cap_n\text{supp}(\tau_n)$  and $\int_N \tau_n dy=1$ (i.e., $\tau_n\rightarrow \delta_q$ in the topology of distributions).
You can see that the compactness of $f$ is required for the integral to be finite if you allow non-compact kernels.
