Schwarz Kernel Theorem states that:
If $A: C_0^\infty(\mathbb{R}^n) \to \mathcal{D}'(\mathbb{R}^n)$ is a linear continuous operator, then there exists a unique kernel $K\in \mathcal{D}'(\mathbb{R}^n\times \mathbb{R}^n)$ such that $\langle Au, v\rangle_{\mathcal{D}', \mathcal{D}} = \langle K, v\otimes u\rangle_{\mathcal{D}',\mathcal{D}}$ for all $u, v\in C_0^\infty(\mathbb{R}^n)$.
My question is:
Suppose we have a sequence of linear continuous operators $A_n: C_0^\infty(\mathbb{R}^n) \to \mathcal{D}'(\mathbb{R}^n)$ with kernels $K_n$. And $A: C_0^\infty(\mathbb{R}^n) \to \mathcal{D}'(\mathbb{R}^n)$ is also continuous and linear with kernel $K$. If $A_nu \to Au$ in $\mathcal{D}'$ for every $u \in C_0^\infty$, can we conclude that $K_n \to K$ in $\mathcal{D}'(\mathbb{R}^n\times \mathbb{R}^n)$?
By definition of kernels in the theorem, one can check that $K_n \to K$ when restricted on $C_0^\infty(\mathbb{R}^n)\otimes C_0^\infty(\mathbb{R}^n)$. So can we use a certain density argument to prove it?