TITLE: Convergence of Schwartz kernels implies convergence of operators
QUESTION [9 upvotes]: Let $K$ be a smoothing operator on $\mathbb{R}^n$, i.e., it defines a map on all Sobolev spaces $K\colon H^r(\mathbb{R}^n) \to H^s(\mathbb{R}^n)$ for all $r, s \in \mathbb{R}$. Now (a variation of) the Schwartz kernel theorem states that it is given by some smooth kernel $k \in C^\infty(\mathbb{R}^n \times \mathbb{R}^n)$, i.e., $(Kf)(x) = \int_{\mathbb{R}^n} k(x, y)f(y) dy$.
Now suppose we have kernels $k_n, k \in C^\infty(\mathbb{R}^n \times \mathbb{R}^n)$ that do define smoothing operator $K_n$ and $K$ (since not every element of $C^\infty(\mathbb{R}^n \times \mathbb{R}^n)$ defines a smoothing operator we have to assume this).

What type of convergence $k_n \to k$ is needed, so that $K_n$ does converge to $K$?

Suppose I want $K_n \to K$ only as maps $L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$. Does it suffice for this that $k_n$ converges uniformly to $k$?
Suppose I want $K_n \to K$ as maps $H^r(\mathbb{R}^n) \to H^s(\mathbb{R}^n)$ for all $r, s \in \mathbb{R}$. Is it sufficient for this that $k_n \to k$ uniformly and also all their derivatives?
I would prefer convergence in the norm topology, but any information about convergence in the strong / weak operator topology would also be nice.

REPLY [7 votes]: Here is a classical theorem. $\newcommand{\bR}{\mathbb{R}}$
Suppose that for $0< a,b<\infty$
$$\sup_x\left(\int_{\bR^n} |k(x,y)|^a dy\right)^{\frac{1}{a}}=M_1(k)<\infty, $$
$$\sup_y\left(\int_{\bR^n} |k(x,y)|^b dx\right)^{\frac{1}{b}}=M_2(k),  $$
and
$$\frac{1}{p}-\left(\frac{b}{a}\right)\frac{1}{q}=1-\frac{1}{a},\;\;1\leq p\leq q <\infty $$
then the integral operator $T=T_k$ defined by the kernel  $k$  defines a  bounded operator $L^p\to L^q$  and  its norm    satisfies the inequality
$$\Vert T_k\Vert \leq M_1(k)^{1-\frac{b}{q}}M_2(k)^{\frac{b}{q}}. $$
For  proofs and many more details, see Section 9.5 of Edwards' book,  Functional Analysis. Theory and Applications, Dover.