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).



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> What type of convergence $k\_n \to k$ is needed, so that $K\_n$ does converge to $K$?
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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.



Convergence of Schwartz kernels implies convergence of operators
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. 


