# Convergence of Schwartz kernels implies convergence of operators

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.

• I posted this question some days ago on Math.StackExchange but did not get any answer. math.stackexchange.com/q/262632/7110 Dec 23 '12 at 15:06
• For the first question, we can try to see what happens when $K_j(x,y)=h_j(x)g(y)$, where $h_j$ and $g$ are smooth and square integrable. Even when $g$ is bounded and $h_j\to 0$ uniformly, we don't necessarily have that $K_j(f)\to 0$ for all $f$. However, in the general case, if we have $\lVert K_n-K\rVert_{L^2(\Bbb R^n\times \Bbb R^n)}\to 0$, there is convergence in the operator norm. Dec 23 '12 at 16:18
• You can certainly get some results just by writing out the definition of the norms and applying Holder's inequality. Do you need something sharper than that? Dec 23 '12 at 16:38
• I know $k_n \to k$ uniformly and tried to deduce convergence of the $K_n$ in the operator norm. But @DavideGiraudo says, that this does not hold in general. So maybe there is some little, small condition (in addition to uniform convergence of the $k_n$), so that I can then deduce $K_n \to K$ in norm? Convergence in $L^2(R^n \times R^n)$ is to strong, since $k_n - k$ is not square-integrable over $R^n \times R^n$. @DeaneYang: I don't see how to apply Hölder here. Dec 23 '12 at 17:49
• >I have a particular case and want to show that it is a compact operator.< Why don't you just tell us the kernel and the space in which you want the compactness then? That may lead to your goal much faster... Dec 25 '12 at 1:42

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.

• Thanks, that answer helps me also with another problem I had. Dec 30 '12 at 17:13