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.

  • $\begingroup$ I posted this question some days ago on Math.StackExchange but did not get any answer. math.stackexchange.com/q/262632/7110 $\endgroup$
    – AlexE
    Dec 23, 2012 at 15:06
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    $\begingroup$ 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. $\endgroup$ Dec 23, 2012 at 16:18
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    $\begingroup$ 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? $\endgroup$
    – Deane Yang
    Dec 23, 2012 at 16:38
  • $\begingroup$ 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. $\endgroup$
    – AlexE
    Dec 23, 2012 at 17:49
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    $\begingroup$ >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... $\endgroup$
    – fedja
    Dec 25, 2012 at 1:42

1 Answer 1


Here is a classical theorem. $\newcommand{\bR}{\mathbb{R}}$

Suppose that for $0< a,b<\infty$

$$\sup_x\left(\int_{\bR^n} |k(x,y)|^a dy\right)^{\frac{1}{a}}=M_1(k)<\infty, $$

$$\sup_y\left(\int_{\bR^n} |k(x,y)|^b dx\right)^{\frac{1}{b}}=M_2(k), $$


$$\frac{1}{p}-\left(\frac{b}{a}\right)\frac{1}{q}=1-\frac{1}{a},\;\;1\leq p\leq q <\infty $$

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.

  • $\begingroup$ Thanks, that answer helps me also with another problem I had. $\endgroup$
    – AlexE
    Dec 30, 2012 at 17:13

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