Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I posted this question now on MathOverflow: – AlexE Dec 23 '12 at 15:07
Do you want convergence as bounded linear operators in operator norm? In this case, first find conditions in order they are well defined, then try to compute the norm in the involved spaces. – Davide Giraudo Dec 23 '12 at 15:24
Yes, operator norm is preferable. But I would also be glad about any information regarding convergence in the strong / weak operator topology. – AlexE Dec 23 '12 at 15:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.