2
$\begingroup$

Let $K$ be an integral kernel of a bounded operator $S:L^2(\mathbb{R}^n) \rightarrow L^2(\mathbb{R}^n) $ defined like

$$(Sf)(x)= \int_{\mathbb{R}^n}K(x,y)f(y)dy.$$

Assume that $K\in C^{\text{bounded}}(\mathbb{R}^n \times \mathbb{R}^n)$ has the following properties. $K$ is symmetric, positive and $K(0,\cdot) \in L^r(\mathbb{R}^n)$ for all $r>2.$

Also $K$ is monotone in the sense that $\left\lVert (x,y) \right\rVert>\left\lVert (x',y') \right\rVert$ then $K(x,y)< K(x',y').$

Then I define the sequence of operators $$(S_nf)(x)=\int_{\mathbb{R}^n}1_{B(0,n)^C\times B(0,n)^C}(x,y) K(x,y)f(y) dy.$$

So the same operator as $S$ but now we throw away everything that is supported close the origin of the kernel.

Obviously $(S_n)$ has the property that $\left\lVert S_n \right\rVert \le \left\lVert S \right\rVert$ and one also checks that $(S_nf) \rightarrow 0$ as $n \rightarrow \infty$ (check this on $C_c^{\infty}$ functions for example.)

I would like to know: Does it follow that $\left\lVert S_n \right\rVert \rightarrow 0$ ? So does pointwise convergence imply uniform convergene in this example?

Edit: It may also be interesting, and for some people maybe more convenient, to look at the analogous problem in $\ell^2(\mathbb{Z}^n),$ so having an operator $$(Tx)(z)= \sum_{y \in \mathbb{Z}^n} L(z,y)x(y)$$ and so on...

$\endgroup$
  • $\begingroup$ I think your monotonicity condition implies that there exists a decreasing function $f$ so that $K(x,y)=f(\|(x,y)\|)$. $\endgroup$ – Anthony Quas Jun 14 '17 at 23:20
  • 1
    $\begingroup$ An afterthought: A general obstacle to the property that you were hoping for is that often it will imply that $S$ is compact (because $S-S_n$ frequently is, for example in my example), so if this is false, your property can't hold. (Parts of my answer could be rephrased in those terms.) $\endgroup$ – Christian Remling Jun 15 '17 at 21:38
1
$\begingroup$

No. Let's take $d=1$ and $$ K=\frac{1}{1+\|(x,y)\|}, \quad f_n(y)=\chi_{(n,\infty)}\frac{1}{y} . $$ Note that this kernel produces a bounded operator on $L^2$; as I learned recently from fedja, one is supposed to use Schur test in these situations, and indeed this works great here with test function $(1+|x|)^{-p}$, $0<p\le 1$.

Since multiplicative constants don't matter here, we may as well use the modified kernel $K_1=1/(1+|x|+|y|)$ to estimate $Tf_n$. Then we can do the integral explicitly: $$ \int_n^{\infty}\frac{dy}{(1+x+y)y} = \frac{1}{1+x}\int_n^{\infty}\left( \frac{1}{y} - \frac{1}{1+x+y}\right) \, dy = \frac{1}{1+x}\, \log\left( 1+\frac{x+1}{n}\right) $$ We now see that even on $x\ge n$, this still has $L^2$ norm $\gtrsim 1/\sqrt{n}=\|f_n\|_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.