Pointwise convergence implies uniform convergence?

Question

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...

Answer 1

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 $\gtrsim 1/\sqrt{n}=\|f_n\|_2$.