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... 



Pointwise convergence implies uniform convergence?
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](https://mathoverflow.net/questions/272085/boundedness-of-integral-operator?rq=1) from fedja, one is supposed to use [Schur test](https://en.wikipedia.org/wiki/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 $\gtrsim 1/\sqrt{n}=\|f\_n\|\_2$.


