Hi all: my issue is to approximate an integral operator with kernel $H\in L^1$, I found a proof but I didn't understand some points.
Let be $A$ an integral operatore in $L^2(Q)$, $Q\subset \mathbb{R}^n$ defined as
$Af=\int_Q H(x,y)f(y)dy $, with $H(x,y)=\frac{K(x,y)}{|x-y|^a}$.
I want to show it can be approximated (in norm) by
$H_M=\begin{cases}H(x,y) & |x-y| > M\\\frac{K(x,y)}{M^a} & |x-y| <M\end{cases}.$
I am interested to $a < n$ , so that $H$ is in $L^1$ but not in $L^2$.
If I do
$|(A-A_M)f|^2=\int_{|x-y_1| <M}\int_{|x-y_2| <M} \ [ H(x,y_1)- H_M(x,y_1)] \ [ H(x,y_2)- H_M(x,y_2)]f(y_1)f(y_2)dy_1 dy_2$
the notes I am following say:
$2|f(y_1)f(y_2)|\le |f(y_1)|^2 + |f(y_2)|^2$ and so
$|(A-A_M)f|^2=\int |H(x,y_1)-H_M(x,y_1)|dy_1 \int |H(x,y_2)-H_M(x,y_2)||f(y_2)|^2dy_2$
(A) why? where is $f(y_1)$???
furthermore it say $\int |H(x,y_1)-H_M(x,y_1)|dy_1 \le 2 \int |H(x,y_1)|dy_1 \le C$
(B) second ineq is because $H \in L^1$, but what about first inequality?
then, for any fixed $y$ I have,
$\int |H(x,y)-H_M(x,y)|dy \le 2 \int_{|x-y| <M} |K(x,y)|\frac{1}{|x-y|^a}dx \le C\frac{1}{M^{n-a}}$
(C) why $\frac{1}{M^{n-a}}$?
finally, integrated by $dx$ one should have
$\|(A-A_M)f\|^2 \le C \int |H(x,y_2)-H_M(x,y_2)||f(y_2)|^2dy_2 dx \le C \frac{1}{M^{n-a}}\|f\|^2$
and this is good, supposing (A-C) right
any help to understand that?