We have the integral operator $$ P:L^2(\Bbb R^n)\to\{\text{meas.functions}\;:\;\Bbb R^n\to\Bbb R\} $$ defined as $$ Tf(x):=\int_{\Bbb R^n}L(x,y)f(y)\,dy $$ where $L$ is a measurable function on $\Bbb R^n\times\Bbb R^n$ such that $$ \int_{\Bbb R^n}|L(x,y)|\,dx\le\beta\;\;\;\forall y\in\Bbb R^n\\ \int_{\Bbb R^n}|L(x,y)|\,dy\le\beta\;\;\;\forall x\in\Bbb R^n\;\;. $$ We know $T$ is bounded on $L^2$; what we have to prove is that the $L^2$ operator norm of $T$ is $\le C\beta$ for some constant $C>0$ (but maybe $C=1$).
I took $f\in L^2$ and I tried (using various Calculus theorems) \begin{align*} \|Tf\|_2^2 &=\int_{\Bbb R^n}\left|Tf(x)\right|^2\,dx\\ &=\int_{\Bbb R^n}\left|\int_{\Bbb R^n}L(x,y)f(y)\,dy\right|^2\,dx\\ &=\int_{\Bbb R^n}\left(\int_{\Bbb R^n}L(x,y)f(y)\,dy\right) \left(\int_{\Bbb R^n}L(x,z)f(z)\,dz\right)\,dx\\ &=\int_{\Bbb R^n}\int_{\Bbb R^n}\left(\int_{\Bbb R^n}L(x,y)L(x,z)\,dx\right)\,dx\,f(y)f(z)\,dydz\\ \end{align*}
but this didn't helped me.
I noticed that if $$ \int_{\Bbb R^n}\left|\int_{\Bbb R^n}L(x,y)f(y)\,dy\right|^2\,dx= \int_{\Bbb R^n}\left|\int_{\Bbb R^n}L(x,y)f(y)\,dx\right|^2\,dy\\ $$
I could conclude, but I don't know how to prove this.
I tried even to use the identity $$ \|Tf\|_p^p=p\int_0^{+\infty}r^{p-1}\lambda_{Tf}(r)\,dr $$ where $$ \lambda_{Tf}(r):=|\{x:|Tf(x)|>r\}| $$ but without success.
Can someone help me please?
SOLVED! \begin{align*} \|Tf\|_2^2 &=\int_{\Bbb R^n}\left|Tf(x)\right|^2\,dx\\ &=\int_{\Bbb R^n}\left|\int_{\Bbb R^n}L(x,y)f(y)\,dy\right|^2\,dx\\ &=\int_{\Bbb R^n}\left|\langle\sqrt{|L(x,\cdot)|},f(\cdot)\sqrt{|L(x,\cdot)|}\rangle\right|^2 dx\\ &\le \int_{\Bbb R^n} \left|\langle\sqrt{|L(x,\cdot)|},\sqrt{|L(x,\cdot)|}\rangle\right|\left| \langle f(\cdot)\sqrt{|L(x,\cdot)|},f(\cdot)\sqrt{|L(x,\cdot)|}\rangle\right| dx\\ &=\int_{\Bbb R^n}\left(\int_{\Bbb R^n}|L(x,y)|\,dy\right)\left(\int_{\Bbb R^n}|L(x,y)||(f(y))|^2\,dy\right)\,dx \end{align*} from which we can easily conclude.
The inequality follows from the Cauchy-Schwarz inequality.