I am currently working trough the chapter 2.10 in the book "Spectral Theory of Self-Adjoint Operators in Hilbert Space" (https://link.springer.com/book/10.1007%2F978-94-009-4586-9).
This chapter works with integral operators of the form
$(Tu)(x)=\int_Yt(x,y)u(y)\mathrm{d}\mu(y)$
where first the following definition is introduced:
If the kernel $t(x,y)$ satisfies the two following conditions
1.The integrand $(Tu,v)=\int_{Y\times Y}t(x,y)u(y)\overline{v(x)}\mathrm{d}\mu(x)\mathrm{d}\mu(y)<\infty$ is summable for all $u,v\in H$
2.The sesqui-linear form defined in the integral above is bounded.
then $T$ is called a regular integral operator.
$\underline{\mathrm{Question\,1}}$
Is condition 2 not just a weaker form of condition 1? Because $||\Phi||=||T||=\sup_{||x||, ||y||=1}|\Phi(x,y)|=\sup_{||x||, ||y||=1}|(Tx,y)|<\infty$ is just a special case of condition 1.
$\underline{\mathrm{Question\,2}}$
In the Proof of Theorem 1:
Let H=$L_2(Y,\mu)$ and let $t(x,y)$ satisfy
$a_1=\sup_y\int_Y|t(x,y)|\mathrm{d}\mu(x)<\infty$ and $a_2=\sup_x\int_Y|t(x,y)|\mathrm{d}\mu(y)<\infty$,
then T is an regular integral operator and $||T||^2\leq a_1a_2$.
the following step occurs
$|(Tu,v)|^2 = |\int_Y \int_Y t(x,y)u(y) \mathrm{d}\mu(y) \overline{v(x)} \mathrm{d}\mu(x)|^2$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\leq \int_{Y \times Y} |t(x,y)|\cdot|u(y)|^2 \mathrm{d}\mu(y)\mathrm{d}\mu(x) \cdot \int_{Y \times Y} |t(x,y)|\cdot|v(y)|^2 \mathrm{d}\mu(y)\mathrm{d}\mu(x)$.
The author sais the Cauchy inequality has been used but I don't see how.
If I use Cauchy the following expression emerges
$|(Tu,v)|^2\leq ||Tu||^2\cdot ||v||^2$. Here the kernel $t(x,y)$ only appears in one of the Integrals not both of them.
I was able to proof that $||T||^2\leq a_1a_2$ in a different way however, it is necessary for me to understand the "trick" (as the author himself calls it on page 54 under equation 11) that is used, since most of the proofs in this chapter use the same one.
The secound problem I have is that I do not see how the $T$ is regular, as Theorem 1 only shows $||T||^2\leq a_1a_2$ which fulfills only the secound part of the definition "regular" but not the first one. But this problem should be closely related to question 1.
Thank you very much in advance! Henrie
You say nothing about what $Y$ nor $H$ are, but it is probably the case that $Y$ is a measure space, and $H=L^2(Y)$.
Condition 2 is stronger than condition 1. They are equivalent, though, because if you start with 1 you can get to 2 via the Uniform Boundedness Principle.
As for Cauchy-Schwarz, let $S$ be the operator with kernel $|t(x,y)|^{1/2}$. Then \begin{align} |(Tu,v)|^2& = \Big|\int_Y \int_Y t(x,y)u(y) \mathrm{d}\mu(y) \overline{v(x)} \mathrm{d}\mu(x)\Big|^2\\[0.3cm] &\leq\bigg[ \int_Y \int_Y |t(x,y)|\, |u(y)|\, |v(x)|\,\mathrm{d}\mu(y) \mathrm{d}\mu(x)\bigg]^2\\[0.3cm] &=\bigg[ \int_Y \int_Y |t(x,y)|^{1/2}\, |u(y)|\,\ |t(x,y)|^{1/2}|v(x)|\,\mathrm{d}\mu(y) \mathrm{d}\mu(x)\bigg]^2\\[0.3cm] &\leq\bigg[ \int_Y \int_Y |t(x,y)|\, |u(y)|^2\,\,\mathrm{d}\mu(y) \mathrm{d}\mu(x)\bigg]\bigg[ \int_Y \int_Y |t(x,y)|\, |v(x)|^2\,\,\mathrm{d}\mu(y) \mathrm{d}\mu(x)\bigg].\\[0.3cm] \end{align}