This is my first question here!
I need to obtain a bound like the following:
Let $P$ be an orthogonal projection and an integral operator defined as
$ Pf(x) = \int\mathop{dy}p(x,y)f(y) $
with $p(x,y)$ the integral kernel of $P$ to be a jointly continuous function and $p(\cdot, y) \in L^2(\Omega)$, where $\Omega \subseteq \mathbb{R}^2$. Then
$ \sup_y \sqrt{\int |p(x,y)|^2 \mathop{dx} } =: \sup_y \| p(\cdot, y) \|_{L^2}^2 \le const. $
I thought I could prove this by using a more general result like
$\|A\| = \sup_y \sqrt{\int |a(x,y)|^2 \mathop{dx}} = \sup_y \| a( \cdot, y) \|^2_{L^2} $ for $A$ a general integral operator with the same hypothesis for its integral kernel (to be jointly continuous and $L^2$-integrable in one variable), so in particular you have the $\ge$ and applying it to the projection case, where $\|P\|=1$, it should be easy to prove the original thesis.
Nevertheless I couldn't arrive much far. I tried to use Jensen Inequality and maybe done something to get to $\|A\| \le \sup_y \| a( \cdot, y) \|^2_{L^2} $ but couldn't find a function $f$ to get the equality or couldn't prove the other inequality to get the equality.
Thank you to everyone who tries!