Let $(X,\Sigma_\mu,d\mu)$ and $(Y,\Sigma_\nu,d\nu)$ be two positive $\sigma$-finite measure space and let $M(d\mu)$ and $M(d\nu)$ be spaces of complex-valued $d\mu$-measurable and $d\nu$-measurable functions. Consider linear integral operator of the form $$\mathcal{K}u(y) = \int k(x,y)u(x)d\mu(x),$$ where $k$ is complex measurable function (with respect to the product $\sigma$-algebra).
I would like to show that $\|\mathcal{K}\|_{B(L^p,L^\infty)}=\|k\|_{L^\infty(L^{p^*})}$, where $B(L^p,L^\infty)$ is the space of boudned operators from $L^p$ to $L^\infty$, $\|.\|_{L^\infty(L^{p^*})}$ denotes iterated norm, and $p^{-1}+p^{^*-1}=1$. The bound $\|\mathcal{K}\|_{B(L^p,L^\infty)}\leq \|k\|_{L^\infty(L^{p^*})}$ follows from Holder's inequality.
But I don't see how to show the reverse inequality. In fact, I'm interested in a simpler case, when $X$ and $Y$ are some bounded and connected subsets of $\mathbb{R}^d$ spaces and $p=p^*=2$, but this does not help me to find a way around it.
Using
Theorem (Hölder's inequality). Let (S, Σ, μ) be a measure space and let $p, q ∈ [1, ∞]$ with $1/p + 1/q = 1.$ Then, for all measurable real- or complex-valued functions f and g on S, ${\displaystyle \|fg\|_{1}\leq \|f\|_{p}\|g\|_{q}.} \|fg\|_{1}\leq \|f\|_{p}\|g\|_{q}.$
you can see that $$|\mathcal{K}u(y)| = |\int k(x,y)u(x)d\mu(x)|\leq \|k(x,y)\|\|u(x)\| \implies (operator \ norm \ of \ \mathcal{K} \leq\|k(.,y)\| $$ where norms are understood, and note that norm of $k(x,y)$ is for fixed $y$, since we integrate with respect to $x$.
Then you take maximum of these norm as $y$ varies. It is looking for supremum of real numbers, so $\|.\|_\infty$ of norm $\|.\|_p$'s is calculated for $k(x,y).$ So, the end result is $\|k(x,y)\|_{L^\infty(L^p)}$.