Strong measurability of operator-valued map induced by a kernel

Question

Let $K\in L^2([0,T]^2)$, and for each $t\in [0,T]$, let $\mathcal{T}_t $ be such that for all $f\in L^2([0,T])$, $\mathcal{T}_t f(s)=\int_0^T K(s,t)K(u,t)f(u)\, d u$ for all $s\in [0,T]$. One can show by the integrability of $K$ that $\mathcal{T}_t $ is a well-defined bounded linear operator acting on $L^2([0,T])$ into itself (i.e., $\mathcal{T}_t\in \mathcal{L}(L^2([0,T]))$). Is it possible to prove that the map $[0,T]\ni t\mapsto \mathcal{T}_t\in \mathcal{L}(L^2([0,T]))$ is strongly measurable?

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I observed that the map $t\mapsto \mathcal{T}_t $ can be decomposed into the map $t\in \mathcal{S}_t\in \mathcal{L}(L^2([0,T]);\mathbb{R})$ such that $\mathcal{S}_t f=\int_0^T K(u,t)f(u)\, d u$ for all $f\in L^2([0,T])$, and the map $t\mapsto K(\cdot, t)\in L^2([0,T])$. But I am not sure whether this observation helps the proof.

Answer 1

It helps to note that $\mathcal{T}_tf(s) = K(s,t)\langle f(\cdot), \overline{K}(\cdot,t)\rangle$.

If $K_n = \sum a_i1_{A_i\times B_i}$ is a finite linear combination of characteristic functions of rectangles then the map $\mathcal{T}_t^n: f \mapsto K_n(s,t)\langle f(\cdot),\overline{K}_n(\cdot, t)\rangle$ is measurable in $t$ with finite range. Find a sequence of such functions $(K_n)$ that converges in $L^2([0,T]^2)$ to $K$. Then for almost every $t$ we have $K_n(\cdot, t)\to K(\cdot, t)$ in $L^2[0,T]$, and from here an easy estimate shows that $\mathcal{T}_t^n f \to \mathcal{T}_tf$ in $L^2[0,T]$ for those values of $t$. So yes, $\mathcal{T}_t$ is Bochner measurable.