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](https://en.wikipedia.org/wiki/Bochner_measurable_function)?




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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.



Strong measurability of operator-valued map induced by a kernel
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.


