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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 $L^2([0,T];\mathbb{R})$ 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?

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.

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?

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.

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 $L^2([0,T];\mathbb{R})$ 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?

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 $L^2([0,T];\mathbb{R})$ 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?

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.

Let $K\in L^2([0,T]^2)$, and for each $t\in [0,T]$, let $\mathcal{T}_t \in \mathcal{L}(L^2([0,T]))$ be such that for all $f\in L^2([0,T])$, $\mathcal{T}_t f(s)=\int_0^T K(s,t)K(u,t)f(u)\, d u$. One can show by the integrability of $K$ that $\mathcal{T}_t $ is well-defined. 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?

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 $L^2([0,T];\mathbb{R})$ 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?