Here comes the context: Let $H_1,H_2$ be Hilbert spaces, $\mathbb{T}$ the unit circle, $M_z$ the operator of multiplication by $z$ on $L^2(\mathbb{T},H_1)$ or $L^2(\mathbb{T},H_2)$ and $T : L^2(\mathbb{T},H_1) \rightarrow L^2(\mathbb{T},H_2)$ a bounded operator such that $M_zT=TM_z$.
I would like to prove that $T=M_F$, more precisely that $T$ is a ''multiplication'' operator for a function $F \in L^{\infty}(\mathbb{T},B(H_1,H_2))$, defined as \begin{equation} M_F: L^2(\mathbb{T},H_1) \rightarrow L^2(\mathbb{T},H_2), h \mapsto (z \mapsto F(z)(h(z)). \end{equation} (I already know that $M_F$ is a well defined, bounded operator). I have seen proofs for the scalar case on mathstackexchange, but none for the vector valued one.
Help is appreciated, Thanks in advance!
Let $A: X\to Y$ be an operator between two reflexiv spaces. By the "matrix" of $A$ denote the map $$Y^*\times X\to \Bbb C,\qquad(\xi, v)\mapsto \xi(A(v)).$$ Note that this map is linear in each argument (ie it is a bilinear map) and continuous in the first argument. Further a bilinear "matrix" map continuous in the first argument uniquely determines a linear operator (here reflexivity is used). Further note that the "matrix" map is jointly continuous iff the operator $A$ is continuous and that two operators agree iff their "matrices" agree.
Let $(\xi,v)$ be a pair in $H_2^*\times H_1$ and $f\in L^2(\Bbb T, \Bbb C)$. Define $$T_{(\xi, v)}(f) = \xi(T(f\cdot v)).$$ Note that $T_{(\xi,v)}$ is a continous linear operator $L^2(\Bbb T, \Bbb C)\to L^2(\Bbb T, \Bbb C)$ commuting with $M_z$ and as such must be equal to multiplication with a function $F_{(\xi,v)} \in L^\infty(\Bbb T,\Bbb Z)$ by the result that you have already seen.
Make liberal use of measure theory to see that since $T_{(\xi_n, v)}\to T_{(\xi,v)}$ if $\xi_n\to \xi$ that then $F_{(\xi_n, v)}\to F_{(\xi,v)}$ in $L^\infty(\Bbb T, \Bbb C)$ and that $F_{(\xi,v)}(z)$ is bilinear almost everywhere. Hence $F_{(\xi,v)}(z)$ defines a linear map $F(z):H_1\to H_2$ almost everywhere for which you have
$$\xi(T( f \cdot v ))\,[z] = T_{(\xi,v)}(f)\,[z]= F_{(\xi, v)}(z)\cdot f(z)= \xi(F(z)[v])\,f(z).$$ Yielding that $T(f\cdot v) = F(z)[v]\,f(z)$. Now use continuity of $T$ and measure theory to check that the linear map $F(z)$ must almost everywhere be a continuous linear operator $H_1\to H_2$.