# Integral operators with operator valued kernels

This is the definition for integral operators I know:

Let $\Omega \subset \mathbb{R}^n$ and $D \subset \mathbb{R}^n$. Let $K : \Omega \times D \to \mathbb{C}$ be measurable. A linear operator $T: A \to B$ where $A, B$ are function spaces is an integral operator iff

$(Tf)(x) = \int_\Omega K(t,x)f(t)dt$

In a book I just read, I saw a kernel defined by

$K : \mathbb{R}^2 \to \mathscr{B}(X)$, $K(t,s)=B_1(t)B_2(s)$

where $\mathscr{B}(X)$ is the space of bounded linear operators on a Hilbert space $X$ and $B_1, B_2: \mathbb{R} \to \mathscr{B}(X)$.

So this kernel is not complex valued but operator valued. The book just gave the kernel and the domain of the integral operator (namely $L^2(\mathbb{R},X)$) but what does the integral operator look like? Maybe:

$(Tf)(x) = \int_\mathbb{R} B_1(t) B_2(s) f(t) dt$

Why can I be sure to integrate over t and not s? Can anyone point me to a definition (in some book) for integral operators with operator valued kernels?

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