Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.