Consider the linear operator $T:L^2(0,1;H) \rightarrow L^2(0,1;H)$ ($H$ is a Hilbert space) given by $$(Tf)(x)=\int_0^x K(x,y)f(y) \, dy,$$ where (C): $K(x,y) \in \mathcal{L}(H)$ such that $||K(x,y)||_{\mathcal{L}(H)} \leq C$ for every $x,y \in [0,1]$.
- What are ``the'' conditions on $K$ with respect to $x,y$ so that $Tf$ is well defined ?
- Under the conditions of the first point, is $T$ compact ?
- Under the conditions of the first point, is $Id-T$ invertible ?
When $H=\mathbb{R}$, if $K$ is measurable then $K \in L^\infty((0,1)\times(0,1))$ by (C) and it is well known that 2) holds.
My main concern is that we deal with vector-valued functions in the general case and I don't want to miss an argument. We need to:
a) check first that $y \in (0,x) \mapsto K(x,y)f(y) \in H$ is Bochner measurable for a.e. $x \in (0,1)$. It will then be clearly integrable thanks to (C) and the Cauchy-Schwarz inequality.
b) check that $x \in (0,1) \mapsto \int_0^x K(x,y)f(y) \, dy \in H$ is Bochner measurable. It will then be clearly in $L^{\infty}(0,1;H) \subset L^2(0,1;H)$ thanks to (C).
c) check that $T$ is compact and $Id-T$ is invertible. Is the proof exactly the same as for the case $H=\mathbb{R}$ or are there any subtleties ?
If someone has a classical book on this, it would be much appreciated.