invertibility of integral operators on the half line Let $Tu=\int_0^\infty u(x)k(x,y)dx$ be an integral operator with kernel $k$. Note that the domain of integration is the half line.   Is there a general theory for invertibility of the operator $I-T$, say on $L^2(\mathbb{R}_+)$, depending on properties of $k$?  Here $I$ is the identity operator.
 A: So, $T:L^2\left(\mathbb R_+ \right) \to L^2\left(\mathbb R_+ \right)$.
A natural setup in this case is Hilbert-Schmidt Operators, that is,
when
$$
\int_{\mathbb R_+ \times \mathbb R_+}  k^2(x,y) dxdy <\infty
$$
In that case,
\begin{eqnarray*}
\left\|Tf \right\|_{L^2( \mathbb R_+)}^2&=& \int_{\mathbb R_+ } (Tf(x))^2 dx \\
&=&\int_{\mathbb R_+ } \left(\int_{\mathbb R_+ } k(x,y) f(y) dy \right)^2 dx \\
&\leq & \int_{\mathbb R_+ \times \mathbb R_+}  k^2(x,y) dxdy \int_{\mathbb R_+ } f(y)^2 dy dx \leq \left\|k \right\|_{L^2(\mathbb R_+ \times \mathbb R_+)}^2 \left\|f \right\|_{L^2( \mathbb R_+)}^2
\end{eqnarray*}
So $T$ is a bounded operator. It is also a compact operator. Thus, the invertibility of $I-T$ is "maybe, maybe not" in the following sense :
There exists a most finitely many linearly independent functions $f$ such that
$$
f = Tf.
$$
There are all sorts of cases when the existence of such eigenvalues can be excluded. The simplest one is when $$\left\|k \right\|_{L^2(\mathbb R_+ \times \mathbb R_+)}<1.$$
In that case,
$$
(I-T)^{-1} = \sum_{n\geq0} T^n,
$$
where $T^n$ means $T$ composed $n$ times with itself.
