Let $C([a,b])$ be equipped with the essential supremum, and $a, b, t_{0} \in \mathbb R$ where $a < t_{0} < b, h \in C([a,b])$ and $A: C([a,b]) \to C([a,b])$ defined as $(Af)(t):=\int_{t_{0}}^{t} h(s)f(s)\,ds$ for all $t \in [a,b].$
We are supposed to use the fact that $I-A$ is invertible and has bounded inverse $\displaystyle\sum_{n=0}^{\infty}A^{n}.$
Show that for any $x_{0}\in \mathbb R$ and $g \in C([a,b])$ there exists a unique solution $f \in C^{1}([a,b])$ of the initial value problem:
$$f'(t)=h(t)f(t) + g(t), t \in [a,b] \quad\text{and}\quad f(t_{0})=x_{0}.$$
My steps: Let $g \in C([a,b])$ and $x_{0} \in \mathbb R$ be arbitrary.
Since $I-A$ is invertible, there exists only one $f \in C([a,b])$ so that $(I-A)f=g\Rightarrow f(t)=Af(t)+g(t),$ where $t \in [a,b].$
How do I show that $f \in C^{1}([a,b]),$ and further how is it I've only got $f(t)=\int_{t_{0}}^{t} h(s)f(s)\,ds+g(t)$ and not $f(t)=\int_{a}^{t} h(s)f(s)\,ds+g(t)?$
$t_{0}$ is not necessarily equal to $a.$
So what must I do? This is the first kind of problem I have encountered on this topic.