How can I show that $\lvert \lvert A\rvert \rvert< 1$ in order to prove that $Id-A$ is invertible For a Banach space $X$, I was able to prove that
for $T \in BL(X)$ with $\lvert \lvert T\rvert \rvert < 1$ then $(Id-T)$ is invertible in $BL(X)$.
Defining $A: C([a,b])\to C([a,b])$ where $(Af)(t)=\int\limits_{t_{0}}^{t}ds h(s)f(s)$ where  $C([a,b])$ is equipped with $\lvert \lvert \cdot \rvert \rvert_{\infty}$ and $h \in C([a,b])$, then I also proved that for any $n \in \mathbb N$:
$$\lvert\lvert A^{n}\rvert\rvert \leq \frac{\lvert b-a\rvert^{n}}{n!}\lvert \lvert h\rvert \rvert_{\infty}(*)$$
Question: For any $G \in C([a,b])$ there is a unique $f \in C([a,b])$ such that $(Id-A)f=G$.
This is clearly a question of proving that $\lvert \lvert A\rvert \rvert < 1$ so that I can then use the previous theorem that $(Id-A)$ is invertible. My problem is I cannot seem to find why $\lvert \lvert T\rvert \rvert < 1$ and I do not see how $(*)$ helps me. Any ideas/hints?
 A: You know the Neumann series $\sum_{i=0}^\infty T^i$ of $T$. It is well-known that if the Neumann series of $T$ converges, then  $Id – T$ is invertible and its inverse is given by the Neumann series of $T$.
The Neumann series certainly converges for $\lVert T \rVert < 1$. However, it also converges if  $\lVert T^n \rVert < 1$ for some $n$. To see this, note that
$$(Id-T)(\sum_{i=0}^{n-1}T^i) = Id -T^n , \tag{1}$$
$$(\sum_{i=0}^{n-1}T^i)(Id-T) = Id -T^n . \tag{2}
$$
Since $Id-T^n$ is invertible in $BL(X)$, it is bijective. Thus (1) implies that $Id-T$ is surjective and (2) that $Id-T$ is injective. Hence $Id-T$ is bijective. This doesn't automatically mean that the algebraic inverse $(Id-T)^{-1} \in L(X)$ is bounded. However, we have $S= (\sum_{i=0}^{n-1}T^i) (Id- T^n)^{-1} \in BL(X)$ and by (1)
$$
(Id-T)S = (Id -T^n)(Id -T^n)^{-1} = Id \tag{3}
$$
which shows that
$$(Id-T)^{-1} = S \in BL(X) . \tag{4}$$
Now you see why $(*)$ helps: $\frac{\lvert b-a\rvert^{n}}{n!}$ becomes arbitrarily small, hence $\lVert A^n \rVert \le \frac{\lvert b-a\rvert^{n}}{n!}\lVert h \rVert_{\infty} < 1$ for large enough $n$.
By the way, you can also show directly via $(*)$ that the Neumann series of $A$ converges. In fact,
$$\left\lVert \sum_{i=n}^m  A^i \right\rVert \le  \sum_{i=n}^m  \lVert A^i \rVert \le \sum_{i=n}^m \frac{\lvert b-a\rvert^{i}}{i!}\lVert h \rVert_{\infty} = \left(\sum_{i=n}^m \frac{\lvert b-a\rvert^{i}}{i!}\right) \lVert h \rVert_{\infty} . \tag{5}$$
But $\sum_{i=n}^m \frac{\lvert b-a\rvert^{i}}{i!}$ is a section of the convergent series $e^{\lvert b-a\rvert} = \sum_{i=0}^\infty \frac{\lvert b-a\rvert^{i}}{i!}$, thus becomes arbitrarily small for large enough $n$.
A: An overkill solution would be to use the Gelfand formula, which states that
$$
\rho(A) =\lim_n \|A^n\|^{1/n}
$$
where the spectral radius $\rho(A)$ is defined as the supremum of the absolute values of all $x$ such that
$$
x\text{Id}-A
$$
does not have a bounded inverse. In particular, for your $A$, the spectral radius vanishes, as per your bound.
Note: if your operator were diagonalizable, then you'd have $\|A^n\|^{1/n}=\|A\|=\rho(A)$. Non exponential decay of the operator norm shows your operator is not diagonalizable. The statement above says that, even more, it does not have any eigenvalue at all.
