Operator equation : $f-Af=g$ 
Problem :
Let $E=\left(\mathcal{C}[0,1],\|.\|_{\infty}\right)$
Defined operator $A$ :
$$A ~~:E\to~~E$$
$$Af(x)=\displaystyle\int\limits_{0}^{x}f(t)dt$$

*

*show that equation $f-Af=g$ has a solution for all $g\in E$ .

Then search operator $(I-A)^{-1}$

I was find $\|A\|=1$ but this dose not implies $(I-A)^{-1}$ exists
$\color{red}{note}~$:
iff $\|A\|<1\implies \left(I-A\right)^{-1}$ exists and $\left(I-A\right)^{-1}=\displaystyle\sum\limits_{n=1}^{\infty}A^{n}$
But I can't use it here ??
Because i see :
$$f-Af=g\implies (I-A)f=g$$
Equation has a solution if and onaly iff $\left(I-A\right)^{-1}$ exists
Can you explain to me where my mistake is?

Thanks
 A: For $n \in \mathbb{N_0}$ and $f \in E$ we have
$$
|A^{n+1}f(x)| = \left| \int_0^x \int_0^{t_1} \dots \int_0^{t_n} f(s) ds dt_n \dots dt_1\right| 
$$
$$
\le \left(\int_0^x \int_0^{t_1} \dots \int_0^{t_n} 1 dsdt_n \dots dt_1 \right) \|f\|_\infty =\frac{x^{n+1}}{(n+1)!}\|f\|_\infty.
$$
Thus $\|A^{n+1}f\|_\infty \le \frac{1}{(n+1)!}\|f\|_\infty$, and we have
$\|A^{n+1}\| \le \frac{1}{(n+1)!}$. In fact $\|A^{n+1}\| = \frac{1}{(n+1)!}$ (check $f=1$). Thus $\sum_{n=0}^\infty A^n$ is (absolut) convergent and equals $(I-A)^{-1}$.
A: Here is a solution without Neumann's series, let me know if something is left unclear, or if I made mistake.

*

*The idea is to build $(\mathrm{I}-A)^{-1}$ on
$\mathcal{C}^0([0,1],\mathbb{C})$ by density of
$\mathcal{C}^1([0,1],\mathbb{C})$ in it.
Hence, let assume first that $g\in \mathcal{C}^1([0,1],\mathbb{C})$,
and consider the problem $$f(x)-Af(x) = g(x),\quad x\in[0,1].$$
It can be given more explicitly as $$f(x)-\int_0^x f(y) \mathrm{d} y
= g(x),\quad x\in[0,1],$$ so that if it admits a solution we have necessarily $f(0)=g(0)$.
Moreover, since $g\in \mathcal{C}^1([0,1],\mathbb{C})$ and $x\mapsto
\int_0^x f(y) \mathrm{d} y \in \mathcal{C}^1([0,1],\mathbb{C})$ (by
the Fundamental Theorem of Calculus), we deduce that any solution
$f$ should belong to $\mathcal{C}^1([0,1],\mathbb{C})$.
Therefore, one may derive to obtain the following Cauchy initial
value problem
\begin{align}\tag{CP}\label{CP}
    \left\{\begin{array}{rl}
            \frac{\mathrm{d}f}{\mathrm{d}x} (x) - f(x)  =& g'(x) \,\text{, } 0 \leqslant x\leqslant 1 \text{, }\\
            f(0) =& g(0)\text{ . }
    \end{array}
    \right.\text{ ,} \end{align}
Existence of uniqueness of solution for such kind of problem is a
direct consequence of Cauchy-Lipschitz Theorem, and notice also that integrating both sides prove that $f$ is a solution of (CP) iff $f-Af = g$.
Duhamel's formula yields :
$$ f(x) = g(0)e^x + \int_{0}^x e^{x-y}g'(y) \mathrm{d}y ,\quad
x\in[0,1].$$
Now, integration by parts shows that, for all $x\in[0,1]$,
\begin{align} f(x) &= g(0)e^{x} + [e^{x-\cdot} g(\cdot)]_{0}^{x} -
\int_{0}^x e^{x-y}g(y) \mathrm{d}y\\ &= g(x) - \int_{0}^x
e^{x-y}g(y) \mathrm{d}y \text{ . } \end{align}
Thus, we can set for all $g\in \mathcal{C}^0([0,1],\mathbb{C})$,
all $x\in[0,1]$
$$Tg(x) := g(x) - \int_{0}^x e^{x-y}g(y) \mathrm{d}y \text{ . } $$
It is not difficult to see that there exists a constant $C>0$
independent of $g$ such that,
$$\lVert Tg \rVert_\infty \leqslant C \lVert g
\rVert_\infty\text{.}$$


*Now, it remains to prove that in fact $T=(\mathrm{I}-A)^{-1}$, i.e. $T-AT = \mathrm{I}$. We are going to argue by density of $\mathcal{C}^1([0,1],\mathbb{C})$ in $\mathcal{C}^0([0,1],\mathbb{C})$ and using above step : for fixed $g\in \mathcal{C}^0([0,1],\mathbb{C})$, we consider $(g_n)_{n\in\mathbb{N}}\subset \mathcal{C}^1([0,1],\mathbb{C})$, such that
$$ \lVert g -  g_n
\rVert_\infty \underset{n\rightarrow + \infty}{\longrightarrow} 0 \text{ . }$$
Finally, by continuity of $A$ and $T$ on $\mathcal{C}^0([0,1],\mathbb{C})$,
$$ Tg - ATg = \lim_{n\rightarrow +\infty} Tg_n - ATg_n = \lim_{n\rightarrow +\infty} g_n =g .$$
