Proving that a function $f$ is a contraction Hello I am solving the following problem and could use some help.
Let (C[0,1],$d_\infty$) be the metric space of continuous functions on [0,1] where the distance function is defined by
Let $d_\infty(f,g)=\sup_{x∈[0,1]}|f(x)−g(x)|. $
Consider the function $T : (C[0, 1], d_\infty)\to (C[0, 1],d_\infty$) defined by
Let $(Tf)(x)=\int_0^xf(t)dt$
Prove:
(a) $T$ has a unique fixed point, i.e. there is a unique $f ∈ C[0,1]$ satisfies $Tf = f$ .
(b) $T^2$ is a contraction.
I know I'd like to use the function $f(x)=c*e^x$, but I don't know how ton put the proof together.
 A: For (a)
$f \in \mathcal C^1([0,1], \mathbb R) =V$ is a fixed point if and only if
$$f(x) = \int_0^x f(t) \ dt$$ for all $x \in [0,1]$. As $f$ is supposed to be continuous, it is also differentiable as $\int_0^x f(t) \ dt$ is differentiable. Therefore, you get the differential equation $$f^\prime(x) = f(x)$$ whose solutions are indeed $f(x) =ce^x$. You also have $f(0)=0$. Hence $c=0$ and $f(x) =0$ is the unique fixed point.
For (b)
For $f,g \in V$ and $x \in [0,1]$
$$\begin{aligned}
\vert (T^2(f) -  T^2(g))(x) \vert &= \left\vert \int_0^x \left(\int_0^t (f(t) - g(t)) \ dt\right) dx\right\vert\\
&\le  \int_0^x \left(\int_0^t \Vert f - g\Vert \ dt\right) dx\\
&\le \left\Vert f - g \right\Vert\int_0^x \left(\int_0^t  \ dt\right) dx\\
&= \left\Vert f - g \right\Vert\int_0^x t \  dt\\
&=\frac{1}{2} \left\Vert f-g \right\Vert
\end{aligned}$$ This implies that $$\Vert T^2(f) - T^2(g) \Vert \le \frac{1}{2} \Vert f - g \Vert$$proving that $T^2$ is indeed a contraction.
A: Answer for the second part: $|T^{2}f(x)|=|\int_0^{x}\int_u^{x}dt du|=|\int_0^{x}(x-u)f(u)du|\leq \|f\|\int_0^{x}f(u)du=\frac 1 2  \|f\|$.
