Proving that a function is not a contraction I am solving the following problem and its parts.
Let (C[0,1],$d_\infty$) be the metric space of continuous functions on [0,1] where the distance function is defined by
$d_\infty(f,g)=\sup_{x∈[0,1]}|f(x)−g(x)|. $
Let $T : (C[0, 1], d_\infty)\to (C[0, 1],d_\infty$) be defined by
$(Tf)(x)=\int_0^xf(t)dt$
Prove that:
T is not a contraction, i.e. there does not exist 0 < K <    1 such that $d_\infty(T f, T g)\leq K · d_\infty(f, g)$
holds for any $f,g ∈ C[0,1]$.
Ive tried an example but I haven't really got anywhere. My work is as follows:
$$f'(t)=\int_0^xtdt=\frac{x^2}{2}=F(x)-F(0)$$
so we pick $t$ & $t^2$ such that
$d$($t$,$t^2$)=sup[$t$-$t^2$] with max value = $\frac{1}{4}$
and $d$(T$t$,T$t^2$)=sup[$\frac{t^2}{2}-\frac{t^3}{3}$] with max value =.167
 A: Take $f \equiv 1$ and $g\equiv 0$. Then,
$$d_\infty(Tf,Tg) = 1 = d_\infty(f,g).$$
A: Let $E:=\mathcal{C}^0([a,b],d_\infty)$ for any $-\infty<a<b<+\infty$. Then $(E,d_\infty)$ is a Banach space, so assuming by contradiction that $T$ is a contraction, there exists an unique fixed point $f$ for $T$ by Banach's fixed point theorem.
Take $\alpha>0$ and let $g_\alpha\in E$ be defined by $g_\alpha(x):=\mathrm{e}^{\alpha x}$. Then, as $Tf=f$, we get:
$$d_\infty(f-Tg_\alpha)=d_\infty(Tf-Tg_\alpha)<d_\infty(f-g_\alpha).$$
By continuity in $\alpha>0$ of the LHS and the RHS above, and by boundedness of $f$ over $[0,1]$, we have
$$d_\infty(f-Tg_\alpha)=\sup_{x\in[0,1]}\left|f(x)-\frac{\mathrm{e}^{\alpha x}}{\alpha}\right|\underset{\alpha\to0^+}{\longrightarrow}+\infty$$
while
$$d_\infty(f-g_\alpha)=\sup_{x\in[0,1]}\left|f(x)-\mathrm{e}^{\alpha x}\right|\underset{\alpha\to0^+}{\longrightarrow}\sup_{x\in[0,1]}\left|f(x)-1\right|<+\infty,$$
a contradiction.
The conclusion is therefore that $T$ can not be a contraction on $E$.
EDIT: In fact, $T$ has an unique fixed point which is the trivial function $f=0_E$, see there: Why $f(x)=e^x$ is a fixed point of $T$?
The above argument can be adapted to situations where the fixed point, if it exists, is not trivial.
