Banach fixed-point theorem and  the map $Tf(x)=\int_0^x f(s)ds $ on $C[0,1]$

First of all recall that for $f\in C[0,1]$ 
\[\| f\| =\sup_{x\in [0,1]} |f|.\]
i) Let $f(x)=0$ and $g(x)=1$. Then
\[\|Tf(x)-Tg(x) \|= \| 0-x\|=1,\]
and 
\[\| f(x)-g(x)\|= \|0-1\|=1.\]
Hence
\[\|Tf(x)-Tg(x) \| \leq \alpha \| f(x)-g(x)\|\]
does not hold for any $0<\alpha <1$. 

ii) Suppose $f(x) \in C[0,1]$ is a fixed point of $T$. Then
\[(1)    Tf(x)=\int_0^{x}f(s)ds=f(x)     \forall x\in [0,1].\]
Substituting $x=0$ in (1) we get $f(0)=0$. Taking a derivative with respect to $x$ in (1) we get 
\[f(x)=f'(x)   \forall x\in [0,1],\]
and hence $f(x)=ce^{x}$. Sinnce $f(0)=0=ce^{0}$. Thus $c=0$ and
$$f(x)\equiv 0,$$
is the only fixed point of $T$. 

(iii) Let $f,g \in C[0,1]$. Then
\[ \|T^2 f -T^2g \|=\sup_{x\in [0,1]} |\int_0^x\int_0^s (f(t)-g(t))dt ds|\]
\[\leq \sup_{x\in [0,1]} |f-g|   \|\int_0^x\int_0^s dt ds|\| = \| f-g\| \| \frac{x^2}{2}\|\] 
\[=\frac{1}{2}\| f-g\|  .\]
Thus 
\[\|T^2 f -T^2g \| \leq \frac{1}{2}\| f-g\| \]
and $T^2$ is a contraction.