Problem on Contraction mapping Here is my question :

Define $T :C[0,1] \rightarrow C[0,1] $ by $T(x)(t) = 1+\int_0 ^t
x(s)ds$. Where the metric in $C[0,1]$ is defined as $d(f,g) = max_{x \in[0.1]}|f(x) - g(x)|$.

a) Is T a contraction?

b) Now we change the space to $C[0,\frac{1}{2}]$ . Is T a contraction? 

Any help will be appreciated.

Thank you.
for the first one answer is no. note that for $f=1$ we have $T(f)=1+t$ and for $g=0, T(g)=1$ so $d(f,g)=1=d(T(f),T(g))$ 

for the second one yes.
\begin{align}
d(T(f),T(g))&amp;=max_{x \in[0.1]}|1+\int_0 ^t
f(s)ds- (1+\int_0 ^t
g(s)ds)|\\
 &amp; =max_{x \in[0.1]}|\int_0 ^t
f(s)ds- 
g(s)ds)| \\ 
 &amp; \leq \frac{1}{2}max_{x \in[0.1]}|f(x)-g(x)| \\
 &amp; =\frac{1}{2}d(f,g)
\end{align}
$d(Tx(t),Ty(t))=\max_{t\in [0,\frac{1}{2}]}|T(x(t)-Ty(t)|=\max_{t\in [0,\frac{1}{2}]}|\int _0^t(x(s)-y(s)) ds|\leq$

$ \max_{t\in [0,\frac{1}{2}]}(\max_{s\in [0,\frac{1}{2}]} |x(s)-y(s)|)\int _0 ^tds=(\max_{s\in [0,\frac{1}{2}]} |x(s)-y(s)|)\max_{t\in [0,\frac{1}{2}]}t$

$\leq \frac{1}{2}(\max_{s\in [0,\frac{1}{2}]} |x(s)-y(s)|)&lt;\max_{s\in [0,\frac{1}{2}]} |x(s)-y(s)|=\max_{t\in [0,\frac{1}{2}]} |x(t)-y(t)|$

$=d(x(t),y(t))$