Consider the space $C([0,1])$ with $\sup$ norm and define $T: C([0,1]) \to C([0,1])$ by $(T(x))(t) = (1-t)x(t) + t \sin \frac{1}{t}$.
Define $B$ the unit closed ball of $C([0,1])$. Show that
$(i)$ $T(B) \subseteq B$
$(ii)$ $\forall x,y \in B, d(T(x),T(y)) \leq d(x,y)$ but $T$ doesn't have fixed point.
My attempt on $(i)$: Let $x \in B$. Then $(T(x)(t)) = (1-t)x(t) + t \sin \frac{1}{t} $ and I need to prove $\sup_{t \in [0,1]} \{ |(T(x))(t)| \} \leq 1$.
$ \sup_{t \in [0,1]} \{ |(T(x))(t)| \} = \sup_{t \in [0,1]} \{ |(1-t)x(t) + t \sin \frac{1}{t}| \} \leq \sup_{t \in [0,1]} \{ |(1-t)x(t)| \} + \sup_{t \in [0,1]} \{ t \sin \frac{1}{t}| \} \leq \sup_{t \in [0,1]} \{ |(x(t)| \} + \sup_{t \in [0,1]} \{ |tx(t)| \} + \sup_{t \in [0,1]} \{ t \sin \frac{1}{t}| \} $
But $\sup_{t \in [0,1]} \{ t \sin \frac{1}{t}| \}$ is already 1, then this wouldn't work...
My attempt on $(ii)$ $d(T(x),T(y)) = d((1-t)x(t) + t \sin \frac{1}{t},(1-t)y(t) + t \sin \frac{1}{t}) = \| (1-t)x(t) + t \sin \frac{1}{t} - (1-t)y(t) - t \sin \frac{1}{t} \| = \| (1-t)(x(t)-y(t)) \| = \| x(t) - y(t) -tx(t) + ty(t) \| \leq \|x(t)-y(t) \| + \|t(y(t)-x(t)) \| \leq \|x(t)-y(t) \| + |t|\|(y(t)-x(t)) \| $
but that would be still bigger than what I need to get:
$d(x,y) = \| x - y \| = \sup_{t \in [0,1]} \{ x(t)-y(t) \}$
Any help would be appreciated. Thanks.
Note that $|t\sin1/t|<1$ on $[0,1] $. In (i), you have $$|(1-t)x (t)|+|t\sin1/t|<(1-t)+t=1. $$ In (2), $$|(1-t)(x (t)-y (t)=(1-t)\,|x (t)-y (t)|\leq|x (t)-y (t)|. $$