I came across this exercise and I am really not sure how to solve it:
1) Prove that for $t_0 > 0$ $A = \{ f \in C[0,t_0]: f(s) \in [0,2] \forall s \in [0,t_0]\}$ is closed in $(C[0,t_0],\|.\|_{\infty})$. I have no problems with that.
2) Find a suitable $t_0 > 0$ such that the map $T: C[0,t_0] \to C[0,t_0]$:
$T(f)(t) = 1 + \int_0^t exp(f(s)) ds$
is a contraction mapping on $A$ int the norm $\|.\|_{\infty}$.
What I understand is that I need to find $t_0$ such that $\forall f,g \in C[0,t_0]$, $\| T(f) - T(g)\|_{\infty} \leq K\|f-g\|_{\infty}$ for some $K \in (0,1)$, but all I managed to get is $| T(f)(t) - T(g)(t)| \leq t\|exp(f)-exp(g)\|_{\infty}$. Can someone help me from there please?
Question: Are your $f,g\in A$? if so you can use the following: What you need is that $exp$ is lipschitz on $[0,2]$, which can be seen by the mean value theorem: $$x,y\in[0,2]: |exp(x)-exp(y)|=|exp(\xi)||x-y|\leq exp(2)|x-y|,$$ since $\xi\in[0,2]$ and $exp$ is monotone. Then you can use this inequality in what you already have: $$t\|exp(f)−exp(g)\|_\infty\leq t \cdot exp(2)\|f-g\|_\infty.$$ Now you just have to choose your $t$ accordingly.
If this is not the case, you may want to check out the proof of the Picard-Lindelöf Theorem.