The conditions for the problem I am currently tackling is as follows:
Let $T, M > 0$. Define :
$X_{T} := \{ u \in C^{1} ([0,T]) : ||u||_{T} \leq 2M \}$
where $||u||_{T} \ = \ ^{\text{sup}}_{t \in (0,T)} [|u(t)| +|u'(t)|]$ is the norm of $C^{1} ([0,T])$
Suppose that $f : \mathbb{R} \rightarrow \mathbb{R}$ belongs to $C^{1} (\mathbb{R})$ and that $f(0) = 0$.
For $a \in \mathbb{R}$ and $u \in C^{1} ([0,T])$ define:
$\psi(u)(t) \ = \ a + \int^{t}_{0} f(u(s)) \text{d}s$
Small note, not sure if this is relevant, but I have previously shown that $X_{T}$ as defined above is a complete metric space under the above norm.
The question for this problem is as follows:
Find $M,T > 0$ (or a suitable condition on $M,T > 0$) such that $\psi$ defines a map from $X_{T}$ to itself.
I have found a minimum value $M$ must take in terms of $a, T,$ and $u$ for a given $u \in X_{T}$ but I can't extend this to a general value that works for all $u$.
I would like to be able to find an upper bound for $^{\text{sup}}_{t \in (0,T)} |f(u(t))|$ over all $u \in X_{T}$ but does such a bound exist? I know there is such a bound for each $u$ individually, but I don't think our conditions on $f$ are strict enough that $f$ is bounded on $\mathbb{R}$. We would need $f$ to be bounded on $\mathbb{R}$ for this to work right?
Can anyone confirm that my hunch is correct, and if so suggest a different approach I might take? Thank you.
It turns out that my professor made a mistake in setting this question. It does not necessarily have a solution unless we are given a bound $|f| \leq 1$ on some $(−2M,2M)$