Let $F : [0,a] \times [b-r,b+r] \rightarrow \mathbb{R}$, $a, b, r$ fixed real numbers, satisfying
(1) for each $k \in [b-r,b+r]$, $g_k : [0,a] \rightarrow \mathbb{R}$ defined by $$g_k(t) = F(t,k)$$ is continuous,
(2) for each $m \in [0,a],$ $$h_m(t) = F(m,t)$$ is continuous
(3) there exists $M > 0$ such that $$|F(s,t)| \leq M$$ for all $s,t.$
If $aM \leq r$, then the initial value problem $$y' = F(u,y(u))$$ with $u \in [0,a]$ and $y(0) = y_0$ has a solution.
$\textbf{Hint :}$ Let $S = \{f \in C[0,a] : |f(t) - y_0| \leq r, t \in [0,a]\}.$ Define $T : S \rightarrow S$ by $$T(f)(x) = y_0 + \int_0^x F(s,f(s)) ds.$$
$\textbf{Attemp}$ Follow that hint, I defined that $T$. I saw one similar problem, and usual trick is to use Banach Fixed point theorem. So that is my plan.
Step 1 : $X$ is complete.
Since $X = \overline{B(y_0,r)}$ a closed ball in $C[0,a]$, it is complete.
Step 2 : $T$ is well-defined ($T(f) \in X$) Seem like the boundedness of $F$ and $aM \leq r$ makes $T$ well-defined (I can already show it.)
So what left is to show that $T$ is a contraction (i.e. there is $C < 1$ such that $||T(f) - T(g)||_\infty \leq C ||f-g||_\infty.$)
Ususally, $T$ has to be $\textit{Lipschitz}$ in the second variable,but not here. I try to use the continuity condition on $F$ too, but not work.
Any help please ?