Let $(X,\Vert \Vert)$ be a Banach space and $f:\mathbb{R} \times X \rightarrow X$ continuous.
If f is linearly bounded in its second argument, i.e. there exist $\alpha, \beta \in C(\mathbb{R} ; \mathbb{R}_0^+) \cap L^1(\mathbb{R})$ such that for arbitrary $(t,v) \in \mathbb{R} \times X$ we have:
$\Vert f(t,v) \Vert \leq \alpha (t) + \beta (t) \Vert v \Vert$
then every solution of $u'(t) = f(t,u(t)), t \in \mathbb{R}$ with $u(t_0) = u_0 \in X, t_0 \in \mathbb{R}$ is bounded.
My attempt:
Given an arbitrary bounded interval, say wlog $V := (t_1, t_2) \subset \mathbb{R}$ we try to show that $u(v) \subset \overline{B(u_0, r)}$.
Let $t \in V$ be arbitrary. The solution u has the form $u(t) = u_0 + \int_{t_0}^t f(x, u(x)) dx$, therefore:
$\Vert u(t) - u_0 \Vert = \Vert u_0 + \int_{t_0}^t f(x, u(x)) dx - u_0 \Vert \leq \int_{t_0}^t f(x,u(x)) dx \Vert \leq \int_{t_0}^t \alpha (x) dx + \Vert u \Vert \int_{t_0}^t \beta (x) dx $
I wanted to use that $\alpha, \beta \in L^1(\mathbb{R})$ and then choose the radius the be the supremum over all radii for all t in $(t_1,t_2)$, but the problem is that I get $\Vert u \Vert $ on the right hand side, which is the term I want to show to be bounded in the first place. I'm also not sure whether I can assume the interval $(t_1, t_2)$ to be of this particular form. Can anybody help out here please?