Linearly boundedness implies boundedness of solution Let $(X,\vert\vert\cdot\vert\vert)$ be a Banach space and $f:\mathbb R \times X \to X$ continuous and linearly bounded in its second variable, i.e. there are functions $\alpha, \beta \in C(\mathbb R; \mathbb R_0^+) \cap L^1(\mathbb R)$ such that $$\vert\vert f(t,v)\vert\vert \leq \alpha(t)+\beta(t)\vert\vert v\vert\vert$$ for any $(t,v) \in \mathbb R \times X$.
How can I show that every solution of $$u'(t)=f(t,u(t)), u(t_o)=u_o \in X$$$t,t_0\in \mathbb R$ is bounded?
 A: First for Hilbert space, let $x(t)=\|u(t)\|^2$. Then
\begin{eqnarray}
 x'(t)&=&2(u'(t),u(t))=2(f(t,u(t)),u(t))\\
&\le&\|f(t,u(t))\|\|u(t)\|\\
&\le&2\alpha(t)\|u(t)\|+2\beta(t)x(t)\\
&\le& \alpha(t)(x(t)+1)+2\beta(t)x(t)\\
&=&\alpha(t)+(\alpha(t)+2\beta(t))x(t)
\end{eqnarray}
from which we have
$$ (x(t)e^{-\int_0^t(\alpha(\tau)+2\beta(\tau))d\tau})'\le \alpha(t)e^{-\int_0^t(\alpha(\tau)+2\beta(\tau))d\tau}. $$
Integrating both sides from $0$ to $t$, one has
$$ x(t)\le x(0)+ \int_0^t\alpha(s)e^{\int_s^t(\alpha(\tau)+2\beta(\tau))d\tau}ds.$$
Since $\alpha,\beta(t)\in L^1(R)$, it is easy to see $x(t)<M$ for some $M>0$ for $\forall t\ge 0$ and hence $u(t)$ is bounded. For Banach space, the situation is almost identical.
A: We have for $t>t_0$
$$\begin{align}
\|u(t)\|=&\Bigl\|\,u(t_0)+\int_{t_0}^tf(s,u(s))\,ds\,\Bigr\|\\
&\le\|u(t_0)\|+\int_{t_0}^t\bigl(\alpha(s)+\beta(s)\,\|u(s)\|\bigr)\,ds\\
&\le \|u(t_0)\|+\int_{t_0}^\infty\alpha(s)\,ds+\int_{t_0}^t\beta(s)\,\|u(s)\|\,ds.
\end{align}$$
Now use Gronwalls's lemma.
The same argument works for $t<t_0$.
A: You get approximately  up to quadratic terms in $h$
\begin{align}
\|u(t+h)\|&\le \|u(t)\|+h·\|f(t,u(t))\|
\\
&\le\|u(t)\|+h·α(t)+h·β(t)||u(t)||
\\
&=(1+h·β(t))\|u(t)\|+h·α(t)
\le e^{h·β(t)}\|u(t)\|+h·α(t)
\end{align}
In the next step you get
$$
\|u(t+2h)\|\le e^{h·β(t+h)}\left(e^{h·β(t)}\|u(t)\|+h·α(t)\right)+h·α(t+h)
$$
so that in continuation this becomes
$$
\|u(t+kh)\|\le e^{h·\sum_{j=0}^{k-1}β(t+jh)}\|u(t)\|+h·\sum_{m=0}^{k-1}α(t+mh)e^{h·\sum_{j=m+1}^{k-1}β(t+jh)}
$$
which obviously transfers in the limit $h\to0$ into integrals
$$
\|u(t_1)\|
\le e^{\int_{t_0}^{t_1}β(s)\,ds}\|u(t_0)\|
    +\int_{t_0}^{t_1}α(r)e^{\int_{r}^{t_1}β(s)\,ds}dr
$$
Now using that $ α$ and $β$ are $L^1$ integrable one easily gets an upper bound for $u$.
Using the monotonously falling functions $A(t)=\int_t^{\infty}α(s)\,ds$ and $B(t)=\int_t^{\infty}β(s)\,ds$ one gets
\begin{align}
\|u(t_1)\|
&\le e^{-B(t_1)}\left(e^{B(t_0)}\|u(t_0)\| + \int_{t_0}^{t_1}α(r)e^{B(r)}dr\right)\\
&\le e^{B(t_0)}\left(\|u(t_0)\|+A(t_0)\right)
\end{align}
