Lipschitz condition and boundedness imply unique 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 $(t,v) \in \mathbb R \times X$.
Thus, every solution of $$u'(t)=f(t,u(t)), u(t_o)=u_o \in X$$$t,t_0\in \mathbb R$ is bounded.
If $f$ now satisfies a Lipschitz condition on $X$, how can I show that the initial value problem is uniquely solvable?
 A: The Picard iteration is
$$
P(y)(t)=y_0+\int_0^tf(s,y(s))\,ds.
$$
The difference of two iterated functions is bounded by
$$
\|P(y)(t)-P(u)(t)\|\le \int_0^t L·\|y(s)-u(s)\|\,ds
$$
The difference of the twice iterated functions is
\begin{align}
\|P^2(y)(t)-P^2(u)(t)\|
&\le \int_0^t L·\|P(y)(r)-P(u)(r)\|\,dr
\\
&\le L^2\int_0^t \int_0^r \|y(s)-u(s)\|\,ds\,dr
\qquad(0\le s\le r\le t)
\\
&= L^2\int_0^t \int_s^t\, dr·\|y(r)-u(r)\|\,dr
\\
&= L^2\int_0^t (t-s)·\|y(s)-u(s)\|\,ds
\end{align}
Continuing this one finds
\begin{align}
\|P^n(y)(t)-P^n(u)(t)\|&\le L^n \int_0^t \frac{(t-s)^{n-1}}{(n-1)!}·\|y(s)-u(s)\|\,ds\\[1em]
\implies \qquad
\|P^n(y)-P^n(u)\|_∞&\le  \frac{L^n ·T^n}{n!}·\|y-u\|_∞
\end{align}
in $C([-T,T],X)$. 
For the function sequence $P^n(y_0)$ this leads to
\begin{align}
\|P^n(y_0)-P^m(y_0)\|_∞&\le \frac{L^m ·T^m}{m!}·\|P^{n-m}(y_0)-y_0\|_∞\\
\text{and}\qquad
\|P^k(y_0)-y_0\|_∞&\le \sum_{j=0}^{k-1}\|P^{j+1}(y_0)-P^{j}(y_0)\|_∞
\\
&\le\|\sum_{j=0}^{k-1}\frac{L^j ·T^j}{j!}\|P(y_0)-y_0\|_∞<e^{LT}\|P(y_0)-y_0\|_∞
\end{align}
From which one concludes that the function sequence is Cauchy, thus converges, and that to a fixed point of the Picard iteration, and thus to a solution of the initial value problem. 
Uniqueness follows because $(LT)^n<n!$ for some large enough $n$ so that for two fixed points $u,y$
$$
\|y-u\|=\|P^n(y)-P^n(u)\|\le \frac{(LT)^n}{n!}\|y-u\|<\|y-u\|
$$
which can only be true for $u=y$.
Since there are no restrictions on $T$, any solution can be extended to all of $\Bbb R$.
