For $0 < \ell < \ell_0$, let $\XX = \map \CC {\closedint {t_0 - \ell_0} {t_0 + \ell_0}; \R^n}$ endowed with the sup norm be the Banach Space of Continuous Functions on Compact Space $\closedint {t_0 - \ell_0} {t_0 + \ell_0} \to \R^n$. By Fixed Point Formulation of Explicit ODE it is sufficient to find a fixed point of the map $T: \XX \to \XX$ defined by: :$\displaystyle \map {\paren {T x} } t = x_0 + \int_{t_0}^t \map f {s, \map x s} \rd s$ {{explain|Notation not clear: what actually does $T x$ mean? And is $x$ that fixed point posited?}} We also have Closed Subset of Complete Metric Space is Complete. Therefore the Banach Fixed-Point Theorem it is sufficient to find a non-empty subset $\YY \subseteq \XX$ such that: :$\YY$ is closed in $\XX$ :$T \YY \subseteq \YY$ :$T$ is a contraction on $\YY$ {{explain|Notation not clear: what does $T \YY$ mean?}} First note that $V$ is closed and bounded, hence compact by the Heine-Borel Theorem. {{questionable|For Heine-Borel Theorem to apply, $V$ needs to be demonstrated to be not only bounded, but also Definition:Totally Bounded Metric Space.}} Therefore since $f$ is continuous, by the extreme value theorem, the maximum $\displaystyle m = \sup_{\tuple {t, x} \mathop \in V} \size {\map f {t, x} }$ exists and is finite. Let $\kappa$ be the Lipschitz constant of $f$. Let: :$\YY = \set {y \in \XX: \norm {\map y t - x_0} \le m \size {t - t_0}, t \in \closedint {t_0 - \ell_0} {t_0 + \ell_0} }$ be the cone in $\XX$ centred at $\tuple {t_0, x_0}$. {{explain|Is this a specialised use of the term "cone" different from the 3-d geometrical construct? If so, it needs to be defined and linked to -- if not, it needs to be explained why it is a cone.}} Clearly $\YY$ is closed in $\XX$. Also for $y \in \YY$ we have: {{begin-eqn}} {{eqn | l = \size {\map {\paren {T y} } t - x_0}       | o = \le       | r = \size {\int_{t_0}^t \map f {s, \map y s} \rd s}       | c =  }} {{eqn | o = \le       | r = m \int_{t_0}^t \rd s       | c = Absolute Value of Definite Integral }} {{eqn | r = m \set {t - t_0}       | c =  }} {{end-eqn}} Therefore $T \YY \subseteq \YY$. Finally we must show that $T$ is a contraction on $\YY$ (we will find that this restricts our choice of $\ell$). Let $y_1, y_2 \in \YY$. We have: {{begin-eqn}} {{eqn | l = \size {\map {\paren {T y_1} } t - \map {\paren {T y_2} } t}       | o = \le       | r = \size {\int_{t_0}^t \map f {s, \map {y_1} s} - \map f {s, \map {y_2} s} \rd s}       | c =  }} {{eqn | o = \le       | r = \int_{t_0}^t \size {\map f {s, \map {y_1} s} - \map f {s, \map {y_2} s} } \rd s       | c = Absolute Value of Definite Integral }} {{eqn | o = \le       | r = \kappa \int_{t_0}^t \size {\map {y_1} t - \map {y_2} t} \rd s       | c = Lipschitz Condition }} {{eqn | o = \le       | r = \kappa \size {t - t_0} \norm {y_1 - y_2}_\sup       | c = Estimation Lemma }} {{end-eqn}} {{explain|Specific meaning of $\norm {y_1 - y_2}_\sup$ needs to be established.}} Taking the supremum over $t$ we have: :$\norm {T y_1 - T y_2}_\sup \le \kappa \ell \norm {y_1 - y_2}_\sup$ for all $y_1, y_2 \in \YY$. Therefore choosing $\ell < \kappa^{-1}$, $T$ is a contraction on $\YY$ as required. This completes the proof. {{qed}} Category:Differential Equations rhsvfu3jx7p7yvofn9wurea5324p5w2