Proof. Take \(t \geq t_0\). For each \(n\), divide \([t_0, t_0 + \alpha ]\) into \(n\) steps of length \(h = \alpha /n\) at \(t_j = t_0 + jh\), and define the Euler polygon \(\phi _n\) by \(\phi _n(t_0) = x_0\) and \[ \phi _n(t) = \phi _n(t_j) + (t - t_j)f\bigl (t_j, \phi _n(t_j)\bigr ), \qquad t_j \leq t \leq t_{j+1} . \] By induction, as in the cone argument, each \(\phi _n\) stays in \(\bar B(x_0, b)\), and it is piecewise linear with slopes of norm at most \(M\), so \(\|\phi _n(t) - \phi _n(s)\| \leq M|t - s|\): the sequence is uniformly bounded and equicontinuous. Writing \(t_{[s]}\) for the grid point at or just below \(s\), \[ \phi _n(t) = x_0 + \int _{t_0}^tf\bigl (t_{[s]}, \phi _n(t_{[s]})\bigr )\,ds = x_0 + \int _{t_0}^tf\bigl (s, \phi _n(s)\bigr )\,ds + \int _{t_0}^t\delta _n(s)\,ds , \] where \(\delta _n(s) = f(t_{[s]}, \phi _n(t_{[s]})) - f(s, \phi _n(s))\). The two points at which \(f\) is evaluated are within \(h\) in \(t\) and \(Mh\) in \(x\), and \(f\) is uniformly continuous on the compact set \(R\), so \(\|\delta _n\|_\infty \to 0\). By Theorem 2.5.1 a subsequence \(\phi _{n_k}\) converges uniformly to a continuous \(\phi \); then \(f(s, \phi _{n_k}(s)) \to f(s, \phi (s))\) uniformly, again by uniform continuity, and passing to the limit gives \(\phi (t) = x_0 + \int _{t_0}^tf(s, \phi (s))\,ds\). By Lemma 2.1.1, \(\phi \) is a solution. □