Theorem (Picard-Lindelof): Suppose we have a smooth function \(f: \mathbb{R} \rightarrow \mathbb{R}\) which is Lipschitz with constant \(L\) and some constant points \(t_0, x_0 \in \mathbb{R}\). Then, for any \(q \in (0, 1)\), there is an unique local solution \(x\) to the initial value problem in the interval \(I_{t_0} = [t_0, t_0 + q/L]\):

\[x'(t) = f(x(t)) \qquad x(t_0) = x_0\]

Proof: The general proof of the Picard-Lindelof theorem leans on the Banach-Caccioppoli fixed point theorem. In particular, if we consider the space of continuous functions \(\mathcal{C}(I_{t_0})\) with domain \(I_{t_0}\), we note that any solution to the initial value problem is a fixed point of the function \(T: \mathcal{C}(I_{t_0}) \rightarrow \mathcal{C}(I_{t_0})\) given by:

\[Tg(t) = x_0 + \int_{t_0}^t f(g(\tau)) \, d\tau\]

It turns out that \(T\) is a contraction mapping under the Chebyshev metric \(d\) on \(I_{t_0}\) for any \(q \in (0, 1)\). The metric space defined by \(d\) on \(\mathcal{C}(I_{t_0})\) happens to be complete so the fixed point theorem implies that such a solution exists and is unique. To make this rigorous, we must prove that \(T\) is well defined as stated, that \(d\) is in fact a complete metric on \(\mathcal{C}(I_{t_0})\), and that \(T\) is a contraction mapping.

Justifying \(T\) is well defined involves simply showing that \(f \circ g\) is integrable and that \(Tg\) is continuous. The first fact is elementary and the second follows immediately from the first Fundamental Theorem of Calculus.

To show that \(d\) defines a valid metric on \(I_{t_0}\), the only nontrivial fact to be shown is that \(d\) adheres to the triangle inequality. Taking arbitrary \(g_1, g_2, g_3 \in \mathcal{C}(I_{t_0})\), we can choose \(t_1, t_2, t_3 \in I_{t_0}\) where these functions achieve their maximal difference in \(I_{t_0}\) (note this makes use of the fact that \(I_{t_0}\) is compact). Then, we have:

\[\begin{align} d(g_1, g_3) &= \vert g_1(t_1) - g_3(t_1) \vert \\ &\leq \vert g_1(t_1) - g_2(t_1) \vert + \vert g_2(t_1) - g_3(t_1) \vert \\ &\leq \vert g_1(t_2) - g_2(t_2) \vert + \vert g_2(t_3) - g_3(t_3) \vert \\ &= d(g_1, g_2) + d(g_2, g_3) \end{align}\]

The completeness of \(d\) on the space of \(\mathcal{C}(I_{t_0})\) follows simply from the uniform convergence theorem. Therefore, \(g\) is continuous and thus lies in the space \(\mathcal{C}(I_{t_0})\).

Finally, what remains to be shown is that \(T\) is a contraction map. This requires use of the Lipschitz regularity of \(f\) and the restricted length of \(I_{t_0}\) as we see in the following:

\[\begin{align} d(T g_1, T g_2) &= \max_{t \in I_{t_0}} \left\vert \int_{t_0}^t f(g_1(\tau)) \, d\tau - \int_{t_0}^t f(g_2(\tau)) \, d\tau \right\vert \\ &= \max_{t \in I_{t_0}} \left\vert \int_0^t f(g_1(\tau)) - f(g_2(\tau)) \, d\tau \right\vert \\ &\leq \max_{t \in I_{t_0}} \int_{t_0}^t \vert f(g_1(\tau)) - f(g_2(\tau)) \vert \, d\tau \\ &\leq \max_{t \in I_{t_0}} \int_{t_0}^t L \vert g_1(\tau) - g_2(\tau) \vert \, d\tau \\ &\leq \frac{q}{L} L \max_{\tau \in I_{t_0}}\vert g_1(\tau) - g_2(\tau) \vert = q d(g_1, g_2) \end{align}\]

We note that the length of the interval \(I_{t_0}\) provided by Picard-Lindelof is independent of our explicit choice of \(t_0\) and \(x_0\) and depends only on the local Lipschitz constant around \(t_0\). Thus, if \(f\) has a global Lipschitz constant as we describe above, we can cover \(\mathbb{R}\) with countably many Picard-Lindelof intervals to obtain a proof for global uniqueness as well.