Given sequence of $L-$Lipschitz functions which converges pointwise, prove uniform convergence

Solution 1:

Let $\varepsilon>0$ be given, and set $\delta=\min\left[\frac{\varepsilon}{3}, \frac{\varepsilon}{3L}\right]$. Since the collection of open balls $\mathcal{B}: = \{B(\, x, \delta) : x \in [a,b] \}$ is a cover for $[a,b]$ we may find a finite subcover, say $\{B(\,x_1, \delta), \, \ldots, \, B(\,x_M, \delta)\}$ (Heine-Borel Theorem). Since $f_n$ converges pointwise on $[a,b]$, for each point $x_j \: \left(\,j=1,\ldots, M \right)$ we may find a positive integer $N_j$ so that \begin{equation} \left|\, f_n(x_j) - f_m(x_j) \right| < \frac{\varepsilon}{3} \text{ whenever } n, m \geq N_j \,. \end{equation} Setting $N = \max [N_1, \ldots, N_M]$ shows that

\begin{aligned} \left|\,f_n(x)- f_m(x) \right| & \leq \left| \,f_n (x)- f_n(x_j) \right| + \left|\, f_n (x_j)- f_m(x_j) \right| + \left|\, f_m(x_j)- f_m(x) \right| \\ & < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon \; \: \text{ whenever } \, n,m \geq N \text{ and } x \in [a,b] . \end{aligned}

Since $\mathbb{R}$ is complete, it follows that the sequence of functions $\{\,f_n\}_{n=1}^\infty$ converges uniformly on $[a,b]$ (Cauchy Criterion).

I prefer uniform convergence first, ask $\,f$ questions later -_-.