Show that if \(\{f_n\}\) is an equicontinuous sequence of functions on a compact set \(K\) and \(f_n\to f\) pointwise on \(K\), then \(f_n\to f\) uniformly on \(K\).
Solution
Goal: Show \(\{f_n\}\) equicontinuous on compact \(K\) with \(f_n \to f\) pointwise implies \(f_n \to f\) uniformly on \(K\).
1. \(f\) is continuous.
Fix \(x_0\) and \(\varepsilon > 0\); by equicontinuity choose \(\delta\) with \(\|x-x_0\| < \delta \Rightarrow |f_n(x) - f_n(x_0)| < \varepsilon/3\) for all \(n\). For \(\|x - x_0\| < \delta\), let \(n \to \infty\) in \(|f_n(x) - f_n(x_0)| < \varepsilon/3\) to get \(|f(x) - f(x_0)| \le \varepsilon/3 < \varepsilon\). So \(f\) is continuous at \(x_0\).
2. Fix \(\varepsilon > 0\).By equicontinuity, for each \(x \in K\) there is \(\delta_x > 0\) with \(\|y - x\| < \delta_x \Rightarrow |f_n(y) - f_n(x)| < \varepsilon/3\) for all \(n\).
This is the definition of equicontinuity: the same \(\delta_x\) works uniformly over all \(n\).
3. Choose finitely many \(x_1, \ldots, x_m\) such that \(K \subseteq \bigcup_i B(x_i, \delta_{x_i}/2)\).
The open cover \(\{B(x, \delta_x/2) : x \in K\}\) of the compact set \(K\) has a finite subcover.
4. For each \(i\), choose \(N_i\) with \(|f_n(x_i) - f(x_i)| < \varepsilon/3\) for all \(n \ge N_i\).
This is pointwise convergence \(f_n(x_i) \to f(x_i)\) at the point \(x_i\).
5. Set \(N = \max_i N_i\).For \(n \ge N\) and any \(x \in K\): pick \(i\) with \(x \in B(x_i, \delta_{x_i}/2)\); then \(|f_n(x) - f(x)| \le |f_n(x) - f_n(x_i)| + |f_n(x_i) - f(x_i)| + |f(x_i) - f(x)|\).
This is the triangle inequality applied to the three differences.
6. Each of the three terms is \(< \varepsilon/3\) for \(n \ge N\).
6.1. \(|f_n(x) - f_n(x_i)| < \varepsilon/3\): \(\|x - x_i\| < \delta_{x_i}/2 < \delta_{x_i}\), equicontinuity.
6.2. \(|f_n(x_i) - f(x_i)| < \varepsilon/3\): \(n \ge N \ge N_i\), by
4.
4.3. \(|f(x_i) - f(x)| < \varepsilon/3\): since \(|f_n(x) - f_n(x_i)| < \varepsilon/3\) for all \(n\) (equicontinuity with \(\|x - x_i\| < \delta_{x_i}\)), let \(n \to \infty\); continuity of \(f\) from 1 also works with a suitable \(\delta\).
Pointwise convergence \(f_n \to f\) at both \(x\) and \(x_i\) gives \(|f(x) - f(x_i)| = \lim_n |f_n(x) - f_n(x_i)| \le \varepsilon/3\).
4.4. Q.E.D.
<2>1–<2>3 sum to \(|f_n(x) - f(x)| < \varepsilon\) for all \(x \in K\), \(n \ge N\): uniform convergence.