Let \((Y, d)\) be a metric space; let \(\mathcal{F}\) be a subset of \(\mathcal{C}(X, Y)\).
Show that if \(\mathcal{F}\) is finite, then \(\mathcal{F}\) is equicontinuous.
Show that if \(f_n\) is a sequence of elements of \(\mathcal{C}(X, Y)\) that converges uniformly, then the collection \(\ts{f_n}\) is equicontinuous.
Suppose that \(\mathcal{F}\) is a collection of differentiable functions \(f: \mathbb{R} \to \mathbb{R}\) such that each \(x \in \mathbb{R}\) lies in a neighborhood \(U\) on which the derivatives of the functions in \(\mathcal{F}\) are uniformly bounded. [This means that there is an \(M\) such that \(\abs{f'(x)} \leq M\) for all \(f\) in \(\mathcal{F}\) and all \(x \in U\).] Show that \(\mathcal{F}\) is equicontinuous.
Solution
1. Part (a): Finite collections are equicontinuous.Proof:
1.1. Let \(\mathcal{F} = \{f_1, \dots, f_k\}\) be finite, and let \(x_0 \in X, \varepsilon > 0\).
1.2. For each \(i \in \{1, \dots, k\}\), continuity of \(f_i\) at \(x_0\) guarantees an open neighborhood \(U_i \subseteq X\) of \(x_0\) such that \(d(f_i(x), f_i(x_0)) < \varepsilon\) for all \(x \in U_i\).
1.3. Define \(U = \bigcap_{i=1}^k U_i\). As a finite intersection of open neighborhoods of \(x_0\), \(U\) is an open neighborhood of \(x_0\).
1.4. For all \(x \in U\) and all \(f \in \mathcal{F}\), \(d(f(x), f(x_0)) < \varepsilon\). Thus \(\mathcal{F}\) is equicontinuous.
2. Part (b): Uniformly convergent sequences \(\{f_n\}\) are equicontinuous.Proof:
2.1. Let \(f_n \to f\) uniformly on \(X\), with \(x_0 \in X\) and \(\varepsilon > 0\). By the uniform limit theorem, \(f: X \to Y\) is continuous.
2.2. Choose an integer \(N \ge 1\) such that \(d(f_n(x), f(x)) < \frac{\varepsilon}{3}\) for all \(n \ge N\) and all \(x \in X\).
2.3. By continuity of \(f\) at \(x_0\), choose an open neighborhood \(U_0 \subseteq X\) of \(x_0\) such that \(d(f(x), f(x_0)) < \frac{\varepsilon}{3}\) for all \(x \in U_0\).
2.4. For each \(n \in \{1, \dots, N-1\}\), choose an open neighborhood \(U_n\) of \(x_0\) such that \(d(f_n(x), f_n(x_0)) < \varepsilon\) for all \(x \in U_n\).
2.5. Set \(U = \bigcap_{n=0}^{N-1} U_n\), which is an open neighborhood of \(x_0\).
2.6. For any \(n < N\) and \(x \in U\), \(d(f_n(x), f_n(x_0)) < \varepsilon\) since \(x \in U_n\).
2.7. For any \(n \ge N\) and \(x \in U\), the triangle inequality gives:\[d(f_n(x), f_n(x_0)) \le d(f_n(x), f(x)) + d(f(x), f(x_0)) + d(f(x_0), f_n(x_0)) < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon.\]
2.8. Thus \(\{f_n\}_{n=1}^\infty\) is equicontinuous at \(x_0\), hence equicontinuous on \(X\).
3. Part (c): Locally uniformly bounded derivatives imply equicontinuity.Proof:
3.1. Let \(x_0 \in \mathbb{R}\) and \(\varepsilon > 0\).
3.2. By hypothesis, there exists an open neighborhood \(U_0 \ni x_0\) and a constant \(M > 0\) such that \(|f'(t)| \le M\) for all \(f \in \mathcal{F}\) and all \(t \in U_0\).
3.3. Choose \(\delta > 0\) such that the interval \(U = (x_0 - \delta, x_0 + \delta) \subseteq U_0\) and \(\delta < \frac{\varepsilon}{M}\).
3.4. For any \(x \in U\) and any \(f \in \mathcal{F}\), by the Mean Value Theorem, there exists a point \(c\) strictly between \(x\) and \(x_0\) such that:\[|f(x) - f(x_0)| = |f'(c)| |x - x_0|.\]
3.5. Since \(c \in U \subseteq U_0\), \(|f'(c)| \le M\).
3.6. Consequently, \(|f(x) - f(x_0)| \le M |x - x_0| < M \delta < \varepsilon\).
3.7. Thus \(\mathcal{F}\) is equicontinuous at \(x_0\), and therefore on \(\mathbb{R}\). Q.E.D.