If \(\{f_n\}_{n\ge1}\) is an equicontinuous sequence of functions on a compact interval and \(f_n\to f\) pointwise, prove that the convergence is uniform.
Let \(\alpha,M>0\) be given, and suppose that \(\{f_n\}_{n\ge1}\) satisfies \(|f_n(x)-f_n(y)|\le M|x-y|^\alpha\) for all \(n\ge1\) and all \(x,y\) in an interval \([a,b]\). Show that \(\{f_n\}_{n\ge1}\) is equicontinuous.
Solution
Goal: (a) Equicontinuous sequence on a compact interval, converging pointwise, converges uniformly; (b) \(|f_n(x) - f_n(y)| \le M|x-y|^\alpha\) implies equicontinuity.
1. (a) \(f_n \to f\) uniformly.
1.1. \(f\) is continuous.Proof: fix \(x_0\); equicontinuity gives \(\delta\) with \(|f_n(x) - f_n(x_0)| < \varepsilon/3\) for all \(n\) and \(|x - x_0| < \delta\); let \(n \to \infty\): \(|f(x) - f(x_0)| \le \varepsilon/3\).
1.2. Fix \(\varepsilon > 0\).By equicontinuity, for each \(x\) there is \(\delta_x\) with \(|f_n(y) - f_n(x)| < \varepsilon/3\) for all \(n\) and \(|y - x| < \delta_x\); finitely many \(\delta_{x_i}/2\)-balls cover the interval (compactness).
1.3. For each \(x_i\) choose \(N_i\) with \(|f_n(x_i) - f(x_i)| < \varepsilon/3\) for \(n \ge N_i\) (pointwise convergence); set \(N = \max N_i\).
1.4. For \(n \ge N\) and any \(x\): pick \(i\) with \(|x - x_i| < \delta_{x_i}/2\); then \(|f_n(x) - f(x)| \le |f_n(x) - f_n(x_i)| + |f_n(x_i) - f(x_i)| + |f(x_i) - f(x)| < \varepsilon\).Proof: three terms each \(< \varepsilon/3\): the first by equicontinuity (\(|x - x_i| < \delta_{x_i}/2 < \delta_{x_i}\)), the second by \(n \ge N_i\), the third by letting \(n \to \infty\) in the equicontinuity bound (or continuity of \(f\) from <2>1).
1.5. Q.E.D. Proof: 4 gives uniform convergence (the standard compactness argument for equicontinuity upgrading pointwise to uniform).
2. (b) \(\{f_n\}\) is equicontinuous on \([a,b]\).
2.1. Given \(\varepsilon > 0\), choose \(\delta = (\varepsilon/M)^{1/\alpha}\).
2.2. For \(|x - y| < \delta\): \(|f_n(x) - f_n(y)| \le M|x - y|^\alpha < M\delta^\alpha = \varepsilon\), for every \(n\) simultaneously.Proof: the Hölder hypothesis with \(\delta\) as chosen; \(\delta\) is independent of \(n\), \(x\), \(y\).
2.3. Q.E.D. Proof: 2 is exactly equicontinuity.