Let \(f_n(x) = \frac{x}{1+nx^2}\) for \(n \in \mathbb{N}\). Let \(\mathcal{F} := \{f_n \, \colon \, n = 1, 2, 3, \ldots\}\) and \([a,b]\) be any compact subset of \(\mathbb{R}\). Is \(\mathcal{F}\) equicontinuous? Justify your answer.
Solution
1. Compute the derivative.
\(f_n(x) = \frac{x}{1+nx^2}\), so \[ f_n'(x) = \frac{(1+nx^2) - 2nx^2}{(1+nx^2)^2} = \frac{1 - nx^2}{(1+nx^2)^2}. \]
2. \(|f_n'(x)| \le 1\) for all \(n\) and all \(x\).
let \(u = nx^2 \ge 0\); then \(|f_n'(x)| = \frac{|1-u|}{(1+u)^2}\). For \(0 \le u \le 1\): \(\frac{1-u}{(1+u)^2} \le \frac{1}{(1+u)^2} \le 1\). For \(u > 1\): \(\frac{u-1}{(1+u)^2} \le \frac{u}{(1+u)^2} \cdot \frac{u}{u} \le 1\) (indeed \(u-1 < u \le (1+u)^2\)). Hence \(|f_n'| \le 1\) everywhere.
3. \(\mathcal{F}\) is equicontinuous on \(\RR\), hence on every compact \([a,b]\).
by the mean value theorem and 2, for all \(n\) and all \(x, y\), \[ |f_n(x) - f_n(y)| \le \sup |f_n'| \cdot |x - y| \le |x - y|. \] So the family is uniformly \(1\)-Lipschitz: given \(\eps > 0\), take \(\delta = \eps\) independently of \(n\) and \(x\). In particular \(\mathcal{F}\) is equicontinuous on any compact \([a,b]\) (and in fact on all of \(\RR\)).
4. Q.E.D.