Let \(f_n: I \to \mathbb{R}\) be the function \(f_n(x) = x^n\). The collection \(\mathcal{F} = \ts{f_n}\) is pointwise bounded but the sequence \((f_n)\) has no uniformly convergent subsequence; at what point or points does \(\mathcal{F}\) fail to be equicontinuous?
Repeat (a) for the functions \(f_n\) of Exercise 9 of §21.
Solution
For \(f_n(x)=x^n\), equicontinuity holds at every \(x_0<1\): choose \(a\) with \(x_0<a<1\); on a sufficiently small neighborhood contained in \([0,a]\), the finitely many small \(n\) are uniformly continuous, while for all sufficiently large \(n\) both \(x^n\) and \(x_0^n\) are arbitrarily small. At \(x_0=1\) equicontinuity fails: for every \(\delta>0\), choose \(x<1\) with \(1-\delta<x<1\) and then \(n\) so large that \(x^n<1/2\), whereas \(f_n(1)=1\).
For the sequence of E-BTI7W, \[ f_n(x)=\frac1{n^3(x-1/n)^2+1}, \] the family fails equicontinuity exactly at \(0\). Indeed \(f_n(0)=1/(n+1)\to0\) but \(f_n(1/n)=1\), and \(1/n\to0\), contradicting equicontinuity at \(0\). If \(x_0\ne0\), choose a neighborhood \(J\) of \(x_0\) whose closure avoids \(0\). For large \(n\), \(1/n\) stays a fixed positive distance from \(J\), so both \(f_n\) and its derivative tend uniformly to \(0\) on \(J\); the finitely many remaining functions are uniformly continuous there. Thus the family is equicontinuous at every \(x_0\ne0\).