Which of the following subsets of \(\mathcal{C}(\mathbb{R}, \mathbb{R})\) are pointwise bounded? Which are equicontinuous?
The collection \(\ts{f_n}\), where \(f_n(x) = x + \sin nx\).
The collection \(\ts{g_n}\), where \(g_n(x) = n + \sin x\).
The collection \(\ts{h_n}\), where \(h_n(x) = \abs{x}^{1/n}\).
The collection \(\ts{k_n}\), where \(k_n(x) = n \sin(x/n)\).
Solution
For fixed \(x\), \(|x+\sin nx|\le |x|+1\), so the family is pointwise bounded. It is not equicontinuous: at \(0\), for any \(\delta>0\) choose \(n\) large and \(x=\pi/(2n)<\delta\); then \(|f_n(x)-f_n(0)|=x+1>1\).
\(g_n(x)=n+\sin x\) is not pointwise bounded at any \(x\). It is equicontinuous since \[ |g_n(x)-g_n(y)|=|\sin x-\sin y|\le|x-y| \] for all \(n\).
The family \(h_n(x)=|x|^{1/n}\) is pointwise bounded: for fixed \(x\), it lies between \(0\) and \(\max\{1,|x|\}\). It is not equicontinuous at \(0\): given \(\delta>0\), choose \(0<x<\min\{\delta,1\}\) and then \(n\) so large that \(x^{1/n}>1/2\), whereas \(h_n(0)=0\).
Since \(|n\sin(x/n)|\le |x|\), the family is pointwise bounded. Also \[ |k_n(x)-k_n(y)|\le n|x/n-y/n|=|x-y|, \] so it is equicontinuous.
Thus (a): bounded, not equicontinuous; (b): equicontinuous, not bounded; (c): bounded, not equicontinuous; (d): both.