Suppose \(\theset{g_n}\) is a uniformly convergent sequence of functions from \(\RR\) to \(\RR\) and \(f:\RR\to \RR\) is uniformly continuous. Prove that the sequence \(\theset{f\circ g_n}\) is uniformly convergent.
Solution
Let \(g_n\to g\) uniformly on \(\mathbb R\). Fix \(\varepsilon>0\). Since \(f\) is uniformly continuous, there exists \(\delta>0\) such that \[ |x-y|<\delta\implies |f(x)-f(y)|<\varepsilon. \] Uniform convergence of \(g_n\) gives \(N\) such that, for \(n\ge N\), \[ \sup_{x\in\mathbb R}|g_n(x)-g(x)|<\delta. \] Therefore for every \(x\in\mathbb R\) and \(n\ge N\), \[ |f(g_n(x))-f(g(x))|<\varepsilon. \] Hence \[ \sup_{x\in\mathbb R}|f(g_n(x))-f(g(x))|\le\varepsilon, \] so \(f\circ g_n\to f\circ g\) uniformly.