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
Uniformly convergent means that \(\norm{g_i - g_j}_{\infty} \to 0\), so \(\sup_{x\in X}\abs{g_i(x)-g_j(x)} \convergesto{i, j\to\infty} 0\). We want to show that given \(\eps\) we can find \(N_0\) such that \(i, j > N_0\) yields \[ \sup_{x\in X}\abs{ f\circ g_i(x) - f\circ g_j(x) } < \eps .\]
Fix \(\eps> 0\), then choose \(\delta_1 = \delta_1(\eps)\) by uniform continuity of \(f\) to guarantee \[ \abs{y_1 - y_2} \leq \delta_1 \implies \abs{f(y_1) - f(y_2) } < \eps \, \forall y_1, y_2\in X .\] Now by uniform convergence of \(\ts{g_n}\), choose \(N_0 = N_0(\delta_1)\) such that \[ i, j \geq N_0 \implies \abs{ g_i(x) - g_j(x) } < \delta_1 \, \forall x\in X .\]
Now writing \(y_1 \da g_i(x), y_2 \da g_j(x)\), choose \(i, j > N_0\) yields \[ \abs{y_1 - y_2} \da \abs{g_i(x) - g_j(x) } < \delta_1 \\ \implies \abs{f(y_1) - f(y_2)} \da \abs{f(g_i(x)) - f(g_j(x))} < \eps ,\] and taking the supremum over \(x\in X\) preserves the inequality since \(\delta_1\) and consequently \(N_0\) only depend on \(\eps\).