Suppose \((g_n)\) is a uniformly convergent sequence of functions from \(\RR\) to \(\RR\), while \(f\colon \RR \to \RR\) is uniformly continuous. Prove that the sequence \((f \circ g_n)\) of composite functions is also uniformly convergent on \(\RR\).
Solution
Let \[ g_n\longrightarrow g \] uniformly on \(\RR\).
1. For every \(\varepsilon>0\) there exists \(\delta>0\) such that\[ \abs{u-v}<\delta \quad\Longrightarrow\quad \abs{f(u)-f(v)}<\varepsilon \]for all \(u,v\in\RR\).
This is exactly the uniform continuity of \(f\).
2. For the \(\delta\) from step 1, there exists \(N\) such that\[ n\geq N \quad\Longrightarrow\quad \abs{g_n(x)-g(x)}<\delta \]for every \(x\in\RR\).
Since \(g_n\to g\) uniformly, the definition of uniform convergence applied to the positive number \(\delta\) gives such an \(N\).
3. For every \(n\geq N\) and every \(x\in\RR\),\[ \abs{f(g_n(x))-f(g(x))}<\varepsilon. \]
Fix \(n\geq N\) and \(x\in\RR\). By step 2, \[ \abs{g_n(x)-g(x)}<\delta. \] Applying step 1 with \[ u=g_n(x), \qquad v=g(x) \] gives \[ \abs{f(g_n(x))-f(g(x))}<\varepsilon. \] This is the claimed estimate.
4. The sequence \((f\circ g_n)\) converges uniformly to \(f\circ g\).
Step 3 gives, for every \(\varepsilon>0\), an \(N\) such that \[ n\geq N \quad\Longrightarrow\quad \abs{(f\circ g_n)(x)-(f\circ g)(x)} < \varepsilon \] for every \(x\in\RR\). This is uniform convergence.
5. Q.E.D.
Step 4 proves the required uniform convergence.