Show that a uniform limit of continuous functions is continuous, and a uniform limit of uniformly continuous functions is uniformly continuous. Show that this is not true if uniform convergence is weakened to pointwise convergence.
Solution
Suppose \(\norm{f_n - f}_\infty\to 0\) and fix \(\eps>0\). For all \(z,w\) and \(n\), \[ \abs{f(z) - f(w)} \leq \abs{f(z) - f_n(z)} + \abs{f_n(z) - f_n(w)} + \abs{f_n(w) - f(w)} \leq 2\norm{f-f_n}_\infty+\abs{f_n(z) - f_n(w)} .\] Choose \(n\) with \(\norm{f-f_n}_\infty<\eps/3\).
- Continuity: fix \(z\). Since \(f_n\) is continuous at \(z\), there is \(\delta>0\) with \(\abs{f_n(z)-f_n(w)}<\eps/3\) for \(\abs{z-w}<\delta\), and then \(\abs{f(z)-f(w)}<\eps\).
- Uniform continuity: if \(f_n\) is uniformly continuous, there is \(\delta>0\) with \(\abs{f_n(z)-f_n(w)}<\eps/3\) for all \(z,w\) with \(\abs{z-w}<\delta\), and then \(\abs{f(z)-f(w)}<\eps\) for all such \(z,w\).
The choice of one \(n\) for all \(z\) uses \(\norm{f-f_n}_\infty<\eps/3\); pointwise convergence gives an \(n\) depending on the point.
Pointwise limits: on \([0,1]\), the uniformly continuous functions \(f_n(x) \da x^n\) converge pointwise to \(\chi_{\theset{1}}\), which is not continuous, hence not uniformly continuous.