Show that a uniform limit of bounded functions is bounded.
Show that a uniform limit of continuous function is continuous.
- I.e. if \(f_n\to f\) uniformly with each \(f_n\) continuous then \(f\) is continuous.
Solution
Let \(f_n \to f\) uniformly on a set \(E\).
1. If each \(f_n\) is bounded, then \(f\) is bounded.
Uniform convergence with \(\eps = 1\) gives \(N\) with \(\|f - f_N\|_\infty \le 1\). By the triangle inequality, \(|f(x)| \le |f_N(x)| + 1 \le \|f_N\|_\infty + 1\) for every \(x \in E\).
2. If each \(f_n\) is continuous, then \(f\) is continuous.
2.1. Fix \(x_0 \in E\) and \(\eps > 0\). There are \(n\) with \(\|f - f_n\|_\infty < \eps/3\) and \(\delta > 0\) with \(|f_n(x) - f_n(x_0)| < \eps/3\) whenever \(x \in E\) and \(|x - x_0| < \delta\).
The first is uniform convergence and the second is continuity of \(f_n\) at \(x_0\).
2.2. Q.E.D.
For \(x \in E\) with \(|x - x_0| < \delta\), the triangle inequality and step 1 give \(|f(x) - f(x_0)| \le |f(x) - f_n(x)| + |f_n(x) - f_n(x_0)| + |f_n(x_0) - f(x_0)| < \eps\). So \(f\) is continuous at the arbitrary point \(x_0\).
3. Q.E.D.
Steps 1 and 2.