Assume that \(\{f_n\}\) is a sequence of continuous functions \(f_n:E\subset\mathbb R\to\mathbb R\) which converges uniformly to \(f\). Prove the results of Theorem 6.1 directly.
Solution
Goal: Prove directly that the uniform limit \(f\) of continuous functions \(f_n: E \to \mathbb R\) is continuous.
1. \(f\) is a function \(E \to \mathbb R\).Proof: \(f_n(x)\) converges for every \(x \in E\) (uniform convergence implies pointwise), so \(f(x) := \lim_n f_n(x)\) is a well-defined real number at each \(x\).
2. Fix \(x_0 \in E\) and \(\varepsilon > 0\).Choose \(N\) with \(\|f_n - f\|_\infty < \varepsilon/3\) for all \(n \ge N\).Proof: uniform convergence of \(f_n\) to \(f\).
3. Choose \(\delta > 0\) such that \(|x - x_0| < \delta \Rightarrow |f_N(x) - f_N(x_0)| < \varepsilon/3\).Proof: \(f_N\) is continuous at \(x_0\).
4. For \(|x - x_0| < \delta\): \(|f(x) - f(x_0)| \le |f(x) - f_N(x)| + |f_N(x) - f_N(x_0)| + |f_N(x_0) - f(x_0)| < \varepsilon/3 + \varepsilon/3 + \varepsilon/3 = \varepsilon\).Proof: the first and third terms are \(< \varepsilon/3\) by 2 (uniform closeness of \(f_N\) to \(f\)), the middle by
3.
5. \(f\) is continuous at \(x_0\).Proof: 4 gives the \(\varepsilon\)-\(\delta\) condition at \(x_0\); \(x_0\) was arbitrary, so \(f\) is continuous on \(E\).
Source Collections
Dependencies
T-RA-WORKSHOP-D7-6-1Uniform limits of continuous functions are continuoususes