(June 2009 #4a) Let \(\{f_n\}\) be a sequence of real-valued continuous functions such that \(f_n\to f\) uniformly on \([0,1]\), and let \(\{x_n\}\subset[0,1]\) be a sequence which converges to \(x\). Show that \[\lim_{n\to\infty}f_n(x_n)=f(x).\]
Solution
Goal: Show \(f_n(x_n) \to f(x)\) given \(f_n \to f\) uniformly on \([0,1]\), \(f_n\) continuous, \(x_n \to x\) in \([0,1]\).
1. \(|f_n(x_n) - f(x)| \le |f_n(x_n) - f(x_n)| + |f(x_n) - f(x)|\).Proof: triangle inequality.
2. \(|f_n(x_n) - f(x_n)| \le \|f_n - f\|_\infty \to 0\).Proof: uniform convergence, and \(x_n \in [0,1]\).
3. \(|f(x_n) - f(x)| \to 0\).Proof: \(f\) is continuous (uniform limit of continuous functions) and \(x_n \to x\), so continuity of \(f\) at \(x\) gives \(f(x_n) \to f(x)\).
4. Q.E.D. Proof: 2 and 3 force the right side of 1 to \(0\), so \(f_n(x_n) \to f(x)\).