Let \(f:[0,1]\to\mathbb{R}\) be continuous with \(f(0)\neq f(1)\) and define \(f_n(x)=f(x^n)\). Prove that \(f_n\) does not converge uniformly on \([0,1]\).
Solution
1. \(f_n \to g\) pointwise, where \(g(x) = f(0)\) for \(x \in [0,1)\) and \(g(1) = f(1)\).
\(x^n \to 0\) for \(0 \le x < 1\) and \(1^n = 1\); \(f\) is continuous, so \(f(x^n) \to f(0)\) for \(x < 1\) and \(f(1^n) = f(1)\).
2. \(g\) is discontinuous at \(1\).
\(g(1) = f(1) \ne f(0) = \lim_{x \to 1^-}g(x)\) by hypothesis \(f(0) \ne f(1)\).
3. If \(f_n\) converged uniformly, its limit would be continuous.
the uniform limit theorem — a uniform limit of continuous functions (\(f_n\) continuous since \(x \mapsto x^n\) and \(f\) are continuous) is continuous.
4. Q.E.D.
the pointwise limit \(g\) of \(f_n\) is unique, and it is discontinuous by 2, contradicting 3; hence the convergence is not uniform.