Define \(f_n: [0,1] \to \mathbb{R}\) by the equation \(f_n(x) = x^n\). Show that the sequence \((f_n(x))\) converges for each \(x \in [0,1]\), but that the sequence \((f_n)\) does not converge uniformly.
Solution
1. \(f_n\to f\) pointwise, where \(f(x)=0\) for \(0\le x<1\) and \(f(1)=1\).
For \(0\le x<1\), \(x^n\to0\); and \(f_n(1)=1\) for every \(n\).
2. \((f_n)\) does not converge uniformly.
Uniform convergence would be to the pointwise limit \(f\). For each \(n\), the point \(x_n=2^{-1/n}\in[0,1)\) has \(\abs{f_n(x_n)-f(x_n)}=\frac12\), so \(\sup_{[0,1]}\abs{f_n-f}\ge\frac12\) for every \(n\). Equivalently, each \(f_n\) is continuous and \(f\) is not continuous at \(1\), while a uniform limit of continuous functions is continuous.
3. Q.E.D.
Steps 1 and 2.