Suppose \(\theset{f_n}_{n\in \NN}\) is a sequence of continuous functions \(f_n: [0, 1]\to \RR\) such that \[ f_n(x) \geq f_{n+1}(x) \geq 0 \quad \forall n\in \NN,\, \forall x\in [0, 1] .\] Prove that if \(\theset{f_n}\) converges pointwise to \(0\) on \([0, 1]\) then it converges to \(0\) uniformly on \([0, 1]\).
Solution
Let \(\eps>0\), we want to show that there exists an \(N_0\) such that \(n\geq N_0\) implies \(\norm{f_n}_\infty<\eps\). Fix \(x\), by pointwise convergence pick \(M_x = M_x(x, \eps)\) so that \(n\geq M \implies \abs{f_n(x)} < \eps\). By continuity, this bound holds in some neighborhood \(U_x \ni x\). Produce a cover \(\ts{U_x}_{x\in [0, 1]}\covers [0, 1]\); by compactness produce a finite subcover \(\ts{U_1, \cdots, U_m} \covers [0, 1]\). Each \(U_i\) corresponds to some \(x_i\) and some \(M_{x_i}\), so choose \(N_0 > \max_{i\leq m} \ts{M_{x_i}}\). Then \(n\geq N_0 \implies N\geq M_{x_i}\) for each \(i\), so \(\abs{f_n(x)} < \eps\) for each \(x\in [0, 1]\) since \(x\in U_i\) for some \(i\). So \(\sup_{x\in X} \abs{f_n(x)} = \norm{f_n}_{\infty } < \eps\).