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
Fix \(\varepsilon>0\) and define \[ U_n=\{x\in[0,1]: f_n(x)<\varepsilon\}. \] Each \(U_n\) is open in \([0,1]\) because \(f_n\) is continuous. Since \(f_{n+1}\le f_n\), the sets are increasing: \[ U_1\subseteq U_2\subseteq\cdots. \] Pointwise convergence to \(0\) implies that every \(x\in[0,1]\) belongs to some \(U_n\), hence \[ [0,1]=\bigcup_{n=1}^\infty U_n. \] Compactness of \([0,1]\) gives a finite subcover. Because the \(U_n\) are nested, there is therefore a single index \(N\) such that \(U_N=[0,1]\). For all \(n\ge N\) and all \(x\in[0,1]\), \[ 0\le f_n(x)\le f_N(x)<\varepsilon. \] Thus \[ \sup_{x\in[0,1]}|f_n(x)|<\varepsilon \qquad(n\ge N), \] which is uniform convergence to \(0\).