Suppose \((f_n)_{n\in\mathbb N}\) is a sequence of continuous functions \(f_n:[0,1]\to\mathbb R\) satisfying \[ f_n(x)\ge f_{n+1}(x)\ge0 \] for all \(n\) and \(x\). Prove that if \(f_n\to0\) pointwise on \([0,1]\), then the convergence is uniform.
Solution
Fix \(\varepsilon>0\) and, for each \(n\), define \[ U_n=\{x\in[0,1]:f_n(x)<\varepsilon\}. \]
1. Each \(U_n\) is open in \([0,1]\), and\[ U_n\subseteq U_{n+1}. \]
Since \(f_n\) is continuous, \[ U_n=f_n^{-1}((-\infty,\varepsilon)) \] is open in the subspace topology on \([0,1]\).
If \(x\in U_n\), then \[ f_{n+1}(x)\leq f_n(x)<\varepsilon, \] so \(x\in U_{n+1}\). Hence the family is increasing.
2. The sets \(U_n\) cover \([0,1]\).
Fix \(x\in[0,1]\). Since \[ f_n(x)\longrightarrow0 \] pointwise, there is an index \(n\) such that \[ f_n(x)<\varepsilon. \] Thus \(x\in U_n\). Since \(x\) was arbitrary, \[ [0,1]=\bigcup_n U_n. \]
3. There is an index \(N\) such that\[ U_N=[0,1]. \]
By steps <1>1–<1>2, the sets \(U_n\) form an open cover of the compact interval \([0,1]\). Choose a finite subcover \[ U_{n_1},\ldots,U_{n_r}. \] Let \[ N=\max\{n_1,\ldots,n_r\}. \] Since the family is increasing by step 1, \[ U_{n_j}\subseteq U_N \] for every \(j\). Therefore \[ [0,1] = \bigcup_{j=1}^r U_{n_j} \subseteq U_N \subseteq [0,1], \] so \(U_N=[0,1]\).
4. For every \(n\geq N\) and every \(x\in[0,1]\),\[ \abs{f_n(x)}<\varepsilon. \]
Step 3 gives \[ f_N(x)<\varepsilon \] for every \(x\in[0,1]\). The hypotheses also give \[ 0\leq f_n(x)\leq f_N(x) \] whenever \(n\geq N\). Hence \[ \abs{f_n(x)} = f_n(x) < \varepsilon. \]
5. The sequence \(f_n\) converges uniformly to \(0\) on \([0,1]\).
For each \(\varepsilon>0\), step 4 supplies an index \(N\) independent of \(x\) such that \[ n\geq N \quad\Longrightarrow\quad \abs{f_n(x)-0}<\varepsilon \] for every \(x\in[0,1]\). This is uniform convergence.
6. Q.E.D.
Step 5 is the required conclusion.