Let \(f_n:[0,1]\to[0,1]\) be nondecreasing functions. Suppose \(f_n(x)\to f(x)\) pointwise and \(f\) is continuous. Prove that \(f_n\to f\) uniformly on \([0,1]\). The functions \(f_n\) need not be continuous.
Solution
1. The limit function \(f\) is nondecreasing.
If \(0\leq x\leq y\leq1\), then \[ f_n(x)\leq f_n(y) \] for every \(n\). Passing to the pointwise limit gives \[ f(x)\leq f(y). \]
2. Given \(\varepsilon>0\), there is a partition\[ 0=x_0<x_1<\cdots<x_m=1 \]such that\[ 0\leq f(x_i)-f(x_{i-1})<\frac{\varepsilon}{2} \]for every \(1\leq i\leq m\).
Since \(f\) is continuous on the compact interval \([0,1]\), it is uniformly continuous. Choose \(\delta>0\) such that \[ \abs{x-y}<\delta \quad\Longrightarrow\quad \abs{f(x)-f(y)}<\frac{\varepsilon}{2}. \] Choose a finite partition with mesh smaller than \(\delta\). Step 1 shows that the endpoint differences are nonnegative, and uniform continuity bounds them above by \(\varepsilon/2\).
3. There is an integer \(N\) such that, for every \(n\geq N\) and every partition point \(x_i\),\[ \abs{f_n(x_i)-f(x_i)}<\frac{\varepsilon}{2}. \]
For each of the finitely many points \(x_0,\ldots,x_m\), pointwise convergence gives an integer after which the displayed inequality holds at that point. Take \(N\) to be the maximum of those finitely many integers.
4. If \(n\geq N\) and \(x\in[0,1]\), then\[ \abs{f_n(x)-f(x)}<\varepsilon. \]
Choose \(i\) such that \[ x_{i-1}\leq x\leq x_i. \] By monotonicity of \(f_n\) and step 1, \[ f_n(x_{i-1})\leq f_n(x)\leq f_n(x_i) \] and \[ f(x_{i-1})\leq f(x)\leq f(x_i). \] Therefore, using steps 2 and 3, \[ \begin{aligned} f_n(x)-f(x) &\leq f_n(x_i)-f(x_{i-1})\\ &= \bigl(f_n(x_i)-f(x_i)\bigr) + \bigl(f(x_i)-f(x_{i-1})\bigr)\\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \end{aligned} \] Similarly, \[ \begin{aligned} f(x)-f_n(x) &\leq f(x_i)-f_n(x_{i-1})\\ &= \bigl(f(x_i)-f(x_{i-1})\bigr) + \bigl(f(x_{i-1})-f_n(x_{i-1})\bigr)\\ &< \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon. \end{aligned} \] Thus \(\abs{f_n(x)-f(x)}<\varepsilon\).
5. The convergence \(f_n\to f\) is uniform on \([0,1]\).
For every \(\varepsilon>0\), step 3 gives an \(N\) such that step 4 holds simultaneously for every \(x\in[0,1]\) whenever \(n\geq N\). This is the definition of uniform convergence.
6. Q.E.D.
Step 5 is the required conclusion.