Theorem 10

Suppose that f_n(x) is a sequence of monotonic functions defined on [a,b] and which are either all increasing or all decreasing. Suppose further that f(x) is a function which is defined and continuous on [a,b] such that

$$\lim_{n\to\infty} f_n (x) = f(x), \qquad \text{for } x \in [a,b].$$

Then, f_n(x) converges uniformly to f(x) on [a,b].

Proof: Suppose first that f_n (x), $n = 1,2,3,\ldots$, are all monotonic increasing. Choose any $\epsilon \gt 0$. Since f(x) is continuous on a closed and bounded set, f(x) is uniformly continuous on [a,b] [Rudin (1953), p. 68, theorem 4.14]. Hence, there exists $\delta \gt 0$ such that for x, $x_0 \in [a,b]$, $\left\vert f(x_0) - f(x) \right\vert < \epsilon/2$ if $\vert x-x_0\vert< \delta$. With each point $x \in [a,b]$, associate the open set $(x-(\delta/4), x+(\delta/4))$. The aggregate of these sets forms an open covering of [a,b]. Then, by the Heine-Borel theorem [Rudin (1953), p. 30, theorem 2.34], there is a finite number of these intervals which cover [a,b]. These intervals will be denoted by $J_1, J_2, \ldots, J_m$. Let $I_i = J_i \cdot [a,b]$ be the set of points common to both J_i and [a,b], where $i = 1,2,\ldots, m$. Define x'_i and x''_i as the greatest lower bound and the least upper bound, respectively, of I_i. Now, x'_i and x''_i are both contained in [a,b], since [a,b] is closed. Lastly, set $I^c_i = \left[ x'_i, x''_i\right]$. Now, if $x \in I^c_i$, then $\left\vert x - x'_i\right\vert < \delta$ and $\left\vert x - x''_i \right\vert < \delta$ from the method of construction. Hence, $\left\vert f(x) - f\left( x'_i\right)\right\vert < \epsilon/2$ and $\left\vert f(x) - f \left( x''_i\right) \right\vert < \epsilon/2$, or

$$\begin{array} {rl} \displaystyle - \frac{\epsilon}2 & \displ... ...le < f(x) - f \left( x'_i \right) < \frac\epsilon2. \end{array}$$

Since $\lim_{n\to\infty} f_n (x) = f(x)$ for every $x \in [a,b]$, it follows that there exist N'_i and N''_i such that

$$\begin{array} {r@{\quad}r} \text{if} \quad n \gt N'_i, \quad... ... \left( x''_i \right) \right\vert < \frac\epsilon2. \end{array}$$

Set $N_i = \max \left( N'_i, N''_i\right)$. Now, since f_n (x) is monotonic increasing on [a,b],

 

$$f_n \left( x'_i \right) \leq f_n (x) \leq f_n \left( x''_i \right)$$ (43)

for $x \in I^c_i$. It follows that

$$\begin{array} {rl} \displaystyle - \epsilon = - \frac\epsilo... ...right] < \frac\epsilon2 + \frac\epsilon2 = \epsilon\end{array}$$

or

$$\vert f_n (x) - f(x) \vert < \epsilon, \qquad \text{for } x \in I^c_i, \quad n \gt N_i.$$

Set $N = \max N_i$, $i = 1,2,\ldots, m$. Then, if n > N,

$$\vert f_n (x) - f(x)\vert < \epsilon$$

for every I^c_i and hence, for every value of x in [a,b]. This proves that f_n (x) converges uniformly to f(x) in [a,b].

If the f_n (x), $n = 1,2,3,\ldots$, are all monotonic decreasing on [a,b], then Eq. (43) becomes

$$f_n \left( x'_i \right) \geq f_n (x) \geq f_n \left( x''_i \right),$$

and as before,

$$\begin{array} {rl} \displaystyle - \epsilon = - \frac\epsilo... ...ght] < \frac\epsilon2 + \frac\epsilon2 = \epsilon,\end{array}$$

$$\vert f_n (x) - f(x) \vert < \epsilon, \qquad \text{for } x \in I^c_i, \quad n \gt N_i.$$

The rest of the proof is the same as previously given for the monotonic increasing case.