Exercise 6.2.10

$f$ is continuous and therefore will map $[a,b]$ to a closed interval $[c,d]$. By the Order Limit Theorem, we also know that $f$ must be increasing - just compare the sequence $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}_n(x_1)$ to $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}_n(x_2)$ when $x_2>x_1$.

The key implication of $f_n$ and $f$ being increasing is that there is a bound to how fast the error $\left|f-f_n\right|$ can grow with respect to $x$, which is proportional to how fast $f$ grows. Intuitively, if we have $f>f_n$ and want the error to grow as fast as possible with respect to $x$, all we can do is hold $f_n$ constant. Alternatively, if we have $f<f_n$, $f_n$ can’t go too far above $f$ since $f$ needs to catch up eventually.

More formally, for $𝜖>0$, define $y_1,y_2,y_3,\dots ,y_n$ to evenly split up $[c,d]$ into intervals of size at most ${𝜖}_1=𝜖∕5$. (In other words, $y_1=c$, $y_n=d$, $y_{k+2}-y_{k+1}=y_{k+1}-y_k<{𝜖}_1$.) Since $f$ is increasing, we can define $x_1,\dots ,x_n$ so that $f(x_k)=y_k$.

Since $f_n\to f$, we can find $M_k$ so that $m_k>M_k$ implies $\left|y_k-f_{m_k}(x_k)\right|<{𝜖}_1$. Let $M=\max <!--\; FUNCTION\; APPLICATION\; -->\{M_1,M_2,\dots <!--\; FUNCTION\; APPLICATION\; -->,M_n\}$. Now, let $m>M$ be arbitrary. Keeping in mind that $f_m$ is increasing, we can bound $f_m(x_{i+1})-f_m(x_i)$ by

Now consider $\left|f(x)-f_m(x)\right|x\in [x_i,x_{i+1}]$ with $i$ arbitrary. Since $f$ is increasing, $y_i\le f(x)\le y_{i+1}$, so $f(x)-y_i<{𝜖}_1$. Similarly $f_m(x)-f_m(x_i)\le f_m(x_{i+1})-f_m(x_i)<3{𝜖}_1$. Finally, we have

Since $m$ and $i$ were arbitrary, this completes the proof.