Exercise 7.25

Following the hint, let $n$ be a positive integer. For $i=0,\dots ,n,$ put $x_i=i∕n$. Let $f_n$ be a continuous, piecewise-linear function on $[0,1]$ such that $f_n(0)=c$ and has slope $f_n^{\prime }(t)=\varphi \left(x_i,f_n(x_i)\right)$ if $x_i<t<x_{i+1}$.

Let $|\varphi |$ be bounded by $M$, so that $|f_n^{\prime }|\le M$. Note that

so that for $0\le x\le 1$, $f_n(x)-c$ is the sum of integrals of $f_n^{\prime }$ over intervals where it is defined. Hence

so that $\{f_n\}$ is uniformly bounded on $[0,1]$. Also, since the continuous, piecewise-linear functions $f_n$ have slopes lying between $-M$ and $M$ on their linear parts, if $𝜀>0$ then for $0\le x\le 1$, $0\le y\le 1$, $|x-y|\le 𝜀∕M$, we have $|f_n(x)-f_n(y)|\le 𝜀$. That is, $\{f_n\}$ is equicontinuous on $[0,1]$. Hence by Theorem 7.25, there is a subsequence $\{f_{n_k}\}$ which converges uniformly to a continuous function $f$ on $[0,1]$.

By Theorem 4.19, $\varphi$ is uniformly continuous on the compact rectangle $R$ given by $0\le x\le 1$, $|y|\le M_1$. That is, if $𝜀>0$, there is a $\delta >0$ such that if the distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ in $R$ is less than $\delta$, then $|\varphi (x_1,y_1)-\varphi (x_2,y_2)|<𝜀$. Since there is a $K$ such that for all $k\ge K$ and all $t\in [0,1]$, we have $|f_{n_k}(t)-f(t)|<\delta$, we have

That is $\varphi \left(t,f_{n_k}(t)\right)$ converges uniformly to $\varphi \left(t,f(t)\right)$ as $k\to \infty$. Hence, if we let

then, since $\varphi \left(x_i,f_{n_k}(x_i)\right)$ converges to $\varphi (x_i,f(x_i))$, and $\varphi \left(t,f_{n_k}(t)\right)$ converges to $\varphi (t,f(t))$, and $\varphi \left(t,f(t)\right)$ is uniformly continuous on $[0,1]$, and the distance between the $x_i$ and the $t$ in the definition of ${\mathrm{\Delta }}_n$ is less than $1∕n$, it’s not hard to see that ${\mathrm{\Delta }}_{n_k}(x)$ will converge uniformly to 0 as $k\to \infty$.

Here are the gory details. Let $𝜀>0$. There is a $\delta >0$ such that if $|t_1-t_2|<\delta$ we have

There is a $K$ such that for $k>K$ we have $1∕n_k<\delta$ and such that for all $t\in [0,1]$,

Then, for $k>K$ and for $t$ such that $x_i<t<x_{i+1}$

By the definition of $f_n$, ${\mathrm{\Delta }}_n(t)=f_n^{\prime }(t)-\varphi \left(t,f_n(t)\right)$ for $x_i<t<t_{i+1}$, so

Since $f_{n_k}(x)$ converges uniformly to $f(x)$ on $[0,1]$, and $\varphi \left(t,f_{n_k}(t)\right)$ converges uniformly to $\varphi \left(t,f(t)\right)$ on $[0,1]$, and ${\mathrm{\Delta }}_{n_k}(t)$ converges uniformly to 0 on $[0,1]$, letting $k\to \infty$ in $(\ast )$, by Theorem 7.16 we have

Hence $f(0)=c$, and by Theorem 6.20, $f^{\prime }(x)=\varphi \left(x,f(x)\right)$ for $0\le x\le 1$.