Exercise 7.26

Repeating the argument of the solution to Exercise 7.25, making changes where necessary, let $n$ be a positive integer. For $i=0,\dots ,n,$ put $x_i=i∕n$. Let $f_n$ be a continuous, piecewise-linear vector-valued function on $[0,1]$ into $R^k$ such that $f_n(0)=c$ and has slope $f_n^{\prime }(t)=\Phi \left(x_i,f_n(x_i)\right)$ if $x_i<t<x_{i+1}$.

Let $||\Phi ||$ 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 $[0,1]$. Let $f_n=(f_{n1},\dots ,f_{nk})$. Then the continuous, piecewise-linear $f_{nm}$ have slopes lying between $-M$ and $M$ on their linear parts, so if $𝜀>0$ then for $0\le x\le 1$, $0\le y\le 1$, $|x-y|\le 𝜀∕Mk$, we have $|f_{nm}(x)-f_{nm}(y)|\le 𝜀∕k$, so $||f_n(x)||\le 𝜀$. That is, $\{f_n\}$ is equicontinuous on $[0,1]$, using the extended definition of “equicontinuous” given in Exercise 17. Hence by the extended version Theorem 7.25 also given in Exercise 25, there is a subsequence $\{f_{n_j}\}$ which converges uniformly to a continuous function $f$ on $[0,1]$.

By Theorem 4.19, $\Phi$ is uniformly continuous on the compact parallelpiped $R$ 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 $||\Phi (x_1,y_1)-\Phi (x_2,y_2)||<𝜀$. 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, $\Phi \left(t,f_{n_k}(t)\right)$ converges uniformly to $\Phi \left(t,f(t)\right)$ as $k\to \infty$. Hence, if we let

then, since $\Phi \left(x_i,f_{n_k}(x_i)\right)$ converges to $\Phi (x_i,f(x_i))$, and $\Phi \left(t,f_{n_k}(t)\right)$ converges to $\Phi (t,f(t))$, and $\Phi \left(t,f(t)\right)$ 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 $f_n$, ${\mathrm{\Delta }}_n(t)=f_n^{\prime }(t)-\Phi \left(t,f_n(t)\right)$ for $x_i<t<t_{i+1}$, so

Since $f_{n_k}(x)$ converges uniformly to $f(x)$ on $[0,1]$, and $\Phi \left(t,f_{n_k}(t)\right)$ converges uniformly to $\Phi \left(t,f(t)\right)$ on $[0,1]$, and ${\mathrm{\Delta }}_{n_k}(t)$ converges uniformly to $0$ on $[0,1]$, letting $k\to \infty$ in $(\ast )$, by the extended version of Theorem 7.16 given in Exercise 17 we have

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