Exercise 8.2.18

Question

A continuous function is called \emph{polygonal} if its graph consists of a finite number of line segments.
\begin{enumerate}[label=(\alph*)] 
\item Show that there exists a polygonal function $p \in C[0,1]$ satisfying $\|f-p\|_\infty < \epsilon / 2$.
\item Show that if $h$ is any function in $C[0,1]$ that is bounded by $1$, then the function
$$g(x) = p(x) + \frac{\epsilon}{2} h(x)$$
satisfies $g \in V_\epsilon(f)$.
\item Construct a polygonal function $h(x)$ in $C[0,1]$ that is bounded by 1 and leads to the conclusion $g \notin A_{m,n}$, where $g$ is defined as in (b). Explain how this completes the argument for Theorem 8.2.12.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item This is Theorem 6.7.3, proved as exercise 6.7.2.
\item Let $q(x) = \epsilon h(x) /2$.
$$\|f - g\|_\infty \leq \|f - p\|_\infty + \|p - (p + q)\|_\infty < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon $$
\item Let $M$ be a bound on $\left|p'\right|$ (for where $p'$ is defined); i.e.\ a bound on the maximum slope of $p$. Then let $h$ zig-zag between $\pm 1$ with slope greater than $2(M + n) / \epsilon$. More formally, let $K > 2(M + n) / \epsilon$, define $i(x) = xK / 2$ , and set
$$h = \begin{cases}
    2 \left(i(x) - \lfloor i(x)\rfloor\right) - 1 & \lfloor i(x) \rfloor \text{ even} \
    1 - 2 \left(i(x) - \lfloor(i(x)\rfloor)\right) & \lfloor i(x) \rfloor \text{ odd}
\end{cases}$$
($\lfloor y \rfloor$ denotes the floor function, the largest integer $z$ with $z \leq y$.) It's clear that $g$ is a polygonal function with each line segment has slope greater than $n$, although proving this formally requires some case work which is not a good use of our time. Thus, $g \notin A_{m,n}$ regardless of $m$.
 \end{enumerate}

Exercise 8.2.18

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Exercise 8.2.18

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