5.10.E: Problems on Regulated Functions
Complete all details in the proof of Theorems \(1-3\).
Explain Examples \((a)-(g)\).
Prove Note \(2 .\) More generally, assuming \(T\) to be complete, prove that if
\[
g_{n} \rightarrow f(\text { uniformly }) \text { on } I=[a, b]
\]
and if the \(g_{n}\) are regulated on \(I,\) so is \(f\).
[Hint: Fix \(p \in(a, b] .\) Use Theorem 2 of Chapter \(4, §11\) with
\[
X=[a, p], Y=N \cup\{+\infty\}, q=+\infty, \text { and } F(x, n)=g_{n}(x) .
\]
Then show that
\[
f\left(p^{-}\right)=\lim _{x \rightarrow p^{-}} \lim _{n \rightarrow \infty} g_{n}(x) \text { exists; }
\]
\(\left.\text { similarly for } f\left(p^{+}\right) .\right]\)
Given \(f, g : E^{1} \rightarrow E^{1},\) define \(f \vee g\) and \(f \wedge g\) as in Problem 12 of Chapter \(4, §8 .\) Using the hint given there, show that \(f \vee g\) and \(f \wedge g\) are regulated if \(f\) and \(g\) are.
Show that the function \(g \circ f\) need not be regulated even if \(g\) and \(f\) are.
[Hint: Let
\[
f(x)=x \cdot \sin \frac{1}{x}, g(x)=\frac{x}{|x|}, \text { and } f(0)=g(0)=0 \text { with } I=[0,1] .
\]
Proceed.]
\(\Rightarrow\) Given \(f : E^{1} \rightarrow(T, \rho),\) regulated on \(I,\) put
\[
j(p)=\max \left\{\rho\left(f(p), f\left(p^{-}\right)\right), \rho\left(f(p), f\left(p^{+}\right)\right), \rho\left(f\left(p^{-}\right), f\left(p^{+}\right)\right)\right\} ;
\]
call it the \(j u m p\) at \(p\).
(i) Prove that \(f\) is discontinuous at \(p \in I^{0}\) iff \(j(p)>0,\) i.e., iff
\[
(\exists n \in N) \quad j(p)>\frac{1}{n} .
\]
(ii) For a fixed \(n \in N,\) prove that a closed subinterval \(J \subseteq I\) contains at most finitely many \(x\) with \(j(x)>1 / n\).
[Hint: Otherwise, there is a sequence of distinct points \(x_{m} \in J, j\left(x_{m}\right)>\frac{1}{n},\) hence a subsequence \(x_{m_{k}} \rightarrow p \in J .\) (Why?) Use Theorem 1 of Chapter \(4,\) §2, \(\left.\text { to show that } f\left(p^{-}\right) \text {or } f\left(p^{+}\right) \text {fails to exist. }\right]\)
\(\Rightarrow\) Show that if \(f : E^{1} \rightarrow(T, \rho)\) is regulated on \(I,\) then it has at most countably many discontinuities in \(I ;\) all are of the "jump" type (Problem 5).
[Hint: By Problem 5, any closed subinterval \(J \subseteq I\) contains, for each \(n,\) at most finitely many discontinuities \(x\) with \(j(x)>1 / n .\) Thus for \(n=1,2, \ldots,\) obtain \(\text { countably many such } x .]\)
Prove that if \(E\) is complete, all maps \(f : E^{1} \rightarrow E,\) with \(V_{f}[I]<+\infty\) on \(I=[a, b],\) are regulated on \(I .\)
[Hint: Use Corollary 1, Chapter 4, §2, to show that \(f\left(p^{-}\right)\) and \(f\left(p^{+}\right)\) exist.
Say,
\[
x_{n} \rightarrow p \text { with } x_{n}<p \quad\left(x_{n}, p \in I\right) ,
\]
but \(\left\{f\left(x_{n}\right)\right\}\) is not Cauchy. Then find a subsequence, \(\left\{x_{n_{k}}\right\} \uparrow,\) and \(\varepsilon>0\) such that
\[
\left|f\left(x_{n_{k+1}}\right)-f\left(x_{n_{k}}\right)\right| \geq \varepsilon, \quad k=1,3,5, \ldots
\]
Deduce a contradiction to \(V_{f}[I]<+\infty .\)
\(\left.\quad \text { Provide a similar argument for the case } x_{n}>p .\right]\)
Prove that if \(f : E^{1} \rightarrow(T, \rho)\) is regulated on \(I,\) then \(\overline{f[B]}\) (the closure \(\text { of } f[B])\) is compact in \((T, \rho)\) whenever \(B\) is a compact subset of \(I .\)
[Hint: Given \(\left\{z_{m}\right\}\) in \(\overline{f[B]}\), find \(\left\{y_{m}\right\} \subseteq f[B]\) such that \(\rho\left(z_{m}, y_{m}\right) \rightarrow 0\) (use \(\text { Theorem } 3 \text { of Chapter } 3, §16) .\) Then "imitate" the proof of Theorem 1 in Chap\(\text { ter } 4, §8\) suitably. Distinguish the cases:
(i) all but finitely many \(x_{m}\) are \(<p\);
(ii) infinitely many \(x_{m}\) exceed \(p ;\) or
(iii) infinitely many \(x_{m}\) equal \(p\).]