5.4.E: Problems on Complex and Vector-Valued Functions on $E^{1}$

Exercise $\PageIndex{1}$

Do the case $g^{\prime}(r)=+\infty$ in Lemma 1.
[Hint: Show that there is $s>r$ with
$$g(x)-g(r) \geq\left(\left|f^{\prime}(r)\right|+1\right)(x-r) \geq|f(x)-f(r)| \text { for } x \in(r, s) .$$
$\text { Such } x \text { are "good." }]$

Exercise $\PageIndex{2}$

Do the case $r=p_{n} \in Q$ in Lemma $1 .$
[Hint: Show by continuity that there is $s>r$ such that $(\forall x \in(r, s))$
$$|f(x)-f(r)|<\frac{\varepsilon}{2^{n+1}} \text { and }|g(x)-g(r)|<\frac{\varepsilon}{2^{n+1}} .$$
Show that all such $x$ are "good" since $x>r=p_{n}$ implies
$$\left.2^{-n}+Q(r) \leq Q(x) . \quad(\text { Why? })\right]$$

Exercise $\PageIndex{3}$

Show that Corollary 3 in §2 (hence also Theorem 2 in §2) fails for complex functions.
$\left.\text { [Hint: Let } f(x)=e^{x i}=\cos x+i \cdot \sin x . \text { Verify that }\left|f^{\prime}\right|=1 \text { yet } f(2 \pi)-f(0)=0 .\right]$

Exercise $\PageIndex{4}$

(i) Verify that all propositions of §4 hold also if $f^{\prime}$ and $g^{\prime}$ are only right derivatives on $I-Q$.
(ii) Do the same for left derivatives. (See footnote 2.)

Exercise $\PageIndex{5}$

(i) Prove that if $f : E^{1} \rightarrow E$ is continuous and finite on $I=(a, b)$ and differentiable on $I-Q,$ and if
$$\sup _{t \in I-Q}\left|f^{\prime}(t)\right|<+\infty ,$$
then $f$ is uniformly continuous on $I$.
(ii) Moreover, if $E$ is complete $\left(\mathrm{e} . g ., E=E^{n}\right),$ then $f\left(a^{+}\right)$ and $f\left(b^{-}\right)$ exist and are finite.
$\text { [Hints: (i) Use Corollary 1. (ii) See the "hint" to Problem 11 (iii) of Chapter } 4, §8 .]$

Exercise $\PageIndex{6}$

Prove that if $f$ is as in Theorem $2,$ with $f^{\prime} \geq 0$ on $I-Q$ and $f^{\prime}>0$ at some $p \in I,$ then $f(a)<f(b) .$ Do it also with $f^{\prime}$ treated as a right derivative (see Problem 4).

Exercise $\PageIndex{7}$

Let $f, g : E^{1} \rightarrow E^{1}$ be relatively continuous on $I=[a, b]$ and have right derivatives $f_{+}^{\prime}$ and $g_{+}^{\prime}$ (finite or infinite, but not both infinite) on $I-Q$.
(i) Prove that if
$$m g_{+}^{\prime} \leq f_{+}^{\prime} \leq M g_{+}^{\prime} \text { on } I-Q$$
for some fixed $m, M \in E^{1},$ then
$$m[g(b)-g(a)] \leq f(b)-f(a) \leq M[g(b)-g(a)] .$$
$\text { [Hint: Apply Theorem } 2 \text { and Problem } 4 \text { to each of } M g-f \text { and } f-m g .]$
(ii) Hence prove that
$$m_{0}(b-a) \leq f(b)-f(a) \leq M_{0}(b-a) ,$$
where
$$m_{0}=\inf f_{+}^{\prime}[I-Q] \text { and } M_{0}=\sup f_{+}^{\prime}[I-Q] \text { in } E^{*} .$$
$\left.\text { [Hint: Take } g(x)=x \text { if } m_{0} \in E^{1} \text { or } M_{0} \in E^{1} . \text { The } \text {infinite case is simple. }\right]$

Exercise $\PageIndex{8}$

(i) Let $f :(a, b) \rightarrow E$ be finite, continuous, with a right derivative on $(a, b) .$ Prove that $q=\lim _{x \rightarrow a^{+}} f_{+}^{\prime}(x)$ exists (finite) iff
$$q=\lim _{x, y \rightarrow a^{+}} \frac{f(x)-f(y)}{x-y} ,$$
i.e., iff
$$(\forall \varepsilon>0)(\exists c>a)(\forall x, y \in(a, c) | x \neq y) \quad\left|\frac{f(x)-f(y)}{x-y}-q\right|<\varepsilon .$$
[Hints: If so, let $y \rightarrow x^{+}$ (keeping $x$ fixed) to obtain
$$(\forall x \in(a, c)) \quad\left|f_{+}^{\prime}(x)-q\right| \leq \varepsilon . \quad \text { (Why?) }$$
Conversely, if $\lim _{x \rightarrow a^{+}} f_{+}^{\prime}(x)=q,$ then
$$(\forall \varepsilon>0)(\exists c>a)(\forall t \in(a, c)) \quad\left|f_{+}^{\prime}(t)-q\right|<\varepsilon .$$
Put
$$M=\sup _{a<t<c}\left|f_{+}^{\prime}(t)-q\right| \leq \varepsilon \quad(\text { why } \leq \varepsilon ?)$$
and
$h(t)=f(t)-t q, \quad t \in(a, b)$.
Apply Corollary 1 and Problem 4 to $h$ on the interval $[x, y] \subseteq(a, c),$ to get
$$|f(y)-f(x)-(y-x) q| \leq M(y-x) \leq \varepsilon(y-x) .$$
Proceed.]
(ii) Prove similar statements for the cases $q=\pm \infty$ and $x \rightarrow b^{-}$. $\text {[Hint: In case } q=\pm \infty, \text { use Problem } 7 \text { (ii) instead of Corollary } 1 .]$

Exercise $\PageIndex{9}$

From Problem 8 deduce that if $f$ is as indicated and if $f_{+}^{\prime}$ is left continuous at some $p \in(a, b),$ then $f$ also has a left derivative at $p .$
If $f_{+}^{\prime}$ is also right continuous at $p,$ then $f_{+}^{\prime}(p)=f_{-}^{\prime}(p)=f^{\prime}(p)$.
$\text { [Hint: Apply Problem } 8 \text { to }(a, p) \text { and }(p, b) .]$

Exercise $\PageIndex{10}$

In Problem $8,$ prove that if, in addition, $E$ is complete and if
$$q=\lim _{x \rightarrow a^{+}} f_{+}^{\prime}(x) \neq \pm \infty \quad \text { (finite) } ,$$
then $f\left(a^{+}\right) \neq \pm \infty$ exists, and
$$\lim _{x \rightarrow a^{+}} \frac{f(x)-f\left(a^{+}\right)}{x-a}=q ;$$
similarly in case $\lim _{x \rightarrow b^{-}} f_{+}^{\prime}(x)=r$.
If both exist, set $f(a)=f\left(a^{+}\right)$ and $f(b)=f\left(b^{-}\right) .$ Show that then $f$ becomes relatively continuous on $[a, b],$ with $f_{+}^{\prime}(a)=q$ and $f_{-}^{\prime}(b)=r$.
[Hint: If
$$\lim _{x \rightarrow a^{+}} f_{+}^{\prime}(x)=q \neq \pm \infty ,$$
then $f_{+}^{\prime}$ is bounded on some subinterval $(a, c), a<c \leq b(\text { why?), so } f \text { is uniformly }$ continuous on $(a, c),$ by Problem $5,$ and $f\left(a^{+}\right)$ exists. Let $y \rightarrow a^{+},$ as in the hint to $\text {Problem } 8 .]$

Exercise $\PageIndex{11}$

Do Problem 9 in §2 for complex and vector-valued functions.
$\text { [Hint: Use Corollary 1 of } §4 .]$

Exercise $\PageIndex{12}$

Continuing Problem $7,$ show that the equalities
$$m=\frac{f(b)-f(a)}{b-a}=M$$
hold iff $f$ is linear, i.e., $f(x)=c x+d$ for some $c, d \in E^{1},$ and then $c=m=M .$

Exercise $\PageIndex{13}$

Let $f : E^{1} \rightarrow C$ be as in Corollary $1,$ with $f \neq 0$ on $I .$ Let $g$ be the real part of $f^{\prime} / f .$
(i) Prove that $|f| \uparrow$ on $I$ iff $g \geq 0$ on $I-Q$.
(ii) Extend Problem 4 to this result.

Exercise $\PageIndex{14}$

Define $f : E^{1} \rightarrow C$ by
$$f(x)=\left\{\begin{array}{ll}{x^{2} e^{i / x}=x^{2}\left(\cos \frac{1}{x}+i \cdot \sin \frac{1}{x}\right)} & {\text { if } x>0, \text { and }} \\ {0} & {\text { if } x \leq 0.}\end{array}\right.$$
Show that $f$ is differentiable on $I=(-1,1),$ yet $f^{\prime}[I]$ is not a convex $\left.\text { set in } E^{2}=C \text { (thus there is no analogue to Theorem } 4 \text { of } §2\right) .$