5.4.E: Problems on Complex and Vector-Valued Functions on \(E^{1}\)
- Page ID
- 23755
Do the case \(g^{\prime}(r)=+\infty\) in Lemma 1.
[Hint: Show that there is \(s>r\) with
\[
g(x)-g(r) \geq\left(\left|f^{\prime}(r)\right|+1\right)(x-r) \geq|f(x)-f(r)| \text { for } x \in(r, s) .
\]
\(\text { Such } x \text { are "good." }]\)
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
\[
\left.2^{-n}+Q(r) \leq Q(x) . \quad(\text { Why? })\right]
\]
Show that Corollary 3 in §2 (hence also Theorem 2 in §2) fails for complex functions.
\(\left.\text { [Hint: Let } f(x)=e^{x i}=\cos x+i \cdot \sin x . \text { Verify that }\left|f^{\prime}\right|=1 \text { yet } f(2 \pi)-f(0)=0 .\right]\)
(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.)
(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.
\(\text { [Hints: (i) Use Corollary 1. (ii) See the "hint" to Problem 11 (iii) of Chapter } 4, §8 .]\)
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).
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
\[
m_{0}=\inf f_{+}^{\prime}[I-Q] \text { and } M_{0}=\sup f_{+}^{\prime}[I-Q] \text { in } E^{*} .
\]
\(\left.\text { [Hint: Take } g(x)=x \text { if } m_{0} \in E^{1} \text { or } M_{0} \in E^{1} . \text { The } \text {infinite case is simple. }\right]\)
(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 \(x \rightarrow b^{-}\). \(\text {[Hint: In case } q=\pm \infty, \text { use Problem } 7 \text { (ii) instead of Corollary } 1 .]\)
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 \(f_{+}^{\prime}(p)=f_{-}^{\prime}(p)=f^{\prime}(p)\).
\(\text { [Hint: Apply Problem } 8 \text { to }(a, p) \text { and }(p, b) .]\)
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 \(\text {Problem } 8 .]\)
Do Problem 9 in §2 for complex and vector-valued functions.
\(\text { [Hint: Use Corollary 1 of } §4 .]\)
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 \(c=m=M .\)
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.
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 \(\left.\text { set in } E^{2}=C \text { (thus there is no analogue to Theorem } 4 \text { of } §2\right) .\)