5.1.E: Problems on Derived Functions in One Variable
- Page ID
- 23751
Prove Theorems 4 and \(5,\) including \(\left(i^{*}\right)\) and \(\left(i i^{*}\right) .\) Do it for dot products as well.
Verify Note 2.
Verify Example (a).
Verify Example (b).
Prove that if \(f\) has finite one-sided derivatives at \(p,\) it is continuous at \(p\).
Restate and prove Theorems 2 and 3 for one-sided derivatives.
Prove that if the functions \(f_{i} : E^{1} \rightarrow E^{*}(C)\) are differentiable at \(p,\) so is their product, and
\[
\left(f_{1} f_{2} \cdots f_{m}\right)^{\prime}=\sum_{i=1}^{m}\left(f_{1} f_{2} \cdots f_{i-1} f_{i}^{\prime} f_{i+1} \cdots f_{m}\right) \text { at } p .
\]
A function \(f : E^{1} \rightarrow E\) is said to satisfy a Lipschitz condition \((L)\) of order \(\alpha(\alpha>0)\) at \(p\) iff
\[
(\exists \delta>0)\left(\exists K \in E^{1}\right)\left(\forall x \in G_{\neg p}(\delta)\right) \quad|f(x)-f(p)| \leq K|x-p|^{\alpha} .
\]
(i) This implies continuity at \(p\) but not conversely; take
\[
f(x)=\frac{1}{\ln |x|}, \quad f(0)=0, \quad p=0 .
\]
\(\text { [Hint: For the converse, start with Problem } 14 \text { (iii) of Chapter } 4, §2 .]\)
(ii) \(L\) of order \(\alpha>1\) implies differentiability at \(p,\) with \(f^{\prime}(p)=0\).
(iii) Differentiability implies \(L\) of order \(1,\) but not conversely. (Take
\[
f(x)=x \sin \frac{1}{x}, f(0)=0, p=0 ;
\]
then even one-sided derivatives fail to exist.)
Let
\[
f(x)=\sin x \text { and } g(x)=\cos x .
\]
Show that \(f\) and \(g\) are differentiable on \(E^{1},\) with
\[
f^{\prime}(p)=\cos p \text { and } g^{\prime}(p)=-\sin p \text { for each } p \in E^{1} .
\]
Hence prove for \(n=0,1,2, \ldots\) that
\[
f^{(n)}(p)=\sin \left(p+\frac{n \pi}{2}\right) \text { and } g^{(n)}(p)=\cos \left(p+\frac{n \pi}{2}\right) .
\]
[Hint: Evaluate \(\Delta f\) as in Example (d) of Chapter \(4, §8 .\) Then use the continuity of \(f\) and the formula
\[
\lim _{z \rightarrow 0} \frac{\sin z}{z}=\lim _{z \rightarrow 0} \frac{z}{\sin z}=1 .
\]
To prove the latter, note that
\[
|\sin z| \leq|z| \leq|\tan z| ,
\]
whence
\[
1 \leq \frac{z}{\sin z} \leq \frac{1}{|\cos z|} \rightarrow 1 ;
\]
similarly for \(g\).]
Prove that if \(f\) is differentiable at \(p\) then
\[
\lim _{x \rightarrow p^{+} \atop y \rightarrow p^{-}} \frac{f(x)-f(y)}{x-y} \text { exists, is finite, and equals } f^{\prime}(p) ;
\]
i.e., \((\forall \varepsilon>0)(\exists \delta>0)(\forall x \in(p, p+\delta))(\forall y \in(p-\delta, p))\)
\[
\left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(p)\right|<\varepsilon .
\]
Show, by redefining \(f\) at \(p,\) that even if the limit exists, \(f\) may not be \(\text { differentiable (note that the above limit does not involve } f(p)) .\)
[Hint: If \(y<p<x\) then
\[
\begin{aligned}\left|\frac{f(x)-f(y)}{x-y}-f^{\prime}(p)\right| & \leq\left|\frac{f(x)-f(p)}{x-y}-\frac{x-p}{x-y} f^{\prime}(p)\right|+\left|\frac{f(p)-f(y)}{x-y}-\frac{p-y}{x-y} f^{\prime}(p)\right| \\ &\left. \leq\left|\frac{f(x)-f(p)}{x-p}-f^{\prime}(p)\right|+\left|\frac{f(p)-f(y)}{p-y}-f^{\prime}(p)\right| \rightarrow 0 .\right] \end{aligned}
\]
Prove that if \(f\) is twice differentiable at \(p,\) then
\[
f^{\prime \prime}(x)=\lim _{h \rightarrow 0} \frac{f(p+h)-2 f(p)+f(p-h)}{h^{2}} \neq \pm \infty .
\]
Does the converse hold (cf. Problem 9)?
In Example \((\mathrm{c}),\) find the three coordinate equations of the tangent line at \(p=\frac{1}{2} \pi .\)
Judging from Figure 22 in §2, discuss the existence, finiteness, and sign of the derivatives (or one-sided derivatives) of \(f\) at the points \(p_{i}\) indicated.
Let \(f : E^{n} \rightarrow E\) be linear, i.e., such that
\[
\left(\forall \overline{x}, \overline{y} \in E^{n}\right)\left(\forall a, b \in E^{1}\right) \quad f(a \overline{x}+b \overline{y})=a f(\overline{x})+b f(\overline{y}) .
\]
Prove that if \(g : E^{1} \rightarrow E^{n}\) is differentiable at \(p,\) so is \(h=f \circ g\) and \(h^{\prime}(p)=f\left(g^{\prime}(p)\right) .\)
[Hint: \(f\) is continuous since \(f(\overline{x})=\sum_{k=1}^{n} x_{k} f\left(\overline{e}_{k}\right) .\) See Problem 5 in Chapter 3, §§4-6.]