陶哲轩实分析（上）10.4及习题-Analysis I 10.4

反函数定理（实数轴上的），第二册中会有在n维欧式空间中的反函数定理。

Exercise 10.4.1

( a ) Since $f(x)=x^n,x∈(0,∞)$ is a continuous and strictly monotone increasing function, by Proposition 9.8.3, $f^{-1} (x)=x^{1/n}=g(x)$ is continuous on $(0,∞)$.
( b ) We already have $f' (x)=nx^{n-1},x∈(0,∞)$, since $g=f^{-1}, f$ is differentiable on $(0,∞)$, and $g$ is continuous by (a), we can use the inverse function theorem to say that $g$ is differentiable at any $x∈(0,∞)$, further for $x∈(0,∞)$, we have $x^{1/n}$ such that $f(x^{1/n} )=x$, thus
$g' (x)=\frac{1}{f'(x^{1/n})}=\frac{1}{n(x^{1/n})^{n-1} }=\frac{1}{n} x^{\frac{1-n}{n}}=\frac{1}{n} x^{1/n-1}$

Exercise 10.4.2

( a ) As $q∈\mathbf Q$, thus $∃m∈\mathbf Z,n∈\mathbf N^+$, s.t. $q=m/n$. By Exercise 10.4.1, we have $g(x)=x^{1/n}$ is differentiable on $(0,∞)$. And $f(x)=[g(x)]^m$, use Theorem 10.1.13 (d) and possibly (g), we can conclude $f$ is differentiable on $(0,∞)$. Use Theorem 10.1.15, we can say
$f' (x)=m[g(x)]^{m-1} g' (x)=mx^{\frac{m-1}{n}} \left(\frac{1}{n} x^{1/n-1} \right)=\frac{m}{n} x^{\frac{m-1}{n}+\frac{1-n}{n}}=qx^{\frac{m-n}{n}}=qx^{q-1}$
( b ) We have $f(1)=1^q=1$, thus by part(a), $f' (1)$ exists, so
$\lim_{x→1;x∈(0,∞)-\{1\} }⁡\frac{x^q-1}{x-1}=\lim_{x→1;x∈(0,∞)-\{1\}}⁡\frac{f(x)-f(1)}{x-1}=f' (1)=q1^{q-1}=q$
during which the second equality comes from Definition 10.1.1
Since the function $(x^q-1)/(x-1)$ is undefined at $x=1$, we conclude
$\lim_{x→1;x∈(0,∞)-\{1\} }⁡\frac{x^q-1}{x-1}=\lim_{x→1;x∈(0,∞)-\{1\} }\frac{x^q-1}{x-1}=q$

Exercise 10.4.3

( a ) Let $ϵ>0$ be an arbitrary small number, then as $α∈\mathbf R$, we can find $p,q∈\mathbf Q$, s.t.
$α-ϵ<p<α<q<α+ϵ$
If $x>1$, then $x^p<x^α<x^q$, thus we have
$\frac{x^p-1}{x-1}<\frac{x^α-1}{x-1}<\frac{x^q-1}{x-1}$
Use Exercise 10.4.2 we know that
$\lim_{x→1;x∈(1,∞) }\frac{x^p-1}{x-1}=p, \quad \lim_{x→1;x∈(1,∞) }⁡\frac{x^q-1}{x-1}=q$
Thus we can say that
$α-ϵ<p≤\lim_{x→1;x∈(1,∞) }⁡\frac{x^α-1}{x-1}≤q<α+ϵ$

Also if $x<1$, then $x^p>x^α>x^q$, but $x-1<0$, thus
$\frac{x^p-1}{x-1}<\frac{x^α-1}{x-1}<\frac{x^q-1}{x-1}$
and similarly we have
$α-ϵ<p≤\lim_{x→1;x∈(1,∞) }⁡\frac{x^α-1}{x-1}≤q<α+ϵ$
So we can conclude
$α-ϵ<\lim_{x→1;x∈(0,∞)-\{1\} }⁡⁡\frac{x^α-1}{x-1}<α+ϵ ⇒\lim_{x→1;x∈(0,∞)-\{1\} }⁡⁡\frac{x^α-1}{x-1}=α$

( b ) For $∀c∈(0,∞)$, by Theorem 10.1.13 $g(x)=(x/c)^α$ is differentiable on $(0,∞)$, thus
$\begin{aligned}\lim_{x/c→1;x/c∈(0,∞)-\{1\} }\frac{(x/c)^α-1}{x/c-1}=α& ⇒ \lim_{x→c;x∈(0,∞)-\{c\} }\frac{(x^α-c^α)/c^α}{(x-c)/c}=α \\&⇒ \lim_{x→c;x∈(0,∞)-\{c\} }\frac{x^α-c^α}{x-c}=αc^{α-1}\\&=\lim_{x→c;x∈(0,∞)-\{c\} }\frac{f(x)-f(c)}{x-c}\\&=f' (c)\end{aligned}$
Thus we can conclude $f$ is differentiable on $(0,∞)$, and $f' (x)=αx^{α-1}$.