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陶哲轩实分析11.1及习题-Analysis I 11.1

2019年5月27日2020年9月11日christangdt4条评论

这一节开始讲Riemann积分了，整个Riemann积分这一章的处理是比较新的，用相对好理解的piecewise constant function来定义上积分和下积分，二者相等即为可积，和以往比较经典的达布和其实是一回事，但接受起来就更容易。第一节主要说分法partition。

Exercise 11.1.1. Prove Lemma 11.1.4.
Lemma 11.1.4. Let $X$ be a subset of the real line. Then the following two statements are logically equivalent:
( a ) $X$ is bounded and connected.
( b ) $X$ is a bounded interval.

Solution: If $X={\emptyset}$ then the statement is obvious, since $X$ is bounded and connected and is an interval for trivial reasons.
Now suppose $X\neq {\emptyset}$ , then (b) implies (a) is clear, by the definition of an interval. Now suppose (a) holds, then $\exists M>0,X\subset [-M,M]$ , since $X$ is non-empty and bounded, we can say $\sup X$ and $\inf X$ both exist in $\mathbf R$ . Now choose any $x\in \mathbf R$ such that

$\inf X<x<\sup X$

By the definition of supremum and infimum, we can find $a,b\in X$ , s.t.

$\inf X\leq a<x<b\leq \sup X$

Since $X$ is connected, $[a,b]\subset X$ , thus $x\in X$ . So we have the relation $(\inf X,\sup X )\subset X$ , also we can easily have $X\subset [\inf X,\sup X ]$ , then $X$ is equal to $(\inf X,\sup X )$ plus possibly two points $\inf X$ and $\sup X$ , anyway $X$ is an interval.

$\blacksquare$

Exercise 11.1.2. Prove Corollary 11.1.6.
Corollary 11.1.6. If $I$ and $J$ are bounded intervals, then the intersection $I\cap J$ is also a bounded interval.

Solution: By Lemma 11.1.4, $I$ and $J$ are bounded and connected, we have $M_1>0,M_2>0$ , s.t.

$(I\subset [-M_1,M_1 ])\wedge (J\subset [-M_2,M_2 ])\implies I\cap J\subset [-\max (M_1,M_2 ),\max (M_1,M_2)]$

Thus $I\cap J$ is bounded.
Let $\forall x,y\in I\cap J$ , then as $x,y\in I$ , we have $[x,y]\subseteq I$ , similarly we have $[x,y]\subseteq J$ , so $[x,y]\subseteq I\cap J$ , this means $I\cap J$ is connected.
We now can conclude, using Lemma 11.1.4 again, that $I\cap J$ is a bounded interval.

$\blacksquare$

Exercise 11.1.3. Let $I$ be a bounded interval of the form $I=(a,b)$ or $I=[a,b)$ for some real numbers $a<b$ . Let $I_1,\dots,I_n$ be a partition of $I$ . Prove that one of the intervals $I_j$ in this partition is of the form $I_j=(c,b)$ or $I_j=[c,b)$ for some $a\leq c\leq b$ .

Solution: First, if $I_j$ is not of the form $(c,b)$ or $[c.b)$ for any $a\leq c\leq b$ , then we can have some $0<{\varepsilon}_j<b-a$ s.t. $b-{\varepsilon}\notin I_j$ , as $I_j$ is an interval, it’s connected, thus we have

$\sup I_j\leq b-{\varepsilon}_j<b$

Now assume all intervals $I_j,1\leq j\leq n$ is not of the form $(c,b)$ or $[c.b)$ for any $a\leq c\leq b$ , then we have $\sup I_j\leq b-{\varepsilon}_j<b,1\leq j\leq n$ . Then we can conclude

$b-\dfrac{\min_{1\leq j\leq n} ({\varepsilon}_j)}{2}\notin I_i,\quad\forall 1\leq i\leq n$

so we have $b-\min_{1\leq j\leq n} ({\varepsilon}_j )/2\notin I$ , this is a contradiction since $I_1,\dots ,I_n$ form a partition of $I$ .
Thus there must exists at least one of $I_j$ in the form of $(c,b)$ or $[c.b)$ . By the definition of partition, such $I_j$ can only appear once if $I_1,\dots ,I_n$ form a partition of $I$ .

$\blacksquare$

Exercise 11.1.4. Prove Lemma 11.1.18.
Lemma 11.1.18. Let $I$ be a bounded interval, and let $\mathbf P$ and $\mathbf P'$ be two partitions of $I$ . Then $\mathbf P \# \mathbf P'$ is also a partition of $I$ , and is both finer than $\mathbf P$ and finer than $\mathbf P'$ .

Solution:

$\mathbf P\#\mathbf P'$ is a partition of $I$ :
Choose $\forall x\in I$ , then there is exactly one $K\in \mathbf P$ and $J\in \mathbf P'$ s.t. $x\in K,x\in J$ , thus $x\in K\cap J$ , and any other element belongs to $\mathbf P\#\mathbf P'$ cannot contain $x$ , otherwise $K$ or $J$ can’t be unique. This means $\mathbf P\#\mathbf P'$ is a partition of $I$ .

$\mathbf P\#\mathbf P'$ is finer than $\mathbf P$ or $\mathbf P'$ :
For every $M\in \mathbf P\#\mathbf P'$ , by definition we shall be able to find $K\in \mathbf P$ and $J\in \mathbf P'$ s.t.

$M=K\cap J$

This means $M\subseteq K,K\in \mathbf P$ and $M\subseteq J,J\in \mathbf P'$ , thus we have $\mathbf P\#\mathbf P'$ is finer than $\mathbf P$ or $\mathbf P'$ .

$\blacksquare$

陶哲轩实分析10.5及习题-Analysis I 10.5

图片 2019年5月24日2020年1月8日christangdt留下评论

洛必达法则，计算极为好使，但使用需小心。

Exercise 10.5.1. Prove Proposition 10.5.1.
Proposition 10.5.1 (L’Hôpital’s rule I). Let $X$ be a subset of $\mathbf R$ , let $f:X\to\mathbf R$ and $g:X\to\mathbf R$ be functions, and let $x_0\in X$ be a limit point of $X$ . Suppose that $f(x_0)=g(x_0)=0$ , that $f$ and $g$ are both differentiable at $x_0$ , but $g'(x_0)\neq 0$ . Then there exists a $\delta >0$ such that $g(x)\neq 0$ for all $x\in (X\cap (x_0-\delta,x_0+\delta))-\{x_0\}$ , and

$\lim\limits_{x\to x_0;x\in (X\cap (x_0-\delta,x_0+\delta))-\{x_0\}}\dfrac{f(x)}{g(x)}=\dfrac{f'(x_0)}{g'(x_0)}$

Solution: By Proposition 10.1.7, since $g'(x_0 )\neq 0$ , we can find a ${\delta}>0$ such that for all $x\in X$ and $|x-x_0 |\leq {\delta}$ , we have

$|g(x)-g(x_0 )-g'(x_0 )(x-x_0 )|\leq \dfrac{|g'(x_0 )|}{2} |x-x_0 |$

Using triangle inequality and $g(x_0 )=0$ , we further get

$\dfrac{|g'(x_0 )|}{2} |x-x_0 |\leq |g(x)|$

Thus if $x\in \left(X\cap (x_0-{\delta},x_0+{\delta})\right)-\{x_0 \}$ , we shall have $|g(x)|>0$ , or $g(x)\neq 0$ . In this case we can safely use Proposition 9.3.14 to have

$\begin{aligned}\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} } \dfrac{f(x)}{g(x)}&=\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} } \frac{f(x)-f(x_0)}{g(x)-g(x_0 )}\\&=\dfrac{\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} }\dfrac{f(x)-f(x_0 )}{x-x_0 }}{\lim\limits_{x\to x_0;x\in (X\cap (x_0-{\delta},x_0+{\delta}))-\{x_0 \} }\dfrac{g(x)-g(x_0)}{x-x_0 }} \\&=\dfrac{f'(x_0)}{g'(x_0 ) }\end{aligned}$

$\blacksquare$

Exercise 10.5.2. Explain why Exercise 1.2.12 does not contradict either of the propositions in this section.

Solution: In the first example, we have $g(x)=1+x\nrightarrow 0$ if $x\to 0$ .
In the second example, notice that to use the L’Hôpital’s rule, we need first the limit

$\lim\limits_{x\to a;x\in (a,b]}\dfrac{f'(x)}{g'(x)}$

exist, then we can safely use the rule, but in this example, the limit

$\lim\limits_{x\to 0+;x\in (0,{\infty}) }\dfrac{(x^2 \sin (x^{-4}))'}{x'}$

does not exist, due to the calculations in the textbook. So it’s not safe to use the L’Hôpital’s rule.

$\blacksquare$

陶哲轩实分析10.4及习题-Analysis I 10.4

2019年5月24日2020年1月7日christangdt5条评论

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. Let $n\geq 1$ be a natural number, and let $g:(0,\infty)\to(0,\infty)$ be the function $g(x):=x^{1/n}$ .
( a ) Show that $g$ is continuous on $(0,\infty)$ .
( b ) Show that $g$ is differentiable on $(0,\infty)$ , and that $g'(x)=\frac{1}{n}x^{\frac{1}{n}-1}$ for all $x\in (0,\infty)$ .

Solution:

( a ) Since $f(x)=x^n,x\in (0,{\infty})$ is a continuous and strictly monotone increasing function, by Proposition 9.8.3, $f^{-1} (x)=x^{1/n}=g(x)$ is continuous on $(0,{\infty})$ .

( b ) We already have $f'(x)=nx^{n-1},x\in (0,{\infty})$ , since $g=f^{-1}, f$ is differentiable on $(0,{\infty})$ , and $g$ is continuous by (a), we can use the inverse function theorem to say that $g$ is differentiable at any $x\in (0,{\infty})$ , further for $x\in (0,{\infty})$ , we have $x^{1/n}$ such that $f(x^{1/n})=x$ , thus

$g'(x)=\dfrac{1}{f'(x^{1/n})}=\dfrac{1}{n(x^{1/n})^{n-1}}=\dfrac{1}{n} x^{\frac{1-n}{n}}=\dfrac{1}{n} x^{1/n-1}$

$\blacksquare$

Exercise 10.4.2. Let $q$ be a rational number, and let $f:(0,\infty)\to\mathbf R$ be the function $f(x)=x^q$ .
( a ) Show that $f$ is differentiable on $(0,\infty)$ and that $f'(x)=qx^{q-1}$ .
( b ) Show that $\lim_{x\to1;x\in (0,\infty)}\frac{x^q-1}{x-1}=q$ for every rational number $q$ .

Solution:

( a ) As $q\in \mathbf Q$ , thus $\exists m\in \mathbf Z,n\in \mathbf N^+$ , s.t. $q=m/n$ . By Exercise 10.4.1, we have $g(x)=x^{1/n}$ is differentiable on $(0,{\infty})$ . And $f(x)=[g(x)]^m$ , use Theorem 10.1.13 (d) and possibly (g), we can conclude $f$ is differentiable on $(0,{\infty})$ . Use Theorem 10.1.15, we can say

$\begin{aligned}f'(x)&=m[g(x)]^{m-1} g'(x)=mx^{\frac{m-1}{n}} \left(\dfrac{1}{n} x^{1/n-1} \right)\\&=\dfrac{m}{n} x^{\frac{m-1}{n}+\frac{1-n}{n}}=qx^{\frac{m-n}{n}}=qx^{q-1}\end{aligned}$

( b ) We have $f(1)=1^q=1$ , thus by part(a), $f'(1)$ exists, so

$\lim\limits_{x\to 1;x\in (0,{\infty})-\{1\}}\dfrac{x^q-1}{x-1}=\lim\limits_{x\to 1;x\in (0,{\infty})-\{1\}}\dfrac{f(x)-f(1)}{x-1}=f'(1)=q1^{q-1}=q$

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

$\lim\limits_{x\to 1;x\in (0,{\infty})-\{1\}}\dfrac{x^q-1}{x-1}=\lim\limits_{x\to 1;x\in (0,{\infty})-\{1\}}\dfrac{x^q-1}{x-1}=q$

$\blacksquare$

Exercise 10.4.3. Let $\alpha$ be a real number, and let $f:(0,\infty)\to\mathbf R$ be the function $f(x)=x^{\alpha}$ .
( a ) Show that $\lim_{x\to 1;x\in (0,\infty)\backslash \{1\}}\frac{f(x)-f(1)}{x-1}=\alpha$ .
( b ) Show that $f$ is differentiable on $(0,\infty)$ and that $f'(x)={\alpha}x^{\alpha -1}$ .

Solution:

( a ) Let ${\varepsilon}>0$ be an arbitrary small number, then as ${\alpha}\in \mathbf R$ , we can find $p,q\in \mathbf Q$ , s.t.

${\alpha}-{\varepsilon}<p<{\alpha}<q<{\alpha}+{\varepsilon}$

If $x>1$ , then $x^p<x^{\alpha}<x^q$ , thus we have

$\dfrac{x^p-1}{x-1}<\dfrac{x^{\alpha}-1}{x-1}<\dfrac{x^q-1}{x-1}$

Use Exercise 10.4.2 we know that

$\lim\limits_{x\to 1;x\in (1,{\infty})}\dfrac{x^p-1}{x-1}=p, \quad \lim\limits_{x\to 1;x\in (1,{\infty}) }\dfrac{x^q-1}{x-1}=q$

Thus we can say that

${\alpha}-{\varepsilon}<p\leq \lim\limits_{x\to 1;x\in (1,{\infty}) }\dfrac{x^{\alpha}-1}{x-1}\leq q<{\alpha}+{\varepsilon}$

Also if $x<1$ , then $x^p>x^{\alpha}>x^q$ , but $x-1<0$ , thus

$\dfrac{x^p-1}{x-1}<\dfrac{x^{\alpha}-1}{x-1}<\dfrac{x^q-1}{x-1}$

and similarly we have

${\alpha}-{\varepsilon}<p\leq \lim\limits_{x\to 1;x\in (1,{\infty}) }\dfrac{x^{\alpha}-1}{x-1}\leq q<{\alpha}+{\varepsilon}$

So we can conclude

${\alpha}-{\varepsilon}<\lim\limits_{x\to 1;x\in (0,{\infty})-\{1\}}\dfrac{x^{\alpha}-1}{x-1}<{\alpha}+{\varepsilon} \implies \lim\limits_{x\to 1;x\in (0,{\infty})-\{1\}}\dfrac{x^{\alpha}-1}{x-1}={\alpha}$

( b ) For $\forall c\in (0,{\infty})$ , by Theorem 10.1.13 $g(x)=(x/c)^{\alpha}$ is differentiable on $(0,{\infty})$ , thus

$\begin{aligned}&\lim\limits_{x/c\to 1;x/c\in (0,{\infty})-\{1\} }\dfrac{(x/c)^{\alpha}-1}{x/c-1}={\alpha} \implies \lim\limits_{x\to c;x\in (0,{\infty})-\{c\}}\dfrac{(x^{\alpha}-c^{\alpha})/c^{\alpha}}{(x-c)/c}={\alpha} \\&\implies \lim\limits_{x\to c;x\in (0,{\infty})-{c} }\dfrac{x^{\alpha}-c^{\alpha}}{x-c}={\alpha}c^{{\alpha}-1}=\lim\limits_{x\to c;x\in (0,{\infty})-{c} }\dfrac{f(x)-f(c)}{x-c}=f'(c)\end{aligned}$

Thus we can conclude $f$ is differentiable on $(0,{\infty})$ , and $f'(x)={\alpha}x^{{\alpha}-1}$ .

$\blacksquare$

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