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陶哲轩实分析（下）1.1及习题-Analysis II 1.1

2019年6月5日2019年6月5日christangdt5条评论

1.1节先从实数的距离引入一个等价的收敛定义（1.1.1），而后定义metric spaces（1.1.2），列出一些例子：实数轴是metric space（1.1.4），metric spaces子集上形成的induced metric spaces（1.1.5），Euclidean空间是metric space（1.1.6），Taxi-cab度量（1.1.7），sup norm度量（1.1.9），三者实际上是等价的；继续举例Discrete metric（1.1.11），Geodesics（1.1.12），shortest paths（1.1.13）等；而后作为（1.1.1）的一般形式，引出在metric spaces上的收敛定义（1.1.14），并举了一个在欧式空间使用 $d_{l^{2}}$ 收敛的序列例子（1.1.17）；最后是几个有用的结论：三种度量的等价性（1.1.18），离散空间的收敛实质是最终恒等于某一个点（1.1.19），极限是唯一的（1.1.20），当然极限的唯一性只在Hausdorff空间中成立；（1.1.21）说明使用不同的metric，收敛的点会不一样。

陶哲轩实分析11.10及习题-Analysis I 11.10

图片 2019年6月4日2020年1月13日christangdt2条评论

Exercise 11.10.1. Prove Proposition 11.10.1.
Proposition 11.10.1 (Integration by parts formula). Let $I=[a,b]$ , and let $F:[a,b]\to\mathbf R$ and $G:[a,b]\to\mathbf R$ be differentiable functions on $[a,b]$ such that $F'$ and $G'$ are Riemann integrable on $I$ . Then we have

$\displaystyle{\int_{[a,b]}FG'=F(b)G(b)-F(a)G(a)-\int_{[a,b]}F'G}$ .

Solution: As $F$ and $G$ are differentiable, they are continuous on $[a,b]$ and thus Riemann integrable by Corollary 11.5.2, then $FG'$ and $F'G$ are Riemann integrable by Theorem 11.4.5. Notice that $(FG)'=F' G+FG'$ and $F'G+FG'$ is Riemann integrable on $[a,b]$ , we can use the second fundamental theorem of calculus to have

$\displaystyle{\int_{[a,b]}(F'G+FG')=(FG)(b)-(FG)(a)}$

By some simple calculation we can have the final result.

$\blacksquare$

Exercise 11.10.2. fill in the gaps marked (why?) in the proof of Lemma 11.10.5.
Lemma 11.10.5 (Change of variables formula I). Let $[a,b]$ be a closed interval, and let $\phi:[a,b]\to [\phi (a),\phi (b)]$ be a continuous monotone increasing function. Let $f:[\phi (a),\phi (b)]\to\mathbf R$ be a piecewise constant function on $[\phi (a),\phi (b)]$ . Then $f\circ \phi:[a,b]\to\mathbf R$ is also piecewise constant on $[a,b]$ , and

$\displaystyle{\int_{[a,b]}f\circ \phi d\phi =\int_{[\phi (a),\phi (b)]}f}$ .

Solution: I will prove Lemma 11.10.5 completely.
Let $\mathbf P$ be a partition of $[{\phi}(a),{\phi}(b)]$ such that $f$ is piecewise constant with respect to $\mathbf P$ . we may assume that $\mathbf P$ does not contain the empty set. For each $J\in \mathbf P$ , let $c_J$ be the constant value of $f$ on $J$ , thus

$\displaystyle{\int_{[{\phi}(a),{\phi}(b)]} f=\sum_{J\in \mathbf P}c_J |J|}$

For each interval $J$ , let ${\phi}^{-1} (J):=\{x\in [a,b]:{\phi}(x)\in J\}$ . Then given any $x,y\in {\phi}^{-1} (J),x<y$ , we have ${\phi}(x),{\phi}(y)\in J$ by the definition of ${\phi}^{-1} (J)$ , and ${\phi}(x)\leq {\phi}(y)$ since ${\phi}$ is monotone increasing, and $\forall z\in [x,y]$ , we have ${\phi}(x)\leq {\phi}(z)\leq {\phi}(y)$ again by ${\phi}$ is monotone increasing, this means ${\phi}(z)\in J$ since $J$ is an interval, so ${\phi}^{-1} (J)$ is connected, and thus is an interval. Furthermore, if $x\in {\phi}^{-1} (J)$ , then ${\phi}(x)\in J$ and $f({\phi}(x))=c_J$ , thus $c_J$ is the constant value of $f\circ {\phi}$ on ${\phi}^{-1}(J)$ .
Thus if we define $\mathbf Q:=\{{\phi}^{-1}(J):J\in \mathbf P\}$ , then for any $c\in [a,b], {\phi}(c)\in [{\phi}(a),{\phi}(b)]$ , so there is a $J\in \mathbf P$ such that ${\phi}(c)\in J$ , which means $c\in {\phi}^{-1}(J)$ . Assume we can find two sets $Q_1,Q_2\in \mathbf Q$ s.t. $c\in Q_1\cap Q_2$ , then it means we can find $J_1,J_2\in \mathbf P$ s.t. ${\phi}(c)\in J_1\cap J_2$ , contradict the fact that $\mathbf P$ is a partition of $[{\phi}(a),{\phi}(b)]$ . We can conclude $Q$ is a partition of $[a,b]$ .
Given any $Q\in \mathbf Q$ , we have $Q={\phi}^{-1}(J)$ for some $J\in \mathbf P$ , so $c_J$ is the constant value of $f\circ {\phi}$ on this $Q$ , which means $f\circ {\phi}$ piecewise constant with respect to $\mathbf Q$ . Thus

$\displaystyle{\int_{[a,b]}f\circ {\phi} d{\phi}=\int_{[\mathbf Q]}f\circ {\phi} d{\phi}=\sum_{J\in \mathbf P}c_J {\phi}[{\phi}^{-1}(J)] }$

To calculate ${\phi}[{\phi}^{-1}(J)]$ , we let $J\in \mathbf P$ , then $J$ is an interval with endpoints $a<b$ , and $|J|=b-a$ , as ${\phi}$ is monotone increasing, we can find $a',b'\in {\phi}^{-1} (J)$ , s.t. ${\phi}(a' )=a,{\phi}(b')=b$ , and ${\phi}^{-1}(J)$ is an interval which has endpoints $a',b'$ , thus ${\phi}[{\phi}^{-1}(J)]={\phi}(b' )-{\phi}(a')=b-a=|J|$ , and the claim follows.

$\blacksquare$

Exercise 11.10.3. Let $a<b$ be real numbers, and let $f:[a,b]\to\mathbf R$ be Riemann integrable function. Let $g:[-b,-a]\to\mathbf R$ be defined by $g(x):=f(-x)$ . Show that $g$ is also Riemann integrable, and $\int_{[-b,-a]}g=\int_{[a,b]}f$ .

Solution: For $\forall {\varepsilon}>0$ , we can find a partition of $[a,b]$ , namely $\mathbf P$ , and piecewise constant function $\overline{f}$ which majorizes $f$ and $\underline{f}$ which minorizes $f$ , both with respect to $\mathbf P$ , such that

$\displaystyle{\int_{[a,b]}f-{\varepsilon}<\int_{[a,b]}\underline{f}\leq \int_{[a,b]}\overline{f}<\int_{[a,b]}f+{\varepsilon}}$

Now we define $\overline{g}(x)=\overline{f}(-x)$ and $\underline{g}(x)=\underline{f}(-x)$ on $[-b,-a]$ , then it is easy to see that $\overline{g}$ majorizes $g$ and $\underline{g}$ minorizes $g$ . For any $J\in \mathbf P$ , we define $K_J=\{-x:x\in J\}$ , then $|K_J |=|J|$ , and $\mathbf P'=\{K_J:J\in \mathbf P\}$ is a partition of $[-b,-a]$ , and the constant value of $\overline{g}$ on $K_J$ is the same as the constant value of $\overline{f}$ on $J$ , the constant value of $\underline{g}$ on $K_J$ is the same as the constant value of $\underline{f}$ on $J$ , so we have

$\displaystyle{\int_{[-b,-a]}\overline{g}=p.c.\int_{[-b,-a]}\overline{g}=\sum_{K_J\in \mathbf P'}c_J |K_J|=\sum_{J\in \mathbf P}c_J |J| =\int_{[a,b]}\overline{f}}$

Similarly we have

$\displaystyle{\int_{[-b,-a]}\underline{g}=\int_{[a,b]}\underline{f}}$

So we have

$\displaystyle{\int_{[a,b]}f-{\varepsilon}<\int_{[-b,-a]}\underline{g}\leq \underline{\int}_{[-b,-a]}g\leq \overline{\int}_{[-b,-a]}g\leq \int_{[-b,-a]}\overline{g}<\int_{[a,b]}f+{\varepsilon}}$

and the conclusion follows.

$\blacksquare$

Exercise 11.10.4. What is the analogue of Proposition 11.10.7 when $\phi$ is monotone decreasing instead of monotone increasing?

Solution: Let $[a,b]$ be a closed interval, and let ${\phi}:[a,b]\to [{\phi}(b),{\phi}(a)]$ be a differentiable monotone decreasing function such that ${\phi}'$ is Riemann integrable. Let $f: [{\phi}(b),{\phi}(a)]\to \mathbf R$ be a Riemann integrable function on $[{\phi}(b),{\phi}(a)]$ . Then $(f\circ {\phi}) {\phi}':[a,b]\to \mathbf R$ is Riemann integrable on $[a,b]$ and

$\displaystyle{\int_{[a,b]}(f\circ {\phi}) {\phi}'=-\int_{[{\phi}(b),{\phi}(a)]}f}$

$\blacksquare$

陶哲轩实分析11.9及习题-Analysis I 11.9

图片 2019年6月4日2020年3月4日christangdt7条评论

Exercise 11.9.1. Let $f:[0,1]\to\mathbf R$ be the function in Exercise 9.8.5. Show that for every rational number $q\in \mathbf Q\cap [0,1]$ , the function $F:[0,1]\to\mathbf R$ defined by the formula $F(x):=\int_0^xf(y)dy$ is not differentiable at $q$ .

Solution: Since $f$ is monotonic increasing, $f$ is Riemann ingegrable and $F$ is well-defined.
We assume $F(x)$ is differentiable at some $q\in Q\cap [0,1]$ , then if $q\neq 0,q\neq 1$ , $F$ is differentiable at $q$ iff we have

$\displaystyle{\lim_{x\to q;x\in [0,q)}\frac{F(x)-F(q)}{x-q}=\lim_{x\to q;x\in (q,1]}\frac{F(x)-F(q)}{x-q}}$

From Exercise 9.8.5 we know that $f$ is not continuous at $q$ and if we locate $q=q(n)$ for some $n\in \mathbf N$ as in Exercise 9.8.5, we shall have $f(x)\geq f(q)+2^{-n},x>q$ , so

$\lim\limits_{x\to q;x\in [0,q)}\dfrac{F(x)-F(q)}{x-q}=\lim\limits_{x\to q;x\in [0,q)}\dfrac{\int_x^qf(y)dy}{x-q}\leq \lim\limits_{x\to q;x\in [0,q)}\dfrac{\int_x^qf(q)dy}{x-q}=f(q) \\ \lim\limits_{x\to q;x\in (q,1]}\dfrac{F(x)-F(q)}{x-q}=\lim\limits_{x\to q;x\in (q,1]}\dfrac{\int_q^xf(y)dy}{x-q}\geq \lim\limits_{x\to q;x\in (q,1]}\dfrac{\int_q^x(f(q)+2^{-n})dy}{x-q}=f(q)+2^{-n}$

thus

$\displaystyle{\lim_{x\to q;x\in (q,1]}\frac{F(x)-F(q)}{x-q}-\lim_{x\to q;x\in [0,q)}\frac{F(x)-F(q)}{x-q}\geq 2^{-n}}$

contradicts the result that the two limits are equal.
If $q=0$ or $q=1$ , then $f$ is discontinuous at 0 or 1 when restricted on $[0,1]$ , due to Exercise 9.8.5, thus $F$ is not differentiable, as the limit $\lim_{x\to 0;x\in (0,1]}\frac{F(x)}{x}$ and $\lim_{x\to 1;x\in [0,1)}\frac{F(x)-F(1)}{x-1}$ do not exist.

$\blacksquare$

Exercise 11.9.2. Prove Lemma 11.9.5.
Lemma 11.9.5 Let $I$ be a bounded interval, and let $f:I\to\mathbf R$ be a function. Let $F:I\to\mathbf R$ and $G:I\to\mathbf R$ be two antiderivatives of $f$ . Then there exists a real number $C$ such that $F(x)=G(x)+C$ for all $x\in I$ .

Solution: When $I$ is empty or a single point then the statement is trivial. Now suppose $I$ is a nonempty interval, choose some $x_0\in I$ , then for any $x\in I,x\neq x_0$ , the function $F-G$ is differentiable on $[x_0,x]$ or $[x,x_0]$ , thus we have some ${\xi}$ between $x$ and $x_0$ , s.t.

$\displaystyle{\frac{(F-G)(x)-(F-G)(x_0)}{x-x_0}=(F-G)'({\xi})=f({\xi})-f({\xi})=0}$

So we have

$\displaystyle{(F-G)(x)=F(x)-G(x)=F(x_0 )-G(x_0 )}$

This means if we let $C=F(x_0)-G(x_0)$ , we will have $F(x)=G(x)+C,\forall x\in I,x\neq x_0$ , since we also have $F(x_0 )=G(x_0 )+C$ valid, the conclusion is true.
Alternatively, we can use the second Fundamental theorem of calculus, since

$\displaystyle{(F-G)'(x)=0,\quad \forall x\in I}$

the function $F-G:I\to \mathbf R$ is an antiderivative of the function $g(x)=0,x\in I$ . Since $g$ is constant, it is Riemann integrable, so choose some $x_0\in I$ , then for any $x\in I,x\neq x_0$ , we can use the second Fundamental theorem of calculus on $[x_0,x]$ or $[x,x_0]$ and have

$\displaystyle{\int_{[x_0,x]}g=0=(F-G)(x)-(F-G)(x_0)}$

$\displaystyle{\int_{[x,x_0]}g=0=(F-G)(x_0 )-(F-G)(x)}$

In either case we have $(F-G)(x_0)=(F-G)(x)$ , This means if we let $C=F(x_0)-G(x_0)$ , we will have $F(x)=G(x)+C,\forall x\in I,x\neq x_0$ . Since we also have $F(x_0)=G(x_0)+C$ valid, the conclusion is true.

$\blacksquare$

Exercise 11.9.3. Let $a<b$ be real numbers, and let $f:[a,b]\to\mathbf R$ be a monotone increasing function. Let $F:[a,b]\to\mathbf R$ be the fucntion $F(x):=\int_{[a,x]}f$ . Let $x_0$ be an element of $[a,b]$ . Show that $F$ is differentiable at $x_0$ if and only if $f$ is continuous at $x_0$ .

Solution: Since $f$ is monotone increasing, $f$ is Riemann integrable, thus if $f$ is continuous at $x_0$ , then $F$ is differentiable at $x_0$ by the first Fundamental theorem of calculus.
Conversely we let $F$ is differentiable at $x_0$ and assume $f$ is not continuous at $x_0$ , then we shall have $f(x_0-)0$ , s.t.

$\displaystyle{f(x_0-)<{\eta}-{\varepsilon}<{\eta}<{\eta}+{\varepsilon}<f(x_0+)}$

As $f$ is monotone increasing, we know $f(x)<{\eta}-{\varepsilon}$ if $x<x_0$ and $f(x)>{\eta}+{\varepsilon}$ if $x>x_0$ . Then for $y>x_0$ , we have

$\displaystyle{F(y)-F(x_0 )=\int_{[x_0,y]}f>\int_{[x_0,y]}({\eta}+{\varepsilon}) =({\eta}+{\varepsilon})(y-x_0 )}$

And for $y<x_0$ , we have

$\displaystyle{F(x_0 )-F(y)=\int_{[y,x_0]}f<\int_{[y,x_0]}({\eta}-{\varepsilon}) =({\eta}-{\varepsilon})(x_0-y)}$

So we have

$\lim\limits_{y\to x_0+}\dfrac{F(y)-F(x_0)}{y-x_0}\geq {\eta}+{\varepsilon}>{\eta} \\ \lim\limits_{y\to x_0-}\dfrac{F(y)-F(x_0)}{y-x_0}=\lim\limits_{y\to x_0-}\dfrac{F(x_0)-F(y)}{x_0-y}\leq {\eta}-{\varepsilon}<{\eta}$

thus $F'(x_0+)>F'(x_0-)$ , a contradiction to $F$ being differentiable at $x_0$ .

$\blacksquare$

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