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陶哲轩实分析9.10及习题-Analysis I 9.10

2019年5月21日2019年12月21日christangdt留下评论

limits at infinity, 也是很短的一节，但其事实上起到了将极限定义从R延拓到R*的作用。

Exercise 9.10.1. Let $(a_n)_{n=0}^{\infty}$ be a sequence of real numbers, then $a_n$ can also be thought of as a function from $\mathbf N$ to $\mathbf R$ , which takes each natural number $n$ to a real number $a_n$ . Show that

$\lim\limits_{n\to\infty;n\in \mathbf N}a_n=\lim\limits_{n\to\infty}a_n$

where the left-hand limit is defined by Definition 9.10.3 and the right-hand limit is defined by Definition 6.1.8. More precisely, show that if one of the above two limits exists then so does the other, and then they both have the same value. Thus the two notions of limit here are compatible.

Solution: First suppose $\lim_{n\to\infty;n\in \mathbf N}a_n=L$ , then $\forall \varepsilon >0,\exists M>0$ , s.t.

$|a_n-L|<\varepsilon,\quad,\forall n\in \mathbf N, n>M$

We let $N=[M]+1>M$ , then if $n>N$ , we shall have $|a_n-L|<\varepsilon$ , this means $\lim_{n\to\infty}a_n=L$ .

Conversely, suppose $\lim_{n\to\infty}a_n=L$ , then $\forall \varepsilon >0,\exists N\in \mathbf N,N>0$ , s.t.

$|a_n-L|<\varepsilon,\quad,\forall n\in \mathbf N, n>N$

We choose $M=N$ in Definition 9.10.3, then we can see $\lim_{n\to\infty;n\in \mathbf N}a_n=L$ .

$\blacksquare$

陶哲轩实分析9.9及习题-Analysis I 9.9

2019年5月20日2019年12月21日christangdt5条评论

这一节讲一致连续，是全局性质或者区间性质而非点的性质。一致连续初看不太好理解，但在后续章节中就会发现一致连续是比逐点连续更强的性质。

Exercise 9.9.1. Prove Lemma 9.9.7.
Lemma 9.9.7. Let $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ be sequences of real numbers (not necessarily bounded or convergent). Then $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ are equivalent if and only if $\lim_{n\to\infty}(a_n-b_n)=0$ .

Solution: If $(a_n)_{n=1}^{\infty}$ and $(b_n)_{n=1}^{\infty}$ are equivalent, then $\forall \varepsilon >0,\exists N>1$ , s.t. $(a_n)_{n=N}^{\infty}$ and $(b_n)_{n=N}^{\infty}$ are $\varepsilon$ -close. Thus

$|a_n-b_n |<\varepsilon ,\quad \forall n>N$

This means $\lim_{n\to {\infty}}(a_n-b_n)=0$ , and all steps above can be reversed.

$\blacksquare$

Exercise 9.9.2. Prove Proposition 9.9.8.
Proposition 9.9.8. Let $X$ be a subset of $\mathbf R$ , and let $f:X\to\mathbf R$ be a function. Then the following two statements are locigally equivalent:
( a ) $f$ is uniformly continuous on $X$ .
( b ) Whenever $(x_n)_{n=0}^{\infty}$ and $(y_n)_{n=0}^{\infty}$ are two equivalent sequences consisting of elements of $X$ , the sequences $f((x_n))_{n=0}^{\infty}$ and $f((y_n))_{n=0}^{\infty}$ are also equivalent.

Solution:

( a ) implies ( b ):
$\forall \varepsilon >0,\exists \delta >0$ , s.t. $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$ . As $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are equivalent, we can find $N$ such that $|x_n-y_n |<\delta ,\forall n>N$ , so we shall get

$|f(x_n )-f(y_n)|<\varepsilon ,\quad \forall n>N$

This means the sequence $(f(x_n))_{n=1}^{\infty}$ and $(f(y_n))_{n=1}^{\infty}$ are equivalent.

( b ) implies ( a ):
Assume $f$ is not uniformly continuous on $X$ , then we shall be able to find a $\varepsilon _0>0$ s.t. for every $n\in \mathbf N,\exists x_n,y_n\in X,|x_n-y_n|<1/n$ , but $|f(x_n )-f(y_n)|\geq \varepsilon_0$ . By Archimedean principle we know the sequence $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ are equivalent, but

$\lim\limits_{n\to {\infty}}(f(x_n )-f(y_n))\geq \varepsilon _0>0$

which means the sequence $(f(x_n))_{n=1}^{\infty}$ and $(f(y_n))_{n=1}^{\infty}$ are not equivalent, a contradiction.

$\blacksquare$

Exercise 9.9.3. Prove Proposition 9.9.12.
Prosition 9.9.12. Let $X$ be a subset of $\mathbf R$ , and let $f:X\to\mathbf R$ be a uniformly continuous function. Let $(x_n)_{n=0}^{\infty}$ be a Cauchy sequence consisting entirely of elements in $X$ . Then $(f(x_n))_{n=0}^{\infty}$ is also a Cauchy sequence.

Solution: $f$ is uniformly continuous on $X$ means $\forall \varepsilon >0, \exists \delta >0$ , s.t. $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$ . As $(x_n )_{n=0}^{\infty}\in X$ is a Cauchy sequence, we can find $N>0$ s.t.

$|x_n-x_m|<\delta ,\quad \forall n,m>N$

This means

$|f(x_m )-f(x_n)|<\varepsilon ,\quad \forall n,m>N$

which shows $(f(x_n ))_{n=0}^{\infty}$ is a Cauchy sequence.

$\blacksquare$

Exercise 9.9.4. Use Proposition 9.9.12. to prove Corollary 9.9.14. Use this corollary to give an alternate demonstration of the results in Example 9.9.10.
Corollary 9.9.14. Let $X$ be a subset of $\mathbf R$ , and let $f:X\to\mathbf R$ be a uniformly continuous function, and let $x_0$ be an adherent point of $X$ . Then the limit $\lim_{x\to x_0;x\in X}f(x)$ exists (in particular, it is a real number).

Solution: Since $x_0$ is an adherent point of $X$ , there is a sequence $(x_n)_{n=1}^{\infty}\in X$ which converges to $x_0$ , in particular, this sequence is a Cauchy sequence by Theorem 6.4.18, thus by Proposition 9.9.12, the sequence $(f(x_n))_{n=1}^{\infty}$ is a Cauchy sequence, by Theorem 6.4.18, $(f(x_n))_{n=1}^{\infty}$ is convergent, let the number being convergent be $L\in \mathbf R$ . To finish the proof, let $(x_n')_{n=1}^{\infty}\in X$ be any sequence which converges to $x_0$ , then $(x_n')_{n=1}^{\infty}$ and $(x_n)_{n=1}^{\infty}$ are equivalent sequence, thus $(f(x_n'))_{n=1}^{\infty}$ and $(f(x_n))_{n=1}^{\infty}$ are also equivalent sequence, thus $(f(x_n'))_{n=1}^{\infty}$ converges to $L$ . By Proposition 9.3.9, $f$ converges to $L$ at $x_0$ . An alternate demonstration of Example 9.9.10: if $f$ is uniformly continuous, then since $0$ is an adherent point of the sequence $(0,2)$ , the limit $\lim_{x\to 0;x\in (0,2)}1/x$ should exist, but in fact $\lim_{x\to 0;x\in (0,2)}1/x=+{\infty}$ .

$\blacksquare$

Exercise 9.9.5. Prove Proposition 9.9.15.
Proposition 9.9.15. Let $X$ be a subset of $\mathbf R$ , and let $f:X\to\mathbf R$ be a uniformly continuous function. Suppose that $E$ is abounded subset of $X$ . Then $f(E)$ is also bounded.

Solution: Assume $f(E)$ is unbounded, then $\forall n\in \mathbf N^+$ , we can find $x_n\in E$ , s.t. $|f(x_n )|>n$ . Since $E$ is bounded, we know the sequence $(x_n)_{n=1}^{\infty}$ is bounded, thus has a convergent subsequence $(x_{n_j})_{j=1}^{\infty}$ , we denote $\lim_{j\to {\infty}}x_{n_j }=x_0$ , by Corollary 9.9.14, $\lim_{j\to {\infty}}f(x_{n_j})$ exists, thus is bounded, but $|f(x_{n_j} )|>n_j>j$ , which is a contradiction.

$\blacksquare$

Exercise 9.9.6. Let $X,Y,Z$ be subsets of $\mathbf R$ . Let $f:X\to Y$ be a function which is uniformly continuous on $X$ , and let $g:Y\to Z$ be a function which is uniformly continuous on $Y$ . Show that the function $g\circ f:X\to Z$ is uniformly continuous on $X$ .

Solution: Let $\forall \varepsilon >0$ , then since $g$ is uniformly continuous, $\exists \delta _1>0$ , s.t. for $\forall y,y'\in Y,|y-y' |<\delta _1$ we have

$|g(y)-g(y')|<\varepsilon$

For this $\delta _1$ , since $f$ is uniformly continuous, $\exists \delta _2>0$ , s.t. for $\forall x,x'\in X,|x-x' |<\delta _2$ we have

$|f(x)-f(x')|<\delta _1$

So for $\forall x,x'\in X,|x-x' |<\delta _2$ we have

$|(g\circ f)(x)-(g\circ f)(x')|=|g(f(x))-g(f(x' ))|<\varepsilon$

Thus $g\circ f$ is uniformly continuous on $X$ .

$\blacksquare$

陶哲轩实分析9.8及习题-Analysis I 9.8

2019年5月16日2019年12月21日christangdt2条评论

这一节讲单调函数，虽然也很短，但单调函数性质很好，易应用，给出的两个例子一个是反函数的存在性（在讲反函数微分时会用到），另一个是构建n次方根的alternative method。习题不错

Exercise 9.8.1. Explain why the maximum principle remains true if the hypothesis that $f$ is continuous is replaced with $f$ being monotone, or with $f$ being strictly monotone.

Solution: If $a<b$ and $f:[a,b]\to \mathbf R$ is a monotonic function on $[a,b]$ , then suppose first $f$ is monotonic increasing, we have $f(b)\geq f(x),\forall x\in [a,b]$ , and $f(a)\leq f(x),\forall x\in [a,b]$ , so $f$ attain its maximum at $b$ and its minimum at $a$ , if $f$ is strictly increasing, the proof is the same.
Suppose next that $f$ is monotonic decreasing, we have $f(a)\geq f(x),\forall x\in [a,b]$ , and $f(b)\leq f(x),\forall x\in [a,b]$ , so $f$ attain its maximum at a and its minimum at $b$ , if $f$ is strictly increasing, the proof is the same.

$\blacksquare$

Exercise 9.8.2. Give an example to show that the intermediate value theorem becomes false if the hypothesis that $f$ is continuous is replaced with $f$ being monotone, or with $f$ being strictly monotone.

Solution: Let $f:[-1,1]\to \mathbf R$ be defined as:

$f(x)=\begin{cases}x,& x\in [-1,0)\\x+1,&x\in [0,1] \end{cases}$

Then $f$ is strictly monotone (thus monotone), $f(-1)=-1$ and $f(1)=2$ , but no element in $[-1,1]$ could make $f(x)=1/2$ .

$\blacksquare$

Exercise 9.8.3. Let $a<b$ be real numbers, and let $f:[a,b]\to\mathbf R$ be a function which is both continuous and one-to-one. Show that $f$ is strictly monotone.

Solution: We shall divide into three cases: $f(a)=f(b),f(a)f(b)$ . The first case is obviously a contradiction to $f$ is one-to-one. So we only consider the last two cases.
Case $f(a)f(y)$ . Since $xf(y)$ , then $f(y)f(b)$ can be similarly proved. In this case $f$ is strictly decreasing.
In conclusion $f$ is strictly monotone.

$\blacksquare$

Exercise 9.8.4. Prove Proposition 9.8.3.Is the proposition still true if the continuity assumption is dropped, or if stricy monotonicity is replaced just by monotonicity? How should one modify the proposition to deal with strictly monotone decreasing functions instead of strictly monotone increasing functions?

Solution: $f$ is strictly monotone increasing means $f(a)<f(b)$ , and for $\forall x,y\in [a,b],f(x)<f(y)$ , thus $f$ is injective. By the intermediate value theorem, $\forall y\in [f(a),f(b)],\exists c\in [a,b],f(c)=y$ , thus $f$ is surjective. In conclusion $f$ is a bijection from $[a,b]$ to $[f(a),f(b)]$ .

To prove $f^{-1}$ is continuous on $[f(a),f(b)]$ , choose some $y_0\in [f(a),f(b)]$ , then $\exists x_0\in [a,b],f(x_0 )=y_0$ . Now for $\forall \varepsilon >0$ , let

$\delta =\begin{cases}\min \{y_0-f(a),f(x_0+\varepsilon )-y_0 \},&a\leq x_0<a+\varepsilon \\ \min \{y_0-f(x_0-\varepsilon ),f(x_0+\varepsilon )-y_0 \},&a+\varepsilon \leq x_0\leq b-\varepsilon \\ \min \{y_0-f(x_0-\varepsilon ),f(b)-y_0 \}, & b-\varepsilon <x_0\leq b\end{cases}$

Then as long as $y\in [f(a),f(b)],|y-y_0 |<\delta$ , we can have

$|f^{-1} (y)-f^{-1} (y_0)|=|f^{-1} (y)-x_0 |<\varepsilon$

Thus $f^{-1}$ is continuous.

To prove $f^{-1}$ is monotonic increasing on $[f(a),f(b)]$ , let $\forall y_1,y_2\in [f(a),f(b)],y_1<y_2$ , then $\exists x_1,x_2\in [a,b]$ , s.t. $f(x_1 )=y_1,f(x_2 )=y_2$ . Now we must have $x_1<x_2$ , otherwise contradicting $f$ being strictly monotone increasing. This means $f^{-1} (y_1 )<f^{-1} (y_2 )$ .

If the continuity assumption is dropped, the proposition is false, a counterexample may be

$f:[0,2]\to [0,3]:=\begin{cases}x,& x\in [0,1)\\x+1,& x\in [1,2] \end{cases}$

which is not a surjection to $[0,3]$ . The inverse $f^{-1}$ exists but is not continuous.

If strict monotonicity is replaced by monotonicity, the proposition is false, a counterexample may be

$f:[0,2]\to \mathbf R:=1,\quad x\in [0,2]$

which is not an injection, and $f^{-1}$ doesn’t exist.

To deal with strict monotone decreasing functions, we shall modify the proposition like this:
Let $a<b$ be real numbers, and let $f:[a,b]\to \mathbf R$ be a function which is both continuous and strictly monotone decreasing. Then $f$ is a bijection from $[a,b]$ to $[f(b),f(a)]$ , and the inverse $f^{-1}:[f(b),f(a)]\to [a,b]$ is also continuous and strictly monotone decreasing.

$\blacksquare$

Exercise 9.8.5. In this exercise we give an example of a function which has a discontiuity at every rational point, but is continuous at every irrational. Since the rationals are countable, we can write them as $\mathbf Q=\{q(0),q(1),q(2),\dots\}$ , where $q:\mathbf N\to\mathbf Q$ is a bijection from $\mathbf N$ to $\mathbf Q$ . Now define a function $g:\mathbf Q\to \mathbf R$ by setting $g(q(n)):=2^{-n}$ for each natrual number $n$ ; thus $g$ maps $q(0)$ to $1$ , $q(1)$ to $2^{-1}$ , etc. Since $\sum_{n=0}^{\infty}2^{-n}$ is absolutely convergent, we see that $\sum_{r\in \mathbf Q}g(r)$ is also absolutely convergent. Now define the function $f:\mathbf R\to \mathbf R$ by

$f(x):=\sum\limits_{r\in \mathbf Q;r<x}g(r)$ .

Since $\sum_{r\in \mathbf Q}g(r)$ is absolutely convergent, we know that $f(x)$ is well-defined for every real number $x$ .
( a ) Show that $f$ is strictly monotone increasing.
( b ) Show that for every rational number $r$ , $f$ is discontinuous at $r$ .
( c ) Show that for every irrational number $x$ , $f$ is continuous at $x$ .

Solution: ( a ) Given $\forall x_1,x_2\in \mathbf R$ , we can find $q\in \mathbf Q,x_1<q<x_2$ , thus $\exists n_0\in \mathbf N,q=q(n_0)$ , so $g(q)=2^{-n_0 }>0$ . We have

$f(x_2 )=\sum\limits_{r\in \mathbf Q:r<x_2}g(r)=\sum\limits_{r\in \mathbf Q:r<x_1}g(r)+\sum\limits_{r\in \mathbf Q:x_1\leq r<x_2}g(r) \geq f(x_1 )+g(q)>f(x_1)$

( b ) $\forall r\in \mathbf Q,\exists n\in \mathbf N,r=q(n)$ . So given this $r$ , for $\forall x>r$ :

$\begin{aligned} f(x)&=\sum\limits_{a\in \mathbf Q:a<x}g(a)=\sum\limits_{a\in \mathbf Q:a<r}g(a)+\sum\limits_{a\in \mathbf Q:r\leq a<x}g(a)\\&=f(r)+\sum\limits_{a\in \mathbf Q:r\leq a<x}g(a)\\&\geq f(r)+g(r)=f(r)+2^{-n}\end{aligned}$

By the comparison principle we have $f(r+)\geq f(r)+2^{-n}$ , thus $f(r+)\neq f(r)$ , use Proposition 9.5.3 we can conclude $f$ is not continuous at $r$ .

( c ) First, we show $f_n (x)$ is continuous at irrational $x$ . Since there is at most $n$ rationals $r$ which could satisfy $g(r)\geq 2^{-n}$ , we name them $r_1,\dots,r_n$ , then it is able to choose

$\delta =\min |x-r_i|>0,\quad 1\leq i\leq n$

We conclude $f_n (y)$ remains constant for any $y\in (x-\delta ,x+\delta )$ , thus $f_n$ is continuous at $x$ . Also, notice that

$|f(x)-f_n (x)|=\sum\limits_{r\in \mathbf Q:r<x;{\ }g(r)<2^{-n}}g(r)\leq \sum\limits_{k=n+1}^{\infty}2^{-k} =2^{-n}$

Now for $\forall \varepsilon >0$ , choose $n$ s.t. $2^{-n}<\varepsilon /2$ , and for this $n$ choose $\delta$ s.t.

$(x-\delta ,x+\delta )\cap \{r\in \mathbf Q:g(r)\geq 2^{-n} \}=\emptyset$

Then if $y\in (x-\delta ,x+\delta )$ , we can conclude $f_n (y)=f_n (x)$ , also

$|f(y)-f_n (y)|=\sum\limits_{r\in \mathbf Q:r<y,{\ }g(r)<2^{-n}}g(r)\leq \sum\limits_{k=n+1}^{\infty} 2^{-k} =2^{-n}$

So we can have

$\begin{aligned} |f(y)-f(x)|&\leq |f(y)-f_n (y)|+|f_n (y)-f(x)|=|f(y)-f_n (y)|+|f_n (x)-f(x)|\\&\leq 2^{-n}+2^{-n}<\frac{\varepsilon }{2}+\frac{\varepsilon }{2}=\varepsilon ,\quad \forall y\in (x-\delta ,x+\delta ) \end{aligned}$

This proves $f$ is continuous at $x$ .

$\blacksquare$

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