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陶哲轩实分析9.4及习题-Analysis I 9.4

2019年5月9日2019年12月20日christangdt2条评论

函数极限性。简单而言，上一节定义的limiting values of functions如果等于函数在这一点的值，就是函数连续。连续性用ε-δ定义和序列定义是等价的，可以证明很多函数都是连续的

Exercise 9.4.1. Prove Proposition 9.4.7.
Proposition 9.4.7 (Equivalent formulations of continuity). Let $X$ be a subset of $\mathbf R$ , let $f:X\to \mathbf R$ be a function, and let $x_0$ be an element of $X$ . Then the following four statements are logically equivalent:
( a ) $f$ is continuous at $x_0$ .
( b ) For every sequence $(a_n)_{n=0}^{\infty}$ consisting of elements of $X$ with $\lim_{n\to\infty}a_n=x_0$ , we have $\lim_{n\to\infty}f(a_n)=f(x_0)$ .
( c ) For every $\varepsilon >0$ , there exists a $\delta >0$ such that $|f(x)-f(x_0)|<\varepsilon$ for all $x\in X$ with $|x-x_0|<\delta$ . ( d ) For every $\varepsilon >0$ , there exists a $\delta >0$ such that $|f(x)-f(x_0)|\leq\varepsilon$ for all $x\in X$ with $|x-x_0|\leq\delta$ .

Solution:
( a ) and ( b ) are equivalent :

$\begin{aligned}(f\text{ is continuous at }x_0) &\iff (\lim_{x\to x_0;x\in X}f(x)=f(x_0 )) \\&\iff ((\forall (a_n )_{n=0}^{\infty} \in E,\lim_{n\to {\infty} }a_n=x_0 )\implies \lim_{n\to {\infty} }f(a_n)=f(x_0))\end{aligned}$

( a ) implies ( c ): Get directly from the definition of $\lim_{x\to x_0;x\in X}f(x)=f(x_0 )$ .
( c ) implies ( d ): Obviously true.
( d ) implies ( a ): Let $\forall \varepsilon >0$ , then due to (d), $\exists \delta '>0$ , s.t. $|f(x)-f(x_0)|\leq \varepsilon /2$ for every $x\in X$ with $|x-x_0 |\leq \delta '$ , so if we let $\delta =\delta '$ , then $|x-x_0 |<\delta$ means $|x-x_0 |\leq \delta '$ , and

$|f(x)-f(x_0)|\leq \varepsilon /2 \implies |f(x)-f(x_0 )|<\varepsilon \implies \lim\limits_{x\to x_0;x\in X}f(x)=f(x_0 )$

$\blacksquare$

Exercise 9.4.2. Let $X$ be a subset of $\mathbf R$ , and let $c\in \mathbf R$ . Show that the constant function $f:X\to \mathbf R$ defined by $f(x):=c$ is continuous, and show that the identity function $g:X\to \mathbf R$ defined by $g(x):=x$ is also continuous.

Solution: First for every $x_0\in X$ , we have

$\lim\limits_{x\to x_0;x\in X}f(x)=\lim\limits_{x\to x_0;x\in X}c=c=f(x_0 )$

So $f$ is continuous on $X$ .
Next, for every $x_0\in X$ , we have

$\lim\limits_{x\to x_0;x\in X}g(x)=\lim\limits_{x\to x_0;x\in X} x=x_0=g(x_0 )$

So $g$ is continuous on $X$ .

$\blacksquare$

Exercise 9.3.3. Prove Proposition 9.4.10.
Proposition 9.4.10 (Exponentiation is continuous, I) Let $a>0$ be a positive real number. Then the function $f:\mathbf R\to \mathbf R$ defined by $f(x):=a^x$ is continuous.

Solution: Choose $\forall x_0\in \mathbf R$ , we prove $f(x)=a^x$ is continuous at $x_0$ . As $a>0$ , we have $a^{x_0 }>0$ .
$\forall \varepsilon >0$ , from Lemma 6.5.3 we can find a $N\in \mathbf N$ , s.t. $|a^{1/n}-1|<\varepsilon /a^{x_0}$ for all $n>N$ . Since

$|a^x-a^{x_0} |=a^{x_0 } |a^{x-x_0}-1|$

We choose $\delta '=1/(N+1)$ , then $\forall x\in \mathbf R$ with $0\leq x-x_0<\delta '$ , we can find a $k>N$ , s.t.

$1/(k+1)<x-x_0<1/k$

Thus $|a^{x-x_0}-1|$ is between $|a^{1/k}-1|$ and $|a^{1/k+1}-1|$ , which means

$|a^{x-x_0}-1|<\varepsilon /a^{x_0 } \implies |a^x-a^{x_0 } |=a^{x_0} |a^{x-x_0}-1|<\varepsilon$

Also, from Lemma 6.5.3 we can find a $N'\in \mathbf N$ , s.t. $|a^{-1/n}-1|<\varepsilon /a^{x_0}$ for all $n>N'$ . Since

$|a^x-a^{x_0} |=a^{x_0} |a^{x-x_0}-1|$

We choose $\delta ''=1/(N'+1)$ , then $\forall x\in \mathbf R$ with $0<x_0-x<\delta ''$ , we can find a $k>N'$ , s.t.

$1/(k+1)<x_0-x<1/k$

Thus $|a^{x-x_0}-1|$ is between $|a^{-1/k}-1|$ and $|a^{-1/k+1}-1|$ , which means

$|a^{x-x_0}-1|<\varepsilon /a^{x_0} \implies |a^x-a^{x_0} |=a^{x_0} |a^{x-x_0}-1|<\varepsilon$

ow let $\delta =\min (\delta ',\delta '')$ , then if $|x-x_0 |<\delta , |a^x-a^{x_0} |<\varepsilon$ , completing the proof.

$\blacksquare$

Exercise 9.4.4. Prove Proposition 9.4.11.
Proposition 9.4.11 (Exponentiation is continuous, II) Let $p$ be a real number. Then the function $f:(0,\infty)\to\mathbf R$ defined by $f(x):=x^p$ is continuous.

Solution: Since $\lim_{x\to 1} x=1$ , we have $\lim_{x\to 1}x^n=(\lim_{x\to 1} x )^n=1$ for all natural numbers $n$ , and since when $x\to 1,1/x\to 1$ , so $\lim_{x\to 1}x^{-n}=1,n\in \mathbf N$ . Use prove by contradiction we can show that $\lim_{x\to 1}x^{1/n}=1, \forall n\in \mathbf N^+$ , thus it’s able to conclude $\lim_{x\to 1} x^q=1,\forall q\in \mathbf Q$ .
Now for $\forall p\in \mathbf R,\exists q_1,q_2, s.t. q_1\leq p<q_2$ , so $x^p$ is between $x^{q_1}$ and $x^{q_2}$ , which means
$\lim_{x\to 1} x^p=1,\quad \forall p\in \mathbf R$

$\lim\limits_{x\to 1} x^p=1,\quad \forall p\in \mathbf R$

Now for $x_0\in (0,+{\infty} )$ , then $x_0^p>0$ . For $\forall \varepsilon >0,\exists \delta >0$ , s.t. for any $x\in (0,+{\infty})$ with $|x-x_0 |<\delta$ , we have $|(x/x_0 )^p-1|<\varepsilon /x_0^p$ , thus

$|x^p-x_0^p |=x_0^p |(x/x_0 )^p-1|<\varepsilon$

$\blacksquare$

Exercise 9.4.5. Prove Proposition 9.4.13.
Proposition 9.4.13 (Composition preserves continuity). Let $X$ and $Y$ be subsets of $\mathbf R$ , and let $f:X\to Y$ and $g:Y\to \mathbf R$ be functions. Let $x_0$ be a point in $X$ . If $f$ is continuous at $x_0$ , and $g$ is continuous at $f(x_0)$ , then the composition $g\circ f:X\to \mathbf R$ is continuous at $x_0$ .

$|g(f(x))-g(f(x_0 ))|<\varepsilon \implies |(g\circ f)(x)-(g\circ f)(x_0)|<\varepsilon$

$\blacksquare$

Exercise 9.4.6. Let $X$ be a subset of $\mathbf R$ , and let $f:X\to \mathbf R$ be a continuous function. If $Y$ is a subset of $X$ , show that the restriction $f|_Y:Y\to \mathbf R$ of $f$ to $Y$ is also a continuous function.

$\blacksquare$

Exercise 9.4.7. Let $n\geq 0$ be an integer, and for each $0\leq i\leq n$ let $c_i$ be a real number. Let $P:\mathbf R\to \mathbf R$ be the function

$P(x):=\sum\limits_{i=0}^nc_ix^i$ ;

such a function is known as polynomial of one variable. Show that $P$ is continuous.

Solution: By Proposition 9.4.9, the function $x^i$ is continuous for each $0\leq i\leq n$ , thus the function $c_i x^i$ is continuous for each $0\leq i\leq n$ , at last their sums is continuous.

$\blacksquare$

陶哲轩实分析9.3及习题-Analysis I 9.3

2019年5月8日2019年12月19日christangdt8条评论

9.3的题目叫limiting values of functions，在Analysis II里有一节同样题目的章节。这一节限制在R上讨论，函数的极限值是后续讨论continuity的基础，函数极限可以和数列极限等价起来，函数极限是唯一的，满足极限算律，极限也是局部的性质（即只和某一点附近的情况有关）

Exercise 9.3.1. Prove Proposition 9.3.9.
Proposition 9.3.9.Let $X$ be a subset of $\mathbf R$ , let $f:X\to \mathbf R$ be a function, let $E$ be a subset of $X$ , let $x_0$ be an adherent point of $E$ , and let $L$ be a real number. Then the following two statements are logically equivalent:
( a ) $f$ converges to $L$ at $x_0$ in $E$ .
( b ) For every sequence $(a_0)_{n=0}^{\infty}$ which consists entirely of elements of $E$ and converges to $x_0$ , the sequence $(f(a_n))_{n=0}^{\infty}$ converges to $L$ .

Solution:
(a) implies (b):
If $f$ converges to $L$ at $x_0$ in $E$ , then $\forall \varepsilon >0,\exists \delta >0, \text{ s.t. } |x-x_0 |<\delta \implies |f(x)-L|<\varepsilon$ . Let $(a_n)_{n=0}^{\infty} \in E$ such that $a_n\to x_0$ , then $\exists N\in \mathbf N, \text{ s.t. } |a_n-x_0 |<\delta ,\forall n>N$ , so if $n>N$ , we have $|f(a_n)-L|<\varepsilon$ , so $(f(a_n))_{n=0}^{\infty} \to L$ .
(b) implies (a):
Assume $f$ does not converge to $L$ at $x_0$ in $E$ , then due to axiom of choice, there is a $\varepsilon _0>0$ , such that for $\forall N\in \mathbf N,\exists x_n\in E, \text{ s.t. } (|x_n-x_0 |<\delta )\wedge (|f(x_n)-L|\geq \varepsilon )$ , the sequence $(x_n)_{n=0}^{\infty} \in E$ , but $(f(x_n))_{n=0}^{\infty}\nrightarrow L$ , a contradiction.

$\blacksquare$

Exercise 9.3.2. Prove the remaining claims in Proposition 9.3.14 (Limit laws for functions).

Solution: Since $x_0$ is an adherence point of $E$ , we can find $(a_n)_{n=0}^{\infty} \in E$ such that $a_n\to x_0$ , then we have $(f(a_n))_{n=0}^{\infty} \to L$ and $(g(a_n ))_{n=0}^{\infty} \to M$ , by Proposition 9.3.9. Thus: by Theorem 6.1.19 (d):

$((f-g)(a_n))_{n=0}^{\infty} \to L-M \implies (f(a_n))_{n=0}^{\infty} -(g(a_n))_{n=0}^{\infty} \to L-M$

by Theorem 6.1.19 (g):

$((\max (f,g))(a_n))_{n=0}^{\infty} \to \max(L,M) \\ \implies \max ((f(a_n ))_{n=0}^{\infty} ,(g(a_n))_{n=0}^{\infty} )\to \max(L,M)$

by Theorem 6.1.19 (h):

$((\min (f,g) )(a_n))_{n=0}^{\infty} \to \min(L,M)\\ \implies \min ((f(a_n ))_{n=0}^{\infty} ,(g(a_n))_{n=0}^{\infty} )\to \min (L,M)$

by Theorem 6.1.19 (b):

$((fg)(a_n))_{n=0}^{\infty} \to LM \\ \implies (f(a_n))_{n=0}^{\infty} (g(a_n))_{n=0}^{\infty} \to LM$

by Theorem 6.1.19 (f):

$((f/g)(a_n))_{n=0}^{\infty} \to L/M \implies (f(a_n ))_{n=0}^{\infty} /(g(a_n))_{n=0}^{\infty} \to L/M$

Then all functions have asserted limits at $x_0$ in $E$ since the above conclusions are true for arbitrary $(a_n)_{n=0}^{\infty} \in E$ such that $a_n\to x_0$ , and thus by Proposition 9.3.9.

$\blacksquare$

Exercise 9.3.3. Prove Lemma 9.3.18.
Proposition 9.3.18 (Limits are local). Let $X$ be a subset of $\mathbf R$ , let $E$ be a subset of $X$ , let $x_0$ be an adherent point of $E$ , let $f:X\to \mathbf R$ be a function, and let $L$ be a real number. Let $\delta >0$ . Then we have

$\lim\limits_{x\to x_0;x\in E}f(x)=L\iff \lim\limits_{x\to x_0;x\in E\cap (x_0-\delta,x_0+\delta)}f(x)=L$

Solution:

$\begin{aligned} &\lim_{x\to x_0;x\in E} f(x)=L \\&\iff \forall \varepsilon >0,\exists \delta '>0,\text{ s.t. } (x\in E)\wedge (|x-x_0 |<\delta ' )\implies |f(x)-L|<\varepsilon \\ &\iff \forall \varepsilon >0,\exists \delta '>0,\text{ s.t. } (x\in E)\wedge (|x-x_0 |<\min (\delta ,\delta')\leq \delta')\implies |f(x)-L|<\varepsilon \\ &\iff \forall \varepsilon >0,\exists \delta '>0,\text{ s.t. } (x\in E\cap (x_0-\delta ,x_0+\delta ))\wedge (|x-x_0 |<\delta ' )\implies |f(x)-L|<\varepsilon \\&\iff \lim_{x\to x_0;x\in E \cap (x_0-\delta,x_0+\delta)} f(x)=L \end{aligned}$

$\blacksquare$

Exercise 9.3.4. Propose a definition for limit superior $\limsup_{x\to x_0;x\in E}f(x)$ and limit inferior $\liminf_{x\to x_0;x\in E}f(x)$ , and then propose an analogue of Proposition 9.3.9 for you definition. (For an additional challenge: prove that analogue.)

Solution:

A definition for limit superior: we say $\limsup_{x\to x_0;x\in E} f(x)=M$ iff for $\forall \varepsilon >0,\exists \delta >0$ , s.t.

$\Big((|x-x_0 |<\delta )\implies (f(x)<M+\varepsilon )\Big)\wedge \Big(\exists x'\in (x_0-\delta ,x_0+\delta ),f(x' )>M-\varepsilon \Big)$

A definition for limit inferior: we say $\liminf_{x\to x_0;x\in E}f(x)=m$ iff for $\forall \varepsilon >0,\exists \delta >0$ , s.t.

$\Big((|x-x_0 |<\delta )\implies (f(x)>m-\varepsilon )\Big)\wedge \Big(\exists x'\in (x_0-\delta ,x_0+\delta ),f(x' )<m+\varepsilon \Big)$

An analogue of Proposition 9.3.9 (limsup case):
Let $E\subseteq X\subseteq R, f:X\to R, x_0\in \overline E$ , then the following two statements are logically equivalent:
(a) $\limsup_{x\to x_0;x\in E}f(x)=M$
(b) For any sequence $(a_n)_{n=0}^{\infty} \in E,a_n\to x_0$ , we have $\limsup_{n\to {\infty} }f(a_n)\leq M$ , and there is a sequence $(b_n)_{n=0}^{\infty} \in E, b_n\to x_0$ , such that $\lim_{n\to {\infty} }f(b_n)=M$ .

An analogue of Proposition 9.3.9 (liminf case):
Let $E\subseteq X\subseteq R, f:X\to R, x_0\in \overline E$ , then the following two statements are logically equivalent:
(a) $\liminf_{x\to x_0;x\in E}f(x)=m$
(b) For any sequence $(a_n)_{n=0}^{\infty} \in E,a_n\to x_0$ , we have $\liminf_{n\to {\infty} }f(a_n)\geq m$ , and there is a sequence $(b_n )_{n=0}^{\infty} \in E, b_n\to x_0$ , such that $\lim_{n\to {\infty} }f(b_n)=m$ .

Proof: I only prove the case for limit superior.
(a) implies (b):
If $\limsup_{x\to x_0;x\in E}f(x)=M$ , then on one hand, $\forall \varepsilon >0$ , the set

$A_n=\{x\in E: |x-x_0 |<1/n\text{ and }M-\varepsilon<f(x)<M+\varepsilon \}$

is non-empty, use axiom of choice to find a $b_n\in A_n$ and form a sequence $(a_n)_{n=0}^{\infty} \in E$ , it’ s easy to see that $b_n\to x_0$ and $\lim{n\to {\infty} }f(b_n)=M$ .
On the other hand, assume $\exists (a_n)_{n=0}^{\infty} \in E, a_n\to x_0$ , and $\limsup_{n\to {\infty} }f(a_n)=M'>M$ , then let

$\varepsilon =(M'-M)/2>0$

We can first have a $\delta >0$ , s.t. $(|x-x_0 |<\delta )\implies (f(x)<M+\varepsilon )$ , and there is a $N\in \mathbf N$ , s.t.

$|a_n-x_0 |<\delta ,\quad \forall n>N$

which means

$f(a_n )<M+\varepsilon ,\quad \forall n>N$

But since $\limsup_{n\to {\infty} }f(a_n)=M'$ , we can find a $K>N$ , s.t. $f(a_K )>M'-\varepsilon$ , now we have

$f(a_K )<M+\varepsilon =M'-\varepsilon <f(a_K )$

This is a contradiction.

(b) implies (a):
Assume the first condition is wrong, then we shall find a $\varepsilon_ 0>0$ , s.t. $\forall N\in \mathbf N$ , the set

$A_n=\{x\in E:|x-x_0 |<1/n\text{ and }f(x)\geq M+\varepsilon _0\}$

is non-empty, choose $a_n\in A_n$ by axiom of choice, then $a_n\to x_0$ as $n\to {\infty}$ , and

$M<M+\varepsilon _0\leq \inf (a_n )\leq \limsup_{n\to {\infty} } f(a_n )\leq M$

This is a contradiction.
Assume the second condition is wrong, then $\forall x\in (x_0-\delta ,x_0+\delta ),f(x)\leq M-\varepsilon 0$ for some $\varepsilon _0>0$ and any $\delta >0$ , then any sequence $(a_n)_{n=0}^{\infty} \in E$ such that $a_n\to x_0$ and $f(a_n)$ converges would have $\lim_{n\to {\infty} }f(a_n)\leq M-\varepsilon_0<M$ , this contradicts the existence of $(b_n )_{n=0}^{\infty}$ .

$\blacksquare$

Exercise 9.3.5. (Continuous version of squeeze test) Let $X$ be a subset of $\mathbf R$ , let $E$ be a subset of $X$ , let $x_0$ be an adherent point of $E$ , and let $f:X\to \mathbf R,g:X\to \mathbf R,h:X\to \mathbf R$ be functions such that $f(x)\leq g(x)\leq h(x)$ for all $x\in E$ . If we have $\lim_{x\to x_0;x\in E}f(x)= \lim_{x\to x_0;x\in E}h(x)=L$ for some real number $L$ , show that $\lim_{x\to x_0;x\in E}g(x) =L$ .

Solution: Let $(a_n)_{n=0}^{\infty} \in E$ such that $a_n\to x_0$ , by Proposition 9.3.9, we know that both $(f(a_n ))_{n=0}^{\infty}$ and $(h(a_n ))_{n=0}^{\infty}$ converge to $L$ , thus for $\forall \varepsilon >0$ , there’ s $N_1,N_2\in \mathbf N$ , such that

$|f(a_n )-L|<\varepsilon ,\quad\forall n>N_1\\|h(a_n )-L|<\varepsilon ,\quad\forall n>N_2$

Let $N=\max(N_1,N_2)$ , then we shall have

$L-\varepsilon <f(a_n )\leq g(a_n )\leq h(a_n )<L+\varepsilon ,\quad\forall n>N$

This means $(g(a_n ))_{n=0}^{\infty}$ converges to $L$ , this is true for arbitrary sequence $(a_n )_{n=0}^{\infty} \in E$ which satisfies $a_n\to x_0$ , so we can conclude $\lim_{x\to x_0;x\in E} g(x)=L$ .

$\blacksquare$

陶哲轩实分析9.2及习题-Analysis I 9.2

2019年5月7日2019年12月19日christangdt

这一节是实值函数算律，相对好理解。

Exercise 9.2.1. Let $f:\mathbf R\to\mathbf R,g: \mathbf R\to\mathbf R, h: \mathbf R\to\mathbf R$ . Which of the following identities are true, and which ones are false? In the former case, give a proof; in the latter case, give a counterexample.

$\begin{aligned}(f+g)\circ h&=(f\circ h)+(g\circ h) \\ f\circ (g+h)&=(f\circ g)+(f\circ h) \\ (f+g)\cdot h&=(f\cdot h)+(g\cdot h) \\ f\cdot (g+h)&=(f\cdot g)+(f\cdot h)\end{aligned}$

Solution: The first one is true.

$\begin{aligned}\big((f+g)\circ h\big)(x)&=(f+g)(h(x))=f(h(x))+g(h(x))\\&=(f\circ h)(x)+(g\circ h)(x)\\&=\big((f\circ h)+(g\circ h)\big)(x)\end{aligned}$

The second one is false.
Let $f(x)=x^2,g(x)=x,h(x)=-x$ , then

$\big(f\circ (g+h)\big)(x)=f(0)=0^2=0\\ \big((f\circ g)+(f\circ h)\big)(x)=(f\circ g)(x)+(f\circ h)(x)=x^2+x^2=2x^2$

The third one is true.

$\begin{aligned}\big((f+g)\cdot h\big)(x)&=\big((f+g)(x)\big)\cdot h(x)\\&=\big(f(x)+g(x)\big)\cdot h(x)\\&=f(x)h(x)+g(x)h(x)\\&=(f\cdot h)(x)+(g\cdot h)(x)\\&=\big((f\cdot h)+(g\cdot h)\big)(x)\end{aligned}$

The fourth one is true.

$\begin{aligned}\big(f\cdot (g+h)\big)(x)&=f(x)\big((g+h)(x)\big)\\&=f(x)\big(g(x)+h(x)\big)\\&=f(x)g(x)+f(x)h(x)\\&=(f\cdot g)(x)+(f\cdot h)(x)\\&=\big((f\cdot g)+(f\cdot h)\big)(x)\end{aligned}$

$\blacksquare$

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