陶哲轩实分析（上）9.4及习题-Analysis I 9.4

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

( a ) ⇔ ( b ):
$\begin{aligned}(f\text{ is continuous at }x_0) &⇔ (\lim_{x→x_0;x∈X}f(x)=f(x_0 )) \\&⇔((∀(a_n )_{n=0}^∞∈E,\lim_{n→∞}a_n=x_0 )⇒\lim_{n→∞}f(a_n)=f(x_0))\end{aligned}$
( a ) implies ( c ):
Get directly from the definition of $\lim_{x→x_0;x∈X}f(x)=f(x_0 )$.
( c ) implies ( d ):
Obviously true.
( d ) implies ( a ):
Let $∀ϵ>0$, then due to (d), $∃δ'>0$, s.t. $|f(x)-f(x_0)|≤ϵ/2$ for every $x∈X$ with $|x-x_0 |≤δ'$, so if we let $δ=δ'$, then $|x-x_0 |<δ$ means $|x-x_0 |≤δ'$, and
$|f(x)-f(x_0)|≤ϵ/2 ⇒|f(x)-f(x_0 )|<ϵ ⇒ \lim_{x→x_0;x∈X}f(x)=f(x_0 )$

Exercise 9.4.2

First for every $x_0∈X$, we have
$\lim_{x→x_0;x∈X}f(x)=\lim_{x→x_0;x∈X}c=c=f(x_0 )$
So $f$ is continuous on $X$.
Next, for every $x_0∈X$, we have
$\lim_{x→x_0;x∈X}g(x)=\lim_{x→x_0;x∈X}⁡x=x_0=g(x_0 )$
So $g$ is continuous on $X$.

Exercise 9.4.3

Choose $∀x_0∈\mathbf R$, we prove $f(x)=a^x$ is continuous at $x_0$. As $a>0$, we have $a^{x_0 }>0$.
$∀ϵ>0$, from Lemma 6.5.3 we can find a $N∈\mathbf N$, s.t. $|a^{1/n}-1|<ϵ/a^{x_0}$ for all $n>N$. Since
$|a^x-a^{x_0} |=a^{x_0 } |a^{x-x_0}-1|$
We choose $δ'=1/(N+1)$, then $∀x∈\mathbf R$ with $0≤x-x_0<δ'$, 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
$|a^{x-x_0}-1|<ϵ/a^{x_0 } ⇒ |a^x-a^{x_0 } |=a^{x_0} |a^{x-x_0}-1|<ϵ$
Also, from Lemma 6.5.3 we can find a $N'∈\mathbf N$, s.t. $|a^{-1/n}-1|<ϵ/a^{x_0}$ for all $n>N'$. Since
$|a^x-a^{x_0} |=a^{x_0} |a^{x-x_0}-1|$
We choose $δ''=1/(N'+1)$, then $∀x∈\mathbf R$ with $0<x_0-x<δ''$, 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
$|a^{x-x_0}-1|<ϵ/a^{x_0} ⇒ |a^x-a^{x_0} |=a^{x_0} |a^{x-x_0}-1|<ϵ$
Now let $δ=\min⁡(δ',δ'')$, then if $|x-x_0 |<δ, |a^x-a^{x_0} |<ϵ$, completing the proof.

Exercise 9.4.4

Since $\lim_{x→1}⁡x=1$, we have $\lim_{x→1}x^n=(\lim_{x→1}⁡x )^n=1$ for all natural numbers $n$, and since when $x→1,1/x→1$, so $\lim_{x→1}x^{-n}=1,n∈\mathbf N$. Use prove by contradiction we can show that $\lim_{x→1}x^{1/n}=1, ∀n∈\mathbf N^+$, thus it’s able to conclude $\lim_{x→1}⁡x^q=1,∀q∈\mathbf Q$.
Now for $∀p∈\mathbf R,∃q_1,q_2, s.t. q_1≤p<q_2$, so $x^p$ is between $x^{q_1}$ and $x^{q_2}$, which means
$\lim_{x→1}⁡x^p=1,\quad ∀p∈\mathbf R$
Now for $x_0∈(0,+∞)$, then $x_0^p>0$. For $∀ϵ>0,∃δ>0$, s.t. for any $x∈(0,+∞)$ with $|x-x_0 |<δ$, we have $|(x/x_0 )^p-1|<ϵ/x_0^p$, thus
$|x^p-x_0^p |=x_0^p |(x/x_0 )^p-1|<ϵ$

Exercise 9.4.5

Let $∀ϵ>0$, as $g$ is continuous at $f(x_0 )$, we can find $δ_1>0$, s.t. $∀y∈Y,|y-f(x_0 )|<δ_1$, we can have $|g(y)-g\big(f(x_0 )\big)|<ϵ$. For this $δ_1>0$ we can find $δ>0$, s.t. $∀x∈X,|x-x_0 |<δ$, we can have $|f(x)-f(x_0)|<δ_1$, thus if $|x-x_0 |<δ$, we get
$|g(f(x))-g(f(x_0 ))|<ϵ ⇒|(g∘f)(x)-(g∘f)(x_0)|<ϵ$

Exercise 9.4.6

Let $y_0∈Y$, so $y_0∈X$, and $f$ is continuous at $y_0$, so $∀ϵ>0,∃δ>0$, s.t. $∀x∈X,|x-y_0 |<δ$, we can have $|f(x)-f(y_0)|<ϵ$.
Use the result above, if $y∈Y,|y-y_0 |<δ$, we can have $|f(y)-f(y_0)|<ϵ$, this proves $f|_Y:Y→\mathbf R$ is continuous at $y_0$, since $y_0$ is arbitrarily chosen, $f|_Y$ is continuous.

Exercise 9.4.7

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