陶哲轩实分析（上）10.1及习题-Analysis I 10.1

介绍了微分的许多基本概念。

Exercise 10.1.1

$f$ is differentiable at $x_0$ means the limit
$\lim_{x→x_0;x∈X-\{x_0 \} }\frac{f(x)-f(x_0)}{x-x_0}$⁡
exists, suppose the limit is $L$, then by definition this means $∀ϵ>0,∃δ>0,$ s.t.
$\left|\frac{f(x)-f(x_0 )}{x-x_0 }-L\right|<ϵ,\quad∀x∈\Big((x_0-δ,x_0+δ)∩X\Big)-\{x_0 \}$
Consider $f|_Y$, if $∀y∈(x_0-δ,x_0+δ)∩Y$ and $y≠x_0$, then since $Y⊂X$, we have $y∈\Big((x_0-δ,x_0+δ)∩X\Big)-\{x_0 \}$, so
$\left|\frac{f|_Y (y)-f|_Y (x_0 )}{y-x_0 }-L\right|=\left|\frac{f(y)-f(x_0 )}{y-x_0 }-L\right|<ϵ$
Which means the limit $\lim_{y→x_0;y∈Y-\{x_0 \} }\frac{f|_Y (y)-f|_Y (x_0 )}{y-x_0}$ ⁡exists, i.e. $f|_Y$ is differentiable at $x_0$.

Exercise 10.1.2

(a) implies (b):
If (a) is valid, then
$\lim_{x→x_0;x∈X-\{x_0 \} }\frac{f(x)-f(x_0)}{x-x_0 }=L$
Thus $∀ϵ>0,∃δ>0$, s.t. whenever $x∈([x_0-δ,x_0+δ]∩X)-\{x_0 \}$
$\left|\frac{f(x)-f(x_0)}{x-x_0 }-L\right|≤ϵ ⇒|f(x)-f(x_0 )-L(x-x_0 )|≤ϵ|x-x_0 |$
Note that if $x=x_0$, then $|f(x)-f(x_0 )-L(x-x_0 )|=0≤0=ϵ|x-x_0 |$. So we can conclude (b) is true.
(b) implies (a):
If (b) is true, then for every $ϵ>0$, we let $x∈([x_0-δ,x_0+δ]∩X)-\{x_0 \}$, obviously $x$ satisfies the condition of (b) and in addition we have $|x-x_0 |≠0$, thus we have
$|f(x)-(f(x_0 )+L(x-x_0 ))|≤ϵ|x-x_0 |$
And from this we can further have
$\left|\frac{f(x)-f(x_0)}{x-x_0 }-L\right|≤ϵ$
which means (a) is true since
$\lim_{x→x_0;x∈X-\{x_0 \}} ⁡\frac{f(x)-f(x_0)}{x-x_0 }=L$

Exercise 10.1.3

If $f$ is differentiable at $x_0$ and we let the derivative be $L$, obviously $|L|<+∞$, by Proposition 10.1.7 we can have $∀ϵ>0,∃δ>0$, s.t.
$|f(x)-(f(x_0 )+L(x-x_0 ))|≤ϵ|x-x_0 |,\quad ∀x∈X,|x-x_0 |≤δ$
This means
$|f(x)-f(x_0 )|≤(L+ϵ)|x-x_0 |,\quad ∀x∈X,|x-x_0 |≤δ$
Now let $δ'=\min⁡(δ,δ/(L+ϵ))$, then if $x∈X,|x-x_0 |≤δ'$, we have $|f(x)-f(x_0 )|≤ϵ$, so $f$ is continuous at $x_0$.
If instead we want to use the limit laws, we can see that
$\begin{aligned} \lim_{x→x_0;x∈X-\{x_0 \}} \frac{f(x)-f(x_0)}{x-x_0}=L &⇒ \lim_{x→x_0;x∈X-\{x_0 \}}(f(x)-f(x_0 ))=\lim_{x→x_0;x∈X-\{x_0 \} } L(x-x_0 ) \\&⇒\lim_{x→x_0 }⁡(f(x)-f(x_0 ))=\lim_{x→x_0 )}L(x-x_0 )=L \lim_{x→x_0}⁡(x-x_0 )=0 \\&⇒ \lim_{x→x_0 )}f(x)=\lim_{x→x_0 }⁡f(x_0)=f(x_0 )\end{aligned}$

Exercise 10.1.4

( a ) We have
$\lim_{x→x_0;x∈X-\{x_0 \}} \frac{f(x)-f(x_0)}{x-x_0}=\lim_{x→x_0;x∈X-\{x_0 \}} \frac{c-c_0}{x-x_0}=\lim_{x→x_0;x∈X-\{x_0 \}}⁡0=0$
( b ) We have
$\lim_{x→x_0;x∈X-\{x_0 \}} \frac{f(x)-f(x_0)}{x-x_0}=\lim_{x→x_0;x∈X-\{x_0 \}} \frac{x-x_0}{x-x_0}=\lim_{x→x_0;x∈X-\{x_0 \}}⁡1=1$
( c ) We have
$\lim_{x→x_0;x∈X-\{x_0 \}} \frac{(f+g)(x)-(f+g)(x_0)}{x-x_0}=\lim_{x→x_0;x∈X-\{x_0 \}} \frac{(f(x)+g(x)-f(x_0 )-g(x_0)}{x-x_0 }\\=\lim_{x→x_0;x∈X-\{x_0 \}}\frac{f(x)-f(x_0)}{x-x_0}+\lim_{x→x_0;x∈X-\{x_0 \}}⁡ \frac{g(x)-g(x_0)}{x-x_0 }=f' (x_0 )+g' (x_0 )$
Thus $f+g$ is differentiable at $x_0$ and $(f+g)' (x_0 )=f' (x_0 )+g' (x_0 )$
( d ) We have, use the “middle-man trick”, that
$\lim_{x→x_0;x∈X-\{x_0 \}}\frac{(fg)(x)-(fg)(x_0)}{x-x_0}=\lim_{x→x_0;x∈X-\{x_0 \}}\frac{f(x)(g(x)-g(x_0 ))+(f(x)-f(x_0 ))g(x_0 )}{x-x_0}\\=\lim_{x→x_0;x∈X-\{x_0 \}}\frac{f(x)(g(x)-g(x_0)}{x-x_0}+\lim_{x→x_0;x∈X-\{x_0 \}}\frac{f(x)-f(x_0 )}{x-x_0} g(x_0 )$
Use Proposition 10.1.10 and limit laws, we can further calculate:
$\begin{aligned}&{\ }{\ }{\ }{\ }{\ }\lim_{x→x_0;x∈X-\{x_0 \}}\frac{f(x)(g(x)-g(x_0)}{x-x_0}+\lim_{x→x_0;x∈X-\{x_0 \}}\frac{f(x)-f(x_0 )}{x-x_0} g(x_0 )\\&=\left(\lim_{x→x_0;x∈X-\{x_0 \}}⁡f(x) \right)\left(\lim_{x→x_0;x∈X-\{x_0 \}}\frac{g(x)-g(x_0 )}{x-x_0 }\right)+g(x_0 ) \lim_{x→x_0;x∈X-\{x_0 \}}\frac{f(x)-f(x_0 )}{x-x_0 }\\&=f(x_0 ) g' (x_0 )+f' (x_0 )g(x_0 )\end{aligned}$

( e ) Combining (a) and (d), let $g(x)=c$, then $g' (x)=0$, and $cf$ differentiable at $x_0$, and
$(cf)' (x_0 )=(gf)' (x_0 )=f(x_0 ) g' (x_0 )+f' (x_0 )g(x_0 )=cf' (x_0 )$
( f ) from (e) we know $–g$ is differentiable at $x_0$, $(-g)' (x_0 )=-g' (x_0 )$, thus by (a) we have
$(f-g)' (x_0 )=\left(f+(-g))' (x_0 \right)=f' (x_0 )+(-g)' (x_0 )=f' (x_0 )-g' (x_0 )$
( g ) We have, use Proposition 10.1.10 and limit laws,
$\lim_{x→x_0;x∈X-\{x_0 \}}\frac{(1/g)(x)-(1/g)(x_0)}{x-x_0 }=\lim_{x→x_0;x∈X-\{x_0 \}}\frac{1/g(x)-1/g(x_0)}{x-x_0}\\=\lim_{x→x_0;x∈X-\{x_0 \}}\frac{-1}{(g(x)g(x_0)} \frac {g(x)-g(x_0)}{x-x_0}=-\frac{g' (x_0 )}{g(x_0 )^2 }$
( h ) by (d) and (g), $f/g$ is differentiable at $x_0$, and
$\left(\frac{f}{g}\right)' (x_0 )=\left(f\left(\frac{1}{g}\right)\right)' (x_0 )=f' (x_0 )\left(\frac{1}{g}\right)(x_0 )+f(x_0 ) \left(\frac{1}{g}\right)' (x_0 )=\frac{f' (x_0 )}{g(x_0 )} -f(x_0 ) \frac{g' (x_0 )}{g(x_0 )^2 }\\=\frac{f' (x_0 )g(x_0 )-f(x_0 ) g' (x_0 )}{(g(x_0 )^2 }$

Exercise 10.1.5

If $n=0$, then $f(x)=x^0=1$, and $f' (x)=0=0x^(-1)=0,x∈\mathbf R-\{0\}$, when $x=0$, use definition we can show $f' (0)=0$. Thus $f$ is differentiable on $\mathbf R$.
Now suppose the case is true for $n$, then in the case of $n+1$, we use Theorem 10.1.13 and have $f(x)=x^{n+1}=x^n⋅x$ is differentiable on $\mathbf R$, and
$f' (x)=(x^n )' x+(x^n ) (x)'=nx^n+x^n=(n+1) x^n,\quad ∀x∈\mathbf R$
Thus the induction hypothesis holds.

Exercise 10.1.6

Use negative induction, first let $n=-1$, so $f(x)=x^{-1}$, then $f' (x)=-x^{-2}$, the induction hypothesis is satisfied.
Now suppose for negative integer $n$ the induction holds, let $f(x)=x^{n-1}=x^n⋅x^{-1}$, by Theorem 10.1.13 $f(x)$ is differentiable on $\mathbf R-\{0\}$, and
$f' (x)=(x^n )' (x^{-1} )+x^n (x^{-1} )'=nx^{n-2}-x^{n-2}=(n-1) x^{n-2}$
Thus the induction hypothesis holds.

Exercise 10.1.7

We have $f' (x_0 )=L_1$ and $g' (y_0 )=L_2$ exists and is finite. Then for $∀ϵ>0, ∃ϵ'>0$ s.t.
$ϵ'^2+|L_1 | ϵ'+|L_2 | ϵ'≤ϵ$
For this $ϵ'$, use Proposition 10.1.7, $∃δ_1>0$, s.t.
$|g(y)-g(y_0 )-L_2 (y-y_0)|≤ϵ' |y-y_0 |,\quad ∀y∈Y,|y-y_0 |≤δ_1$
For this $δ_1,∃δ_2>0$, s.t.
$|f(x)-f(x_0)|<δ_1,\quad ∀x∈X,|x-x_0 |<δ_2$
We can also have a $δ_3>0$, s.t.
$|f(x)-f(x_0 )-L_1 (x-x_0)|≤ϵ' |x-x_0 |,\quad ∀x∈X,|x-x_0 |≤δ_3$
Let $δ=\min⁡(δ_2,δ_3)$, then when $x∈X,|x-x_0 |≤δ$, we shall have $|f(x)-f(x_0)|<δ_1$, and
$|g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 ))|≤ϵ' |f(x)-f(x_0)|$
Use the triangle inequality we can further deduce
$\begin{aligned}&{\ }{\ }{\ }{\ }{\ }|g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 ))|\\&=|g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 )-L_1 (x-x_0 ))-L_2 L_1 (x-x_0)|\\&≥|g(f(x))-g(f(x_0 ))-L_2 L_1 (x-x_0 )|-|L_2 (f(x)-f(x_0 )-L_1 (x-x_0 ))|\end{aligned}$
Thus
$\begin{aligned}&{\ }{\ }{\ }{\ }{\ }|g(f(x))-g(f(x_0 ))-L_2 L_1 (x-x_0 )|\\&≤|g(f(x))-g(f(x_0 ))-L_2 (f(x)-f(x_0 ))|+|L_2 (f(x)-f(x_0 )-L_1 (x-x_0 ))|\\&≤ϵ' |f(x)-f(x_0 )|+|L_2 | ϵ' |x-x_0 |\\&≤ϵ'^2 |x-x_0 |+|L_1 | ϵ' |x-x_0 |+|L_2 | ϵ' |x-x_0 |\\&=(ϵ'^2+|L_1 | ϵ'+|L_2 | ϵ' )|x-x_0 |≤ϵ|x-x_0 |\end{aligned}$
Use proposition 10.1.7, we can say the function $g∘f$ is differentiable at $x_0$, and
$(g∘f)' (x_0 )=L_2 L_1=g' (y_0 ) f' (x_0 )$