陶哲轩实分析（上）10.3及习题-Analysis I 10.3

单调函数和导数的关系，容易理解。

Exercise 10.3.1

Since $f$ is differentiable at $x_0$, we have
$f' (x_0 )=\lim_{x→x_0;x∈X-\{x_0 \}}⁡\frac{f(x)-f(x_0)}{x-x_0}$
If $f$ is monotone increasing, then we have
$f(x)≤f(x_0 ),x<x_0,\quad f(x)≥f(x_0 ),x>x_0$
Thus in both $x<x_0$ and $x>x_0$ we shall have
$\frac{f(x)-f(x_0)}{x-x_0}≥0 ⇒ f' (x_0 )=\lim_{x→x_0;x∈X-\{x_0 \} }⁡\frac{f(x)-f(x_0)}{x-x_0}≥0$
The case when $f$ is monotone decreasing can be similarly proved.

Exercise 10.3.2

Define
$f(x)=\begin{cases}x,&x∈(-1,0]\\2x,&x∈(0,1) \end{cases}$
This doesn’t contradict Proposition 10.3.1 since Proposition 10.3.1 requires $f$ to be differentiable at the point (0 in this case).

Exercise 10.3.3

Define
$f(x)=x^3,\quad x∈(-1,1)$
Then $f$ is differentiable at 0, $f' (0)=0$, but $f$ is monotone increasing.
This doesn’t contradict Proposition 10.3.1 since $f'>0$ is a necessary but not sufficient condition for $f$ to be strictly monotone increasing.

Exercise 10.3.4

For any $x≠y∈[a,b]$, without loss of generality we can suppose $x<y$, then $f|_{[x,y]}$ is continuous and differentiable on $[x,y]$, thus by mean value theorem, we can find a $c∈(x,y)⊂(a,b)$ such that
$f' (c)=\frac{f(y)-f(x)}{y-x}=\frac{f(x)-f(y)}{x-y}$
If $f' (x)>0,∀x∈[a,b]$, then $f' (c)>0$ and $f(x)<f(y)$, so $f$ is strictly monotone increasing.
If $f' (x)<0,∀x∈[a,b]$, then $f' (c)<0$ and $f(x)>f(y)$, so $f$ is strictly monotone decreasing.
If $f' (x)=0,∀x∈[a,b]$, then $f' (c)=0$ and $f(x)=f(y)$, so $f$ is a constant function.

Exercise 10.3.5

Define
$f(x)=\begin{cases}x+1,&x∈(-1,0)\\x-1,&x∈(0,1)\end{cases}$
Then if $X=(-1,0)∪(0,1)$, then $f$ is differentiable on $X$ and $f' (x)=1>0,∀x∈X$, but we have $f(1/2)=-1/2<f(-1/2)=1/2$, thus $f$ in not strictly monotone increasing.
The key condition which is different from Proposition 10.3.3 is that $X$ is allowed to be a disconnected set.