陶哲轩实分析（上）9.8及习题-Analysis I 9.8

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

If $a<b$ and $f:[a,b]→\mathbf R$ is a monotonic function on $[a,b]$, then suppose first $f$ is monotonic increasing, we have $f(b)≥f(x),∀x∈[a,b]$, and $f(a)≤f(x),∀x∈[a,b]$, 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 $f(a)≥f(x),∀x∈[a,b]$, and $f(b)≤f(x),∀x∈[a,b]$, so $f$ attain its maximum at a and its minimum at $b$, if $f$ is strictly increasing, the proof is the same.

Exercise 9.8.2

Let $f:[-1,1]→\mathbf R$ be defined as:
$f(x)=\begin{cases}x,& x∈[-1,0)\\x+1,&x∈[0,1] \end{cases}$
Then $f$ is strictly monotone (thus monotone), $f(-1)=-1$ and $f(1)=2$, but no element in $[-1,1]$ could make $f(x)=1/2$.

Exercise 9.8.3

We shall divide into three cases: $f(a)=f(b),f(a)<f(b),f(a)>f(b)$. The first case is obviously a contradiction to $f$ is one-to-one. So we only consider the last two cases.
Case $f(a)<f(b)$. Assume we can find $x<y∈[a,b]$ s.t. $f(x)≥f(y)$, since $f$ is one-to-one we can’t have $f(x)=f(y)$, so we can only assume $f(x)>f(y)$. Since $x<y$, we have $y≠a$. So if $f(a)>f(y)$, then $f(y)<f(a)<f(b)$, by the intermediate value theorem, $∃c∈[y,b]$, s.t. $f(c)=f(a)$, this contradicts the fact that $f$ is one-to-one. If $f(a)<f(y)$, then $f(a)<f(y)<f(x)$, by the intermediate value theorem, $∃c∈[a,x]$, s.t. $f(c)=f(y)$, this again contradicts the fact that $f$ is one-to-one. So $∀x<y∈[a,b],f(x)<f(y)$, thus $f$ is strictly increasing.
Case $f(a)>f(b)$ can be similarly proved. In this case $f$ is strictly decreasing.
In conclusion $f$ is strictly monotone.

Exercise 9.8.4

$f$ is strictly monotone increasing means $f(a)<f(b)$, and for $∀x,y∈[a,b],f(x)<f(y)$, thus $f$ is injective. By the intermediate value theorem, $∀y∈[f(a),f(b)],∃c∈[a,b],f(c)=y$, 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 $y_0∈[f(a),f(b)]$, then $∃x_0∈[a,b],f(x_0 )=y_0$. Now for $∀ϵ>0$, let
$δ=\begin{cases}\min⁡\{y_0-f(a),f(x_0+ϵ)-y_0 \},&a≤x_0<a+ϵ\\ \min⁡\{y_0-f(x_0-ϵ),f(x_0+ϵ)-y_0 \},&a+ϵ≤x_0≤b-ϵ\\ \min⁡\{y_0-f(x_0-ϵ),f(b)-y_0 \}, & b-ϵ<x_0≤b\end{cases}$
Then as long as $y∈[f(a),f(b)],|y-y_0 |<δ$, we can have
$|f^{-1} (y)-f^{-1} (y_0)|=|f^{-1} (y)-x_0 |<ϵ$
Thus $f^{-1}$ is continuous.

To prove $f^{-1}$ is monotonic increasing on $[f(a),f(b)]$, let $∀y_1,y_2∈[f(a),f(b)],y_1<y_2$, then $∃x_1,x_2∈[a,b]$, 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
$f:[0,2]→[0,3]≔\begin{cases}x,& x∈[0,1)\\x+1,& x∈[1,2] \end{cases}$
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
$f:[0,2]→\mathbf R≔1,\quad x∈[0,2]$
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 $f∶[a,b]→\mathbf R$ 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 $f^{-1} ∶[f(b),f(a)]→[a,b]$ is also continuous and strictly monotone decreasing.

Exercise 9.8.5

( a ) Given $∀x_1,x_2∈\mathbf R$, we can find $q∈\mathbf Q,x_1<q<x_2$, thus $∃n_0∈\mathbf N,q=q(n_0)$, so $g(q)=2^{-n_0 }>0$. We have
$f(x_2 )=∑_{r∈\mathbf Q:r<x_2}g(r)=∑_{r∈\mathbf Q:r<x_1}g(r)+∑_{r∈\mathbf Q:x_1≤r<x_2}g(r) ≥f(x_1 )+g(q)>f(x_1)$

( b ) $∀r∈\mathbf Q,∃n∈\mathbf N,r=q(n)$. So given this $r$, for $∀x>r$:
$\begin{aligned} f(x)&=∑_{a∈\mathbf Q:a<x}g(a)=∑_{a∈\mathbf Q:a<r}g(a)+∑_{a∈\mathbf Q:r≤a<x}g(a)=f(r)+∑_{a∈\mathbf Q:r≤a<x}g(a)\\&≥f(r)+g(r)=f(r)+2^(-n)\end{aligned}$
By the comparison principle we have $f(r+)≥f(r)+2^{-n}$, thus $f(r+)≠f(r)$, 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’s at most $n$ rationals $r$ which could satisfy $g(r)≥2^{-n}$, we name them $r_1,…,r_n$, then it’s able to choose
$δ=\min⁡|x-r_i |>0,\quad 1≤i≤n$
We conclude $f_n (y)$ remains constant for any $y∈(x-δ,x+δ)$, thus $f_n$ is continuous at $x$. Also, notice that
$|f(x)-f_n (x)|=∑_{r∈\mathbf Q:r<x;{\ }g(r)<2^{-n}}g(r)≤∑_{k=n+1}^∞2^{-k} =2^{-n}$

Now for $∀ϵ>0$, choose $n$ s.t. $2^{-n}<ϵ/2$, and for this $n$ choose $δ$ s.t.
$(x-δ,x+δ)∩\{r∈\mathbf Q:g(r)≥2^{-n} \}=∅$
Then if $y∈(x-δ,x+δ)$, we can conclude $f_n (y)=f_n (x)$, also
$|f(y)-f_n (y)|=∑_{r∈\mathbf Q:r<y,{\ }g(r)<2^{-n}}g(r)≤∑_{k=n+1}^∞ 2^{-k} =2^{-n}$
So we can have
$\begin{aligned} |f(y)-f(x)|&≤|f(y)-f_n (y)|+|f_n (y)-f(x)|=|f(y)-f_n (y)|+|f_n (x)-f(x)|\\&≤2^{-n}+2^{-n}<\frac{ϵ}{2}+\frac{ϵ}{2}=ϵ,\quad ∀y∈(x-δ,x+δ) \end{aligned}$
This proves $f$ is continuous at $x$.