I am reading Zorich's book "Analysis I" and ran into the following statement:
Let $f:[a,b]\to \mathbb{R}$ and $f$ is continuous on $[a,b]$. Then $f$ is injective if and only if $f$ is strictly monotonic.
Proof: $\Leftarrow$ is quite obvious.
The most interesting part is $\Rightarrow$. Suppose that $f$ is not strictly monotonic. Then we can find three points $x_1<x_2<x_3$ from $[a,b]$ such that $f(x_2)$ does not lie in between $f(x_1)$ and $f(x_3)$. In this case $f(x_3)$ lies in between $f(x_1)$ and $f(x_2)$ or $f(x_1)$ lies in between $f(x_2)$ and $f(x_3)$.
Question: Suppose $f$ is not strictly monotonic, i.e. $f$ is not strictly increasing and is not strictly decreasing. Then we can find points $x_1,x_2,x_3,x_4\in [a,b]$ such that $x_1<x_2$ with $f(x_1)\geq f(x_2)$ and $x_3<x_4$ with $f(x_3)\leq f(x_4)$.
I have spent smth like couple of hours trying to derive that exists three points with the above properties as in Zorich's book but I failed.
I'd be thankful if you can help me, please!
Best Answer
Note that I am using $(a,b)_* = (\min(a,b), \max(a,b))$ in the following.
You are trying to prove that $f:[a,b] \to \mathbb{R}$ is strictly monotonic iff for all $x_1,x_2,x_3$ such that $a \le x_1<x_2<x_3\le b$ we have $f(x_2) \in (f(x_1),f(x_3))_*$.
If $f$ is strictly monotonic then we can presume that $f(a) < f(b)$ (otherwise take $-f$) and if $x_1<x_2<x_3$ then since $f(x_1) < f(x_2)< f(x_3)$ we see that $f(x_2) \in (f(x_1),f(x_3))_*$.
For the other direction suppose for all $x_1<x_2<x_3$ we have $f(x_2) \in (f(x_1),f(x_3))_*$. We would like to show that $f$ is strictly monotonic.
Since $f({1 \over 2}(a+b)) \in (f(a),f(b))_* $ we see that $f(a) \ne f(b)$. We can presume that $f(a)<f(b)$ (otherwise take $-f$). Note that for $x \in (a,b)$ we have $f(x) \in (f(a),f(b))$.
Now suppose $a \le x < y \le b$. If $x=a$ (or $y=b$) then the previous statement shows that $f(a)< f(x)$ (or $f(y)< f(b)$), so we can assume $a<x<y<b$. In this case we have $f(x) \in (f(a),f(y)$, in particular $f(x)<f(y)$.