Continuous injection of $[a,b]$ is strictly monotonic from Zorich's book

Question

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!

Answer 1

Zorich seemingly uses the definition that $f$ is strictly monotonic that whenever $x_2$ lies between $x_1$ and $x_3$ then $f(x_2)$ lies between $f(x_1)$ and $f(x_3)$.

Answer 2

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 $f(x)<f(y)$.

Answer 3

Proceed by contradiction and assume $f:[a,b]\to\Bbb R$ is a continuous injection but $f$ is not strictly monotonic. Choose $x_1,x_2,x_3\in [a,b]$ such that $x_1<x_2<x_3$ but $f(x_1)< f(x_2)>f(x_3)$. Now choose any point $$c\in (f(x_1),f(x_2))\cap (f(x_2),f(x_3))$$ so that the Intermediate Value Theorem gaurantees there exists $x_4\in(x_1,x_2)$ and $x_5\in(x_2,x_3)$, necessarily distinct because $(x_1,x_2)\cap(x_2,x_3)=\emptyset$, and such that $f(x_4)=f(x_5)=c\;$ which contradicts the fact that $f$ is an injection. $\it{reductio\;ad\;absurdum}$