Different approach for showing that if $f:[a,b]\to\mathbb{R}$ is continuous and injective, then the maximum value of $f$ must occur at $a$ or $b$.

Question

Trying to take a different approach in proving if $f:[a,b] \rightarrow \mathbb{R}$ is continuous and injective, then the maximum value of $f$ must occur at $a$ or $b$. trying to prove this by using the intermediate value theorem.

Here is what I have so far:

Let $a$,$b \in \mathbb{R}$ Assume to the contrary that the max value occurs at an interior point of the interval $[a,b]$. Let $x_\circ$ equal this interior point.

The first case would occur when $f(a) < f(b)$ then $f(a) < f(b) < f(x_\circ)$ If we consider $y_\circ \in ( f(b), f(x_\circ))$ is there a way to show that $y_\circ$ has at least 2 preimages by applying the intermediate value theorem?

Of course the second case would be $f(b) < f(a)$...

Answer 1

Yes! You're very close. Notice first that you only really need to show this for the first case by symmetry, since the second case is virtually the same. Or instead, just say $c = \max(f(a), f(b))$ and deal with both at once.

Now, take $z \in (c, f(x_0))$. We have that $f(a) < z < f(x_0)$, and $f(b) < z < f(x_0)$, so what can we say about the preimage of $z$?

Answer 2

Then of course $y_0\in(f(a),f(x_0))$ as well! (Remember $f(a)\lt f(b)\lt f(x_0)$.)

Thus there is $\alpha$ between $a$ and $x_0$ and another $\beta$ between $x_0$ and $b$ such that $f(\alpha)=f(\beta)=y_0$, so $f$ is not injective.

You were really close to the proof.