# Prove that $f: [a,b] \rightarrow \mathbb{R}$ is strictly monotone

I'm a 1st year mathematics student, and in my analysis class I'm having trouble with proving the following:

Let $a < b$ be real numbers, and let $f: [a,b] \rightarrow \mathbb{R}$ be a function which is continuous and injective. Prove that $f$ is strictly monotone.

So far, I only think I have to divide this into three cases:
Case 1: $f(a) = f(b)$. With the fact that $f$ is injective, this leads to a contradiction.
Case 2: $f(a) < f(b)$. In this case, I think I have to use the previous contradiction, and in some way the intermediate value theorem, to show that $f$ is strictly monotone increasing but I don't know how.
Case 3: $f(a) > f(b)$. Here I think I have to show that $f$ is strictly monotone decreasing. But again, I have no idea on how to do it.

Could anyone help me with this?

As you mentioned in your post, $f(a) = f(b)$ is not possible, because $f$ is injective. Let's assume $f(a) < f(b)$. Now let $c \in (a,b)$. Suppose $f(c) > f(b)$. By the intermediate value theorem there exists a $d_1 \in (a,c)$, such that $f(d_1) = \frac{f(b) + f(c)}{2}$, and there exists a $d_2 \in (c,b)$, such that $f(d_2) = \frac{f(b) + f(c)}{2}$. Since $d_1 \neq d_2$, this is a contradiction to the injectivity of $f$. So we have $f(c) < f(b)$. Indeed, $f(c) = f(b)$ is also not possible, because of the injectivity of $f$. Use the same argument, to show that $f(a) < f(c)$. Now we know that $$f(a) < f(c) < f(b) \quad \text{for each } c \in (a,b) \; .$$ Now let $x,y \in (a,b)$ with $x < y$. We have to show that $f(x) < f(y)$. For an arbitrary point $c \in (x,y)$ we have $f(a) < f(c) < f(b)$. Now by considering $f_{|[a,c]}$ (which is also injective and continuous) and the reasoning above, we deduce $f(a) < f(x) < f(c)$. By considering $f_{|[c,b]}$ and the reasoning above, we deduce $f(c) < f(y) < f(b)$, so all in all we have $$f(a) < f(x) < f(c) < f(y) < f(b) \; ,$$ which is what we wanted to show. So $f$ is strictly increasing.

You can use a similiar argumentation for the case $f(a) > f(b)$, and you will find, that $f$ is strictly decreasing in this case.

Hint:

Suppose that $f$ is not strictly monotone, continuous and injective.

Then there exist $x,y,z$ such that $a < x < y < z < b$ and $f(x) < f(y) > f(z)$ or $f(x) > f(y) < f(z)$. (Try to show that if not, the function is either strictly increasing or strictly decreasing)

But because $f$ is continuous, $f$ takes all values in $[f(x),f(y)]$ and $[f(z), f(y)]$ in the first case. Can you get a contradiction form here?

Let us assume that $f(a) < f(b)$. We will prove that $f(a)$ is minimum value of $f$ and $f(b)$ is maximum value of $f$ in the interval $[a, b]$.

Let us state the argument for maximum value of $f$. Let $M = f(c)$ be the maximum value of $f$. Clearly since $f(a) < f(b)$ so $a < c \leq b$. If $c = b$ we are done with the proof. So let us assume that $a < c < b$ so that the maximum value $M$ is attained at an interior point $c$. Clearly we then have $f(a) < f(b) < M = f(c)$ and hence if we take a number $k$ with $f(b) < k < f(c) = M$ then we also have $f(a) < k < f(c) = M$ and hence by intermediate value theorem there is an $x_{1} \in (a, c)$ and an $x_{2} \in (c, b)$ such that $f(x_{1}) = f(x_{2}) = k$. This is not possible because $f$ is one-one (injective). It follows that that $c = b$ and hence the maximum value of $f$ is $f(b)$. Similarly we can show that $f(a)$ is the minimum value of $f$.

Now it is easy to see that we are done. Let there be points $x, y \in [a, b]$ with $a \leq x < y \leq b$. Since $f(a)$ is minimum value of $f$ we must have $f(a) < f(y)$ and clearly the function $f$ is continuous and injective in sub-interval $[a, y]$ it follows from the argument in preceding paragraph that $f(y)$ is maximum value of $f$ in $[a, y]$ so that $f(x) < f(y)$. Thus $f$ is strictly increasing in $[a, b]$.

The same argument can be continued almost word for word if $f(a) > f(b)$ and in this case $f(a)$ is the maximum value and $f(b)$ is the minimum value of $f$ and the function is strictly decreasing.