A continuous, injective function $f: \mathbb{R} \to \mathbb{R}$ is either strictly increasing or strictly decreasing.

Question

I would like to prove the statement in the title.

Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. So suppose $x < y$.

And that's pretty much how far I got. Help will be appreciated.

Answer 1

Prove the contrapositive instead: if $f$ is not strictly increasing and not strictly decreasing, then it is not one-to-one.

For example, say there are points $a\lt b\lt c$ such that $f(a)\lt f(b)$ and $f(b)\gt f(c)$. Either $f(a)=f(c)$ (in which case $f$ is not one-to-one), or $f(a)\lt f(c)$, or $f(c)\lt f(a)$.

If $f(a)\lt f(c)\lt f(b)$, then by the Intermediate Value Theorem there exists $d\in (a,b)$ such that $f(d)=f(c)$; hence $f$ is not one-to-one.

Now, there are other possibilities (I made certain assumptions along the way, and you should check what the alternatives are if they are not met).

Answer 2

Consider $g\colon \{(x,y)\mid x<y\} \to \mathbb R$, defined by $g(x,y):=f(x)-f(y)$. Clearly $g$ is continuous. Since the domain of $g$ is connected and $g$ has no zeroes, the image of $g$ is an interval not containing $0$.

Answer 3

Suppose $f : \mathbb{R} \rightarrow \mathbb{R}$ is continuous and not strictly increasing. Then there exists two points such that $f(a) = f(b)$, or there exists three points $a < b < c$ such that $f(a) < f(b)$ and $f(b) < f(c)$. The first case contradicts injectivity. Suppose the second, without loss of generality, suppose that $f(b) - f(a) \leq f(c) - f(b)$. Then $f(b) \leq f(b) - (f(b) - f(a)) = f(a) \leq f(c)$. By the intermediate value theorem, there exists $d$ such that $b < d < c$ such that $f(d)= f(a)$. This contradicts injectivity.

Answer 4

Since $f$ is one-to-one, for $a<b$ we have $f(a)\neq f(b)$. We first consider the case when $f(a)<f(b)$. I claim that in this case $f$ is strictly increasing.

First, note that for any $x\in(a,b), f(a)<f(x)$. If not, then since $f$ is 1-1, must have $f(x)<f(a)$. But then by the IVT there is some $c\in(x,b)$ so that $f(c)=f(a)$, contradicting $f$ being 1-1.

Now suppose for contradiction that $f$ is not strictly increasing. So there is some $x,y\in I$, $x<y$ with $f(y)<f(x)$. By the previous paragraph, we also have $f(a)<f(x)$. So by the IVT, there is some $c\in (a,x)$ with $f(c)=f(y)$, contradicting $f$ being 1-1.

Thus $f$ is strictly increasing if $f(a)<f(b)$. If $f(b)<f(a)$, a similar argument (with all inequalities reversed) shows $f$ is strictly decreasing.

Answer 5

Lemma: Let $f:\mathbb{R}\to\mathbb{R}$ be a function with the following property: For any $x,y,z\in\mathbb{R}$ such that $x<y<z$, $f(y)$ is strictly between $f(x)$ and $f(y)$. Then $f$ is strictly monotonic.

Proof: $f$ should injective. For, if $x,y\in \mathbb{R}$ and $x<y$, then there exists $z\in \mathbb{R}$ such that $x<z<y$, and either $f(x)<f(z)<f(y)$ or $f(x)>f(z)>f(y)$, i.e., $f(x)\neq f(y)$.

Next, let $x\in \mathbb{R}$. There exists $y\in \mathbb{R}$ such that $x\neq y$. First suppose that $x<y$. Since $f$ is injective, either $f(x)<f(y)$ or $f(x)>f(y)$. We will show that $f$ is strictly increasing in the former case, and strictly decreasing in the latter case. Suppose that $f(x)<f(y)$. Let $z\in \mathbb{R}$. Note the following:

• If $x<z<y$, then necessarily $f(x)<f(z)<f(y)$.
• If $x<y<z$, then necessarily $f(x)<f(y)<f(z)$.
• If $z<x<y$, then necessarily $f(z)<f(x)<f(y)$.

It follows that if $x<z$, then $f(x)<f(z)$, and if $z<x$, then $f(z)<f(x)$. This shows that $f$ is strictly increasing.

Now suppose that $f(x)>f(y)$. Let $z\in \mathbb{R}$. Note the following:

• If $x<z<y$, then necessarily $f(x)>f(z)>f(y)$.
• If $x<y<z$, then necessarily $f(x)>f(y)>f(z)$.
• If $z<x<y$, then necessarily $f(z)>f(x)>f(y)$.

It follows that if $x<z$, then $f(x)>f(z)$, and if $z<x$, then $f(z)>f(x)$. This shows that $f$ is strictly decreasing.

Next, suppose that $y<x$. Since $f$ is injective, either $f(y)<f(x)$ or $f(y)>f(x)$. We will show that $f$ is strictly increasing in the former case, and strictly decreasing in the latter case. Suppose that $f(y)<f(x)$. Let $z\in \mathbb{R}$. Note the following:

• If $y<x<z$, then necessarily $f(y)<f(x)<f(z)$.
• If $y<z<x$, then necessarily $f(y)<f(z)<f(x)$.
• If $z<y<x$, then necessarily $f(z)<f(y)<f(x)$.

It follows that if $x<z$, then $f(x)<f(z)$, and if $z<x$, then $f(z)<f(x)$. This shows that $f$ is strictly increasing.

Now suppose that $f(y)>f(x)$. Let $z\in \mathbb{R}$. Note the following:

• If $y<x<z$, then necessarily $f(y)>f(x)>f(z)$.
• If $y<z<x$, then necessarily $f(y)>f(z)>f(x)$.
• If $z<y<x$, then necessarily $f(z)>f(y)>f(x)$.

It follows that if $x<z$, then $f(x)>f(z)$, and if $z<x$, then $f(z)>f(x)$. This shows that $f$ is strictly decreasing.

Now let $f:\mathbb{R}\to \mathbb{R}$ be a continuous and injective function. Suppose that $f$ is not strictly monotonic. Then by the lemma above, there exist $x,y,z\in \mathbb{R}$ such that $x<y<z$ and $f(x)>f(y)<f(z)$ or $f(x)<f(y)>f(z)$, either of which contradicts the intermediate value theorem.