19
$\begingroup$

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.

$\endgroup$
19
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Thanks, I knew you went for the kill with the IVT; just wasn't sure how to set it up. $\endgroup$ – WacDonald's Jul 13 '12 at 3:19
  • $\begingroup$ Since it is strictly increasing and the domain is unbounded hence the function should be unbounded also? $\endgroup$ – Upstart Jul 17 at 18:40
  • $\begingroup$ @Upstart: Certainly not. $\arctan(x)$ is continuous, strictly increasing, with unbounded domain, but its values always lie between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. $\endgroup$ – Arturo Magidin Jul 17 at 19:27
7
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The third paragraph is a non-sense. $\endgroup$ – user26857 Oct 9 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.