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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.

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3 Answers 3

up vote 6 down vote accepted

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

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Thanks, I knew you went for the kill with the IVT; just wasn't sure how to set it up. –  WacDonald's Jul 13 '12 at 3:19

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.

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Since $f$ is one-to-one, $f(a)\neq f(b)$. 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 $f(x)=f(a)$, contradicting $f$ being 1-1.

Now suppose for contradiction that $f$ is not strictly increasing. So there is some $x,y\in I$ with $f(y)\leq f(x)$. However, since f is 1-1, we cannot have $f(y)=f(x)$, so $f(y)<f(x)$. 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.

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The first sentence makes no sense. In any event, correct answers were given over a year ago, and I'm not sure how you propose to improve on them. –  user43208 Oct 24 '13 at 19:26
    
@user43208 Which part of the first sentence doesn't make sense to you? –  user87274 Oct 24 '13 at 19:30
    
To say $f$ is one-to-one means distinct arguments $a, b$ are carried to distinct values $f(a), f(b)$. So it makes no sense to conclude $f(a) = f(b)$ (which you then proceed to gainsay in the very next sentence). –  user43208 Oct 24 '13 at 19:33
    
Oh my mistake. It should be $f(a)\neq f(b)$. –  user87274 Oct 24 '13 at 19:39

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