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Topic: monotonic function
Replies: 14   Last Post: Nov 29, 2012 6:53 PM

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David Bernier

Posts: 3,892
Registered: 12/13/04
Re: monotonic function
Posted: Nov 29, 2012 6:53 PM
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On 11/29/2012 12:15 PM, David C. Ullrich wrote:
> On Thu, 29 Nov 2012 05:01:18 -0500, David Bernier
> <> wrote:

>> On 11/28/2012 09:27 AM, Timothy Murphy wrote:
>>> quasi wrote:

>>>> Any increasing function is monotonic.
>>>> Any decreasing function is monotonic.

>>> ...
>>> Also, maybe: a continuous function f: R->R is strictly monotonic
>>> if and only if it is injective?

>> My deductions show that:

Here above, I should have written

"My work shows that Timothy Murphy's guess is correct."

>> If a continuous function f: R -> R isn't strictly monotonic,
>> it might or might not be injective.

> What's an example of a continuous f : R -> R which is
> not strictly monotonic but is injective?

I find that there are no such functions.


I was trying to prove one direction of the equivalence
f injective <==> f strictly monotonic,
using proof by contradiction.
So I redo the beginning in corrected form.


Suppose f: R->R is continuous.

We will show that
f injective <===> f strictly monotonic.

The first implication we prove is that
f is injective ==> f is strictly monotonic. (A)

The implication (A) is equivalent to the implication (A'),

f is not strictly monotonic ==> f is not injective (A').

We prove the implication (A') through a proof by

Proof by contradiction of (A'):

So, in order to get a contradiction, starting from
f is not strictly monotonic, we'll assume
that f _is_ injective .

f not strictly monotonic and yet f injective over an
interval [a, b] with a< b would give one of two graph shapes:

(f continuous)

x = a/\
\x = b

/x = b

There's a strict maximum at some point between a and b (1st case),
and a strict minimum of f at some point between a and b (2nd case).
This is because f is continuous and [a, b] is a compact interval.

Then, by the intermediate value theorem (whether it be the 1st or
2nd case), it's easy to see that some value in the range of
f on [a, b] is attained at at least two distinct points
in [a, b]. So f is not injective on [a, b].
So f: R -> R isn't injective. (contradiction)

End of Proof.

(A') has been proved.
So (A) is proved, i.e.:

Suppose f: R->R is continuous.
Then f is injective ==> f is strictly monotonic.

The other direction, that
f is strictly monotonic ==> f is injective,

is trivial.

David Bernier

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