The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
David Bernier

Posts: 3,892
Registered: 12/13/04
Re: monotonic function
Posted: Nov 29, 2012 5:47 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 11/29/2012 05:01 AM, 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:
> If a continuous function f: R -> R isn't strictly monotonic,
> it might or might not be injective.
> If f is not injective, we're done.
> --
> Then the remaining case is where 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
> or:
> /x = b
> x=a\/

Upon further reflection, it seems that this paragraph here:

> 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.


can be omitted.

In the 1st case, there is a 'c' with a< c < b such that:
f(a), f(b) and f(c) are distinct and
f(c)> f(a) and f(c) > f(b).

In the 2nd case skectched above, there is once again a 'c'
with a< c < b , but in this instance, 'c' is such that:
f(a), f(b), f(c) distinct and
f(c) < f(a), and f(c) < f(b).

Them we can use the intermediate value theorem
on [a, c] as well as on [c, b], since f is

> 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.
> This contradicts the assumption that f is injective.
> --
> So, assuming f: R->R is continuous,
> f not strictly monotonic ==> f not injective.
> By the contrapositive, under the same assumptions on f,
> f injective ===> f strictly monotonic.
> On the other hand, assuming f: R->R is continuous,
> if f is strictly monotonic, a<b ==> f(a) < f(b) without
> loss of generality. That is seen to rule out f being
> non-injective.
> So on the other hand, assuming f: R->R is continuous,
> f strictly monotonic ==> f is injective.
> Taken together, we get:
> Under the assumption that f: R -> R is continuous, then:
> f is injective <==> f is strictly monotonic.


David Bernier

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.