On 11/29/2012 12:15 PM, David C. Ullrich wrote: > On Thu, 29 Nov 2012 05:01:18 -0500, David Bernier > <email@example.com> 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 contradiction.
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:
x = a/\ \x = b
or: /x = b x=a\/
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,