
Re: monotonic function
Posted:
Nov 29, 2012 5:01 AM


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\/
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. 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 noninjective.
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

