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 continuous:
> > 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. [...]