Prove Surjective function (R:reals) with |f(x)-f(y)|>k|x-y|

[math8]

(R:reals)
Let f:R-->R be continuous and satisfy |f(x)-f(y)|>or eq. to k|x-y| for all x, y in R and some k>0. Show that f is surjective.

I can show that f is injective: let f(x) = f(y), hence k|x-y|< or eq. to 0, thus x=y.

I had a suggestion that it might be helpful to show that f has closed image. But I don't see how to work with that.
I know that f is surjective if for each y in R there is an x in R such that f(x)=y.
I don't see how to proceed.

 

[jgens]

Well, assuming the function is injective (I'm not great with these proofs so I'm not sure yours is correct) and f is also surjective, then f is a bijection; hence, f must have a have a one to one correspondance. Now suppose f is not bijective, then there must be some value f(x) = f(y) where y > x (for definiteness), can you show how this contradicts your initial conditions?

 

[Dick]

This isn't a very difficult proof. Consider f(-x), f(0) and f(x) while x goes to infinity.

 

[jgens]

Crap, I mixed my definitions up. Sorry math8!

Well, here's my idea for a proof for surjection (really this time). Suppose f is not surjective. Then, since f is continuous, as x -> infinity, f(x) -> N (or at least varies between values in some finite interval). Now we fix y such that y = 0 and f(y) = C. Therefore, abs(N - C) >= k lim (x -> infinity) abs(x) which cannot be true. Use a similar approach to completely show f is surjective (I think it should work, though my approach is a quite informal). Again, I'm really sorry!