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

In summary, the conversation discusses proving that a continuous function f:R-->R is surjective if it satisfies the inequality |f(x)-f(y)|>or eq. to k|x-y| for all x, y in R and some k>0. The participants propose various methods for proving this, including showing that f is injective and has a closed image, and using a proof by contradiction.
  • #1
math8
160
0
(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.
 
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  • #2
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?
 
  • #3
This isn't a very difficult proof. Consider f(-x), f(0) and f(x) while x goes to infinity.
 
  • #4
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!
 

1. What is a surjective function?

A surjective function is a function in which every element in the output range has at least one corresponding input value. This means that every possible output value is "hit" or reached by the function.

2. What does it mean to prove a function is surjective?

To prove that a function is surjective, we must show that every element in the output range has at least one corresponding input value. This can be done by demonstrating that for every possible output value, there exists at least one input value that produces that output.

3. What does the notation |f(x)-f(y)|>k|x-y| mean?

This notation is known as the triangle inequality and it states that the absolute value of the difference between f(x) and f(y) is greater than or equal to k multiplied by the absolute value of the difference between x and y. This is often used in proofs involving surjective functions.

4. How do you prove a function is surjective using the given inequality?

In order to prove that a function is surjective using this inequality, we must show that for any arbitrary output value, there exists an input value that produces that output. This can be done by using the triangle inequality to manipulate the given inequality and then finding a value of x that satisfies the new inequality.

5. Can a function be both surjective and injective?

Yes, a function can be both surjective and injective. This type of function is known as a bijective function. It means that every element in the output range has exactly one corresponding input value, and every possible output value is "hit" or reached by the function.

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