Continous function in interval problem

[nuuskur]

Homework Statement

Let f: [0,2]\to\mathbb{R} be continuous and f(0) = f(2). Show that there exist x,y\in [0,2] with the following property:
(*)\ y-x = 1 and f(x) = f(y)\ (*)

Homework Equations

Bolzano-Cauchy theorem: If a function f is continuous in some interval [a,b] and f(a) <0, f(b) > 0 (or vice versa) then there exists c\in (a,b) such that f(c) = 0

The Attempt at a Solution

If f was constant, then it's trivial. Fix x\in [0,1] and the condition f(x+1)-f(x) = 0 is satisfied.
Hence, assume f is not constant.

Let us observe function g(x) := f(x+1)-f(x), 0\leq x\leq 1. The objective is to show that g(x)=0 is possible with which we will have proven the existence of the required x,y (is this correct to say? )

Let us note that:
g(0) = f(1) - f(0) and g(1) = f(2) - f(1) = f(0) - f(1). If g(0) > 0, then g(1) <0 (or vice versa), we can therefore conclude that:
Per Bolzano-Cauchy theorem there exists c\in (0,1) such that g(c) = f(c+1) - f(c) = 0 from which we can establish x = c and y = c+1 and the condition (*) is satisfied

If g(0) = 0 then also g(1) = 0 and again the condition (*) is satisfied. Q.E.D

 

[fresh_42]

Looks nice. And what exactly do you expect as replies here? Does your proof still hold for arbitrary c ∈[0,d] with y - x = c and arbitrary continuous functions f : [a,b] → ℝ with f(a) = f(b) and how big can d be at most?

 

[nuuskur]

Most of all expect criticism on presenting proof, generalizing the problem is welcome. Also would like the proof to be challenged if there is something I might have missed.

 

[fresh_42]

Most of all expect criticism on presenting proof, generalizing the problem is welcome. Also would like the proof to be challenged if there is something I might have missed.

I've seen nothing wrong. And your presentation reveals that you work carefully and think about the special cases. I'ld let it go through. (Bluster me if I'm wrong!) There is only one little, tiny, small remark from my side: Don't shout QUOD ERAT DEMONSTRANDUM. A simpe qed will do and a ◊ or box is even more pleasant

 

[fresh_42]

Most of all expect criticism on presenting proof, generalizing the problem is welcome. Also would like the proof to be challenged if there is something I might have missed.

I've seen nothing wrong. And your presentation reveals that you work carefully and think about the special cases. I'ld let it go through. (Bluster me if I'm wrong!) There is only one little, tiny, small remark from my side: Don't shout QUOD ERAT DEMONSTRANDUM. A simpe qed will do and a ◊ or box is even more pleasant.

- sorry, probs with the connection and a failed search for a remove button