For a continuous function on [0,2] , f(0)=f(2)

[kbgregory]

Homework Statement

Suppose f is a continuous function on [0,2] with f(0) = f(2). Show that there is an x in [0,1] where f(x) = f(x+1).

Homework Equations

By the Intermediate Value Theorem, we know that any values between sup{f(x)} and inf {f(x)} over x in [0,2] will be repeated, since the function returns to where it began and must retrace, as it were, its steps (or perhaps it is a constant function).

The Attempt at a Solution

What I have so far is:

Let A = {f(x) : x in [0,1]} and let B={f(x) : x in [1,2]} , then (A $$\cap$$ B) contains all the values of f(x) that are repeated in these intervals. So f$$^{-1}$$(A$$\cap$$ B) contains all the values of x in [0,2] for which f(x) repeats itself. It remains to be shown that there is x1 and x2 in f$$^{-1}$$(A $$\cap$$B) for which x2 = x1 + 1, and I am unsure of how to do this, as I am unsure of my entire approach.

Any suggestions would be greatly appreciated.

 

[Office_Shredder]

You might want to look at the function f(x)-f(x+1) on the interval [0,1].

See if you can prove it has a zero using the IVT

 

[kbgregory]

Wow, thanks a lot! Got it.