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I am reviewing for a midterm and this a problem from a previous year's final.

Assume that $f \in C ([0,2])$ and $f (0) = f (2)$ Prove that there exist $x_1$ and $x_2$ in $[0,2]$ such that $x_2 -x_1 = 1$ and $f (x_2) = f (x_1)$

I am not even sure where to begin with this problem. I can't use the intermediate value theorem, and I know nothing about whether or not the function is differentiable so the mean value theorem doesn't apply either. If someone could possibly solve the problem or give some hints so I can continue reviewing I would appreciate it.

user97549
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1 Answers1

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Consider the function $g(x)=f(x)-f(x+1)$, defined on $[0,1]$. Note that $$g(0)=f(0)-f(1)=f(2)-f(1)=-g(1).$$ By continuity (and the intermediate value theorem), for some $t\in[0,1]$ we must have $g(t)=0$, that is, $f(t)=f(t+1)$.

Andrés E. Caicedo
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