Suppose $f$ is continuous on $[0,2]$ and that $f(0) = f(2)$. Then $\exists$ $x,$ $y \in [0,2]$ such that $|y – x| = 1$ and that $f(x) = f(y)$.
Let $g(x) = f(x+1) – f(x)$ on $[0,1]$. Then $g$ is continuous on $[0,1]$, and hence $g$ enjoys the intermediate value property! Now notice $$g(0) = f(1) – f(0)$$ $$g(1) = f(2) – f(1)$$
Therefore $$ g(0)g(1) = -(f(0) – f(1))^2 < 0$$
since $f(0) = f(2)$. Therefore, there exists a point $x$ in $[0,1]$ such that $g(x) = 0$ by the intermediate value theorem. Now, if we pick $y = x + 1$, i think the problem is solved.
I would like to ask you guys for feedback. Is this solution correct? Is there a better way to solve this problem?
Best Answer
Very good!
It is enough to write $g(1)=-g(0)$..
And, as Hagen commented, you only forgot to mention the case when already $f(0)=f(1)$, but then it is readily done.