seupeljewj

2022-06-24

Suppose $f$ is continuous on [0,2] and that $f\left(0\right)=f\left(2\right)$. Then $\mathrm{\exists }$$x,$$y\in \left[0,2\right]$ such that $|y-x|=1$ and that $f\left(x\right)=f\left(y\right)$.

Let $g\left(x\right)=f\left(x+1\right)-f\left(x\right)$ on [0,1]. Then $g$ is continuous on [0,1], and hence $g$ enjoys the intermediate value property! Now notice
$g\left(0\right)=f\left(1\right)-f\left(0\right)$
$g\left(1\right)=f\left(2\right)-f\left(1\right)$
Therefore
$g\left(0\right)g\left(1\right)=-\left(f\left(0\right)-f\left(1\right){\right)}^{2}<0$
since $f\left(0\right)=f\left(2\right)$. Therefore, there exists a point $x$ in [0,1] such that $g\left(x\right)=0$ 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?

alisonhleel3

Expert

It is enough to write $g\left(1\right)=-g\left(0\right)$..

And, as Hagen commented, you only forgot to mention the case when already $f\left(0\right)=f\left(1\right)$, but then it is readily done.

Do you have a similar question?