Jaxon Hamilton

2022-07-23

I am not sure if the IVT should be applied here. I am try to do this problem and am stuck an how proceed:
Suppose $f:\left[-1,1\right]\to \mathbb{R}$ is continuous and satisfies $f\left(-1\right)=f\left(1\right)$. Prove that there exists $y\in \left[0,1\right]$ such that $f\left(y\right)=f\left(y-1\right)$.
So far, I have consideblack a new function $g\left(x\right)=f\left(x\right)-f\left(x-1\right)$, $x\in \left[0,1\right]$, but am stuck after this.

Expert

Using the fact that $f\left(1\right)=f\left(-1\right)$, we have
$g\left(0\right)=f\left(0\right)-f\left(-1\right)=f\left(0\right)-f\left(1\right)$
$g\left(1\right)=f\left(1\right)-f\left(0\right)=-\left[f\left(0\right)-f\left(1\right)\right]$
So, $g\left(0\right)$ and $g\left(1\right)$ have opposite signs. By continuity of $g$, it must have a root in [0,1], so there exists $y\in \left[0,1\right]$ such that $g\left(y\right)=0$, i.e. $f\left(y\right)=f\left(y-1\right)$.

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