aurelegena

2022-09-06

Show that if $f$ is continuous on [0,1] with $f(0)=f(1)$, there must exist $x,y\in [0,1]$ with $|x-y|=\frac{1}{2}$ and $f(x)=f(y)$

I've been working on this for a while, and can't seem to figure out where to start. Any hints would be appreciated

I've been working on this for a while, and can't seem to figure out where to start. Any hints would be appreciated

farbhas3t

Beginner2022-09-07Added 6 answers

Let $g$ be the function defined at $[0,\frac{1}{2}]$ by

$g:t\mapsto f(t)-f(t+\frac{1}{2})$

we have

$g$g is continuous at $[0,\frac{1}{2}]$

and

$g(0).g(\frac{1}{2})=-(f(0)-f(\frac{1}{2}){)}^{2}\le 0$ since $f(0)=f(1)$.

then

$\mathrm{\exists}x\in [0,\frac{1}{2}]\phantom{\rule{thickmathspace}{0ex}}$ such that $g(x)=0$ or

$f(x)=f(x+\frac{1}{2})=f(y)$

with $y=x+\frac{1}{2}$ satisfying

$|y-x|=\frac{1}{2}$

$g:t\mapsto f(t)-f(t+\frac{1}{2})$

we have

$g$g is continuous at $[0,\frac{1}{2}]$

and

$g(0).g(\frac{1}{2})=-(f(0)-f(\frac{1}{2}){)}^{2}\le 0$ since $f(0)=f(1)$.

then

$\mathrm{\exists}x\in [0,\frac{1}{2}]\phantom{\rule{thickmathspace}{0ex}}$ such that $g(x)=0$ or

$f(x)=f(x+\frac{1}{2})=f(y)$

with $y=x+\frac{1}{2}$ satisfying

$|y-x|=\frac{1}{2}$

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