2
$\begingroup$

Let $f: [0,1] \to \mathbb{R}$ be continuous and such that $f(0) = f(1)$. For each $n \in \mathbb{N}$, prove that there exists $x \in [0,1]$ such that $x + 1/n \in [0,1]$ and $f(x + 1/n) = f(x)$.

Don't even know where to start.

$\endgroup$
3
  • 1
    $\begingroup$ Edit your question title $\endgroup$
    – Gyanshu
    Jan 16, 2017 at 16:32
  • $\begingroup$ what do you suggest? @Gyanshu $\endgroup$ Jan 16, 2017 at 16:35
  • $\begingroup$ It is fine now. $\endgroup$
    – Gyanshu
    Jan 16, 2017 at 17:56

1 Answer 1

3
$\begingroup$

Denote $$g(x):=f(x+\frac{1}{n})$$ and $$h(x):=g(x)-f(x)$$

Then, we have $$h(0)=g(0)-f(0)=f(\frac{1}{n})-f(0)$$ $$h(\frac{1}{n})=f(\frac{2}{n})-f(\frac{(1}{n})$$ if $n>1$ and so on.

We have to show that $h$ has a root in $[0,1]$. If $h(\frac{k}{n})=0$ for some $k<n$, we are done. If we assume that the signs of the values are all equal, we have $f(0)<f(\frac{1}{n})<\cdots <f(1)$ or $f(0)>f(\frac{1}{n})>\cdots >f(1)$. But this is a contradiction to $f(0)=f(1)$.

Hence, we have at least once a different sign and can apply the intermediate value theorem because $h$ is continous.

$\endgroup$
1
  • 1
    $\begingroup$ Ohhh, I didn't think of that. Thanks a lot sir! $\endgroup$ Jan 16, 2017 at 16:54

Not the answer you're looking for? Browse other questions tagged .