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I've been trying to prove this statement:

Let $f$ be a continuous function on $[0,1]$ such that $f(0)=f(1):=c$, and let $n$ be a natural constant. Prove that there exists $xin[0,1-frac{1}{n}], xin mathbb R$ such that $f(x+frac{1}{n})=f(x).$

I tried to use the Intermediate Value Theorem by defining $g(x)=f(x+frac{1}{n})-f(x)$. I could see that $g(0)=f(frac{1}{n})-c$ and $g(1-frac{1}{n})=c-f(1-frac{1}{n})$ yet I could not figure out how to prove that $mathrm smathrm imathrm gmathrm n (g(0)cdot g(1-frac{1}{n}))=-1.$

Edit: Please notice that this statement is not (necessarily) true for every n. The statement is about some constant n.

Thank you and have a good day!.

Let
$$g(x)=fleft(x+frac{1}{n}right)-f(x)$$
Then we need to prove that $exists x$ so that $g(x)=0$.
$$g(0)=fleft(frac{1}{n}right)-c$$
$$gleft(frac{1}{n}right)=fleft(frac{2}{n}right)-fleft(frac{1}{n}right)$$
$$gleft(frac{2}{n}right)=fleft(frac{3}{n}right)-fleft(frac{2}{n}right)$$
$$dots$$
$$gleft(frac{n-2}{n}right)=fleft(frac{n-1}{n}right)-fleft(frac{n-2}{n}right)$$
$$gleft(frac{n-1}{n}right)=c-fleft(frac{n-1}{n}right)$$
If one of the above values are $0$, then we are done. So suppose that all of them are different from $0$. But then we can sum them:
$$gleft(frac{0}{n}right)+gleft(frac{1}{n}right)+dots+gleft(frac{n-1}{n}right)=0$$
Because none of the values on the LHS are zero, then there must be at least $1$ negative and at least $1$ positive. But this means that we are done: if $g(a)<0$ and $g(b)>0$ then $exists d$ between $a$ and $b$ with $g(d)=0$, because $g$ is continuous.

Remark1: And this statement is true for all $n in mathbb{N}_+$

Remark2: I had the same exercise 1 or 2 months ago. I wanted to make a question about it, but I figured out the solution while I was writing the question :)

An extra exercise: Construct a function $f in C[0,1]$, $f(0)=f(1)$ which does not have a chord with length $2/5$: So $nexists x in [0, 1-2/5]$ so that $f(x)=f(x+2/5)$. This was an additional excercise in my worksheet, but I could not solve it.