# Proving that there exists a solution to ##f(x+\frac1n)-f(x)=0##

archaic
Homework Statement:
Given ##f:[0,1]\to\mathbb R## which is continuous, ##f(0)=f(1)##, and ##n\in\mathbb N^*## (doesn't include ##0##), show that ##f(x+\frac1n)-f(x)=0## has a solution.
Relevant Equations:
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This was shown to me by an acquaintance who told me that if we consider ##h(x)=f(x+\frac1n)-f(x)##, then the fact that ##\sum_{k=0}^{n-1}f(\frac kn+\frac1n)-f(\frac kn)=0## somehow tells us that there are two opposite terms in that sum, and, as such, we can apply the IVT to ##h(x)##.

Clearly the sum is null because it is in the form ##f(1) + (f(a)-f(a)\text{ same argument a})+(f(b)-f(b)\text{ same argument b})+...-f(0)## and ##f(1)=f(0)##, not because there somehow exists two different in argument terms that cancel ##f(a)+f(b)=0##.
In either case, we require a ##h(a)## and a ##h(b)## such that ##h(a)h(b)<0##, not that but with ##f##.

But, from what I have understood, this is the correction given by their professor, so maybe I am missing something?

EDIT: Ignore the following nonsense. I thought of going about it by contradiction, but the use of a limit leaves me undecided as to whether this counts as a solution.
The negative of the proposition is ##\exists n\in\mathbb N^*\,|\,\forall x\in\mathbb R,\,f(x+\frac1n)-f(x)\neq0##, i.e ##f(x+\frac1n)<f(x)## or ##>##. In both cases, if we take ##n\to\infty##, we get an impossible inequality.

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• Delta2

Mentor
What about ##f(x) = 2## for ##x \in [0, 1]##? f is continuous and f(0) = f(1).

What about ##f(x) = 2## for ##x \in [0, 1]##? f is continuous and f(0) = f(1).

Isn't every ##x \in [0, 1- \frac{1}{n} ]## a solution for that example?

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