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.
Any comments? Thank you!

 

[Mark44]

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

 

[etotheipi]

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?

 

[Infrared]

You know that the sum must have a non-negative term and a non-positive term for it to be zero. This means that the function $h(x)=f(x+1/n)-f(x)$ is somewhere non-negative and somewhere non-positive. By IVT, this means that it is somewhere zero.

 

[archaic]

You know that the sum must have a non-negative term and a non-positive term for it to be zero. This means that the function $h(x)=f(x+1/n)-f(x)$ is somewhere non-negative and somewhere non-positive. By IVT, this means that it is somewhere zero.

Yes.. I got that as I was going to sleep. It's so obvious, I don't know what happened.