# Prove that there exists $x_n$ such that $0 \leq x_n \leq 1-\frac{1}{n}$ and $f(x_n)=f(x_n+\frac{1}{n})$. [duplicate]

Suppose that the function $f:[0,1] \to \mathbb{R}$ is continuous on $[0,1]$ and $f(0)=f(1)$. Prove that for each natural number $n$, there exists $x_n \in \mathbb{R}$ such that $0 \leq x_n \leq 1-\frac{1}{n}$ and $f(x_n)=f(x_n+\frac{1}{n})$.

Though I don't know how the proof would look like, I have a strong feeling that it has something to do with the Intermediate Value Theorem, judging by the continuity of $f$ and the existence of such a $x_n$. So I guess I'm supposed to define a function $g(x)=f(x)-f(x+\frac{1}{n})$ on $[0,\frac{1}{n}]$ and try to claim that $g(x_n)=0$ for some $x_n \in [0,\frac{1}{n}]$. Unfortunately I don't know how to proceed further from here, maybe because I haven't made use of the fact that $f(0)=f(1)$.

Any hint and suggestion is much appreciated. Thank you!

• Oh my bad, sorry for asking a duplicated question, I did a full search and couldn't find any similar problems so I decided to post it here. Anyway, thanks to all involved, for taking your time answering my question. One more vote and it should be closed. Commented Apr 24, 2013 at 16:27
• drawar, it is hard to find that, you don't have to apologize. Commented Apr 24, 2013 at 17:26

This problem is tricky, by which I mean there's a proof that doesn't use the usual toolkit (IVT, for instance).

Suppose not. Then there exists $n \geq 1$ for which, for every $x \in [0,1-1/n]$, $f(x) \neq f(x + 1/n)$. The left and right hand side are continuous functions, and so $\neq$ is either a $<$ or a $>$ identically on the domain $[0,1-1/n]$. Choose $>$ without loss.

Plugging in $x = k/n$, we have $$f(k/n) > f((k+1)/n)$$ Applying in succession, we obtain $$f(0) > f(1/n) > \cdots > f((n-1)/n) > f(1)$$ which is a contradiction.

• The IVT comes in, for example, in the assertion that $\neq$ is either a $<$ or a $>$. I guess I meant that the IVT comes up in a tricky way. Commented Apr 24, 2013 at 15:33
• I think I see your point here. Did you mean if there exist $x_1$ and $x_2$ s.t $f(x_1) > f(x_1+\frac{1}{n})$ and $f(x_2) < f(x_2+\frac{1}{n})$ then there is an $x$ between $x_1$ and $x_2$ s.t $f(x) = f(x+\frac{1}{n})$, which is a contradiction? Commented Apr 24, 2013 at 16:12
• @drawar Yeah, that's one way to see it. Be careful where the contradiction hypothesis comes in, though. Commented Apr 24, 2013 at 16:24

Consider $g_n(x)=g(x+\frac{1}{n})-g(x)$ for $x\in [0,\frac{n-1}{n}]$.

Now $0=g(1)-g(0) = g_n(0) + g_n(\frac{1}{n}) + \dots g_n(\frac{n-1}n)$. If $g_n(\frac{k}k)=0$ for any $k$ you are done. On the other hand, if not, since their sum is $0$, at least one of the values $g_n(k/n)$ must be positive and at least one must be negative.

Now use the intermediate value theorem.

• Very nice, thank you! Commented Apr 24, 2013 at 15:59