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Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous with $f(1)=f(0)$. If $h\in (0, 1/2)$ is not of the form $1/n$, there does not necessarily exist $|x − y| = h$ satisfying $f(x) = f(y)$. Provide an example that illustrates this using h = 2/5.

I'm really mind blown with this question. I could share with you what I've done but its just some squiggles of a graph and it feels like there is absolutely no way I can prevent a gap of $2/5$ sneaking in. I'm taking $|x-y|=2/5$ to be, "Can I fit a ruler under my curve of length 2/5 so that it touches both sides" and I just cannot see a situation where that doesn't happen. I'm guessing that my intuition has been flawed by my drawing attempt and in fact the solution is a very non-differentiable curve.

The same question has been "answered" here Using Intermediate Value Theorem for continuous functions but unfortunately this poor fellow has been entirely mislead by a false example.

sequencesandseries
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    This is the subject of the Universal Chord theorem. In the accepted answer to this question there is a sketch of how to achieve this. – Ross Millikan Mar 27 '21 at 15:12
  • I don't think the question as clearly stated as it might be. As I understand it, you are required to provide an example of a continuous function $f(x)$ with $f(1) = f(0) $ and $f(x_1) = f(x_2) \implies |x_1 - x_2| \ne 2/5$ – Tom Collinge Mar 27 '21 at 15:20
  • @Tom: The question asks you to think of a function $f$ where $f$ is continuos, $f(0) = f(1)$, and forall $x$ in domain, $f(x + 0.4) \neq f(x)$. – Anon Mar 27 '21 at 15:23
  • @Kaind I think that is just the negation of what I wrote (i.e. the same thing) ? – Tom Collinge Mar 27 '21 at 15:26
  • @Kaind As I understand the query, $(x + 0.4)$ must also be in the domain, which means that $0 \leq x \leq 0.6$. – user2661923 Mar 27 '21 at 15:26
  • @Tom: that's correct. user2661923: Yes I know, i was feeling a bit lazy to write $[0,0.6]$, hence I just wrote 'domain' which isn't entirely correct. – Anon Mar 27 '21 at 15:28
  • Here are examples as well: https://math.stackexchange.com/q/16374/42969 – Martin R Mar 27 '21 at 15:52

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I made a picture following the sketch of Brian Rushton at the post linked by Ross Millikan. More generally, if you want to avoid having $f(x)=f(x+a)$ where $0<a<1$ is not of the form $\frac{1}{n}$, the idea would be to take the quotient of the $x$-axis by $a \mathbb{Z}$ and draw the graph on the resulting cylinder.

enter image description here

Mike F
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I just had a huge chat with user @user21820 on this problem. Do see his answer at: Intuition for the Universal Chord Theorem

Either way I'll just write an explicit function as an answer for this problem: $f(x) = \begin{cases} -10x &, x \in [0,0.2) \\ 1 + 15(x-0.4) &, x \in [0.2,0.4) \\ 1 -10(x - 0.4) &, x \in [0.4,0.6) \\ 2 + 15(x-0.8) &, x \in [0.6,0.8) \\ -10(x - 1) &, x \in [0.8,1] \\ \end{cases} $

The above function $f(x)$ is just a bunch of alternatively parallel line segments joined at the endpoints. It satisfies the property: $$x \in [0,0.6] \Rightarrow f(x + 0.4) = f(x) + 1$$ A plot for f(x) is given by: enter image description here

Anon
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    I gave +1, but this is pretty much the same plot I put in my answer! Observation that $f(x+0.4)=f(x)+1$ is nice. – Mike F Mar 28 '21 at 02:25
  • Hey, I realize you have the 'cylinder ' approach to plotting the curve. But user21820's answer to that problem on the link in my answer describes a slightly different method of construction which allows us to find $f$ s.t. $f(x+0.4) = f(x) + c$ for any constant $c$. – Anon Mar 28 '21 at 03:44