# Prove: if $f(x)$ is continuous with $f(0)=f(1)=0$ and $0<k<1/2$ then there exists $0\leq x\leq1$ such that $f(x+k)=f(x)$

It is true (even if $$k>1/2$$) if $$f(x)\geq0$$ by constructing $$g(x)=f(x+k)-f(x)$$ and using the IVT since $$g(0)=f(k)\geq0$$ and $$g(1-k)=-f(1-k)\leq0$$. I'm OK with relaxing the assumptions to $$f$$ being smooth, if it helps.

Why $$k\leq1/2$$? Build $$f(x)$$ such that $$f(x)>0$$ in $$(0,0.5)$$ and $$f(x)<0$$ in $$(0.5,1)$$.

• Isn't there a paper by R. P. Boas on this? True for some $k$, false for others. Sorry for such a vague recollection... Dec 31, 2023 at 19:26
• Why are you assume $f(k)$ and $f(1-k)$ are positive? Dec 31, 2023 at 20:11
• @fleablood The first paragraph of the question shows that the sought claim holds if $f$ is nonnegative. Dec 31, 2023 at 20:22
• I'd appreciate if you can point me to the paper above (by R. P. Boas) Dec 31, 2023 at 21:07
• The conclusion holds iff $k={1\over n}$ for a positive integer $n.$ Jan 1 at 1:31

Here is a counterexample for $$k = \frac27$$. Let $$f \colon [0, 1] \to \mathbb R$$ be the piecewise function consisting of the quadratic functions

• from $$(0, 0)$$ through $$(\frac17, \frac12)$$ to $$(\frac27, -\frac1{10})$$;
• from $$(\frac27, -\frac1{10})$$ through $$(\frac{11}{28}, \frac15)$$ to $$(\frac12, 0)$$; and
• the images of the previous two functions under a $$180^\circ$$ rotation around $$(\frac12, 0)$$.

Here is a plot of blue $$f(x)$$ and yellow $$f(x + \frac27)$$.

Here's a proof sketch:

Extend $$f(x)$$ to be zero outside of $$[0,1]$$ and define $$g(x)=f(x+k)-f(x)$$. Then, $$\sum_{r=0}^{m} g(r\cdot k)$$ for $$m>1/k$$ is zero. Either $$g(r\cdot k)$$ is zero for all $$r$$ or there are $$r_0$$ and $$r_1$$ such that $$g(r_1\cdot k)<0$$ and $$g(r_0\cdot k)>0$$ and the theorem follows from the IVT.

• In "the theorem follows from the IVT," what happens if the interval between $r_0 \cdot k$ and $r_1 \cdot k$ is not a subset of $[0, 1 - k]$? Dec 31, 2023 at 18:56
• Agree. There is only one point along the $0,k,2k,...,r_0\cdot k,...,r_1\cdot k,...,mk$ series that this can happen (if $r_1$ is the minimum such that $r_1>1/k$). How would you handle this edge case? Dec 31, 2023 at 19:14
• I believe I found a counterexample. I will edit my answer to include a graph of $f$ and $g$ later on. Dec 31, 2023 at 19:19