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 $(0.5,1)$.

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

2 Answers 2


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 $f(x + \frac27)$.

Plot of f(x) and f(x + 2/7)


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.

  • $\begingroup$ 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 $[0, 1 - k]$? $\endgroup$
    – L. F.
    Dec 31, 2023 at 18:56
  • $\begingroup$ 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? $\endgroup$
    – MDman
    Dec 31, 2023 at 19:14
  • $\begingroup$ I believe I found a counterexample. I will edit my answer to include a graph of $f$ and $g$ later on. $\endgroup$
    – L. F.
    Dec 31, 2023 at 19:19

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