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My coauthor and I have run into the following problem in a research project involving normal numbers. We suspect that the following question may be resolved using standard techniques in analysis. We take the empty product to be $1$.

Question

  1. Let $m\ge 2$ and let $c_1,c_2,\cdots,c_m$ be integers $\geq 2$. Suppose that $f:[0,1] \to [0,1]$ is continuous and non-decreasing, f(0)=0, f(1)=1, and satisfies \begin{equation}\tag{1} \frac{1}{m} \sum_{r = 0}^{m-1} \sum_{j = 0}^{c_1 \ldots c_{r} - 1} \left(f\left(\frac{j+x}{c_1 \ldots c_{r}}\right) - f\left(\frac{j}{c_1 \ldots c_{r}}\right)\right)=x \end{equation} for all $x \in [0,1]$. Note that $f(x) = x$ satisfies these conditions. Is it true that this is the only function that satisfies these conditions?

    Since $f$ is non-decreasing, the derivative exists almost everywhere. Differentiating the previous equation, we have $$\tag{2}\frac{1}{m} \sum_{r=0}^{m-1} \sum_{j=0}^{c_1 \cdots c_r -1} \frac{1}{c_1 \cdots c_r} f'\left(\frac{j+x}{c_1 \cdots c_r}\right) = 1$$ for a.e. $x \in [0,1]$. Does this imply that $f'(x)=1$ almost everywhere? If this is true, then we can prove that the answer to our question is yes.

  2. In case the answer to question $1$ is no, consider the following conditions. Let $m \geq 2$ and let $(c_1, c_2, \cdots)$ be a sequence of integers $\geq 2$ with $c_{i} = c_{i+m}$ for all $i$. Suppose that $f: [0,1] \to [0,1]$ is continuous, non-decreasing, f(0)=0, f(1)=1, and satisfies \begin{equation}\tag{3} \frac{1}{mt} \sum_{r = 0}^{mt-1} \sum_{j = 0}^{c_1 \ldots c_{r} - 1} \left(f\left(\frac{j+x}{c_1 \ldots c_{r}}\right) - f\left(\frac{j}{c_1 \ldots c_{r}}\right)\right)=x \end{equation} for all $x \in [0,1]$ and all positive integers $t$. Is it true that $f(x) =x$ for all $x \in [0,1]$? Note that a similar equation holds for the derivative as well.

What we've tried

Note that the left side of Equation 2 is an average of $m$ Perron-Frobenius operators (acting on $L^1([0,1])$) applied to $f'$. We have that $f' \in L^1([0,1])$ by applying the Lebesgue decomposition theorem to the distributional derivative of $f$. We do not have much experience with the study of Perron-Frobenius operators, so we are not sure if this observation is helpful.

The left sides of both Equation 1 and 3 can be regarded as averages of variations of $f$ over specific types of partitions.

Taking the Fourier transform of both sides of Equation 1 gives (writing $e_n(x) = e^{2\pi i n x}$ and $a_n = \int_0^1 f(x) e_n(x) dx$) $$\tag{4} \frac{1}{m} \sum_{r=0}^{m-1} c_1 \cdots c_r a_{c_1 \cdots c_r n} = \frac{i}{2\pi n} \hbox{ for } n\neq 0 $$ and $$\tag{5} \frac{1}{m} \sum_{r=0}^{m-1} c_1 \cdots c_r \left( a_0 - \sum_{j=0}^{c_1 \cdots c_r -1} \frac{1}{c_1 \cdots c_r} f\left(\frac{j}{c_1 \cdots c_r}\right) \right) = \frac{1}{2}. $$

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  • $\begingroup$ Since this is part of a research project, I might suggest MathOverflow. However, people might have an answer on this site. $\endgroup$ – Clarinetist Jul 17 '14 at 1:51
  • $\begingroup$ We wanted to try here first, but if we don't get any answers, we'll post to overflow. $\endgroup$ – Dylan Airey Jul 17 '14 at 1:54

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