Final update: on 11/29/2019: I have worked on this a bit more, and wrote an article summarizing all the main findings. You can read it here.

This functional equation appears in the following context. Let $\alpha\in[0,1]$ be an irrational number (called seed) and consider the sequence $x_n=\{2^n \alpha\}$. Here the brackets represent the fractional part function. In particular, $\lfloor 2x_n\rfloor$ is the $n$-th digit of $\alpha$ in base $2$.

The values $x_n$ are distributed in a certain way due to the ergodicity of the underlying process. The density associated with this distribution is the function $f$, and for the immense majority of seeds $\alpha$ that density is uniform on $[0, 1]$, that is, $f(x) = 1, x \in [0, 1]$. Such seeds $\alpha$ producing the uniform density are sometimes called normal numbers; their digit distribution is also uniform.

However, the functional equation $f(x) = \frac{1}{2}\Big(f(\frac{x}{2}) + f(\frac{1+x}{2})\Big)$ may have plenty of other solutions. Such solutions are called non-standard solutions. Can you find a seed $\alpha$ producing a non-standard solution, with an explicit form for $f$? Maybe a step-wise uniform function? The set of seeds producing non-standard solutions is known to have Lebesgue measure zero, but there are infinitely many such seeds.

All rational seeds $\alpha$ work, but they produce a discrete distribution. Thus their density is of the discrete type. Here however, I am interested in a continuous function $f$, even if it has infinitely many points of discontinuity (that is, a function $f$ continuous almost everywhere: the set of discontinuity points has Lebesgue measure zero.)


I am looking for a function $f$ that is a density on $[0, 1]$, so there are additional constraints here: $\int_0^1 f(x)\,dx = 1$ and $f(x) \geq 0$. However, note that if $f$ is a solution, then $cf$ is also a solution regardless of the constant $c$. So any solution can be normalized to integrate to one. Also, $cf+d$ is also a solution ($c, d$ constants).

Second update

Below is a density satisfying all the requirements. Actually, the plot below represents its percentile distribution. It was produced with a seed $\alpha$ built as follows: its $n$-th binary digit is $1$ if $\mbox{Rand}(n) < 0.75$, and $0$ otherwise, using a pseudo random number generator. Note that $P._{25} = 0.5$ and corresponds to a dip ($P._{25}$ denotes the $25$-th percentile.) Dips are everywhere, only the big ones are visible. By contrast, the percentile distribution for the uniform case (if you replace $0.75$ by $0.50$ in $\mbox{Rand}(n) < 0.75$) is a straight line, with no dips.

enter image description here

Note: I eventually answered my question, see the second answer.

  • $\begingroup$ $f(x)=0$ is a solution, but it's not clear to me how that would relate to the seed mechanism. $\endgroup$ Commented Aug 21, 2019 at 16:57
  • 1
    $\begingroup$ One thing we know: $f(0)=f(1/2)=f(1),$ regardless of the solution elsewhere. $\endgroup$ Commented Aug 21, 2019 at 17:12
  • $\begingroup$ $f(x)=\text{const}$ works for any constant actually. $\endgroup$ Commented Aug 21, 2019 at 17:17
  • 1
    $\begingroup$ Have you looked at any of the numbers whose digits are known not to be normal or equidistributed, e.g. Liouville numbers? $\endgroup$ Commented Aug 21, 2019 at 17:28
  • $\begingroup$ @StevenStadnicki: it is on my "to do" list, thanks. $\endgroup$ Commented Aug 21, 2019 at 17:39

3 Answers 3


Not an answer, just some thoughts.

Let's try applying Fourier transform:

$$f(x)=\frac{1}{\sqrt{2 \pi}}\int_{-\infty}^\infty e^{i k x} g(k) dk$$

The equation becomes:

$$\int_{-\infty}^\infty \left(2 e^{i k x}-e^{i k x/2} (1+e^{i k /2}) \right) g(k) dk=0$$

Making a simple substitution:

$$\int_{-\infty}^\infty e^{i k x} \left(g(k)- (1+e^{i k }) g(2k) \right) dk=0$$

We can consider a particular case (or is this the general case?):

$$g(k)=(1+e^{i k }) g(2k)$$



For any $a \neq 0$:


$$g(2a)= \frac{C}{1+e^{ia}}$$

$$g(2^n a)= C \prod_{q=1}^{n-1} \frac{1}{1+e^{i q a}}$$

In the same way we can find $g(2^{-n} a)$.

For $a$ not being a rational multiple of $\pi$ the infinite product doesn't converge, and the function itself is not nice.

If we call:

$$G_n= -\log g(2^n), \qquad g(1)=1$$

Then the plot $G_n$ looks like this:

enter image description here

Not sure if this could lead to a non-trivial solution to the original equation though.

  • $\begingroup$ I will check this out. Also the constraint that $\int_0^1 f(x) dx = 1$ is not important after all: if $f$ is a solution, then $c\cdot f$ is also a solution regardless of the constant $c$. $\endgroup$ Commented Aug 21, 2019 at 18:46
  • $\begingroup$ Sounds very interesting and inspiring. I updated my post and included an example of a special function (continuous everywhere it seems, but maybe differentiable nowhere) that satisfies the equation. $\endgroup$ Commented Aug 22, 2019 at 5:53

Here I discuss in more detail the case mentioned in the section "second update". This special distribution was produced with a seed $\alpha$ built as follows: its $n$-th binary digit is $1$ if $\mbox{Rand}(n)< p$, and $0$ otherwise, using a pseudo random number generator. I used $p=0.75$ in my example.

Now $x_n$ (introduced in the first paragraph in my question) is a random variable, and we have:

$$x_n=\sum_{k=1}^\infty \frac{d_{n+k}}{b^k}.$$

Here $b$ is the base ($b=2$), and $d_{n+k}$ is the $(n+k)$-th digit of $\alpha$ in base $b$. Furthermore, by construction, these digits are identically and independently distributed with a Bernouilli distribution of parameter $p$. Thus, using the convolution theorem, the characteristic function of $x_n$ is

$$\phi(t; p, b) = \prod_{k=1}^\infty \Big(1-p(1-\exp \frac{it}{b^k})\Big).$$

Take the derivative of the inverse Fourier transform (see section inverse formula here) and you obtain

$$f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ity} \phi(t; p,b) dy.$$

If $p=0.5$ and $b=2$ we are back to the uniform case. If $p\neq 0.5$ then the solution is quite special: the density $f$ is nowhere continuous it seems. See picture below for $p=0.55, b=2$.

enter image description here

Now we should prove that this case is ergodic, for the functional equation to apply. I also tried to check with some sampled values of $x$ to see whether $f(x) = \frac{1}{2}\Big(f(\frac{x}{2}) + f(\frac{1+x}{2})\Big)$, but the function being discontinuous everywhere, and since I got its value approximated probably to no more than two decimals, it is not easy. The distribution attached to this density has the following moments:

  • Expectation: $\frac{p}{b-1}.$
  • Variance: $\frac{p(1-p)}{b^2-1}.$

An Excel spreadsheet with the computation of $x_n$ up to $n=200,000$ and with a precision of 14 decimals, is available upon request. You can interactively change $b$ or $p$ and see the result on the chart. Despite dealing with more than 200K digits, the computations are done very efficiently. Finally, the functional equation can be adapted to base $b$, provided $b$ is an integer. It becomes:

$$f(x) = \frac{1}{b}\sum_{m=0}^{b-1}f\Big(\frac{x+m}{b}\Big).$$


We are told that $f\in L^1([0,1])$ and that the equation $$ f(x)=\frac 1 2 \left( f\left(\frac x 2 \right)+f\left(\frac{1+x} 2\right)\right)\tag{*}$$ holds, for almost all $t\in[0,1]$.

Let $a_n=\int_0^1\exp(2\pi i n x) f(x)dx$. Formula (*) gives: $$ a_n = \frac1 2 \int_0^1 \exp(2\pi i n x) f(x/2)dx + \frac 1 2 \int_0^1 \exp(2\pi i n x)f((1+x)/2) dx$$ $$ = \int_0^{1/2}\exp(2\pi i 2u ) f(u) du + \int_{1/2}^1 \exp(2\pi i 2u ) f(u) du $$ by substituting $u=2x$ and so $$a_n = \int_0^1 \exp(2\pi i n 2u) f(u)du = a_{2n},$$ and so $a_n=a_{2n}=a_{4n}=a_{8n}\cdots.$

If $n\ne0$ we have $a_{2^kn} = a_n$ for all $k$, so for such an $n$, $$a_n = \lim_{k\to\infty} a_{2^kn} = 0$$ by the Riemann Lebesgue lemma. This implies $f$ is a constant, (the constant $a_0$) almost everywhere.

  • $\begingroup$ There clearly are many distributions other than the uniform one (corresponding to the case that you describe, with $f(x)$ being a constant). However, the case that you describe also clearly seems to be the only one for which a density exists. In short, in all other cases, the distribution appears to be nowhere differentiable. To put it differently, there is only one solution if $f\in L^1([0, 1])$, and it must be a constant. Any other solution don't belong to that space. $\endgroup$ Commented Nov 24, 2019 at 21:48
  • $\begingroup$ Interestingly, the constant case corresponds to a normal number, with its binary digits behaving exactly like i.i.d. Bernouilli($\frac{1}{2}$) variables. The other cases correspond to non-normal numbers. Let $S$ be the set of non-normal numbers in $[0, 1]$, in base $2$. Its Lebesgue measure is known to be zero. $\endgroup$ Commented Nov 24, 2019 at 21:53
  • $\begingroup$ In short, the characteristic function in my question only makes sense if $p=\frac{1}{2}$. In all other cases, it does not make sense, or you need to invent a non-standard concept of characteristic functions, using non-standard analysis, and the classical theory of Fourier transforms does not apply. I can elaborate more on this if you are interested. $\endgroup$ Commented Nov 24, 2019 at 22:00
  • $\begingroup$ I have answered the question you actually asked, but your comments make it clear you are interested in some other question. (The graph you give in your "second update", for example, is not a graph of the solution of your functional equation, but yet is interesting to you.) Why don't you figure out just what it is, and post it as a separate question. $\endgroup$ Commented Nov 24, 2019 at 22:18
  • $\begingroup$ True, I am interested in something different, I guess related to normal numbers and attrators in stochastic/chaotic dynamical systems, with a strong probabilistic number theory / experimental math flavor (like David Bailey's work) hoping to build solid foundations in what I am interested in (a moving target by itself.) Yet many of the issues that I have translate into various problems that are of interest to a broader audience. And actually that graph in my second update is related to the functional equation if it was expressed in terms of a distribution rather than in terms of a density. $\endgroup$ Commented Nov 24, 2019 at 22:35

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