Non standard solution to $f(x) = \frac{1}{2}\Big(f(\frac{x}{2}) + f(\frac{1+x}{2})\Big)$

Solution 1:

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?):

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

Examples:

$$g(0)=0$$

For any $a \neq 0$:

$$g(a)=C$$

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

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

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:

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

Solution 2:

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 $p=0.55, b=2$.

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:

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

Solution 3:

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.