# The function $f (x) = f \left (\frac x2 \right ) + f \left (\frac x2 + \frac 12\right)$

The function $f: [0,1] → \mathbb R$ satisfies the equation $$f (x) = f \left (\frac x2 \right ) + f \left (\frac x2 + \frac 12\right)$$ for every $x$ in $[0,1]$.

Can we assert that $f (x) = c (1-2x)$ for some real $c$ if:

a) $f$ twice continuously differentiable on $[0,1]$;

b) $f$ continuously differentiable on $[0,1]$;

c) $f$ continuous on [0,1]?

• the only thing i can see is that $$f(x+1)-f(x) = f(x/2+1)-f(x/2)$$ and $f(1/2)=0$ – gt6989b Feb 23 '16 at 19:28
• There is no much sense writing $f(x+1)$ due to $x$ range. By the way we can see that $f(1/4)=-f(3/4)$ and $f'(0)=f'(1/2)=f'(1)$. – Maffred Feb 23 '16 at 20:03
• Where is this question from? Have you tried the usual "tricks" (e.g., dyadic numbers -- not sure it works, but worth trying) – Clement C. Feb 23 '16 at 21:00
• Well, there seem to be highly pathological solutions existing to the original functional equation. This can be seen by partitioning the rationals such that each set of the partition contains rationals whose binary expansions (eventually periodic) have identical periodic cycles. – user175531 Feb 23 '16 at 21:49

This is a partial answer that proves the result when $f$ is asumed continuously differentiable. Let $M$ be the maximum of $f'$ on $[0,1]$. Define $A=\{a\in[0,1]:f'(a)=M\}$. Since $$2f'(a)=f'\left(\frac{a}{2}\right)+f'\left(\frac{1+a}{2}\right)$$ We see immediatly that $$a\in A\Longrightarrow(\frac{a}{2}\in A)\quad\hbox{and}\quad (\frac{1+a}{2}\in A)$$ Now, $A$ is not empty, and if $\alpha$ is an element from $A$ then the preceding property implies that $$\forall\,n\ge 1, k\in\{0,\ldots,2^{n-1}\},\quad \frac{k+a}{2^n}\in A$$ Now, $A$ is closed and contains a dense subset of $[0,1]$ it must be All the interval $[0,1]$, i.e. $A=[0,1]$.

This proves that $f'$ is constant, so $f(x)= m x+ c$ for some constants $m$ and $c$. Replacing back in the functionsl equation we conclude that we must have $m+2c=0$, that is $f(x)= -2c x+ c$ as desired.

The case of just continous solutions to the functionl equation seems difficult. The example of Julian-Rosen is remarkable.

Here is a continuous counterexample.

Write $\lfloor x\rfloor$ for the floor of $x$, and $\{x\}:=x-\lfloor x\rfloor$ for the fractional part. Let $g:[0,1]\to \mathbb{R}$ be a continuous function satisfying $g(0)=g(1)$, and define $$f(x)=\sum_{n\geq 1} \frac{(-1)^{\lfloor2^n x\rfloor}}{2^n} g(\{2^n x\}).$$ It is an exercise to check that $f$ is continuous. It can also be checked that $$f\left(\frac{x}{2}\right)=\frac{g(x)}{2}+\frac{1}{2}f(x),\hspace{1cm}f\left(\frac{x}{2}+\frac{1}{2}\right)=-\frac{g(x)}{2}+\frac{1}{2}f(x),$$ so that $f$ satisfies the functional equation.

If $g(x)=c$ is constant, then $f(x)=c(2x-1)$. If we take $g(x)$ to be non-constant, we get a different $f$. I have included a plot of $f$ when $g(x)$ is the distance from $x$ to the nearest integer.

Here is a way to find Julian's example. This functional equation is very well behaved under the Fourier transform. Let $f$ be an integrable solution of the equation. For any integer $n$, let:

$$\hat{f} (n) := \int_0^1 f(x) e^{-2\pi inx} \ dx.$$

Then:

$$\hat{f} (n) = \int_0^1 f \left( \frac{x}{2}\right) e^{-2\pi inx} \ dx + \int_0^1 f \left( \frac{x+1}{2}\right) e^{-2\pi inx} \ dx \\ = 2\int_0^{\frac{1}{2}} f (u) e^{-2\pi in (2u)} \ dx + 2 \int_{\frac{1}{2}}^1 f (u) e^{-2\pi in (2u-1)} \ dx \\ = 2 \hat{f} (2n).$$

Necessarily, $\hat{f} (0) = 0$. To get continuous solutions, take any sequence $(a_{2k+1})_{k \in \mathbb{Z}}$ which is summable and defined on the odd integers. Then extend this sequence to the integers by $b_0 = 0$ and:

$$b_{2^n (2k+1)} := 2^{-n} a_{2k+1},$$

and put $f(x) := \sum_{n \in \mathbb{Z}} b_n e^{2 \pi i nx}$. Then $f$ is continuous (its Fourier coefficients are summable), and satisfies the functional equation. If you want real-valued solutions, take the imaginary or real part (or choose $a_{-k} = a_k$).

More generally, the Fourier transform is well-defined when the coefficients are square-integrable. So you can choose a sequence $(a_{2k+1})_{k \in \mathbb{Z}}$ whose square is summable, extend it in the same way to a sequence $(b_n)_{n \in \mathbb{Z}}$, which will still be square-summable, and take the inverse Fourier transform. For instance, with $a_{2k+1} = -i(2k+1)^{-1}$, you get $b_n = -\delta_{0n} i n^{-1}$, so a solution to the functional equation is:

$$f(x) = 2\sum_{n=1}^{+ \infty} \frac{\sin (2 \pi n x)}{n},$$

which, up to a multiplicative constant, is $1-2x$. If you try to make this function periodic, it will have a discontinuity at the integers (which is hidden here by the fact that we worked on $[0,1]$), which explain that the coefficients of the Fourier transform are not summable.

It can be shown that any continuous solution $f$ to the functional equation such that $f(0) = f(1)$ must at best have a modulus of continuity $\omega_f (h) \simeq h |\ln (h)|$ (and I think that this modulus of continuity is optimal almost everywhere), so they cannot be $\mathcal{C}^1$, and they look indeed pathological. This is because their Fourier coefficients do not decay very quickly.

With this method, you can solve other similar functional equations, say, for instance,

$$f(x) = \frac{2}{3} \left[ f \left( \frac{x}{2}\right) + f \left( \frac{x+1}{2}\right) \right],$$

or:

$$f(x) = f \left( \frac{x}{3}\right) + f \left( \frac{x+1}{3}\right) + f \left( \frac{x+2}{3}\right),$$

as well as some similar functional equations with more variables.