# Continuous $f$ satisfying $f(2x)=f(x-1/4)+f(x+1/4)$ on $(-1/2,1/2)$

What are the continuous functions $f\colon (-\frac{1}{2},\frac{1}{2}) \to \mathbb{C}$ that satisfy the following functional equation, and how are they derived? $$f(2x)=f(x-\frac{1}{4})+f(x+\frac{1}{4})\,\,\,\,\,\,\,\,\text{for }x \in (-\frac{1}{4},\frac{1}{4})$$

I think this could be interesting because in addition to the obvious solutions $f(x)=zx$, this is also satisfied by $f(x)=\ln (2\cos(\pi x))$.

• Yes. I'll edit to clarify. – Malper Dec 30 '13 at 4:05
• @coffeemath Oops, I meant to write $f(x)=\ln(2\cos(\pi x))$ – Malper Dec 30 '13 at 4:32
• Yep. just checked that and it works numerically. (too busy to check via identities...). – coffeemath Dec 30 '13 at 4:34
• Assuming integrability, applying $\int_{-1/4}^{1/4} \ dx$ to both sides of the functional equation, and changing variables gives $\frac{1}{2} \int_{-1/2}^{1/2} f(x) \ dx = \int_{-1/2}^{1/2} f(x) \ dx$, so it is at least true that $\int_{-1/2}^{1/2} f(x) \ dx = 0$ i.e. the function averages to $0$ over its domain. – Mike F Dec 30 '13 at 10:14
• For real analytic solutions of this functional equation, see this question. – Malper Jan 2 '14 at 0:53

If $f$ is as in the problem statement, then, from the functional equation and continuity of $f$ at $x=0$, we see that $$\lim_{x \ \to -\frac{1}{4}} f(2x) - f(x-\frac{1}{4}) = f(0).$$

Now, suppose $f_0$ is any continuous function on $(-\frac{1}{2},0]$ such that \begin{align*} \lim_{x \ \to -\frac{1}{4}} f_0(2x) - f_0(x-\frac{1}{4}) = f_0(0). && && (*) \end{align*} A simple sufficient condition on $f_0$ implying the above condition is $\lim_{x \to - \frac{1}{2}} f_0(x)$ exists and $f_0(0) = 0$.

I claim that any $f_0$ as above extends uniquely to an $f$ as in the problem statement.

We partition $(-\frac{1}{2},\frac{1}{2})$ into subintervals so that $f$ can be defined on one interval at a time. Put \begin{align*} x_{-1}=-\frac{1}{2} && x_0 = 0 && x_1 = \frac{1}{4} && x_2 = \frac{3}{8} && x_3 = \frac{7}{16} && \ldots \end{align*} \begin{align*} I_0 = (x_{-1},x_0] && I_1 = (x_0, x_1] && I_2 = (x_1,x_2] && I_3 = (x_2,x_3] && \ldots \end{align*} Generally: \begin{align*} x_n = \frac{1}{2} - \frac{1}{2^{n+1}} && n \geq -1 \end{align*} \begin{align*}I_n = (x_{n-1}, x_n ] && n \geq 0. \end{align*} Now, whenever $x \in I_n$ for $n \geq 1$, a quick calculation shows that $2x -1/2 \in I_{n-1}$ and $x - 1/2 \in I_0$. So, we get a function $f_n$ on $I_n$ for all $n \geq 0$ by the recursive definition \begin{align*} f_n(x) = f_{n-1}(2x -\frac{1}{2}) - f_0(x-\frac{1}{2}) && x \in I_n && n \geq 1. \end{align*} By construction, the function $f$ on $(-\frac{1}{2},\frac{1}{2})$ obtained by glueing all the $f_n$ together satisfies the functional equation, this recursive definition having been derived directly from the functional equation. Since $f_0$ is continuous, it is clear that each $f_n$ is continuous. It remains to see the $f_n$ agree at the endpoints so that $f$ is also continuous. We check inductively that \begin{align*} \lim_{x \to x_n} f_{n+1}(x) = f_n(x_n) && n \geq 0. && && (**)\end{align*} When $n=0$, this says $$\lim_{x \to 0} f_0(2x - \frac{1}{2}) - f_0(x-\frac{1}{2}) = f_0(0)$$ which is just a restatement of $(*)$, so the induction starts. Suppose the claim holds when for some $n \geq 0$ look at $$f_{n+2}(x) = f_{n+1}(2x - \frac{1}{2}) - f_0(x - \frac{1}{2})$$ with $x \in I_{n+2}$ converging to $x_{n+1}$. Then $2x - \frac{1}{2}$ is converging to $x_n$ and $x - \frac{1}{2}$ is converging to $x_{n+1} - \frac{1}{2} \in I_0$. So, by induction and continuity of $f_0$, the whole expression above is converging to $$f_n(x_n) - f_0(x_{n+1} - \frac{1}{2}) = f_n(2x_{n+1} - \frac{1}{2}) - f_0(x_{n+1} - \frac{1}{2}) = f_{n+1}(x_{n+1})$$ and the claim follows by induction.

We have obtained the continuous $f$ extending $f_0$ as claimed and by design it is the unique extension satisfing the functional equation.

• Very nice! I'll add that if we want $f$ to be $\mathcal{C}^n$ (including $n = \infty$), then we must require that $f_0 \in \mathcal{C}$ and that $f_0$ satisfies $\lim_{x \to -\frac{1}{4}}(2^k f_0^{(k)}(2x)-f_0^{(k)}(x-\frac{1}{4}))=f^{(k)}_0(0)$ for all positive integers $k \leq n$. Then each derivative of $f$ is continuous by the same inductive logic that Mike used in his answer. – Malper Dec 31 '13 at 23:23

example

Using Mike's method, with $f_0(x) = x^2$

Here is an unbounded one, $f_0(x) = \log(4x+2)$

So now, what about differentiable, or real-analytic solutions? Or even solutions continuous at both endpoints.

• Neat, thanks for making these! – Mike F Dec 30 '13 at 19:31