periodical function $f$ with period $1$ satisfying $f(x)+f(x+1/2)=f(2x)$ is zero.

Question

Let $f$ be continuously differentiable on $\Bbb R$, periodical with period $1$, and $f(x)+f(x+1/2)=f(2x)$ for all $x\in\Bbb R$. Show that $f\equiv 0$.

A natural attempt is to use Fourire series. Let $f(x)=a_0/2+\sum_{n=1}^\infty(a_n\cos 2 n\pi x+b_n\sin 2n\pi x)$. Then after checking both side of $f(x)+f(x+1/2)=f(2x)$ , we find $a_0=0$, $2a_{2n}=a_n$, $2b_{2n}=b_n$. Then I have on idea to proceed on.

Answer 1

You can inductively prove that

$$ f(x) = \sum_{k=0}^{2^n - 1} f\left(\frac{x}{2^n} + \frac{k}{2^n}\right) $$

holds for all $x \in \Bbb{R}$ and $n \geq 1$. Since $f \in C^1$, this implies

$$ f'(x) = \sum_{k=0}^{2^n - 1} f'\left(\frac{x}{2^n} + \frac{k}{2^n}\right) \frac{1}{2^n} \xrightarrow[n\to\infty]{} \int_{0}^{1} f'(t) \, dt = f(1) - f(0) = 0. $$

So $f$ is constant, which then implies $f \equiv 0$ by the functional equation.

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Edit (2017/08/20). Here are two remarks:

1. We can give an alternative solution using Fourier series: Differentiating the functional equation, we have $f'(x) + f'(x+\frac{1}{2}) = 2f'(2x)$. So it follows that

   $$ \hat{f'}(k) := \int_{0}^{1} f'(x)e^{-2\pi i k x} \, dx = \int_{0}^{1} \frac{f'(\frac{x}{2}) + f'(\frac{x+1}{2})}{2} \, e^{-2\pi i k x} \, dx = \hat{f'}(2k). $$

   So by the Parseval's identity, for any $k \neq 0$ and $m \geq 1$ we have

   $$ |\hat{f'}(k)|^2 = \frac{1}{m}\sum_{j=0}^{m-1} |\hat{f'}(2^j k)|^2 \leq \frac{1}{m} \sum_{n \in \mathbb{Z}} |\hat{f'}(n)|^2 = \frac{1}{m} \int_{0}^{1} |f'(x)|^2 \, dx $$

   and letting $m\to\infty$ gives $\hat{f'}(k) = 0$. This tells that $f'$ is constant, from which we easily deduce $f' \equiv 0$ and consequently $f \equiv 0$.

2. Without differentiability, we have non-trivial solutions such as

   $$ f(x) = \sum_{n=1}^{\infty} \frac{\cos(2^n \pi x)}{2^n}. $$

   Indeed, Weierstrass $M$-test tells that $f$ is continuous. Also, it follows that

   $$ f(x) + f(x + \tfrac{1}{2}) = \sum_{n=1}^{\infty} \frac{\cos(2^n \pi x) + \cos(2^n \pi x + 2^{n-1}\pi)}{2^n} = \sum_{n=2}^{\infty} \frac{2\cos(2^n \pi x)}{2^n} = f(2x).$$

   So the differentiability condition is essential.

Answer 2

We are given that $f$ is $C^1$. Then $f'(x)+f'(x+\frac12)=2f'(2x)$. As a continuous preriodic function, $f'$ reaches its maximum at some $a\in[0,1)$. Then so it does also at $\frac a 2$ (and $\frac a2+\frac12$). By induction, this also happens at $2^{-n}a$, by continuity also at $0 $. By the same argument, $f'$ reaches its minimum at $0$, hence is constant. Then $f(x)=mx+b$, and only $m=0,b=0$ work.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{Lets}\quad\mrm{f}\pars{x} = \sum_{n\ \in\ \mathbb{Z}}A_{n}\exp\pars{2n\pi\ic x} \\[5mm] &\mrm{f}\pars{x} + \,\mrm{f}\pars{x + 1/2} = \mrm{f}\pars{2x}\! \implies\! \pars{~\substack{\ds{\exp\pars{2n\pi\ic x} + \exp\pars{2n\pi\ic x}\exp\pars{n\pi\ic} = \exp\pars{4n\pi\ic x}}~} \\[2mm] \,\,\,\mbox{if}\quad \ds{A_{n} \not= 0}} \\[5mm] & \implies \left.\vphantom{\Large A}\exp\pars{2n\pi\ic x} = 1 + \pars{-1}^{n}\right\vert_{\ A_{n}\ \not=\ 0}\quad \pars{~\substack{\mbox{There isn't any integer}\ \ds{n\ in\ \mathbb{Z}} \\[2mm] \mbox{which satisfies, $\ds{\forall\ x}$, this relation.} \\[2mm] \mbox{Then,}\ds{\ A_{n} = 0\,,\ \forall\ n \in \mathbb{Z}.}}~} \\[5mm] & \implies \bbx{\mrm{f}\pars{x} = 0\,,\ \forall\ x} \end{align}