4

Let $f:\mathbb R \to \mathbb R$ have the properties

(1): for any prime number $p$ and any real number $x$, $$\sum_{j=0}^{p-1}f\left(x+\dfrac{j}{p}\right)=0$$

(2): there exist real numbers $a$ and $b(>a)$ such that: $x\in (a,b) \implies f(x)=0$

Show that $f(x)\equiv 0$

Perhaps we can use this well known fact: $$1+a+a^2+\cdots+a^{n-1}=0,a=\exp \left (\dfrac{k\pi\cdot i}{n}\right )$$ but I don't know how to use it here.

math110
  • 93,304
  • 2
    Two orienting remarks for potential solvers: First, if condition (1) for an integer $p$, then it also holds for any multiple $kp$, by summing it applied to $x,x+\frac1{kp},\dots,x+\frac{p-1}{kp}$. So one can assume condition (1) for every integer $p\ge2$. (2) The functions $f(x)=\sin\pi x$ and $f(x)=\cos\pi x$ satisfy condition (1) alone, so condition (2) is definitely necessary. - Is $f$ assumed to have any regularity properties, such as continuity? – Greg Martin Nov 08 '14 at 19:48

1 Answers1

1

Denote by $p_k$ the $k$-th prime. Fix an integer $r\geq 1$. Let $P=p_1p_2p_3 \ldots p_r$ and $g(x)=f(\frac{x}{P})$. Then $g$ satisfies :

$$ \sum_{j=0}^{p_k-1} g\left(y+j\prod_{l\neq k}p_l\right)=0 \tag{1} $$

for any $k\in [1,r]$ and $y\in{\mathbb R}$. For $P\in{\mathbb Q}[X]$, $P=\sum_{k=0}^d a_kX^k$, define

$$ \psi(P)=\sum_{k=0}^d a_k g(y+k) \tag{2} $$

Then $\psi$ is a ring homomorphism, and (1) tell us that the polynomial $Q_k=\sum_{j=0}^{p_k-1} X^{j\prod_{l\neq k}p_l}$ is in ${\sf Ker}(\psi)$. Now $Q_k=\Phi_{p_k}(X^{\prod_{l\neq k}p_l})$, so that $z\in{\mathbb C}$ is a root of $Q_k$ iff $z^{\frac{P}{p_k}}$ is a primtive $p_k$-th root of unity.

It follows that $z$ is a common root of all the $Q_k (1\leq k \leq r)$ iff $z$ is a primitive $P$-th root of unity. The gcd of the $Q_k (1\leq k \leq r)$ is therefore $\Phi_{P}$. But since ${\sf Ker}(\psi)$ is an ideal of $P\in{\mathbb Q}[X]$, we have $\Phi_P \in {\sf Ker}(\psi)$. We know that the degree of $\Phi_P$ is exactly $\phi(P)=\prod_{k=1}^r (p_k-1)=d_k$, so we can write $\phi_P=X^{d_k}+ \sum_{i=0}^{d_k-1} a_i X^i$ where the $a_i$ are integers. We then have :

$$ g(y+d_k)=\sum_{i=0}^{d_k-1}a_ig(y+i) \tag{3} $$ for any $y\in{\mathbb R}$. Iterating (3), we see that if $g$ is zero on an interval of length $\geq d_k$ then it must be zero everywhere. We deduce that if $f$ iz zero on an interval of length $\geq \frac{d_k}{P}=e_k$ then it must be zero everywhere. But

$$ e_k=\prod_{k=1}^r \left(1-\frac{1}{p_k}\right) \tag{4} $$

tends to zero when $r\to\infty$ (by Euler’s product formula for example), so we are done.

Ewan Delanoy
  • 61,600