IMC 2010 · Day 2 · P10

Let $N > 1$ be some integer to be defined later, and consider set of

real polynomials $$J_N = \left\{ c_0 + c_1 x + \dots + c_n x^n \in \mathbb{R}[x] \ \middle|\ \forall x \in \mathbb{R}\ \sum_{k=0}^{n} c_k f \left( x + \frac{k}{N} \right) = 0 \right\}.$$ Notice that $0 \in J_N$, any linear combinations of any elements in $J_N$ is in $J_N$, and for every $P(x) \in J_N$ we have $x P(x) \in J_N$. Hence, $J_N$ is an ideal of the ring $\mathbb{R}[x]$.

By the problem's conditions, for every prime divisors $p$ of $N$ we have $\dfrac{x^N - 1}{x^{N/p} - 1} \in J_N$. Since $\mathbb{R}[x]$ is a principal ideal domain (due to the Euclidean algorithm), the greatest common divisor of these polynomials is an element of $J_N$. The complex roots of the polynomial $\dfrac{x^N - 1}{x^{N/p} - 1}$ are those $N$th roots of unity whose order does not divide $N/p$. The roots of the greatest common divisor is the intersection of such sets; it can be seen that the intersection consist of the primitive $N$th roots of unity. Therefore, $$\gcd \left\{ \frac{x^N - 1}{x^{N/p} - 1}\ \middle|\ p \mid N \right\} = \Phi_N(x)$$ is the $N$th cyclotomic polynomial. So $\Phi_N \in J_N$, which polynomial has degree $\varphi(N)$.

Now choose $N$ in such a way that $\dfrac{\varphi(N)}{N} < b - a$. It is well-known that $\liminf\limits_{N \to \infty} \dfrac{\varphi(N)}{N} = 0$, so there exists such a value for $N$. Let $\Phi_N(x) = a_0 + a_1 x + \dots + a_{\varphi(N)} x^{\varphi(N)}$ where $a_{\varphi(N)} = 1$ and $|a_0| = 1$. Then, by the definition of $J_N$, we have $\sum\limits_{k=0}^{\varphi(N)} a_k f \left( x + \frac{k}{N} \right) = 0$ for all $x \in \mathbb{R}$.

If $x \in \left[ b, b + \frac{1}{N} \right)$, then $$f(x) = -\sum_{k=0}^{\varphi(N)-1} a_k f \left( x - \frac{\varphi(N) - k}{N} \right).$$ On the right-hand side, all numbers $x - \frac{\varphi(N) - k}{N}$ lie in $(a, b)$. Therefore the right-hand side is zero, and $f(x) = 0$ for all $x \in \left[ b, b + \frac{1}{N} \right)$. It can be obtained similarly that $f(x) = 0$ for all $x \in \left( a - \frac{1}{N}, a \right]$ as well. Hence, $f = 0$ in the interval $\left( a - \frac{1}{N}, b + \frac{1}{N} \right)$. Continuing in this fashion we see that $f$ must vanish everywhere.