Zero function

The following is an exercise proposed by one of our readers.

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $f\left( \frac{m}{2^n} \right)=0 \; \text{forall} \; m \in \mathbb{Z} \; \text{and} \; n \in \mathbb{N}$. Prove that $f=0$ forall $x \in \mathbb{R}$.

Solution

Let us begin with a very well known lemma.

Lemma: The set $\mathcal{S}= \left \{ \frac{m}{2^n} , \; m ,n \in \mathbb{Z}\right \}$ is dense in $\mathbb{R}$.

Proof: Since $2^n \rightarrow +\infty$ then we can find an $n$ such that $2^n > \frac{1}{b-a}$. This in return means that $2^n b - 2^n a >1$ and since the difference is greater than $1$ , then in the interval $(2^n a, 2^n b)$ there exists an integer $m$ such that $a< \frac{m}{2^n}<b$. This concludes the proof.

Note: Numbers of this form are called dyadic numbers.

Now due to density we can pick up a sequence $x_n \in \mathcal{S}$ such that $x_n \rightarrow x \in \mathbb{R}$ . Due to continuity of $f$ we have that $\lim f(x_n)=0 \Rightarrow f(x)=0$. The result now follows.