It is known that
$$
\cos(\frac{x}{2})\cos(\frac{x}{4})\cos(\frac{x}{8})\dots = \frac{\sin x}{x} = O\_{x \rightarrow \infty}(x^{-1})
$$


Is it true that
$$
f(x) = \cos(\frac{x}{3})\cos(\frac{x}{9})\cos(\frac{x}{27})\dots = o\_{x \rightarrow \infty} (1) ?
$$
If so, what is the rate of convergence?


It seems to me that $f(x)$ converges to zero, but very slowly. For example $f(1081882100) \approx 0.27$. I guess the reason is that $f(x)$ is the Fourier transform of the uniform distribution on the Cantor set $C$ supported in $[-1/2, 1/2]$, which is highly irregular. To see this, let $X \sim \mathcal U (C)$, then by the self-similarity of $C$, we have
$$
X \stackrel{(d)}{=} X/3 + Y
$$
where $Y \sim \mathcal U(\pm 1/3)$ is independent of $X$. So
$$
\mathbb E[e^{itX}] = \mathbb E[e^{itX/3}]\cos(x/3) 
$$
From which we obtain
$$
\mathbb E[e^{itX}] = \cos(\frac{x}{3})\cos(\frac{x}{9})\cos(\frac{x}{27})\dots
$$


**Update.** The answer is no by Noam Elkies. However now I want to ask the same question for
$$
f\_a(x) = \prod\_{n \geq 1} \cos(\frac{x}{a^k})
$$
for $a>1$ and $a \neq 2$.



$\cos(\frac{x}{3})\cos(\frac{x}{9})\cos(\frac{x}{27})\dots$ as $x \rightarrow \infty$
No. If $x = 3^n \pi$ then $|f(x)| = f(\pi) \neq 0$
(numerically it's about $0.466$), and $3^n \pi$ can be arbitrarily large.


Such a construction fails for $\prod\_{k=1}^\infty \cos(x/2^k)$
because if some $x/2^k$ is nearly $\pi$,
or more generally some *odd* multiple of $\pi$, then
$x/2^{k+1}$ is nearly a half-integral multiple of $\pi$
and thus has a very small cosine.


[**added later**] For $f\_a(x) := \prod\_{k=1}^\infty \cos(x/a^k)$
with $a > 2$, the same construction does work if $a$ is an integer
(consider $x = a^n \pi$), and more generally if $a$ is a
Pisot-Vijayaraghavan number (which need not exceed $2$,
e.g. if $a = (1+\sqrt5)/2$ then there are arbitrarily large $x$
such that $f\_a(x)$ remains bounded away from zero).


