$\cos(\frac{x}{3})\cos(\frac{x}{9})\cos(\frac{x}{27})\dots$ as $x \rightarrow \infty$

Question

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 $a \neq 2$.

Answer 1

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).