Evaluating $f(x) f(x/2) f(x/4) f(x/8) \cdots$ Let $f : \mathbb R \to \mathbb R$ be a given function with $\lvert f(x) \rvert \le 1$ and $f(0) = 1$. Is there a nice simplified expression for $$\begin{align}F(x) &= f(x) f(x/2) f(x/4) f(x/8) \cdots \\ &= \prod_{i=0}^\infty f(x/2^i)?\end{align}$$ If there isn't a general solution, as seems likely, can anything useful be said about the case when $f(x) = \operatorname{sinc} x = \frac{1}{x} \sin x$?
This question arose when idly wondering about the limit of convolving an infinite number of dyadically(?) scaled versions of a kernel $g$ together. Taking the Fourier transform of $g(x) * 2g(2x) * 4g(4x) * 8g(8x) * \cdots$ yields the above expression.
 A: Assuming analyticity of $f(x)$ it seems more tractable to work with the form involving sums after taking logarithms:
$$ \log F(x) = \sum_{i=0}^\infty \log f(x/2^i)$$
Now if $\log f(x)$ has convergent power series expansion in a neighborhood of zero:
$$ \log f(x) = \sum_{k=1}^\infty a_k x^k$$
Note that since $f(0) = 1$ the constant term $a_0$ of $\log f(x)$ is zero and thus omitted.
Then:
$$ \log F(x) = \sum_{k=1}^\infty a_k (\sum_{i=0}^\infty 2^{-ik}) x^k$$
Note that the inner summations are just $\sum_{i=0}^\infty 2^{-ik} = 1/(1 - 2^{-k})$ and thus uniformly bounded by $2$.
A: For what it's worth, I found a partial answer to the thought that motivated my original question. When $f(x) = \operatorname{sinc}(x)$, it is the Fourier transform of a rectangular function, $g(t) = 1/2$ for $t \in [-1,1]$ and $g(t) = 0$ otherwise (or some scaling thereof, depending on convention). The infinite product $f(x) f(x/2) f(x/4) f(x/8) \cdots$ corresponds to the convolution which I wrote as $g(t) * 2g(2t) * 4g(4t) * 8g(8t) * \cdots$ in a small abuse of notation. Each term in the latter convolution is the probability distribution function of a uniformly distributed random variable, and the convolution is the pdf of their sum. This distribution is given by (some scaling of) the Fabius function: a compactly supported, infinitely differentiable function that is not analytic anywhere in its support. So the original infinite product falls off exponentially fast but probably cannot be expressed in any nice form.
A: I've uploaded here plots of some approximation of limit distributions  which are defined by inverse Fourier transform:
$$
\mathrm{h}_a(x):=\mathcal{F^{-1}}\biggl[\prod\limits_{k=1}^{\infty}\mathrm{sinc}(t\cdot a^{-k})\biggr](x).
$$
