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.
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 $2$.