4 added 403 characters in body

Formally

I am going to address the question for $\mathrm{Bernoulli}(1/2)$ measures, using probabilistic language. This is not a complete answer, but I am trying to relate your question to the properties of the distribution of $f(x)$. Clearly, for $x<1/2$ we never even reach zero, but my guess is that for $x>1/2$ this distribution is absolutely continuous, though I am unable to prove this at the moment.

So formally, at least,

$\displaystyle \mathsf{E} \, \sum_{f(x)=0} \mathsf{1}\{|f^\prime(x)| < \epsilon\} = \intop_0^1 \mathsf{E} \, \delta(f(x)) \mathsf{1}\{|f(x)|<\epsilon\} |f^\prime(x)| dx \le \epsilon \intop_0^1 \mathsf{E} \, \delta(f(x)) dx$.

$\mathsf{E} \, \delta$ is the density at zero, and it can be made perfect sense of, provided that the law of $f(x)$ has continuous density at zero. I don't know whether it has continuous density, but if we manage to prove that $f(x)$ has at least bounded density for $x>1/2$, then we can write inequalities with approximations of $\delta$ to get the same results...

3 deleted 624 characters in body; added 3 characters in body

Here is a well-known idea that should work

Formally, but probably it takes some time to elaborate.

Using probabilistic languageat least,

$Z := \displaystyle \{x mathsf{E} \mid f(x)=0\}$ is a point process, so the first thing that we can actually calculate is the expectation of its counting measure. The argument is a combination of Fubini and a calculus-style change of variables\sum_{f(x)=0} \mathsf{1}\{|f^\prime(x)| < \epsilon\} = \intop_0^1 \mathsf{E} \, but the technical conditions are cumbersome. There are nice explicit formulas\delta(f(x)) \mathsf{1}\{|f(x)|<\epsilon\} |f^\prime(x)| dx \le \epsilon \intop_0^1 \mathsf{E} \, see e.g. the Metatheorem in Adler & Taylor 'Random fields and geometry'\delta(f(x)) dx$.The same ideas lead to a definition of conditional expectation given$x \mathsf{E} \in Z$- this was done by the guys working on stationary processes and extreme value theory, probably it \delta$ is explained in Cramer & Leadbetter, but I'm not surethe density at zero, and I have no time to search for it now. Calculation of this conditional distribution of $f^\prime(x)$ given $f(x)=0$ (as $x$ varies!) should require nothing more than knowing the joint distribution can be made perfect sense of$(f(x),f^\prime(x),f^{\prime\prime}(x))$ for a fixed $x$, which is a convolution. In particular, my guess is provided that if the conditional distribution law of $f^\prime(x)$ given $f(x)=0$ (as $\omega$ varies) f(x)$has no atom continuous density at zero, then you won't have double zeroes. Perhaps I'll try I don't know whether it has continuous density, but if we manage to prove that$f(x)$has at least bounded density for$x>1/2$, then we can write more precisely a bit later..inequalities with approximations of$\delta$to get the same results... 2 added 30 characters in body; added 45 characters in body Here is a well-known idea that should work, but probably it takes some time to elaborate. Using probabilistic language,$Z := \{x \mid f(x)=0\}$is a point process, so the first thing that we can actually calculate is the expectation of its counting measure. The argument is a combination of Fubini and a calculus-style change of variables; there , but the technical conditions are cumbersome. There are nice explicit formulas, see e.g. the Metatheorem in Adler & Taylor 'Random fields and geometry'. The same ideas lead to a definition of conditional expectation given$x \in Z$- this was done by the guys working on stationary processes and extreme value theory, probably it is explained in Cramer & Leadbetter, but I'm not sure, and I have no time to search for it now. Calculation of this conditional distribution of$f^{\prime\prime}(x)$f^\prime(x)$ given $f(x)=0$ (as $x$ varies!) should require nothing more than knowing the joint distribution of $(f(x),f^\prime(x),f^{\prime\prime}(x))$ for a fixed $x$, which is a convolution. In particular, my guess is that if the conditional distribution of $f^\prime(x)$ given $f(x)=0$ (as $\omega$ varies) has no atom at zero, then you won't have double zeroes.

Perhaps I'll try to write more precisely a bit later...

1