I am interested in the derivative of a function defined on a subset $S$ of $[0, 1]$. The subset in question is dense in $[0, 1]$ but has Lebesgue measure zero. My actual question can be found at the bottom of this post.

There has been a few questions on the subject, but none leading to something interesting as far as I am concerned. See here, here (concept of arithmetic derivative) and also here (Minkowski's question mark function, related to the material discussed here.)

Generally these discussions lead to some kind of non-sense math. Here it is the opposite. I have a framework that does work as far as applications and computations are concerned, but I have a hard time putting it into some sound mathematical framework. It would have to be some kind of non-standard calculus.

Perhaps the simplest example (though I have plenty of other similar cases) is as follows. Let $Z$ be a random variable defined as follows: $$Z = \sum_{k=1}^\infty \frac{X_k}{2^k}$$

where the $X_k$'s are identically and independently distributed with a Bernouilli$(p)$ distribution. Thus $P(X_k = 1) = p$ and $P(X_k = 0) = 1-p$. Here $0 < p < 1$. In short, the $X_k$'s are the binary digits of the random number $Z$.

There are two cases.

Case $p=\frac{1}{2}$

In this case, $Z$ has a uniform distribution on $S$, where $S$ is the set of normal numbers in $[0, 1]$. It is known that $S$ has Lebesgue measure $1$, and that $S$ is dense in $[0, 1]$. Yet it is full of holes (no rational number is is a normal number due to their periodicity, thus the $X_k$'s are not independent for rational numbers.)

This is the simplest case. One might wonder if the density $f_Z$ (the derivative of the distribution $F_Z$) exists. Yet $f_Z(z) = 1$ if $z \in S$ works perfectly well for all purposes. It can easily be extended to $f_Z(z) = 1$ if $z \in [0, 1]$. Let us denote the extended function as $\tilde{f}_Z$. You can compute all the moments using the extended $\tilde{f}_Z$ and get the right answer. If $s$ is a positive real number, then $$E(Z^s) = \int_0^1 z^s \tilde{f}_Z(z) dz = \frac{1}{s+1}.$$

You could argue that $\tilde{f}_Z$ (and thus $f_Z$) can be obtained by inverting the above functional equation, using some kind of Laplace transform. So we can bypass the concept of derivative entirely, it seems.

Case $p\neq \frac{1}{2}$

Now we are dealing with a hard nut to crack, and a wildly chaotic system: $Z$'s support domain is a set $S'$ that is a subset of non-normal numbers in $[0, 1]$. This set $S'$ has now Lebesgue measure zero, yet it is dense in $[0, 1]$. For the distribution, it is not a problem: even discrete random variables have a distribution $F_Z$ defined for all positive real numbers: $F_Z(z) = P(Z \leq z)$.

The issue is with the density $f_Z = dF_Z/dz$. It sounds either it should be zero everywhere or not exist. My guess is that you might be able to define a new, workable concept of density. In the neighborhood of every point $z \in S'$, it looks like $g(z,h) = (F_Z(z+h) - F_Z(z))/h$ oscillates infinitely many times with no limit as $h\rightarrow 0$, yet these oscillations are bounded most of the time, perhaps leading to the fact that averaging $g(z, h)$ around $h = 0$, using smaller and smaller values of $h$, could provide a sound definition for the density $f_Z$.

Again, despite the chaotic nature of the system (see how the the would-be density could potentially look like in the picture below) all the following quantities exist and can be computed exactly and then confirmed by empirical evidence, even though the integrals below may not make sense:

$$E(Z) = \int_{0}^{1} z f_Z(z) dz = p \\ E(Z^2) = \int_{0}^{1} z^2 f_Z(z) dz =\frac{p}{3}(1+2p)\\ E(Z^3) = \int_{0}^{1} z^3 f_Z(z) dz =\frac{p}{7}(1+4p+2p^2)\\ E(Z^4) = \int_{0}^{1} z^4 f_Z(z) dz =\frac{p}{105}(7+46p + 44p^2+8p^3) $$ Indeed, a general formula for $E(Z^s) = \int_0^1 z^s f_Z(z)dz$ is available for $s \geq 0$, defined by the following functional equation (see here): $$E(Z^s) = \frac{p}{2^s-1+p}\cdot E((1+Z)^s) .$$

In other words, we would have, under some appropriate calculus theory with a sound definition of integral and derivative: $$\int_{S'}z^s f_Z(z)dz = \frac{p}{2^s-1+p}\cdot\int_{S'}(1+z)^s f_Z(z) dz .$$

Here is how the density $f_Z$, if properly defined, could look like for $p=0.75$ (see here and here):

enter image description here

Below is the empirical percentile distribution, for this particular $Z$:

enter image description here

Other related problems

If instead, we consider the model $Z = X_1 + X_1 X_2 + X_1 X_2 X_3 + \cdots$ with $X$ Bernouilli$(p)$, then $Z$ has a geometric distribution of parameter $1-p$, see section 2.2 in this article. This system is also equivalent to the binary numeration system discussed so far: see section 5 in the same article. It results in $Z$ having a standard, well-known discrete distribution. But if this time $P(X=-0.5) = 0.5 = P(X=0.5)$ then $Z$ is uniform on a subset $S$ of $[-1, 1]$, with $S$ also full of holes.

Here is another interesting model:


The distribution for $X$ is as follows: $$P(X=0) = \frac{1}{2}, P(X=1) = \frac{1}{1 + \sqrt{5}}, P(X=2) = \frac{3 - \sqrt{5}}{4} \mbox{ } (\star)$$

This corresponds to a different numeration system, and the choice for $X$'s distribution is not arbitrary, see here: in short, it makes the system smoother and possibly easier to solve. To the contrary $P(X=0) = P(X=1)=$ $P(X=2)= \frac{1}{3}$ yields a far more chaotic system, and the case $P(X=0) =$ $P(X=1) = \frac{1}{2}$ is so wild that the support domain of $Z$ has huge gaps, very visible to the naked eye.

Normal numbers in the nested square root system are very different from normal numbers in the binary numeration system. The successive digits have a very specific auto-correlation structure, and the digits $0, 1, 2$ are not evenly distributed for normal numbers in that system. It is clear that if we assume that the $X_k$'s are i.i.d, then $Z$ is not a normal number in that system. Yet we get a very good approximation for $F_Z$, much smoother (at least visually) than in the binary numeration system investigated earlier, with $p\neq \frac{1}{2}$. In particular, $F_Z$ is very well approximated by a log function, see chart below.

enter image description here

Here the blue line is the empirical distribution for $Z$, the red line is the log approximation. And below is a spectacular chart, featuring the approximation error $\epsilon(z) = F_Z(z) -\log_2(z)$. It's a fractal! (source: see section 2.2 in this article). In short, it is no more differentiable than a Brownian motion, and technically, the derivative $f_Z$ does not exist. Yet all moments of $Z$ can be computed exactly from the functional equation attached to that system ($F_{Z^2}=F_{X+Z}$) and confirmed empirically. Even though the distribution looks smooth to the naked eye, we are dealing here with a very chaotic system in disguise. Again we need non-standard calculus to handle the density, whose support is a dense set of Lebesgue measure zero in $[1, 2]$.

Since fractals are nowhere differentiable, $f_Z$ does not exist. Yet one could imagine a "density" that would look like $f_Z(z)=\frac{1}{z\log 2}$ for $z\in [1,2]$.

enter image description here

My question

Is there an existing theory to handle this type of density-like-substance? Following some advice, I also posted this question on MathOverflow, here.

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    $\begingroup$ First, this is good, well-written, and of interesting content. Second (and I hope this doesn't sound insulting, because I don't mean it to be): what's your question? As someone pointed out in the comment of one of your links, the fact that we don't have a traditional density w.r.t. Lebesgue measure shouldn't be terribly surprising; the same is true for any discrete random variable, for instance. My best guess at your question is: "Is there an existing theory to handle this type of density-like-substance?" -- did I miss the mark? $\endgroup$ – Aaron Montgomery Dec 6 '19 at 16:03
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    $\begingroup$ You are right, and I will add that in my question. The issue is probably the use of the Lebesgue measure, which should be avoided in this context. Thank you. $\endgroup$ – Vincent Granville Dec 6 '19 at 16:05
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    $\begingroup$ This is quite interesting. Two remarks: I don't see why you make a detour through the normal numbers for $p=\frac12$ instead of just saying that the density is uniform on $[0,1]$. Non-normal numbers do have probability $0$, but so does every other number. There's nothing special about them in this regard; you could just as well exclude any other set with measure $0$ because it has probability $0$. In a similar vein, it makes no sense to say that "the $X_k$'s are not independent for rational numbers"; independence is a property of the distribution, not of individual numbers. $\endgroup$ – joriki Dec 7 '19 at 7:03
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    $\begingroup$ @VincentGranville: That's all true, but it doesn't establish your conclusion that "the $X_k$'s are not independent for rational numbers". That's like saying "I throw two dice independently, and then I exclude the cases where they show the same face, because in those cases they're not independent." $\endgroup$ – joriki Dec 7 '19 at 8:11
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    $\begingroup$ @VincentGranville Your problem evokes to my mind Jenny Harrison's differential chain/chainlet calculus which is a rather original take on geometric measure theory. I don't know how far along she is today in the publication of her ideas. She basically constructs a form of discrete calculus and calculus over fractals using limits of polyhedral chains, that turn into something like Dirac measures. I wouldn't be qualified to guide you more than that, but I do believe you might find some insight in her work that could be useful. $\endgroup$ – Tristan Duquesne Dec 7 '19 at 16:46

It really looks to me that you're after the concept of a generalized function which allow you to manipulate things that are decidedly not functions in a way that really looks like they are functions. The basic premise of a generalized function is that you know how they're supposed to integrate against nice functions, but they clearly are not functions themselves - which is exactly the path you seem to be going down. You can also think of them as what happens if you keep differentiating a continuous function - much in the same way that "a measure" is a reasonably good answer to the question of what happens if you differentiate an increasing function.

More formally, to define a generalized function, we start by defining some functions which are very well behaved, known as test functions. A function $f:\mathbb R\rightarrow\mathbb R$ is a test function if and only if it is smooth and compactly supported. The set of test functions is known as $T$. We also put a topology on $T$ by the rule that $f_1,f_2,\ldots$ converges to $f$ if and only if, for each $n$, the $n^{th}$ derivatives of this sequence converge uniformly to the $n^{th}$ derivative of $f$.

A generalized function is then just a continuous map from $T$ to $\mathbb R$. The idea is that we would associate a continuous function $g:\mathbb R\rightarrow\mathbb R$ to the map $$f\mapsto \int f(x)g(x)\,dx.$$ More generally, we can associate to each measure $\mu$ the generalized function $$f\mapsto \int f(x)\,d\mu.$$ The really neat thing about this definition is that we can use tricks to start to reason about distributions the same way we would reason about functions; let me focus on your second example as it's a nice example of how this reasoning works out without too much difficulty.

So, first, we would like to define the derivative of a generalized function $g$. You can either think about this either as a convenient extension of the map $g\mapsto g'$ from the subset of generalized functions which are actually differentiable to the entire set, or as, noting that differentiation is linear, taking the transpose of the map that takes $g\mapsto g'$ in the set of test functions taken as an inner product space - or you can just think about integration by parts as justification. In any case, you should come across the following formula (where we now abuse the integral sign $\int g' \cdot f$ to formally mean $g'(f)$ - but avoid this notation because it is more confusing than abusing notation!): $$\int g'\cdot f = -\int g\cdot f'.$$ This is, of course, actually true if $g$ and $f$ are differentiable and their product is compactly supported - being a consequence of integration by parts - and one can (and should) use it to differentiate distributions in general.

You can also define compositions of distributions with nice functions in a similar manner; let's stick with linear functions for simplicity. In particular, if $g(x)$ is a distribution, lets define $g(2x)$ in the same way as before (abusing notation even more now - just remember that $g(x)$ and $g(2x)$ are not truly functions, but symbols of their own right): $$\int g(2x)\cdot f(x)\,dx = \frac{1}2\int g(x)\cdot f\left(\frac{x}2\right)\,dx$$ which is essentially the result of a $u$-substitution. Similarly, we could define $$\int g(2x-1)\cdot f(x)\,dx = \frac{1}2\int g(x)\cdot f\left(\frac{x+1}2\right)\,dx.$$

So, even though we can't plot these things (though, if we're at least still within the realm of measure theory - which these density functions absolutely are - we can make histograms which look like what you've done), we can reason about them by analogies to functions that are rooted formally in a theory that will let us extract results via integration later. In particular, our theory lets us, without reservation, differentiate any increasing function - and thus fills in what your values $f_Z$ have to be.

So, let's let $G_Z$ be the cumulative distribution given by randomly selecting the bits of a number in $[0,1]$ from a Bernoulli distribution with parameter $p$. While we could notice some properties of $G_Z$ directly and infer things from differentiation, let's instead look at some more interesting reasoning directly on $g_Z$ enabled by this.

So, as a first step, let's let $Z_n$ be a discrete random variable given as a base two value $0.b_1b_2b_3\ldots b_n$ of bits chosen from the given Bernoulli distribution. Then, $g_{Z_n}$ will represent our probability mass function and we will take it to satisfy: $$\int g_{Z_n}(x) \cdot f(x) = \mathbb E[f(Z_n)].$$ This is, of course, just summing up some of the values of $f$. However, it is helpful, because to produce the distribution of $Z_{n+1}$, all we need to do is randomly pick the first bit, then add half of a value chosen as in $Z_n$. This works out to the following equation: $$g_{Z_{n+1}}(x) = 2(1-p)g_{Z_n}(2x) + 2p\cdot g_{Z_n}(2x-1).$$ which can be explicitly verified in terms of the expectations, if desired. However, the final distribution you want ought to be (and is, in pretty much any topology one might put on the generalized functions) is the limit of $g_{Z_n}$ as $n$ goes to $\infty$ - and the operations here are continuous, so we will get $$g_{Z}(x) = 2(1-p)\cdot g_Z(2x) + 2p\cdot g_Z(2x-1)$$ which is exactly the sort of algebraic equation that would lead to the sort of structures you are seeing - and gives you the iterative formula that leads to the sort of density plots you've made. Of course, as a measure, this is supported on the set of numbers $x$ in $[0,1]$ where the asymptotic density of the set of indices where the corresponding bit of $x$ in binary is $1$ equals $p$ - so can't be converted to a function (even via tools like the Radon-Nikodym derivative which converts from measures to functions where possible) - but, nonetheless, these generalized functions can still provide some of the framework you seem to be after.

As an aside, you can work with measures this way too; if you let $\mu$ be the probability measure associated to the process we've been discussing, you can write $$\mu(S) = p\mu\left(\frac{S}2\right) + (1-p)\mu\left(\frac{S+1}2\right)$$ which has the same structure. If you're happy to work with measures and don't want to try differentiating the measure any further, then they're a good answer to your search as well.

I might note that these absolutely cannot see the sort of holes that you are talking about - but every finite measure on $\mathbb R$ is supported on a totally disconnected set (because only countably many points can have positive finite mass, and any countable set of points not including such a point has no measure). Generally, with probability, the "blurring" effect of using smooth test functions is actually desirable, because the measure of an interval is not the sum of the measures of its points. This is also why one rarely every works with functions when dealing with measures, but instead works with $L^p$ spaces (whose members are not functions, but rather almost-everywhere equivalence classes of functions).

To say this more generally: a function is defined by the fact that it can be evaluated at any point. This is not really something you desire when talking about probability distributions, because the quantity is too spread out to register evaluation. A measure, more usefully, is defined by the fact that it can integrate against continuous functions (e.g. as in the Riesz representation theorem) and then a distribution is a generalization which can integrate against smooth functions. The latter two objects tend to be more useful in these situations.

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  • $\begingroup$ Very useful, great answer. I was a afraid to start spending time re-inventing the wheel, and your answer is definitely what I needed. It will take me a bit of time to digest this material, but it is worth it! $\endgroup$ – Vincent Granville Dec 11 '19 at 16:20
  • $\begingroup$ Sorry, I only now realize that I accidentally downvoted while trying to upvote yesterday and my vote is now locked in unless an edit is made. $\endgroup$ – ajs3 Dec 12 '19 at 21:21

Measure-theoretic probability provides a formalism in which all quantities in your question are rigorously defined. In my opinion this is the natural framework to use (instead of introducing an ad-hoc approach) since you have already invoked measure theory in your question!

Even better, there is substantial (and beautiful) literature studying the exact same fractal systems you are interested in - read to the bottom of this answer (skimming the pedantic bits you probably already know) to find out more.

TL;DR The answer to your question

Is there an existing theory to handle this type of density-like-substance?

is a resounding Yes!

Some background

In measure-theoretic probability, one starts with a sample space $(\Omega,\mathcal F)$ which is an arbitrary set equipped with a $\sigma$-algebra. In this case, it is natural to take $\Omega=\{0,1\}^{\mathbb N}$ and use the product $\sigma$-algebra. Equip $(\Omega,\mathcal F)$ with the measure $\mathbb P_p=\textrm{Ber}_p^{\mathbb N}$ which is the product of $p$-Bernoulli measures as you have described.

In this formalism, a random variable is nothing other than a measurable function $f\colon \Omega\to\mathbb R$ where $\mathbb R$ is equipped with the Borel $\sigma$-algebra. In particular, we have a random variable given by $$ Z(\omega_1,\omega_2,\ldots)=\sum_{n=1}^{\infty}\frac{\omega_n}{2^n}. $$ It is a measurable function since it is a limit of measurable functions (namely, each partial sum is straightforwardly seen to be measurable).

Since $\mathbb P_p(0\leq Z\leq 1)=1$, the random variable $Z$ is seen to be integrable, so that the mean $\mathbb EZ$ is well-defined. By the same token, $\mathbb E[Z^k]$ is well-defined for all integers $k\geq 0$.

This is not just abstract nonsense: it leads to an operational calculus of the kind you seek. For instance, we have the well-known tail integral for moments of a random variable given by $$ \mathbb E[Z^k]=\int_0^{\infty}kZ^{k-1}\mathbb P_p(Z>t)\ dt,\qquad k\in\mathbb N, $$ which is a special case of Tonelli's theorem.

The distribution of the random variable $Z$, let's call it $\mu_p$, is defined to be the pushforward of the measure $\mathbb P_p$ under the mapping $Z\colon\Omega\to\mathbb R$. In symbols, $$\mu_p(A)=\mathbb P_p\bigl(Z^{-1}(A)\bigr),$$for all Borel subsets $A$ of $\mathbb R$. It is a probability measure on $\mathbb R$, with support contained in $[0,1]$. As you have pointed out, the density does not exist in general. In fact, by the Radon-Nikodym theorem it exists if and only if the measure $\mu_p$ is absolutely continuous with respect to Lebesgue measure, in which case the density is precisely the Radon-Nikodym derivative of $\mu_p$ with respect to the Lebesgue measure. The power of measure-theoretic probability is that all techniques and formulas that one is accustomed to working with carry over to the language of probability measures and random variables, ceasing to rely upon the existence of a density. (Of course, complications arise and a little adjustment is required - but this forms the core of the modern probabilist's arsenal of tools.) Thus one has the Lebesgue integral $$ \mathbb E[Z^k]=\int_0^{\infty}t^{k}\ d\mu_p(t),\qquad k\in\mathbb N, $$ regardless of whether $\mu_p$ possesses a density.

Modern research

With all this being said, it is still a very interesting question (related to open lines of current research) to understand when the density exists, and what are its fractal properties. I refer you to the survey paper by Peres, Schlag, and Solomyak entitled "Sixty Years of Bernoulli Convolutions", where a closely related generalization of your $p=\tfrac12$ case is considered, namely to the random variable $$ Z_\lambda(\omega_1,\omega_2,\ldots)=\sum_{n=1}^{\infty}\omega_n\lambda^n,\qquad \lambda\in[0,1]. $$ The distribution $\nu_\lambda$ of the random variable $Z_\lambda$ has fascinating properties when $\lambda\in(\tfrac12,1)$, and the study of these properties leads to beautiful connections with "harmonic analysis, the theory of algebraic numbers, dynamical systems, and Hausdorff dimension estimation" (direct quote from the abstract).

The fractal nature of these measures is a prominent motif throughout this line of inquiry, and appears at the forefront in the article Self-similar measures and intersections of Cantor sets written by a subset of the authors of the survey paper, which goes beyond the $p=\tfrac12$ case to study the measure $\nu_{\lambda}^{p}:=(Z_\lambda)_*(\mathbb P_p)$, of which the special case $\lambda=\tfrac12$ is the distribution of your random variable $Z$.

The aforementioned survey article is featured prominently in the proceedings of an international conference on fractal geometry, collected in the (unfortunately non-freely available) book Fractal Geometry and Stochastics II, which serves as a jumping off point to the literature on dynamical systems leading to fractal geometries closesly related to your list of "other related problems" as well. Note that many of the individual articles collected in this book (and its bibliography) are freely available at either the authors' webpages or on preprint servers.

Broader picture

To make contact with the answer posted by Milo here, let me point out that the theory of measures may be regarded as a special case of the theory of generalized functions - i.e., every measure is a generalized function with respect to the class of bounded continuous test functions (with the Lebesgue integral supplying the pairing between measure and test function).

My perspective is that measure theory is (broadly speaking) a more developed field than that of generalized functions, for the same reason that continuous functions are better understood than measures. What I mean is that one can group the objects of study in real analysis by how "well-behaved" the objects are. After linear functions, the best behaved functions are polynomials, after which comes (in order of increasing generality) analytic functions, then smooth functions, then continuous functions, then measurable functions, then measures, and finally generalized functions. (Everything function $f$ that appears before measures in this list can be canonically associated with the measure $f\cdot \textrm{Lebesgue}$; and measures can be regarded as generalized functions as already explained - so everything is on the same footing, just with varying degrees of regularity.)

Fortunately the problems you are interested in (and all the different flavors of dynamical systems, including ergodic theory) live at (or below) the "measure theory" rung on the ladder of abstractions.

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    $\begingroup$ Fantastic answer, just like Milo. I wish I could accept both of them but Math.StackExchange won't allow it. I believe there is a way to give 200 points to both of you. I'd like you to also receive these points, if you know how to do it.. $\endgroup$ – Vincent Granville Dec 11 '19 at 16:23
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    $\begingroup$ The case $\lambda\in ]\frac{1}{2},1[$ is fascinating and I am familiar with it. It corresponds, if I understand it correctly, to the representation of a number in a base $b\in ]1, 2[$. I will read all your references. I also invite you to read what I wrote on this very topic (non-integer bases), see my article "Fascinating New Results in the Theory of Randomness", at dsc.news/2HLUjTO $\endgroup$ – Vincent Granville Dec 11 '19 at 16:30

This is not really an answer, but rather a path to build a potential density (derivative) on a support domain dense in $[1, 2]$ yet of Lebesgue measure zero. The purpose of this exercise, after all, was pure math to define a derivative on sets such as rational numbers.

Let's look at the last example involving a fractal as a component of the distribution $F_Z$. Clearly no derivative $f_Z$ can exist, yet $f_Z^{[1]}(z)=\frac{1}{z \log 2}$ (with $z \in [1, 2]$) is in some way a first-order approximation to any would-be density.

If you smooth out the fractal (using splines) in the last picture, you would easily end up with a second order approximation for $f_Z(z)$:

$$f_Z^{[2]}(z) = \frac{1}{z\log 2} + \frac{d\epsilon^*(z)}{dz}$$

where $\epsilon^*(z)$ is the smoothed version of $\epsilon(z) = F_Z(z) - \log_2(z)$.

You can repeat the process and get third or fourth order approximations. From a mathematical viewpoint, the splines can be chosen such as that the $n$-th order approximation (denoted as $f_Z^{[n]}$) to the non-existing theoretical density $f_Z$, has all its first $n$ moments exactly correct, that is, $$E(Z^k) = \int_1^2 z^k f_Z^{[n]}(z) dz, \mbox{ for } k = 0, 1, \cdots, n-1.$$

My guess is that the sequence $f_Z^{[n]}$ won't converge, but could be just as good as traditional asymptotic approximations such as enter image description here

which are known to diverge yet provide great, useful approximations.

Another path to a potential solution

The density $f_Z$ may be viewed as a function whose value at any point is undefined, a bit like the exact space/time position of any tiny particle in the universe, according to quantum physics, can not be determined exactly. Instead, that position has some kind of discrete distribution. The same is true for quantum computers. While in those cases, the distribution's support domain (e.g. the energy levels) have tiny gaps, yet gaps with a length strictly different from zero, in my case (the fractal-like distribution) has gaps -- infinitely many and everywhere -- but the gap between two adjacent values, just like the gap between "closest" rational numbers, is exactly zero. That is, a potential solution would be to define $f_Z(z)$ not as a single value, but as a continuous set of values, each having its own density depending on $z$, (in the standard meaning of the word density) as being a potential value of $f_Z(z)$. The astute reader will easily find a connection to Bayesian probability theory and singular distributions (see here). Does this make sense?

I believe that my concept, if properly formalized, could help set some foundations for some of the 7 millennium problems (see here, each one coming with a $1 million award), especially the Navier-Stokes equations in fluid dynamics involving viscosity. In some sense, my "density" concept involves a blend of smooth solutions mixed with turbulence, thus the analogy.

Yet all the time I spent on this (and I am by no means done) is connected to what I consider to be the most challenging, unsolved mathematical problems of all times: proving that $\sqrt{2}$ is a normal number. I offer $500,000 to anyone who solves it. I came close to a solution, but I probably have 5 more years before I get it finalized. See here how far I got.

It's kind of funny that the systems that I have described here are all solutions of extremely elementary stochastic functional equations (elementary in the sense that their writing appears as simple as quadratic equations), yet incredibly hard to handle. To summarize, I presented three of them:

  • $F_Z = F_{(X+Z)/2}$ - the binary numeration system
  • $F_Z = F_{X(1+Z)}$ - another one-to-one mapping to the binary numeration system
  • $F_{Z^2} = F_{X+Z}$ - the nested square roots system

But after all, the unsolved Navier-Stokes equations have also a very simple / short expression.

Philosophical comment

Pretty much all functions are nowhere continuous, much less differentiable. It's just that the ones mathematicians are dealing with, behave incredibly nicely for the most part. Any perfect physical model of the universe will have the same features as the last example in my discussion: a smooth component (Newton's laws), a discrete component (quantum physics) and some kind of chaotic/fractal behavior involving support domains full of holes.

And just like Newton's theory is just an approximation, if you ignore the fractal component in my last example, you also get a very good approximation.

The mathematical equations that fully govern the universe will eventually have this nasty feature discussed here: densities (more precisely, solutions of stochastic functional equations) that technically do not exist, that are acting like mysterious objects or black holes, yet in some way exist if we had the right mathematical framework to describe them.

In some sense, it might not be different from the invention of complex numbers and the imaginary $i$ satisfying $i^2=-1$. I will call the non-existent densities featured here as imaginary densities. A better word could be quantum densities, or singular densities.

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