A family of singular functions and its relation to harmonic fractal analysis and fuzzy logic

The mapping

(a, b, a₂, a₃ ∈ ]0,1 [) is a contraction for the Hausdorff metric in 𝒦 ([0, 1]²).

The graph of f_a,b, is the unique invariant set for the iterated function system defined by (1), that is the fixed point of C. The next result follows as a consequence of the definition of f_a,b, and its graph is a compact set.

The function f_a,b is monotonic and continuous.

Now, since each affine contraction represents a functional equation, according to the Banach fixed point theorem, we have f_{a ,b} determined by the following functional equations.

The function f_a,b is the unique bounded solution of the system of functional equations

The area under the graph of f_a,b, is∫01fa,b(x)dx=2a1+a22.

In [22], for Okamoto’s one-parameterized function it is proved that f_a = f_{a, 1-a} is non-differentiable almost everywhere at the critical parameter value. In [19], the performance of fa′ is explored almost everywhere for different values of a. For f_a,b we have the following general result.

The derivativefa,b′(x)when it exists, can only vanish.

If fa,b′(0+) exists, then it must vanish. To prove it, we take into account that fa,b13n=an and fa,b23n=ban−1. This implies that the slope of the segments joining the point (0, 0) with (13n,an),(23n,ban−1) and 13n−1,an−1 are, repectively, (3a)n,3b(3a)n−12and(3a)n−1.Iffa,b′(0+) exists and does not vanish, then the

limits of the quotients (3a)n(3a)n−1=3aand3b(3a)n−12(3a)n−1=3b2 exist and are equal to 1. That is, (a,b)=13,23 which is a contradiction.

A similar argument allows us to obtain fa,b′(1−)=0, if it exists.

By repeting this reasoning or using the system of functional equations (2), it follows that if exist, the numbers fa,b′m3k−andfa,b′m3k+ are zero.

Let us suppose that x cannot be expanded in the form m3k. If fa,b′(x) exists and does not vanish,

for sequences (x_k) and (y_k) that converge to x such that x_k ≤ x < y_k.

Let us take

and

then we have

Thus, if 3a_i ≠ 1 for the three values, this is a contradiction. If some of them equal 1, then the other two differ from 1, because we have avoided the case a = 1/3 and b = 2/3. Therefore, if we obtain 3a_i = 1, we shall change that quotient by

or by

This quotient is 3(ai+aj)2, with i ≠ j, and differs from 1.

For each k we have described the possibility of taking the quotient on a finite set (where 1 is not included). Consequently, the limit fa,b′(x) must be zero.

Note that the theorem above provides a new proof for the singularity of f_a,b.

([3, Th. 4]). If F₂(F_n) is finite for every n, then the convergence of two series∑n=1∞E(Fn) and∑n=1∞σ2(Fn) implies the weak convergence of H_n(x) = F₁ * F₂ * ...* F_n(x) (i.e. H_n(x) → H(x) for each x at which H is continuous).

([3, Th. 35]). If F = F₁ * F₂ * · · · is a convergent infinite convolution of distribution functions F_n each of which is purely discontinuous, then F is either purely discontinuous, or singular, or absolutely continuous.

The next result can be found in [23, pg. 46].

Let γ_n = max_z{F_n(z+) – F_n(z–)}, then the infinite convolution F(x) = (F₁ * F₂ * ...) (x) iscontinuous if and only if the series ∑n≥1(1−γn) diverges.

Finally, in [26] the following useful result is proven.

If F_n and F have characteristic functions F~n and F~,, respectively, and (F_n) is weakly convergent to F, then F~nx→F~x for each x.

Let be the infinite convolution function S := F₁ *F₂*F₃*..., where F₁, F₂, F₃,... are distribution functions given, for each positive integeer n, by

This function S is well defined. In fact, let us note that, for every n:

The convergence of the corresponding series ensures, by Lemma 2.6, the weak convergence of the convolution, that is, S(x) exists for every x.

With the notation as in the above definition, we have:

1. The characteristic function of S isS~(t)=∏n=1Q(a1+a2eit3n+a3ei2t3n).

2. The sequence of Fourier coefficients of S~,, i.e. Sp=∫01e2πpixdS(x),, does not converge to zero.

3. S is a singular function.

1. The characteristic function of F_n is a1+a2eit3n+a3ei2t3n. If H = F * G, then their respective characteristic functions satisfy the relation H~=F~G~, and as a consequence, by Lemma 2.6, we obtain that the characteristic of S is S~(t)=∏n=1∞(a1+a2eit3n+a3ei2t3n).

2. Let us denote by Sp:=S~(2πp) the Fourier coefficients of dS. The expression of S~ allows us to obtain that S_{3q p}= S_p for all q ≥ 0. Since the measure is not equal to the Lebesgue measure, then there exists n > 0 such that Sp ≠ 0, and, as a consequence, the Fourier coefficients do not converge to zero.

3. The functions in the convolution are purely discontinuous, thus S is also pure. But it is neither discontinuous, by Theorem 2.8, nor absolutely continuous, because by b) it does not satisfy the Riemann-Lebesgue theorem. Therefore, it must be singular.

For a nonnegative integer m < 3ⁿ, let us consider its expansion in the base 3 m = r₀3⁰ + r₁3^l + r₂3² + ... + r_n-13^n-1, with r_i ∈ {0, 1, 2}. Set c(m) (resp., u(m) and d(m)) the number of times that the digit 0 appears (resp., 1s and 2s) in the above expansion of m.

The distribution function F₁ * F₂ * · · · * F_n has the following associated probability

The proof is reached by induction. The statement is true for the first values. Let us suppose that it is true for n — 1, and we will prove it for n. (Let us denote by Pn¯ the associated probability with F_n.)

If m is congruent with zero module 3, i.e. m = 3k, then

and

Taking into account that the expansion of m has one less 0 than the expansion of k, and the same number of 1s and 2s, by applying the induction hypothesis we obtain the desired result.

For the cases m = 1 (mod 3) and m = 2 (mod 3) we proceed in a similar way.

Applying the above lemma and making an inductive hypothesis we obtain the following result.

a) F1∗F2∗⋯∗Fn3n−1−13n=a.

b) F1∗F2∗⋯∗Fn2⋅3n−1−13n=b.

Now, we show the relationship between f_a,b and S.

f_a,b is the infinite convolution function S.

We have to prove that the graph of G_n = F₁ * F₂ *...* F_n and Cⁿ(d) are the same set, where d denotes the segment (0,1)(1,1)¯ and C is given by (2.1), but for a finite set of points. Namely, the intersection of Cⁿ(d) and the line x=k3nfor1≤k≤3n−1 has two points: k3n,Gnk3nandk3n,Gnk3n− where the minus sign means left-side limit. For the other values of x, the graph of G_n and the set Cⁿ(d) coincide.

Set Cⁿ(x) := max{y : (x, y) ∈ Cⁿ(d)}. It is immediate that G113=C113,G123=C123. Thus we will show that for n, it follows that Cnk3n=Gnk3nfor0≤k<3n. Let us suppose that it is also true for n – 1.

For 0 ≤ m < 3^n-1, we consider these possibilities:

a) Gnm3n=∑k=0mPnk3n=∑k=0ma1Pn−1k3n−1=a1Gn−1m3n=a1Cn−1m3n=Cnm3n, where the second equality is true because when we write k with one more digit, this must be zero.

b)If Gnm+3n−13n=a1+a2Gn−1m3n−1=a1+a2Cn−1m3n−1=Cnm3n−1, the proof is similar to the above.

c)If Gnm+2.3n−13n=a1+a2+a3Gn−1m3n−1=a1+a2+a3Cn−1m3n−1=Cnm3n−1, the proof also follows similarly.

This ensures that the statement is also true for n.

If G_n are step functions that coincide with f_a,b at points in the form m3n with 0 ≤ m < 3ⁿ, then limn→∞Gn(x)=fa,b(x)for allx ∈[0,1].

Theorems 2.11and 2.14 give further proof of Okamoto and Wunsch’s result [18, Th. 5] using exclusively probability theory methods. In fact, f_a,b is a singular function and additionally we know its characteristic function, as well as the fact that the sequence of its Fourier coefficients does not converge to zero.

A pair (E, f_α) is a Katok foliation if:

1. E ⊂ [0, l]²is a set of measure 1;

2. f_α is a family of analytic functions from ]0,1[ to [0, 1];

3. the graphs of the functions f_α fill the interior of [0, 1]²;

4. the graphs of the functions f_α are pairwise disjoint;

5. the graph of each function f_α intersects with E at one point at most.

For each x ∈ ]0, 1 [, the function gx(t):=ft2,t(x)is analytic at t ∈ ]0,1 [ and g_x(0) = 0, g_x (1) = 1.

If x = d₁ + s₁d₂ + s₁s₂d₃ + s₁s₂s₃d₄ +.... then g_x(t) is obtained by the substitution of d_i,s_i with one of the values 0, t, t², 1 — t, f (1 — f), depending on d¡. For each substitution, we obtain an expansion series in powers of t. It is clear that g_x(0) = 0 and g_x(1) = 1 : the series expansion begins with a term in the tⁿ power, thus g_x (0) = 0. For the latter, we can write in the form tⁿ + a series of positive terms, with 0 ≤ g_x (t) ≤ 1. Hence, by continuity, the equality follows.

For x, x’ ∈ ]0, 1[, x≠ x’, the graphs of g _x and g_x’ in ]0, l[ are disjoint.

If (t, g_x(t)) and (t, g_x’(t)) are equal, the series expansion in the t², t -representation system of x and x’ is the same. But this would imply x = x’, a contradiction.

The functions ft2,t are singular, and their graphs fill the interior of the unit square [0,1].

Singularity follows from the above results. Let x ∈ ]0,1[. The function g_x is continuous with g_x(0) = 0 and g_x(1) = 1, and as a consequence, it takes all the values 0 < y < 1.

The pair (E,g_x), where

E is a set of measure 1 by Fubini’s theorem. The analyticity was already obtained above. To deduce that the graph of each function in the foliation intersects at most at one point is a consequence of the fact that the variables d¡ (that only depend on x) appear in the same proportion in the image of g_x. Because this proportion is a limit, it is possible that it does not exist. If it exists, then we find a value t with the desired proportions.

In exchange of t²,t for t^u ,t^v with u > v, one can obtain a bi-parametric family of foliations.

For r such that 0 < r < 1/2, let us set functions:

The only fixed point of G will be denoted by N_r.

N_r is a fractal set, and we will study harmonic functions on it. This type of sets has already been used to show the first example of copulas with fractal support (see for instance [30] or [31]). Let us recall that the fixed point N_r in the Banach contraction mapping is obtained as the limit by iterations of G starting at any choosen point. Taking as the starting point the compact unit square, for r = 1/3, the first two are as Figure 2 shows.

Fig. 2

The two first iterations of the unit square in the case r = 1/3

We name the vertices of [0,1]² : p₁ = (0, 1), p₂ = (1, 1), p₃ = (0,0) and p₄, = (1,0), and for the vertices in the first iteration:

(see Figure 3).

Fig. 3

Vertices considered in the unit square

For the sake of our paper, we state that a function h defined in C is harmonic if

implies that

And the same occurs on each of the squares that appear at each stage. G can be extended to C by continuity, because the set of vertices of the squares form a dense set. These functions will be denoted by hs1s2s3s4. These functions and the rest that follows in this section, depend on the parameter r, but we omit it for simplicity. We will study their performance on the diagonal of the set, and we will do it for the particular case of h_α10α, which we denote by h_α for short. Actually, the only condition we impose is that s₁ = s₄ = α, because it is necessary to take it into account:

When we are restricted to the diagonal, it is necessary to study the subsets G₁ ([0, 1]²), G₃ ([0, 1]²) and G₅([0, 1]²) : they are the new squares having a non-empty intersection with it. The self-similarity of the set and the way the function takes the values at the vertices of the new squares, allow us to write the following equations:

that can be rewritten as:

Set g_α : [0,1] → [0,1], g_α(x) := h_α ((x, x)) for all x ∈ [0,1]. The above equalities for these functions give:

and if α = 1/2, then they become the functional equations for g12:

Now attending to the functional equations of f_r,1-r (with 0 < r < 1/2) we have the following result.

fr,1−r−1andg12 coincide on the unit interval.

g_α is a singular function.

Proof. The equalities already established for the family of functions g_α ensure the way to divide the unit interval [0,1] into subintervals where the corresponding function is the scale copy of g12. Thus, we complete the proof, because the inverse of a singular function is also singular.

There exists a set of measure zero and Hausdorff dimensionln⁡27−ln⁡r2(1−2r)that is mapped on a set of measure one by g_α.

There exists a set of measure one that is mapped on a set of measure zero by g_α,r and Hausdorff dimension

This study can be generalized to the set C’ determined as the fixed point of the mapping generated by the following functions:

The results are analogous, but the functions we now obtain are the inverses of f_{r, r’}, and the fractal dimensions of the related fractal sets areln⁡27−ln⁡rr′(1−r′−r)and−rln⁡r+r′ln⁡r′+(1−r−r′)ln⁡(1−r−r′)ln⁡3 respectively.

For 0 < a < b < 1, the function n_a,b : [0, 1] → [0, 1] given by

for all x in [0, 1] is a strong negation.

Proof. Continuity and monotony are evident. Through functional equations (3), we deduce that n_a,b, is the only function defined in [0,1] that is bounded and satisfies the relations:

By the definition of n_a,b and properties of f_ab(1-b)(1-a):

The others follow is an analogous way. Finally, its uniqueness follows from that of f_{ab(1—b)(1—a)} in (3) or from the fixed point Banach theorem.

To show that na,b2(x)=x it is sufficient to use equations (4). From them, we deduce that na,b2 satisfies the functional equations