# Analysis of a Fractal Boundary: The Graph of the Knopp Function

Authors: Mourad Ben Slimane; Clothilde Mélot
Journal: Abstract and Applied Analysis
 (2015)

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## Abstract

A usual classification tool to study a fractal interface is the computation of its fractal dimension. But a recent method developed by Y. Heurteaux and S. Jaffard proposes to compute either weak and strong accessibility exponents or localLp regularity exponents (the so-called p-exponent). These exponents describe locally the behavior of the interface. We apply this method to the graph of the Knopp function which is defined for x∈0,1 as Fx=∑j=0∞2-αjϕ2jx, where 0<α<1 and ϕx=distx,z. The Knopp function itself has everywhere the same p-exponent α. Nevertheless, using the characterization of the maxima and minima done by B. Dubuc and S. Dubuc, we will compute the p-exponent of the characteristic function of the domain under the graph of F at each point (x,F(x)) and show that p-exponents, weak and strong accessibility exponents, change from point to point. Furthermore we will derive a characterization of the local extrema of the function according to the values of these exponents.


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## Body

## 1. Introduction

At the beginning of the century several examples of nondifferentiable functions were studied, such as the Weiertrass function or the example we will focus on in the following, that is, the Takagi-Knopp or so called Knopp function (see [1] and references therein for a review). The issue was the study of the regularity.Indeed, in 1918, Knopp [2] introduced a new family of nondifferentiable functions defined on the interval [0,1]. Going beyond the construction of Weierstrass of a continuous nondifferentiable function, his goal was to build examples of continuous functions for which one-sided limits of the difference quotient at all points do not exist. He considered the function Fa,b given by the series, for x∈[0,1],
(1)Fa,bx=∑j=0∞‍ajϕbjx,
where ϕ(x)=dist⁡(x,Z), 0<a<1, and b is an integer such that ab>4.Forb=2 and a=2-α, this function can be seen as a series expanded in the Faber-Schauder basis Λj,k:x↦2j/2Λ(2jx-k), j∈N, k=0,…,2j-1, where Λ is the Schauder function defined by Λ(x)=inf⁡(x,1-x) if x∈[0,1] and 0 elsewhere. In fact
(2)F2-α,2(x)=∑j=0∞‍∑k=02j-1‍2-αjΛ(2jx-k).
We will write F for F2-α,2 in the following.Thus, for example, using the characterization of Lipschitz spaces with the help of coefficients in the Schauder-basis [3], one gets immediately the fact that F belongs to Cα([0,1]).A further step to study the regularity of this function can be to follow the ideas developed in multifractal analysis. The goal in multifractal analysis is to study the sets of points where the function has a given pointwise regularity, and doing so to check if the regularity changes from point to point and quantify these changes. Recall the definition of Hölder pointwise regularity and localLp regularity.Definition 1.
Letx0∈Rd and α≥0. A locally bounded function f:Rd→R belongs to Cα(x0) if there exists C>0 and a polynomial P=Px0 with degree deg⁡(P)≤[α] (integer part of α), such that on a neighborhood of x0,
(3)fx-Px0x≤Cx-x0α.
The pointwise Hölder exponent of f at x0 is hf(x0)=sup⁡{α:f∈Cα(x0)}.Definition 2 (see [4]).
Letx0∈Rd. Let p∈[1,∞] and u such that u≥-d/p. Let f be a function in Lloc⁡p. The function f belongs to Tup(x0) if there exists R>0, a polynomial P with deg⁡(P)≤u, and C>0 such that
(4)∀ρ≤R:1ρd∫|x-x0|≤ρ‍|f(x)-P(x)|pdx1/p≤Cρu.
Thep-exponent of f at x0 is ufp(x0)=sup⁡{u:f∈Tup(x0)}.Then again with the help of the Faber-Schauder basis one can prove that, for allx0∈[0,1], F is in Cα(x0) (details for this technique can be found in [5]). It is then easy to check that actually upF(x0)=hF(x0)=α at all x0∈[0,1]. Thus, from the point of view of various notions of regularity, even if it is not differentiable, the function F is rather “regular” since one can compute at each point x0 the same regularity exponent. This remark was actually the starting point of this work.Indeed obviously the graph of the function has a very irregular behavior, and it has also some self-similarity properties. What can we say on the domainΩ={X=(x,y):y≤F(x)} under the graph of F?Denote in the following by
I
Ω
the characteristic function of
Ω
, which takes the value
1
onΩand
0
outsideΩ.A first reflex is to compute fractal dimensions of the boundary∂Ω. The box dimension of the graph can be derived by standard methods (see Tricot [6]) and is exactly dim⁡B(∂Ω)=2-α. Let us mention that Ciesielski [7, 8] proved results of this type for Schauder and Haar bases expansions in the case of more general families of functions. Jaffard [9] and Kamont and Wolnik [10] obtained then general formulas that allow deriving the box dimensions of the graphs of arbitrary functions from their wavelet expansions.For what concerns the Hausdorff dimension of the graph ofF, as far as we know, the question is not solved yet in its all generality. It was proved by Ledrappier [11] in 1992 to be 2-α in the special case where a=2α-1 is an Erdös number. By the results of Solomyak [12] on Erdös numbers this amounts to have the computation for almost every α in [0,1].Beside the computation of the box and Hausdorff dimension, which provide global quantities to describe the graph of the function, several methods were recently developed to classify fractal boundaries with the help of pointwise exponents. The idea was to be able to give a finer description of the geometry of the boundary, since the pointwise behavior was studied. In [13], Jaffard and Mélot focused on the computation of the dimension of the set of points where IΩ has a given p-exponent in the sense of Definition 2. In [14], Heurteaux and Jaffard studied pointwise exponents more related to the geometry. These are the exponents we are actually interested in.Indeed denote bymeas⁡ the Lebesgue measure in Rd and B(X,r) the d dimensional open ball of center X and radius r>0. Heurteaux and Jaffard [14] gave the following definitions.Definition 3.
LetΩ be a domain of Rd and let X0∈∂Ω. The point X0 is weak α-accessible in Ω if there exist C>0 and r0>0 such that
(5)∀r≤r0meas⁡Ω⋂‍BX0,r≤Crα+d.
The supremum of all the values of α such that (5) holds is called the weak accessibility exponent in Ω at X0. We denote it by EΩw(X0).Example 4.
Let0<β<1 and Ω={X=(x,y)∈R2:|y|≤|x|β}. Denote by Ωc the complement of Ω. Then one can easily check that at each point X1≠(0,0) of the boundary ∂Ω we have EΩw(X1)=0=EΩcw(X1) and at X0=(0,0) we have EΩcw(X0)=(1/β)-1 and EΩw(X0)=0.Definition 5.
LetΩ be a domain of Rd and let X0∈∂Ω. The point X0 is strong α-accessible in Ω if there exist C>0 and r0>0 such that
(6)∀r≤r0meas⁡Ω⋂‍BX0,r≥Crα+d.
The infimum of all the values of α such that (6) holds is called the strong accessibility exponent in Ω at X0. We denote it by EΩs(X0).The following proposition is given in [14].Proposition 6.
LetX0∈∂Ω. Then
(7)d+EΩw(X0)=liminf⁡r→0log⁡meas⁡Ω∩BX0,rlog⁡r,d+EΩs(X0)=limsup⁡r→0log⁡meas⁡Ω∩BX0,rlog⁡r.ObviouslyEΩs(X0)≥EΩw(X0). We will see that thanks to our result one can prove that these two exponents can be different.Tricot [15] proved that these exponents are related to local dimension computation. Let us mention, without giving too much details, the relationship of this work [15] with these exponents. Indeed the author focuses on the formula
(8)Hϕ(E)=liminf⁡ɛ→0∑i≥0‍ϕEi:E⊂⋃i≥0‍Ei,diam⁡Ei≤ɛ
with ϕ:BE→(0,∞) some “set function” and BE the set of closed balls centered on E.Given an open setV such that E⊂∂V the special choice of
(9)ϕα(B)=Vol(B∩V)Vol(B)diam⁡Bα
leads to definitions of Hausdorff, exterior and interior dimensions, and Packing, exterior and interior dimensions.The following characterization, written for the setting we are interested in, holds.Theorem 7 (see [15]).
LetΩ be a bounded open set in Rd with boundary ∂Ω such that meas⁡(∂Ω)=0. Let X0∈Ω. Let r>0 and
(10)αBX0,r=d-log⁡meas⁡Ω⋂‍BX0,rlog⁡(r).
Then(11)liminf⁡r→0αBX0,r=dim⁡intX0,limsup⁡r→0αBX0,r=Dimint({X0})
with dim⁡int the Hausdorff interior dimension and Dimint the Packing Hausdorff dimension.We clearly havedim⁡int⁡({X0})=-EΩw(X0) and Dimint⁡({X0})=-EΩs(X0). Let us stress that in the setting of Tricot dim⁡ext({X0})=-EΩcw(X0) and Dimext({X0})=-EΩcs(X0) with Ωc the complementary of Ω in Rd. We rather refer to [15] for more details on local dimensions in their all generality.We will compute these quantities at the points of the boundary∂Ω of Ω={(x,y)∈R2:0≤x≤1,0≤y≤F(x)}, where F is the function defined by (2). For that we will use the characterization of the maxima and minima done in [16]. This will yield the p-exponent at each point of IΩ. We will actually derive the fact that the set of local extrema of the function is fully characterized by the set of points where this p-exponent has a given value.We will also prove that the weak and strong accessibility exponents inΩ and Ωc change from point to point on the graph ∂Ω of F. They also help to provide exact characterization of the sets of local maxima and local minima. Finally we will prove that there is a set of nontrivial Hausdorff dimensions such that the strong accessibility exponents in Ω and Ωc are the same and the weak and strong accessibility exponents are different.Let us emphasize that this is to our knowledge the first time that the computation of these exponents was done in a nearly exhaustive study on a given example. The characterization we get for the set of extrema raise several questions: is it a general property? Do other functions share it? Could it lead to a finer classification of functions in Hölder classes? We would like to address them in future works.Let us come back now to our work. The outline of the paper is the following. In Section2 we set our main result. In Section 3 some notations, preliminary remarks and technical lemmas, help us to prepare Section 4 where are the main proofs.


## 2. Main Results

### 2.1. Statement of Our Main Result

Our goal is to prove the following Theorem.Theorem 8 (the main theorem).
Leta=2-α with 0<α<1 and b=2. Let F be the function defined by (2).
LetΩ={X0=(x,y)∈R2:0≤x≤1,0≤y≤F(x)} and let f=IΩ.
Then, at each pointX0 of ∂Ω, the graph of F, one has(1)
ufp(X0)=(1/p)(1/α)-1 if and only if F(X0) is a local extremum of F. Furthermore
(a)
EΩw(X0)=EΩs(X0)=(1/α)-1 if and only if F(X0) is a local maximum of F. And in this case EΩcw(X0)=EΩcs(X0)=0.(b)
EΩcw(X0)=EΩcs(X0)=(1/α)-1 if and only if F(X0) is a local minimum of F. And in this case EΩw(X0)=EΩs(X0)=0.(2)
In the other cases whereF(X0) is not a local extremum of F, one has EΩw(X0)=EΩcw(X0)=0.(3)
Furthermore one can find a subsetDα⊂∂Ω such that EΩs(X0)=EΩcs(X0)=(1/α)-1 and EΩw(X0)=EΩcw(X0)=0 for all X0∈Dα. The orthogonal projection of Dα on [0,1] has the Hausdorff dimension α.


## 2.1. Statement of Our Main Result

Our goal is to prove the following Theorem.Theorem 8 (the main theorem).
Leta=2-α with 0<α<1 and b=2. Let F be the function defined by (2).
LetΩ={X0=(x,y)∈R2:0≤x≤1,0≤y≤F(x)} and let f=IΩ.
Then, at each pointX0 of ∂Ω, the graph of F, one has(1)
ufp(X0)=(1/p)(1/α)-1 if and only if F(X0) is a local extremum of F. Furthermore
(a)
EΩw(X0)=EΩs(X0)=(1/α)-1 if and only if F(X0) is a local maximum of F. And in this case EΩcw(X0)=EΩcs(X0)=0.(b)
EΩcw(X0)=EΩcs(X0)=(1/α)-1 if and only if F(X0) is a local minimum of F. And in this case EΩw(X0)=EΩs(X0)=0.(2)
In the other cases whereF(X0) is not a local extremum of F, one has EΩw(X0)=EΩcw(X0)=0.(3)
Furthermore one can find a subsetDα⊂∂Ω such that EΩs(X0)=EΩcs(X0)=(1/α)-1 and EΩw(X0)=EΩcw(X0)=0 for all X0∈Dα. The orthogonal projection of Dα on [0,1] has the Hausdorff dimension α.


## 3. Useful Notations and Results

### 3.1. Lemmas for Practical Computation of the Exponents

From the computation of the weak accessibility exponent inΩ and Ωc it is easy to derive the p-exponent. In [13], Jaffard and Mélot proved that IΩ∈Tα/pp(X0) if and only if either X0 is weak α-accessible in Ω or X0 is weak α-accessible in Ωc. As a consequence we have
(12)puIΩp(X0)=max⁡EΩwX0,EΩcwX0.We will also need the following lemma.Lemma 9.
Letf:R→R be in Cα(x0) with 0<α<1 and Ω the domain below (resp., above) the graph of f. Consider X0=(x0,f(x0)). Then X0 is strong (1/α)-1, accessible in both Ω and Ωc.Proof.
Suppose thatΩ is the domain below the graph of f. Without any loss of generality, we can assume that X0=(0,0). Let r>0. Since f is in Cα(0) and 0<α<1 then there exists a constant C≥0 such that in neighborhood of 0(13)fx≤Cxα.
Thus
(14)-Cxα≤fx≤Cxα.
Obviously meas⁡(Ωc⋂B(X0,r)) (resp., meas⁡(Ω⋂B(X0,r))) is greater than the area A=C′∫0r‍y1/αdy=C′′r1+1/α above (resp., below) the graph of x↦C|x|α and below (resp., above) the square of side r and center X0.
The same results hold ifΩ is the domain above the graph of f (we have just to replace Ω by Ωc).One of our goals for the points which are not extrema ofF will be to find sequences of local maxima or minima such that the following key lemma holds.Lemma 10.
Letf:R→R be in Cα(R) and Ω the domain below the graph of f. Consider X0=(x0,f(x0)). Suppose that there exists cα>0, a sequence rn of positive numbers, such that rn→0 as n→+∞, xn∈]x0-rn,x0+rn[, and n0∈N, such that
(15)∀n≥n0f(xn)=f(x0)-cαrnα.
Then EΩcw(X0)=0.Proof.
We can suppose thatxn∈]x0-rn,x0[(the case xn∈]x0,x0+rn[ is similar). Then by the mean value theorem we can find bn∈]xn,x0[ such that f(bn)=f(x0)-rn. Let bn′=inf⁡{bn∈]xn,x0[;f(bn)=f(x0)-rn}. Since f is continuous we get f(bn′)=f(x0)-rn and bn′∈]xn,x0[. It follows from the definition of bn′ and the mean value theorem that
(16)∀t∈xn,bn′f(t)<f(x0)-rn.
Thus
(17)xn,bn′×[f(x0)-rn,f(x0)+rn]⊂B(X0,rn)⋂‍Ωc.
Therefore
(18)meas⁡BX0,rn⋂‍Ωc≥2bn′-xnrn.
Since
(19)fxn-fbn′=fx0-cαrnα-fx0-rn=rn-cαrnα≃-rnα,
in the sense that there exists a constant C≥1 such that for every n we have (1/C)rnα≤cαrnα-rn≤Crnα.
Sincef belongs to Cα(R) we get
(20)Cbn′-xnα≥fbn′-fxn≃rnα.
Thus
(21)bn′-xn≥C′rn.
Following (18) and (21) we get
(22)meas⁡BX0,rn⋂‍Ωc≥Crn2.
Since meas⁡(B(X0,r)⋂Ωc)≤meas⁡(B(X0,r))≤r2 for all r≥0 we get
(23)lim⁡n→+∞log⁡(meas⁡(B(X0,rn)⋂‍Ωc))log⁡(rn)=2.
Thus thanks to Proposition 6 we have EΩcw(X0)≤0 which yields EΩcw(X0)=0.
By replacingf by -f we also have the following result.Lemma 11.
Letf:R→R be in Cα(R) and Ω the domain below the graph of f. Consider X0=(x0,f(x0)). Suppose that there exist cα>0, rn→0 as n→+∞, xn∈]x0-rn,x0+rn[, and n0∈N, such that
(24)∀n≥n0f(xn)=f(x0)+cαrnα.
Then EΩw(X0)=0.


### 3.2. Dyadic Expansions and Approximation by Dyadics

We give some properties of the approximation of a point by the dyadics. Such properties will be used later.Letx∈[0,1]. Set i1(x),…,ij(x),…, the binary digits of x; that is,
(25)x=∑l=1∞‍il(x)2l.(i)
Note that dyadic points, that is, pointsx=2-NK with K∈2N+1, are characterized by the fact that one can find N>0 such that iN(x)=1 and in(x)=0 for n>N or equivalently iN(x)=0 and in(x)=1 for n>N.Furthermore forn>N the number x-2-n is dyadic. Since 2-n=∑j=n+1∞‍2-j, then x-2-n=(∑j=1N-1‍ij2-j)+2-(N+1)+2-(N+2)+⋯+2-n.On the other handx+2-n has the simple expansion ∑j=1N(ijx/2j)+(1/2n).We will denote byD the set of all dyadic points in [0,1].(ii)
Let us come back to the general case withx any point in [0,1].For eachj∈N, define Kj (=Kj(x)) by
(26)Kj2-j-x=inf⁡k∈Nk2-j-x.Set(27)rj(x)=log⁡|Kj2-j-x|log⁡2-j.Define the rate of approximation ofx by dyadics as
(28)r(x)=limsup⁡j↦∞rj(x).Since|Kj2-j-x|≤2-j, then, for every x, we have r(x)≥1. If x is dyadic then r(x)=∞ (by taking the convention log⁡0=-∞). If x is normal (i.e., the frequency of ones (or zeros) in the binary expansion of x is equal to 1/2) then r(x)=1.(iii)
Ifr(x)>1, following the definition of r(x), then for any δ>0 such that r(x)-δ>1 one can find a subsequence Jn→+∞ for n→+∞ such that
(29)rJn(x)≤2-Jn(r(x)-δ).LetJn′=[Jn(r(x)-δ)]. We then have
(30)KJn2-Jn-2-Jn′≤x≤KJn2-Jn+2-Jn′.Thus, eitherx belongs to the dyadic interval [KJn2-Jn-2-Jn′,KJn2-Jn] and in this case it satisfies iJn+1(x)=⋯=iJn′-1(x)=1, or it belongs to the other interval [KJn2-Jn,KJn2-Jn+2-Jn′] and in this case it satisfies iJn+1(x)=⋯=iJn′-1(x)=0.In both cases let us notice that the binary expansion ofx contains chains of 0 or 1 whose length Jn′-Jn~Jn′ increases when n→+∞.


### 3.3. Approximation by Sequences of Maxima ofF

We will see in the following that points in[0,1] of the set
(31)S=k∈N,N0∈N,k2N0+1312N0,k2N0+2312N0
will play a big role in this work, since they actually are the locations of the local maxima of the function F (see below). Remark that they are characterized by the fact that, for each x∈S, one can find j0∈N such that for j≥j0 we have ij(x)+ij+1(x)=1.As in the case of dyadic approximation we can define a rate of approximation by this kind of points.Indeed let forx∈[0,1](32)|mj-x|=inf⁡k∈Nk2j+1312j-x,k2j+2312j-x.
Define
(33)sj(x)=log⁡mj-xlog⁡2-j.
Then the rate of approximation of x by elements of S is given by
(34)s(x)=limsup⁡j↦∞sjx.
Since |mj-x|≤2-j, then, for every x, we have s(x)≥1.In the case of dyadic numbers, we haves(x)=1. But remark that in other nontrivial cases there is no obvious relationship between s(x) and r(x). Indeed one can check on the following examples that s and r can independently take any value.(i)
Letx=∑j=1+∞(ij(x)/2j) with i3k+1(x)=0=i3k+2(x) and i3k+2(x)=1 for all k∈N. Then we have r(x)=s(x)=1.(ii)
Letu>1. Then x=∑n=0+∞2-[un] with [un] the integer part of un. We have r(x)=u whereas s(x)=1.(iii)
Letu>1. Then x=∑n=1+∞2-2n-∑n=0+∞2-2[un]. We have s(x)=u whereas r(x)=1.(iv)
Letu>1 and s>1. Let x=∑n=1+∞2-2[sn-1un]+∑k=[sn-1un]+1[snun+1]-12-2k. Then r(x)=u and s(x)=s.


### 3.4. The Shift Operator

Since0<α<1 it is easy to check that we obtain from (2) with a=2-α and b=2(35)Fx=∑j=0∞‍∑i=(i1,…,ij)∈{0,1}j‍2-αjΛ2jx-2j-1i1-⋯-2ij-1-ij.
The term of (35) corresponding to j=0 is Λ(x). But, the function Λ is supported in [0,1], and therefore F vanishes outside [0,1] and for x∈[0,1](36)Fx=∑j=0∞‍2-αjΛ2jx-2j-1i1x-⋯-2ij-1x-ijx.
For dyadic rationals x, x=2-NK with K∈2N+1, as we already said it, there exist two binary expansions, one such that iN(x)=1 and in(x)=0 for n>N and another one such that iN(x)=0 and in(x)=1 for n>N. The two right-hand sides of (36) corresponding to the two choices of i(x) give identical results.Denote byτ the shift operator
(37)τx=∑l=2∞‍il(x)2-l+1=∑l=1∞‍il+1(x)2-l.
Observe that
(38)τx=2x,ifx∈0,12,2x-1,ifx∈12,1.
Hence
(39)Fx=∑j=0∞‍2-αjΛτjx,(40)τjx=∑l=1∞‍il+j(x)2-l.Our self-similar function is of the formF(x)=∑j=0∞‍∑k=02j-1‍Cj,kΛ(2jx-k) with
(41)Cj,k=2-αjifj≠0,k≠0,C0,0=1.Forn≥1, denote
(42)Fn(x)=∑j=0n‍2-αjΛτjx.
Remark that Fn is affine on intervals of type In+1=k/2n+1,(k+1)/2n+1. Remark also that if t∈[0,1], then Λ(t)=(-1)i1(t)t+i1(t). So, if t′∈[0,1] and i1(t)=i1(t′), then Λ(t)-Λ(t′)=(-1)i1t(t-t′).It follows that ifk/2n+1=∑j=1n+1‍(ij/2j) then the slope of Fn at any point x of the interval ]k/2n+1,(k+1)/2n+1[ is exactly
(43)Cn=Cn(x)=∑j=0n‍-1ij+1(x)2(1-α)j=∑j=0n‍-1ij+12(1-α)j.


### 3.5. Extrema ofF

We will need the following characterization of the extrema ofF proved in [16, 17]. Let us start with the local and global minima.Proposition 12.
Let0<α<1 and let F be the function defined by (39); then (i)
0 and 1 are the abscissas of the global minima ofF;(ii)
the dyadic points are the abscissas of the minima ofF and furthermore
(44)min⁡x∈INF(x)=min⁡FN-1k2N,FN-1k+12NwithIN=k/2N,(k+1)/2N.In the case of the maxima, the statement of the result is slightly more technical. We need the following proposition of [17] using the same notations used previously.Proposition 13.
Let0<α<1 and let F be the function defined by (39). Let t=21-α and X(p) the list of positions where F(x)+px attains its maximum on [0,1]. Let M(p) be the maximum on [0,1] of F(x)+px. Then (i)
X-tN-1/(t-1)={(1/3)(1/2N),(2/3)(1/2N)} for N=0,1,….(ii)
Xp={(1/3)(1/2N)} if -(tN+1-1)/(t-1)<p<-(tN-1)/(t-1).(iii)
X(p)=1-X(-p) for all p≠0.(iv)
max⁡x∈IN⁡F(x)=FN-1(k/2N)+2-NαM(CN-1((2k+1)/2N+1)/tN).The following proposition is a consequence of the previous one.Proposition 14.
Let0<α<1 and let F be the function defined by (39). Then (i)
1/3 and 2/3 are the abscissas of the global maxima of F;(ii)
the abscissas of the local maxima ofF are the points of S.


### 3.6. Approximation of Slopes ofFn

Suppose first we have some information about the dyadic expansion ofx. Then we have the following lemma, which helps to control the behavior of the slopes of the affine function Fn-1.Lemma 15.
(
1
) Let x=K/2N be a dyadic number. Then one can find N0>N, A>0, and B>0 depending only on x such that if n≥N0 then
(45)∀y∈x,x+2-nA21-αn≤Cn-1y≤B21-αn,(46)∀y∈x-2-n,x-A2(1-α)n≥Cn-1(y)≥-B2(1-α)n.
(
2
) Let x be the abscissa of a local maximum of F. Then one can find J0>0, A>0, and B>0 such that for n≥J0(47)in-1(x)+in(x)=1,Cn-1(x)Cn(x)<0,A2(1-α)n≤-1in+1(x)Cn-1(x)≤B2(1-α)n.
(
3
) Let x be a nondyadic point such that r(x)>1. Then one can find two subsequences Jn and Jn′ with Jn′/Jn>1 for all n, such that iJn(x)=iJn′+1(x) and ij(x)+iJn(x)=1 for Jn<j<Jn′+1. Furthermore one can find J0>0, A>0, and B>0 such that for n>J0(48)A21-αJn′≤-1iJn+1(x)CJn′-1(x)≤B21-αJn′.
(
4
) Let x be a nondyadic point such that s(x)>1. Then one can find two subsequences Jn and Jn′ with Jn′/Jn>1 for all n, such that ij(x)+ij+1(x)=1 for Jn<j<Jn′ and iJn′(x)=iJn′+1(x). Furthermore one can find J0>0, A>0, and B>0 such that for n>J0(49)A21-αJn′≤-1iJn′(x)CJn′-1(x)≤B21-αJn′.Proof.
Consider the following cases.
Case 1. The idea is very simple since it is a direct computation.
Indeed following (43) we have for y∈]x,x+2-n[(50)Cn-1y=∑j=0n-1‍-1ij+1y2-αj2j=∑j=0N-2‍-1ij+1y2j1-α-2N-11-α+∑j=Nn-1‍2j1-α=∑j=0N-2‍-1ij+1y2j1-α+2N1-α(1-2α-1)+2(N+1)(1-α)2(n-N-1)(1-α)-12(1-α)-1.
The second equation with y∈]x-2-n,x[ can be computed in the same way, up to a change of signs.
Thus one can findN0>N, A>0, and B>0 such that (46) holds for n>N0.
Case 2. It is enough to remark that x has the following binary expansion:
(51)x=kN02N0+∑l=0∞‍122l+1+N0.
As a consequence of Proposition 14, the same kind of computation yields Case 2.
Case 3. Since r(x)>1, for any δ>0 one can find two subsequences Jn and Jn′ such that iJn+1(x)=⋯=iJn′-1(x) and |x-KJn2-Jn|≤2-Jn′ with Jn′=[(r(x)-δ)Jn].
Suppose first eventually up to a small change of definition ofJn′ that iJn(x)=1=iJn′+1(x) and iJn+1(x)=⋯=iJn′-1(x)=iJn′(x)=0.
Then with the same kind of computation as in Case 1 one gets(52)A21-αJn′≤CJn′-1(x)≤B21-αJn′.
In the other case iJn+1(x)=⋯=iJn′-1(x)=iJn′(x)=1, the sign of the slope will be changed.
Case 4. This follows exactly the same ideas than previously. Since s(x)>1 for any δ>0 one can find two subsequences Jn and Jn′ such that for all Jn<j<Jn′ij(x)+ij+1(x)=1 and iJn(x)=iJn-1(x), iJn′(x)=iJn′+1(x). Then with the same kind of computation as in Case 1 one gets
(53)A21-αJn′≤-1iJn′(x)CJn′-1(x)≤B21-αJn′.
Hence Lemma 15 holds.
If we do not have any further information onx, the following lemma will be useful.Lemma 16.
Letx∈[0,1] be a nondyadic number. (1)
Then there existsδ>0 and δ′=1/(21-α-1) such that for all n∈N one can find Jn≥n such that
(54)δ′2Jn(1-α)>CJn-1(x)>δ2Jn(1-α).(2)
Ifx∉S then there exists δ>0 such that for all n≥0 there exists Jn≥n such that (54) holds and
(i)
either[CJn-1x>0  and iJn+1(x)=0],(ii)
or[CJn-1(x)<0  and iJn+1(x)=1].Proof.
(1) The upper bound is a straightforward computation.
Suppose the contrary; that is, for allδ>0 one can find N∈N such that for all n≥N(55)-δ2(1-α)n≤Cn-1(x)≤δ2(1-α)n.
If we suppose without loss of generality that in+1(x)=0 then at step n(56)-δ2(1-α)n≤Cn-1(x)≤δ2(1-α)n-δ2(1-α)n+2(1-α)n≤Cn(x)≤δ2(1-α)n+2(1-α)n-δ21-α+121-α≤Cn(x)2(1-α)(n+1)≤δ21-α+121-α.
It is enough to choose δ such that -(δ/21-α)+(1/21-α)>δ to have a contradiction.
(2) Suppose the contrary; that is, there existsx∉S and that, for all β>0, there exists N∈N, such that for all n≥N(a)
either|Cn-1(x)|≤β2Jn(1-α),(b)
or[Cn-1(x)<0andin+1(x)=0],(c)
or[Cn-1(x)>0andin+1(x)=1].
Remark first that the points ofS satisfy exactly (2b) and (2c). Indeed for x∈S and assuming that in0+1(x)=1, x has a binary expansion (51).
Thus following (43) the slope Cn-1(x) satisfies
(57)Cn-1x=∑j=0N0-1‍-1ij+1x21-αj+∑j=N0n-1‍21-αj-1ij+1x=∑j=0N0-1‍-1ij+1x21-αj+2N0(1-α)-21-αn-N0-11+21-α.
Hence, for n being large enough (2b) and (2c) are satisfied.
Our goal is thus to prove that if we chooseβ being small enough then only (2b) and (2c) can be satisfied, which will lead to the fact that x∈S and thus to a contradiction.
Let us start by the following special cases.
(i) We claim that if one can findk being large enough such that Ck-1(x)=0 then x∈S, which is a contradiction.
Let us prove this claim.
We will need the following sequence: let forn∈N⋆(58)dn=2-1-α∑j=0n‍-1j2-j1-α=2-(1-α)1+2-(1-α)1--1n+12-(n+1)(1-α).
We clearly have dn≥d1>0 for all n≥1.
Chooseβ≤d1/2 and N such that the hypotheses are satisfied.
Suppose thatk≥N+1 is such that Ck-1(x)=0. Then |Ck(x)|=2k(1-α)=2-(1-α)2(k+1)(1-α).
Remark thatβ<d1≤2-(1-α) and thus |Ck(x)|>β2(k+1)(1-α).
Suppose without loss of generality thatCk(x)>0 (the case Ck(x)<0 is symetrical and can be proved in exactly the same way). Thus ik+2(x)=1 and
(59)Ck+1x=-2k+11-α+2k1-α=-2k+21-α2-1-α-2-21-α<-β2(k+2)(1-α).
Let us prove by induction on n that for all n∈N⋆(P)|Ck+n(x)|=dn2(k+n+1)(1-α),-1nCk+n(x)>0.
We just proved that (P) is true for n=1.
Suppose that forn∈N⋆(P) is true. Suppose without loss of generality that Ck+n(x)>0 (the case Ck+n(x)<0 is symetrical and can be proved in exactly the same way). Thus ik+n+2(x)=1 and
(60)Ck+n+1x=Ck+nx-2n+k+11-α=dn2(k+n+1)(1-α)-2n+k+11-α=-(-2-(1-α)dn+2-(1-α))2(k+n+2)(1-α).
Since dn satisfies exactly dn+1=-2-(1-α)dn+2-(1-α), we have the result and property (P) is satisfied at level n+1.
Thus for alln∈N⋆(P) is true. Recall that since β<d1≤dn for all n∈N⋆, this implies that for all n∈N⋆ik+n+2(x)+ik+n+3(x)=1, which is exactly the characterization of the points in S and is in contradiction with the hypothesis x∉S.
In the following we will always keep the hypothesis0<β≤d1/2 so that for n large enough we always have Cn(x)≠0.
(ii) We now consider the case where|Cn-1(x)| is close to the value of β2n(1-α) and prove that this yields that x∈S, thus a contradiction.
Let0<β≤d1/2 and β′>0 whose value will be precised later on.
Supposen≥N is such that β′2n(1-α)>Cn-1(x)>β2n(1-α) (the case Cn-1(x)<-β2n(1-α) can be done exactly in the same way). Then in+1(x)=1 and Cn(x)=Cn-1(x)-2n(1-α); hence
(61)β2n(1-α)-2n1-α≤Cn(x)≤β′2n(1-α)-2n(1-α)β-12-1-α2n+11-α≤Cnx≤β′-12-(1-α)2(n+1)(1-α).
Choose β′ such that (β′-1)2-(1-α)<-β; hence β<β′<(1-β)/2-(1-α), which is possible since β≤d1/2<2-(1-α)/(1+2-(1-α)).
This yieldsin+2(x)=0. Thus
(62)Cn+1x=Cn-1x+-2n1-α+2n+11-α>2-1-α-2-21-α2n+21-α>β2(n+2)(1-α).
Let us prove by induction on k that, for all k∈N,
(Q)-1k+1Cn+k(x)>β2(n+k+1)(1-α).
We just prove that the case k=0 is true.
Suppose one can findk∈N such that (Q) is true for all 0≤k′≤k.
Let us prove that it is true atk+1. Without loss of generality suppose Cn+k(x)>0; thus in+k+2(x)=1.
We have(63)Cn+k+1x=Cn+k-1x+2n+k1-α-2n+k+11-α<-2-1-α-2-21-α2n+k+21-α<-β2(n+k+2)(1-α).
This proves that (Q) is true at k+1.
Thus by induction(Q) is true for all k∈N. This means that for all k∈Nin+k+2(x)+in+k+3(x)=1, and thus x∈S, hence the contradiction.
(iii) We now study the case where|Cn-1(x)|≤β2n(1-α) and prove that if we choose β being small enough then it will lead to x∈S.
Indeed letβ≤inf⁡(d1/2,d1′/2) with
(64)d1′=2-(1-α)-2-2(1-α)1+2-2(1-α).
And suppose n≥N such that |Cn-1(x)|≤β2n(1-α). Suppose in+1(x)=0 without lost of generality. Thus we have
(65)Cnx=Cn-1x+2n1-α,-β+12-1-α2n+11-α≤Cnx≤(β+1)2-(1-α)2(n+1)(1-α).
Remark that with the choice of β we made, we have on one hand β<(-β+1)2-(1-α) and on the other hand β<(β+1)2-(1-α)<(1-β)/2-(1-α). Thus following the previous result using β′=(β+1)2-(1-α), x∈S and we have a contradiction.
(iv) We consider the case whereCn-1(x)>0 and Cn(x)<0 for n being large enough under the previous range of values of β.
Letβ≤inf⁡(d1/2,d1′/2) (recall that d1 is defined by (58) and d1′ by (64)).
And suppose that forn≥N we have Cn-1(x)>0 and Cn(x)<0.
Following the previous case we haveCn-1(x)>β2n(1-α) and Cn(x)<-β2(n+1)(1-α). Thus in+1(x)=1 and in+2(x)=0.
Then(66)Cn+1x=Cn-1x-2n1-α+2n+11-α>β2(n+2)(1-α)
since by definition of β and d1 we have 2-(1-α)-2-2(1-α)≥d1>β. Thus in+3(x)=1.
We haveCn+2(x)=Cn(x)+2(n+1)(1-α)-2(n+2)(1-α)<-β2(n+3)(1-α).
A proof by induction exactly in the same way as previously proved yields that fork≥0 we have Cn+2k+1(x)>0 and Cn+2k(x)<0; thus in+2k+2(x)+in+2k+3(x)=1 for all k∈N and we have x∈S, hence a contradiction.
We will now go for the main proof, taking into account what we just proved.
In the following we will considerδ>0 and Jn defined as in Point 1, β=inf⁡(d1/2,δ/2,d1′/2), and n such that Jn≥N. Thus for all n∈N|CJn-1(x)|/2Jn(1-α)>δ>β.
SupposeCJn-1(x)>0. This means that CJn(x)=CJn-1(x)-2Jn(1-α)<CJn-1(x). The only case we want to consider is CJn(x)>β2(Jn+1)(1-α) since for all the other cases the previous points yield x∈S.
ThusiJn+2(x)=1. It is clear that one can find k∈N such that for all 0≤k′≤kCJn+k′(x)>β2(Jn+k′+1)(1-α) and iJn+k′+2(x)=1 and CJn+k+1(x)<β2(Jn+k+2)(1-α).
Hence either|CJn+k+1(x)|≤β2(Jn+k+2)(1-α) and x∈S or CJn+k+1(x)<-β2(Jn+k+2)(1-α) and since CJn+k(x)>β2(Jn+k+1)(1-α) we get the desired result.
In all cases we proved that Points (2a), (2b), and (2c) lead tox∈S, which is a contradiction. Hence Lemma 16 holds.


## 3.1. Lemmas for Practical Computation of the Exponents

From the computation of the weak accessibility exponent inΩ and Ωc it is easy to derive the p-exponent. In [13], Jaffard and Mélot proved that IΩ∈Tα/pp(X0) if and only if either X0 is weak α-accessible in Ω or X0 is weak α-accessible in Ωc. As a consequence we have
(12)puIΩp(X0)=max⁡EΩwX0,EΩcwX0.We will also need the following lemma.Lemma 9.
Letf:R→R be in Cα(x0) with 0<α<1 and Ω the domain below (resp., above) the graph of f. Consider X0=(x0,f(x0)). Then X0 is strong (1/α)-1, accessible in both Ω and Ωc.Proof.
Suppose thatΩ is the domain below the graph of f. Without any loss of generality, we can assume that X0=(0,0). Let r>0. Since f is in Cα(0) and 0<α<1 then there exists a constant C≥0 such that in neighborhood of 0(13)fx≤Cxα.
Thus
(14)-Cxα≤fx≤Cxα.
Obviously meas⁡(Ωc⋂B(X0,r)) (resp., meas⁡(Ω⋂B(X0,r))) is greater than the area A=C′∫0r‍y1/αdy=C′′r1+1/α above (resp., below) the graph of x↦C|x|α and below (resp., above) the square of side r and center X0.
The same results hold ifΩ is the domain above the graph of f (we have just to replace Ω by Ωc).One of our goals for the points which are not extrema ofF will be to find sequences of local maxima or minima such that the following key lemma holds.Lemma 10.
Letf:R→R be in Cα(R) and Ω the domain below the graph of f. Consider X0=(x0,f(x0)). Suppose that there exists cα>0, a sequence rn of positive numbers, such that rn→0 as n→+∞, xn∈]x0-rn,x0+rn[, and n0∈N, such that
(15)∀n≥n0f(xn)=f(x0)-cαrnα.
Then EΩcw(X0)=0.Proof.
We can suppose thatxn∈]x0-rn,x0[(the case xn∈]x0,x0+rn[ is similar). Then by the mean value theorem we can find bn∈]xn,x0[ such that f(bn)=f(x0)-rn. Let bn′=inf⁡{bn∈]xn,x0[;f(bn)=f(x0)-rn}. Since f is continuous we get f(bn′)=f(x0)-rn and bn′∈]xn,x0[. It follows from the definition of bn′ and the mean value theorem that
(16)∀t∈xn,bn′f(t)<f(x0)-rn.
Thus
(17)xn,bn′×[f(x0)-rn,f(x0)+rn]⊂B(X0,rn)⋂‍Ωc.
Therefore
(18)meas⁡BX0,rn⋂‍Ωc≥2bn′-xnrn.
Since
(19)fxn-fbn′=fx0-cαrnα-fx0-rn=rn-cαrnα≃-rnα,
in the sense that there exists a constant C≥1 such that for every n we have (1/C)rnα≤cαrnα-rn≤Crnα.
Sincef belongs to Cα(R) we get
(20)Cbn′-xnα≥fbn′-fxn≃rnα.
Thus
(21)bn′-xn≥C′rn.
Following (18) and (21) we get
(22)meas⁡BX0,rn⋂‍Ωc≥Crn2.
Since meas⁡(B(X0,r)⋂Ωc)≤meas⁡(B(X0,r))≤r2 for all r≥0 we get
(23)lim⁡n→+∞log⁡(meas⁡(B(X0,rn)⋂‍Ωc))log⁡(rn)=2.
Thus thanks to Proposition 6 we have EΩcw(X0)≤0 which yields EΩcw(X0)=0.
By replacingf by -f we also have the following result.Lemma 11.
Letf:R→R be in Cα(R) and Ω the domain below the graph of f. Consider X0=(x0,f(x0)). Suppose that there exist cα>0, rn→0 as n→+∞, xn∈]x0-rn,x0+rn[, and n0∈N, such that
(24)∀n≥n0f(xn)=f(x0)+cαrnα.
Then EΩw(X0)=0.


## 3.2. Dyadic Expansions and Approximation by Dyadics

We give some properties of the approximation of a point by the dyadics. Such properties will be used later.Letx∈[0,1]. Set i1(x),…,ij(x),…, the binary digits of x; that is,
(25)x=∑l=1∞‍il(x)2l.(i)
Note that dyadic points, that is, pointsx=2-NK with K∈2N+1, are characterized by the fact that one can find N>0 such that iN(x)=1 and in(x)=0 for n>N or equivalently iN(x)=0 and in(x)=1 for n>N.Furthermore forn>N the number x-2-n is dyadic. Since 2-n=∑j=n+1∞‍2-j, then x-2-n=(∑j=1N-1‍ij2-j)+2-(N+1)+2-(N+2)+⋯+2-n.On the other handx+2-n has the simple expansion ∑j=1N(ijx/2j)+(1/2n).We will denote byD the set of all dyadic points in [0,1].(ii)
Let us come back to the general case withx any point in [0,1].For eachj∈N, define Kj (=Kj(x)) by
(26)Kj2-j-x=inf⁡k∈Nk2-j-x.Set(27)rj(x)=log⁡|Kj2-j-x|log⁡2-j.Define the rate of approximation ofx by dyadics as
(28)r(x)=limsup⁡j↦∞rj(x).Since|Kj2-j-x|≤2-j, then, for every x, we have r(x)≥1. If x is dyadic then r(x)=∞ (by taking the convention log⁡0=-∞). If x is normal (i.e., the frequency of ones (or zeros) in the binary expansion of x is equal to 1/2) then r(x)=1.(iii)
Ifr(x)>1, following the definition of r(x), then for any δ>0 such that r(x)-δ>1 one can find a subsequence Jn→+∞ for n→+∞ such that
(29)rJn(x)≤2-Jn(r(x)-δ).LetJn′=[Jn(r(x)-δ)]. We then have
(30)KJn2-Jn-2-Jn′≤x≤KJn2-Jn+2-Jn′.Thus, eitherx belongs to the dyadic interval [KJn2-Jn-2-Jn′,KJn2-Jn] and in this case it satisfies iJn+1(x)=⋯=iJn′-1(x)=1, or it belongs to the other interval [KJn2-Jn,KJn2-Jn+2-Jn′] and in this case it satisfies iJn+1(x)=⋯=iJn′-1(x)=0.In both cases let us notice that the binary expansion ofx contains chains of 0 or 1 whose length Jn′-Jn~Jn′ increases when n→+∞.


## 3.3. Approximation by Sequences of Maxima ofF

We will see in the following that points in[0,1] of the set
(31)S=k∈N,N0∈N,k2N0+1312N0,k2N0+2312N0
will play a big role in this work, since they actually are the locations of the local maxima of the function F (see below). Remark that they are characterized by the fact that, for each x∈S, one can find j0∈N such that for j≥j0 we have ij(x)+ij+1(x)=1.As in the case of dyadic approximation we can define a rate of approximation by this kind of points.Indeed let forx∈[0,1](32)|mj-x|=inf⁡k∈Nk2j+1312j-x,k2j+2312j-x.
Define
(33)sj(x)=log⁡mj-xlog⁡2-j.
Then the rate of approximation of x by elements of S is given by
(34)s(x)=limsup⁡j↦∞sjx.
Since |mj-x|≤2-j, then, for every x, we have s(x)≥1.In the case of dyadic numbers, we haves(x)=1. But remark that in other nontrivial cases there is no obvious relationship between s(x) and r(x). Indeed one can check on the following examples that s and r can independently take any value.(i)
Letx=∑j=1+∞(ij(x)/2j) with i3k+1(x)=0=i3k+2(x) and i3k+2(x)=1 for all k∈N. Then we have r(x)=s(x)=1.(ii)
Letu>1. Then x=∑n=0+∞2-[un] with [un] the integer part of un. We have r(x)=u whereas s(x)=1.(iii)
Letu>1. Then x=∑n=1+∞2-2n-∑n=0+∞2-2[un]. We have s(x)=u whereas r(x)=1.(iv)
Letu>1 and s>1. Let x=∑n=1+∞2-2[sn-1un]+∑k=[sn-1un]+1[snun+1]-12-2k. Then r(x)=u and s(x)=s.


## 3.4. The Shift Operator

Since0<α<1 it is easy to check that we obtain from (2) with a=2-α and b=2(35)Fx=∑j=0∞‍∑i=(i1,…,ij)∈{0,1}j‍2-αjΛ2jx-2j-1i1-⋯-2ij-1-ij.
The term of (35) corresponding to j=0 is Λ(x). But, the function Λ is supported in [0,1], and therefore F vanishes outside [0,1] and for x∈[0,1](36)Fx=∑j=0∞‍2-αjΛ2jx-2j-1i1x-⋯-2ij-1x-ijx.
For dyadic rationals x, x=2-NK with K∈2N+1, as we already said it, there exist two binary expansions, one such that iN(x)=1 and in(x)=0 for n>N and another one such that iN(x)=0 and in(x)=1 for n>N. The two right-hand sides of (36) corresponding to the two choices of i(x) give identical results.Denote byτ the shift operator
(37)τx=∑l=2∞‍il(x)2-l+1=∑l=1∞‍il+1(x)2-l.
Observe that
(38)τx=2x,ifx∈0,12,2x-1,ifx∈12,1.
Hence
(39)Fx=∑j=0∞‍2-αjΛτjx,(40)τjx=∑l=1∞‍il+j(x)2-l.Our self-similar function is of the formF(x)=∑j=0∞‍∑k=02j-1‍Cj,kΛ(2jx-k) with
(41)Cj,k=2-αjifj≠0,k≠0,C0,0=1.Forn≥1, denote
(42)Fn(x)=∑j=0n‍2-αjΛτjx.
Remark that Fn is affine on intervals of type In+1=k/2n+1,(k+1)/2n+1. Remark also that if t∈[0,1], then Λ(t)=(-1)i1(t)t+i1(t). So, if t′∈[0,1] and i1(t)=i1(t′), then Λ(t)-Λ(t′)=(-1)i1t(t-t′).It follows that ifk/2n+1=∑j=1n+1‍(ij/2j) then the slope of Fn at any point x of the interval ]k/2n+1,(k+1)/2n+1[ is exactly
(43)Cn=Cn(x)=∑j=0n‍-1ij+1(x)2(1-α)j=∑j=0n‍-1ij+12(1-α)j.


## 3.5. Extrema ofF

We will need the following characterization of the extrema ofF proved in [16, 17]. Let us start with the local and global minima.Proposition 12.
Let0<α<1 and let F be the function defined by (39); then (i)
0 and 1 are the abscissas of the global minima ofF;(ii)
the dyadic points are the abscissas of the minima ofF and furthermore
(44)min⁡x∈INF(x)=min⁡FN-1k2N,FN-1k+12NwithIN=k/2N,(k+1)/2N.In the case of the maxima, the statement of the result is slightly more technical. We need the following proposition of [17] using the same notations used previously.Proposition 13.
Let0<α<1 and let F be the function defined by (39). Let t=21-α and X(p) the list of positions where F(x)+px attains its maximum on [0,1]. Let M(p) be the maximum on [0,1] of F(x)+px. Then (i)
X-tN-1/(t-1)={(1/3)(1/2N),(2/3)(1/2N)} for N=0,1,….(ii)
Xp={(1/3)(1/2N)} if -(tN+1-1)/(t-1)<p<-(tN-1)/(t-1).(iii)
X(p)=1-X(-p) for all p≠0.(iv)
max⁡x∈IN⁡F(x)=FN-1(k/2N)+2-NαM(CN-1((2k+1)/2N+1)/tN).The following proposition is a consequence of the previous one.Proposition 14.
Let0<α<1 and let F be the function defined by (39). Then (i)
1/3 and 2/3 are the abscissas of the global maxima of F;(ii)
the abscissas of the local maxima ofF are the points of S.


## 3.6. Approximation of Slopes ofFn

Suppose first we have some information about the dyadic expansion ofx. Then we have the following lemma, which helps to control the behavior of the slopes of the affine function Fn-1.Lemma 15.
(
1
) Let x=K/2N be a dyadic number. Then one can find N0>N, A>0, and B>0 depending only on x such that if n≥N0 then
(45)∀y∈x,x+2-nA21-αn≤Cn-1y≤B21-αn,(46)∀y∈x-2-n,x-A2(1-α)n≥Cn-1(y)≥-B2(1-α)n.
(
2
) Let x be the abscissa of a local maximum of F. Then one can find J0>0, A>0, and B>0 such that for n≥J0(47)in-1(x)+in(x)=1,Cn-1(x)Cn(x)<0,A2(1-α)n≤-1in+1(x)Cn-1(x)≤B2(1-α)n.
(
3
) Let x be a nondyadic point such that r(x)>1. Then one can find two subsequences Jn and Jn′ with Jn′/Jn>1 for all n, such that iJn(x)=iJn′+1(x) and ij(x)+iJn(x)=1 for Jn<j<Jn′+1. Furthermore one can find J0>0, A>0, and B>0 such that for n>J0(48)A21-αJn′≤-1iJn+1(x)CJn′-1(x)≤B21-αJn′.
(
4
) Let x be a nondyadic point such that s(x)>1. Then one can find two subsequences Jn and Jn′ with Jn′/Jn>1 for all n, such that ij(x)+ij+1(x)=1 for Jn<j<Jn′ and iJn′(x)=iJn′+1(x). Furthermore one can find J0>0, A>0, and B>0 such that for n>J0(49)A21-αJn′≤-1iJn′(x)CJn′-1(x)≤B21-αJn′.Proof.
Consider the following cases.
Case 1. The idea is very simple since it is a direct computation.
Indeed following (43) we have for y∈]x,x+2-n[(50)Cn-1y=∑j=0n-1‍-1ij+1y2-αj2j=∑j=0N-2‍-1ij+1y2j1-α-2N-11-α+∑j=Nn-1‍2j1-α=∑j=0N-2‍-1ij+1y2j1-α+2N1-α(1-2α-1)+2(N+1)(1-α)2(n-N-1)(1-α)-12(1-α)-1.
The second equation with y∈]x-2-n,x[ can be computed in the same way, up to a change of signs.
Thus one can findN0>N, A>0, and B>0 such that (46) holds for n>N0.
Case 2. It is enough to remark that x has the following binary expansion:
(51)x=kN02N0+∑l=0∞‍122l+1+N0.
As a consequence of Proposition 14, the same kind of computation yields Case 2.
Case 3. Since r(x)>1, for any δ>0 one can find two subsequences Jn and Jn′ such that iJn+1(x)=⋯=iJn′-1(x) and |x-KJn2-Jn|≤2-Jn′ with Jn′=[(r(x)-δ)Jn].
Suppose first eventually up to a small change of definition ofJn′ that iJn(x)=1=iJn′+1(x) and iJn+1(x)=⋯=iJn′-1(x)=iJn′(x)=0.
Then with the same kind of computation as in Case 1 one gets(52)A21-αJn′≤CJn′-1(x)≤B21-αJn′.
In the other case iJn+1(x)=⋯=iJn′-1(x)=iJn′(x)=1, the sign of the slope will be changed.
Case 4. This follows exactly the same ideas than previously. Since s(x)>1 for any δ>0 one can find two subsequences Jn and Jn′ such that for all Jn<j<Jn′ij(x)+ij+1(x)=1 and iJn(x)=iJn-1(x), iJn′(x)=iJn′+1(x). Then with the same kind of computation as in Case 1 one gets
(53)A21-αJn′≤-1iJn′(x)CJn′-1(x)≤B21-αJn′.
Hence Lemma 15 holds.
If we do not have any further information onx, the following lemma will be useful.Lemma 16.
Letx∈[0,1] be a nondyadic number. (1)
Then there existsδ>0 and δ′=1/(21-α-1) such that for all n∈N one can find Jn≥n such that
(54)δ′2Jn(1-α)>CJn-1(x)>δ2Jn(1-α).(2)
Ifx∉S then there exists δ>0 such that for all n≥0 there exists Jn≥n such that (54) holds and
(i)
either[CJn-1x>0  and iJn+1(x)=0],(ii)
or[CJn-1(x)<0  and iJn+1(x)=1].Proof.
(1) The upper bound is a straightforward computation.
Suppose the contrary; that is, for allδ>0 one can find N∈N such that for all n≥N(55)-δ2(1-α)n≤Cn-1(x)≤δ2(1-α)n.
If we suppose without loss of generality that in+1(x)=0 then at step n(56)-δ2(1-α)n≤Cn-1(x)≤δ2(1-α)n-δ2(1-α)n+2(1-α)n≤Cn(x)≤δ2(1-α)n+2(1-α)n-δ21-α+121-α≤Cn(x)2(1-α)(n+1)≤δ21-α+121-α.
It is enough to choose δ such that -(δ/21-α)+(1/21-α)>δ to have a contradiction.
(2) Suppose the contrary; that is, there existsx∉S and that, for all β>0, there exists N∈N, such that for all n≥N(a)
either|Cn-1(x)|≤β2Jn(1-α),(b)
or[Cn-1(x)<0andin+1(x)=0],(c)
or[Cn-1(x)>0andin+1(x)=1].
Remark first that the points ofS satisfy exactly (2b) and (2c). Indeed for x∈S and assuming that in0+1(x)=1, x has a binary expansion (51).
Thus following (43) the slope Cn-1(x) satisfies
(57)Cn-1x=∑j=0N0-1‍-1ij+1x21-αj+∑j=N0n-1‍21-αj-1ij+1x=∑j=0N0-1‍-1ij+1x21-αj+2N0(1-α)-21-αn-N0-11+21-α.
Hence, for n being large enough (2b) and (2c) are satisfied.
Our goal is thus to prove that if we chooseβ being small enough then only (2b) and (2c) can be satisfied, which will lead to the fact that x∈S and thus to a contradiction.
Let us start by the following special cases.
(i) We claim that if one can findk being large enough such that Ck-1(x)=0 then x∈S, which is a contradiction.
Let us prove this claim.
We will need the following sequence: let forn∈N⋆(58)dn=2-1-α∑j=0n‍-1j2-j1-α=2-(1-α)1+2-(1-α)1--1n+12-(n+1)(1-α).
We clearly have dn≥d1>0 for all n≥1.
Chooseβ≤d1/2 and N such that the hypotheses are satisfied.
Suppose thatk≥N+1 is such that Ck-1(x)=0. Then |Ck(x)|=2k(1-α)=2-(1-α)2(k+1)(1-α).
Remark thatβ<d1≤2-(1-α) and thus |Ck(x)|>β2(k+1)(1-α).
Suppose without loss of generality thatCk(x)>0 (the case Ck(x)<0 is symetrical and can be proved in exactly the same way). Thus ik+2(x)=1 and
(59)Ck+1x=-2k+11-α+2k1-α=-2k+21-α2-1-α-2-21-α<-β2(k+2)(1-α).
Let us prove by induction on n that for all n∈N⋆(P)|Ck+n(x)|=dn2(k+n+1)(1-α),-1nCk+n(x)>0.
We just proved that (P) is true for n=1.
Suppose that forn∈N⋆(P) is true. Suppose without loss of generality that Ck+n(x)>0 (the case Ck+n(x)<0 is symetrical and can be proved in exactly the same way). Thus ik+n+2(x)=1 and
(60)Ck+n+1x=Ck+nx-2n+k+11-α=dn2(k+n+1)(1-α)-2n+k+11-α=-(-2-(1-α)dn+2-(1-α))2(k+n+2)(1-α).
Since dn satisfies exactly dn+1=-2-(1-α)dn+2-(1-α), we have the result and property (P) is satisfied at level n+1.
Thus for alln∈N⋆(P) is true. Recall that since β<d1≤dn for all n∈N⋆, this implies that for all n∈N⋆ik+n+2(x)+ik+n+3(x)=1, which is exactly the characterization of the points in S and is in contradiction with the hypothesis x∉S.
In the following we will always keep the hypothesis0<β≤d1/2 so that for n large enough we always have Cn(x)≠0.
(ii) We now consider the case where|Cn-1(x)| is close to the value of β2n(1-α) and prove that this yields that x∈S, thus a contradiction.
Let0<β≤d1/2 and β′>0 whose value will be precised later on.
Supposen≥N is such that β′2n(1-α)>Cn-1(x)>β2n(1-α) (the case Cn-1(x)<-β2n(1-α) can be done exactly in the same way). Then in+1(x)=1 and Cn(x)=Cn-1(x)-2n(1-α); hence
(61)β2n(1-α)-2n1-α≤Cn(x)≤β′2n(1-α)-2n(1-α)β-12-1-α2n+11-α≤Cnx≤β′-12-(1-α)2(n+1)(1-α).
Choose β′ such that (β′-1)2-(1-α)<-β; hence β<β′<(1-β)/2-(1-α), which is possible since β≤d1/2<2-(1-α)/(1+2-(1-α)).
This yieldsin+2(x)=0. Thus
(62)Cn+1x=Cn-1x+-2n1-α+2n+11-α>2-1-α-2-21-α2n+21-α>β2(n+2)(1-α).
Let us prove by induction on k that, for all k∈N,
(Q)-1k+1Cn+k(x)>β2(n+k+1)(1-α).
We just prove that the case k=0 is true.
Suppose one can findk∈N such that (Q) is true for all 0≤k′≤k.
Let us prove that it is true atk+1. Without loss of generality suppose Cn+k(x)>0; thus in+k+2(x)=1.
We have(63)Cn+k+1x=Cn+k-1x+2n+k1-α-2n+k+11-α<-2-1-α-2-21-α2n+k+21-α<-β2(n+k+2)(1-α).
This proves that (Q) is true at k+1.
Thus by induction(Q) is true for all k∈N. This means that for all k∈Nin+k+2(x)+in+k+3(x)=1, and thus x∈S, hence the contradiction.
(iii) We now study the case where|Cn-1(x)|≤β2n(1-α) and prove that if we choose β being small enough then it will lead to x∈S.
Indeed letβ≤inf⁡(d1/2,d1′/2) with
(64)d1′=2-(1-α)-2-2(1-α)1+2-2(1-α).
And suppose n≥N such that |Cn-1(x)|≤β2n(1-α). Suppose in+1(x)=0 without lost of generality. Thus we have
(65)Cnx=Cn-1x+2n1-α,-β+12-1-α2n+11-α≤Cnx≤(β+1)2-(1-α)2(n+1)(1-α).
Remark that with the choice of β we made, we have on one hand β<(-β+1)2-(1-α) and on the other hand β<(β+1)2-(1-α)<(1-β)/2-(1-α). Thus following the previous result using β′=(β+1)2-(1-α), x∈S and we have a contradiction.
(iv) We consider the case whereCn-1(x)>0 and Cn(x)<0 for n being large enough under the previous range of values of β.
Letβ≤inf⁡(d1/2,d1′/2) (recall that d1 is defined by (58) and d1′ by (64)).
And suppose that forn≥N we have Cn-1(x)>0 and Cn(x)<0.
Following the previous case we haveCn-1(x)>β2n(1-α) and Cn(x)<-β2(n+1)(1-α). Thus in+1(x)=1 and in+2(x)=0.
Then(66)Cn+1x=Cn-1x-2n1-α+2n+11-α>β2(n+2)(1-α)
since by definition of β and d1 we have 2-(1-α)-2-2(1-α)≥d1>β. Thus in+3(x)=1.
We haveCn+2(x)=Cn(x)+2(n+1)(1-α)-2(n+2)(1-α)<-β2(n+3)(1-α).
A proof by induction exactly in the same way as previously proved yields that fork≥0 we have Cn+2k+1(x)>0 and Cn+2k(x)<0; thus in+2k+2(x)+in+2k+3(x)=1 for all k∈N and we have x∈S, hence a contradiction.
We will now go for the main proof, taking into account what we just proved.
In the following we will considerδ>0 and Jn defined as in Point 1, β=inf⁡(d1/2,δ/2,d1′/2), and n such that Jn≥N. Thus for all n∈N|CJn-1(x)|/2Jn(1-α)>δ>β.
SupposeCJn-1(x)>0. This means that CJn(x)=CJn-1(x)-2Jn(1-α)<CJn-1(x). The only case we want to consider is CJn(x)>β2(Jn+1)(1-α) since for all the other cases the previous points yield x∈S.
ThusiJn+2(x)=1. It is clear that one can find k∈N such that for all 0≤k′≤kCJn+k′(x)>β2(Jn+k′+1)(1-α) and iJn+k′+2(x)=1 and CJn+k+1(x)<β2(Jn+k+2)(1-α).
Hence either|CJn+k+1(x)|≤β2(Jn+k+2)(1-α) and x∈S or CJn+k+1(x)<-β2(Jn+k+2)(1-α) and since CJn+k(x)>β2(Jn+k+1)(1-α) we get the desired result.
In all cases we proved that Points (2a), (2b), and (2c) lead tox∈S, which is a contradiction. Hence Lemma 16 holds.


## 4. Computation of Weak and Strong Accessible Exponents

### 4.1. Case of Dyadic Points

We will prove the following proposition.Proposition 17.
Ifx is a dyadic point, and X=(x,F(x)) then
(67)EΩcw(X)=1α-1,EΩw(X)=0,EΩcs(X)=1α-1,EΩs(X)=0,ufpX=1p1α-1.Ifx is a dyadic point, that is, x=2-NK with K∈2N+1, we consider its binary expansion in which iN(x)=1 and in(x)=0 for n>N. For n>N the number x-2-n is dyadic. Since 2-n=∑j=n+1∞‍2-j then x-2-n=(∑j=1N-1‍ij2-j)+2-(N+1)+2-(N+2)+⋯+2-n. On the other hand x+2-n has the simple expansion (∑j=1N(ijx/2j))+(1/2n).Remark thatFN-1(x)=Fn-1(x)=F(x) for n>N and Fn-1(x+2-n)=F(x+2-n).Any pointy in the interval ]x,x+2-n[ satisfies the expansion iN(y)=iN+1(y)=⋯=in-1(y)=in(y)=0.It follows that(68)Fx+2-n-Fx=Fn-1x+2-n-Fn-1x=2-nCn-1(y)
with y any of the points of the interval ]x,x+2-n[.Following Lemma15 and Case 1 there exist two constants A>0 and B>0 and J0≥N (which depend only on the given dyadic point x) such that
(69)∀n≥J0A2-αn≤F(x+2-n)-F(x)≤B2-αn.
Thus we have F(x+2-n)-F(x)≥A.2-αn.On the other hand, following remarks of Section3.2, for any y∈]x-2-n,x[ we have iN(y)=1=⋯=in(y). Thus
(70)Fx-2-n-Fx=Fn-1x-2-n-Fn-1x=-Cn-1(y)2-n.Whence, following Lemma15 and Case 1 we have for n≥J0(71)∀n≥J0A2-αn≤-F(x-2-n)+F(x)≤B2-αn.Letρ>0 and J≥J0 such that 2-J-1≤ρ≤2-J.SinceF≥FJ, then Ωj⊂Ω, where ΩJ is the domain below the graph of FJ. So
(72)meas⁡(B(X,ρ)∩Ωc)≤meas⁡(B(X,ρ)∩ΩJc).Butmeas⁡(B(X,ρ)∩ΩJc) is smaller than the area hb/2 of a triangle with altitude h issued from X and a corresponding hypotenuse b (see Figures 1 and 2).Figure 1
Overview of the function.Figure 2
Zoom at the pointx=1/22.Clearly, we can takeh~2-J. On the other hand, if we write 2-(j+1)<b/2<2-j with j≥J, then, using properties (69) and (71) (in which we replace nα by J), we get b/2~2-J/α. Since α<1, (69) and (71) are valid with n=J/α≥J0.Whence(73)meas⁡(B(X,ρ)∩Ωc)≤Cρ1+(1/α).
We conclude that
(74)EΩcw(X)≥1α-1.SinceEΩcw(X)≤EΩcs(X)≤(1/α)-1 this yields
(75)EΩcs(X)=1α-1.Sincef=1Ω then
(76)ufpX≥1p1α-1.Sincemeas⁡(B(X,r))=meas⁡(B(X,r)⋂Ω)+meas⁡(B(X,r)⋂Ωc) we get EΩs(X)=0; hence EΩw(X)=0.Whence Proposition17 holds.


### 4.2. Case of a Local Maximum ofF

We will prove the following proposition.Proposition 18.
LetX=(x,F(x)).
Ifx is a local maximum,
(77)EΩw(X)=1α-1,EΩcw(X)=0,(78)EΩs(X)=1α-1,EΩcs(X)=0,(79)ufp(X)=1p1α-1.Letx be a local maximum of F. There is an interval I containing x such that for all x′∈I, F(x)≥F(x′). Let N be such that the dyadic interval kN/2N,(kN+1)/2N which contains x is contained in I.Following Proposition14, we know that x has the binary expansion (51); that is, x=(kN0/2N0)+∑l=0∞(1/22l+1+N0).As a consequence of Lemma15, and following Case (2), one can find J0 and two constants A and B such that for n≥J0  (47) holds. Remark that it implies clearly that for n≥J0in(x)=1 if n is odd, and in(x)=0 if n is even.Our goal now is to evaluateF(x)-F(x′) with x′ in the interval kn/2n,(kn+1)/2n⊂kN/2N,kN+1/2N and x′≠x. If x′ is a dyadic then we take its expansion of type ij(x′)=0 for j to be large enough.Letm≥n be the smallest integer such that im(x)=im(x′) and im+1(x)≠im+1(x′). To fix the ideas, suppose that im+1(x)=1 and im+1(x′)=0. Thus
(80)132m-1≥x-x′≥12m+1+12m+3-∑j=m+2∞‍12j≥12m+3.Sinceim(x)=0 we have Cm-1(x)>0.Thus(81)Fx-Fx′=Cm-1xx-x′+∑k=m+∞‍2-kαΛτkx-Λτkx′.We have(82)A2m1-α2-m-3=C12-mα≤Cm-1xx-x′≤B2m1-α2-m-13=C22-mα.Since fork≥m-1 we have Λ(τkx)=1/3, this yields
(83)0≤∑k=m+∞‍2-kαΛτkx-Λτkx′≤C32-mα.Thus we have(84)C12-mα≤F(x)-Fx′≤C3+C22-mα.Let us compute the weak and strong exponents atx.Letρ and J≥J0 such that 2-J-1≤ρ≤2-J. Thus obviously
(85)meas⁡(B(X,ρ)⋂‍Ω)≤meas⁡(B(X,2-j)⋂‍Ω).Remark first that if(x′,y′)∈B(X,2-J)⋂Ω then |x-x′|<2-J, |y′-F(x)|<2-J, and y′≤F(x′). Since x is a local maximum on the interval kJ/2J,(kJ+1)/2J, then y′≤F(x′)≤F(x) and so 0≤F(x)-F(x′)<2-J. Hence (x′,F(x′))∈B(X,2-J)⋂Ω.Furthermore sinceF(x′) satisfies 0≤F(x)-F(x′)≤2-J, and following (84)  x′ belongs to [x-C2-J/α,x+C2-J/α] with C depending only on C3+C2.ThusB(X,2-J)⋂Ω is contained in a rectangle of length 2-J and width C2-J/α.This yields(86)meas⁡(B(X,ρ)⋂‍Ω)≤C2-J(1+(1/α))≤C′ρ1+(1/α).
We can conclude that
(87)EΩw(X)≥1α-1.
Since EΩw(X)≤EΩs(X)≤(1/α)-1 this yields
(88)EΩw(X)=EΩs(X)=1α-1.Sincemeas⁡(B(X,ρ))=meas⁡(B(X,ρ)⋂Ω)+meas⁡(B(X,ρ)⋂Ωc) we get
(89)EΩcs(X)=EΩcw(X)=0.And finally(90)ufp(X)=1p1α-1.


### 4.3. Case ofx∉D⋃S

Ifx∉D⋃S then we will compute separately the weak and strong exponents. We will first prove that for any point x in [0,1] which is not a maximum or a minimum of F the two weak exponents vanish.Proposition 19.
Letx∉D⋃S and X=(x,f(x)). Then EΩw(X)=EΩcw(X)=0.Proof.
We will prove first that we always haveEΩcw(X)=0 but will separate the proofs in cases r(x)>1 and r(x)=1. Then we will prove that EΩw(X)=0 and prove it separately for s(x)>1 and s(x)=1.
(i) Case
r
(
x
)
>
1
. We follow the notations of Case  3 of Lemma 15; that is, one can find two subsequences Jn and Jn′ such that Jn′/Jn>1 and iJn(x)=iJn′+1(x), ij(x)+iJn(x)=1 for Jn<j<Jn′+1. Suppose without lose of generality that iJn+1(x)=0. Let x~n=KJn/2Jn=∑j=1Jnij(x)/2j. Thus we have
(91)2-Jn′-1≤x-x~n≤2-Jn′.
Since Case  3 of Lemma 15 holds, we get
(92)A21-αJn′2-Jn′-1≤FJn′-1x-FJn′-1x~n≤B21-αJn′2-Jn′,A′2-αJn′≤FJn′-1(x)-FJn′-1(x~n)≤B′2-αJn′.
We have F(x)=FJn′-1(x)+∑k=Jn′+∞2-kαΛ(τkx) and FJn′-1(x~n)=F(x~n). Clearly ∑k=Jn′+∞2-kαΛ(τkx)≥0. Thus following (92) we have
(93)A21-αJn′2-Jn′-1≤Fx-Fx~n≤B21-αJn′2-Jn′+2-Jn′α∑k=0+∞‍2-kαΛτk+Jn′xA′2-αJn′≤F(x)-F(x~n)≤B′2-Jn′+2-αJn′F(τJn′x)A′2-αJn′≤F(x)-F(x~n)≤B′2-Jn′+2-αJn′FτJn′13A′2-αJn′≤F(x)-F(x~n)≤C2-αJn′
because the global maxima of F are reached at points 1/3 and 2/3.
We can now apply the mean value theorem and get that for eachn≥J0 we can find yn∈]x-2-Jn′,x+2-Jn′[ such that F(x)-F(yn)=A′·2-αn/2.
Thus using Lemma10 we can conclude that EΩcw(X)=0.
(ii) Case
r
(
x
)
=
1
. Let Jn be defined just as in Lemma 16, that is, that one can find δ>0, δ′>0, and Jn such that (54) is satisfied.
Following the definition ofr(x), for all γ>0, there exists n0 such that for all j≥Jn0Kj2-j-x>2-j(1+γ). Thus in particular for all n≥n0 we have
(94)2-Jn>KJn2-Jn-x>2-Jn(1+γ).
Suppose on one hand CJn-1(x)≥0. Then choose x~n=KJn2-Jn if x∈]KJn2-Jn,2-Jn+KJn2-Jn[(resp.x~n=KJn2-Jn-2-Jn if x∈]KJn2-Jn-2-Jn,KJn2-Jn[).
We have obviously(95)FJn-1(x)-FJn-1(x~n)=CJn-1(x)(x-x~n)≥0.
If we suppose on the other hand CJn-1(x)≤0, then we can choose in the same way a dyadic number x~n=k/2Jn such that
(96)FJn-1(x)-FJn-1(x~n)=CJn-1(x)(x-x~n)≥0.
Together with (54) this yields in any of these cases that
(97)δ′2-αJn≥FJn-1x-FJn-1x~n≥δ2-Jn1+γ21-αJn,δ′2-αJn≥FJn-1(x)-FJn-1(x~n)≥δ2-Jn(1+γ)2(1-α)Jn.
Since ∑k=Jn′+∞2-kαΛ(τkx)≥0 and FJn-1(x~n)=F(x~n) we get
(98)δ′2-αJn+2-αJnF13≥Fx-Fx~n≥FJn-1x-FJn-1x~n≥δ2-Jn1+γ21-αJn,C2-αJn≥F(x)-F(x~n)≥δ2-Jn(γ+α).
To get EΩcw(X) we only have to adapt the proof of Lemma 10 to the case rn=2-Jn. Suppose without loss of generality that x<x~n (the other case can be treated in a similar way) and let rn=2-Jn for n≥n0.
Indeed, since forγ being small enough and for n being large enough 2-Jn is negligible in front of 2-(α+γ)Jn (what we denote 2-(α+γ)Jn≫2-Jn), following the mean value theorem we can find bn∈]min⁡(x,x~n),max⁡(x,x~n)[ such that bn=sup⁡{un∈]x,x~n[,f(un)=f(x)-rn}. For all t∈]bn,x~n[, we have f(t)<f(x)-rn. Thus following the same method as in Lemma 10 we can find C>0 such that
(99)C′rn2≥meas⁡B(X,rn)⋂‍Ωc≥Crn(1+(γ/α))+1.
This yields
(100)2≤liminf⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωclog⁡⁡rn≤limsup⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωclog⁡⁡rn≤2+γα.
Since γ>0 is arbitrary and rn is independent of γ, we have the result and EΩcw(X)=0.
(iii) Case
s
(
x
)
>
1
. Following Case 4, then one can find two subsequences Jn and Jn′ with Jn′/Jn>1 for all n, such that ij(x)+ij+1(x)=1 for Jn<j<Jn′ and iJn′(x)=iJn′+1(x).
Suppose without losing generality thatiJn′(x)=0.
LetX~n such that X~n=(kJn/2Jn)+(2/32Jn)=∑j=1Jn(ijx/2j)+(2/3(2Jn)). We have clearly
(101)2-Jn′-1≤-x+X~n≤2-Jn′.
Following the same sketch as in the proof with r(x)>1 we can say that using Case  4 of Lemma 15(102)A′2-αJn′≤FJn′-1(X~n)-FJn′-1(x)≤B′2-αJn′
and since 2-αJn′F(τJn′1/3)≥∑k=Jn′+∞2-kαΛ(τkX~n)-∑k=Jn′+∞2-kαΛ(τkx)≥0 we have indeed
(103)A′2-αJn′≤F(X~n)-F(x)≤C2-αJn′.
Thus using the mean value theorem and Lemma 11 as in the previous case we conclude that EΩw(X)=0.
(iv) Case
s
(
x
)
=
1
. Let Jn be defined just as in Lemma 16, that is, that one can find δ>0, δ′>0, and Jn such that (54) is satisfied as well as Point 2 of Lemma 16.
Following the definition ofs(x), for all γ>0 there exists n0 such that for all j≥Jn0mj-x>2-j(1+γ). Thus in particular for all n≥n0 we have
(104)2-Jn>mJn-x>2-Jn(1+γ).
Suppose on one hand that CJn-1(x)>0. Then take X~n=(KJn/2Jn)+(2/3(2Jn)). Since iJn+1(x)=0 we have FJn-1(X~n)-FJn-1(x)>0.
Together with (54) this yields that
(105)δ′2-αJn≥FJn-1(x)-FJn-1(X~n)≥δ2-Jn(1+γ)2(1-α)Jnδ′2-αJn≥-FJn-1(x)+FJn-1(X~n)≥δ2-Jn(1+γ)2(1-α)Jn.
The same computation used previously yields
(106)δ′2-αJn+2-αJnF13≥-Fx+FX~n≥-FJn-1x+FJn-1X~n≥δ2-Jn1+γ21-αJn,C2-αJn≥-F(x)+F(X~n)≥δ2-Jn(γ+α).
To get EΩw(X) we only have to adapt the proof of Lemma 11 in the same way we adapt the one of Lemma 10 in the case r(x)=1.
Thus takingrn=2-Jn and following the same method used previously we can find C>0 such that
(107)C′rn2≥meas⁡B(X,rn)⋂‍Ω≥C2-Jn(1+(γ/α))2-Jn.
This yields
(108)2≤liminf⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωlog⁡⁡rn≤limsup⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωlog⁡⁡rn≤2+γα.
Since γ>0 is arbitrary, we have the result and EΩw(X)=0.
Hence Proposition19 holds.For what concerns the strong accessibility exponent we have the following result.Proposition 20.
Supposex∉D⋃‍S and let X0=(x,F(x)). Then we have the following. (1)
Ifr(x)>1/α then EΩcs(X0)=(1/α)-1.(2)
Ifs(x)>1/α then EΩs(X0)=(1/α)-1.(3)
LetDα the set of x∉D⋃S such that r(x)>1/α and s(x)>1/α. Then the Hausdorff dimension of Dα is α.Proof.
(1) Let us prove Point 1. Sincer(x)>1/α and following Point 3 of Lemma 15, for δ>0 such that r(x)-δ>1/α, we can find Jn′ and Jn such that (i)
xn=KJn2-Jn and |x-xn|≤2-Jn′;(ii)
|F(x)-F(xn)|≤2-αJn′.
Since we can chooseJn′ such that 2-Jn′≤C2(r(x)-δ)Jn (see the proof of Point 3 of Lemma 15), then |x-xn| is negligible in front of 2-Jn (what we denote |x-xn|≪2-Jn) and |F(x)-F(xn)|≤2-αJn′≪2-Jn.
Thus we can choose a constantC such that with ρn=C2-Jn and B(X0,ρn/2)⊂B((xn,F(xn)),ρn). Following the proof of Proposition 17 and more precisely (73) we have
(109)meas⁡BX,ρn2⋂‍Ωc≤meas⁡BKJn2-Jn,FKJn2-Jn,ρn⋂‍Ωc≤Cρn1+1/α.
This yields EΩcs(X)≥(1/α)-1.
SinceEΩcs(X)≤(1/α)-1, we get EΩcs(X)=(1/α)-1.
(2) We follow exactly the same proof as previously replacingxn by X~n the sequence of local maxima defined in the proof of Point 4 of Lemma 15.
(3) We follow here the results proved by [18] and summarized in [19] for our special case. Indeed recall the definition given in [18] of an ubiquitous system in a real interval of R.Definition 21.
LetU be a real open interval. Let (xi)i≥1 be points in U and let (ri)i≥1 be a sequence of positive real numbers such that lim⁡i→∞⁡ri=0. The family (xi,ri)i≥1 is a homogeneous ubiquitous system in U if the set limsup⁡iB(xi,ri) is of full Lebesgue measure in U.Theorem D of [19] proved in [18] yields the following result.Theorem 22.
Letτ be a real number with τ≥1. With the above notations if the families (xi,ri)i≥1 and (xi′,ri′)i≥1 are two homogeneous ubiquitous systems in U, then the Hausdorff dimension of the set limsup⁡B(xi,riτ)⋂limsup⁡B(x~j,r~jτ) is at least equal to 1/τ.LetU=]0,1[ and consider K1={(k/2j,2-j),k∈N,0<k<2j,j≥1}. It is a countable set and can be written as K1={(xi,ri),i≥1} with xi a dyadic number for all i≥1. Let K2={(x,r),x∈S,r=2-j/3forj≥1}. It is again a countable set and we can rewrite it as K2={(x~i,r~i),i≥1} with x~i∈S for all i≥1.It is clear thatlimsup⁡B(xi,ri) and limsup⁡B(x~i,r~i) are of full Lebesgue measure.Remark then thatDα=limsup⁡B(xi,riτ)⋂limsup⁡B(x~i,r~iτ) with τ=1/α. Since Dα⊂⋃j≥J,0≤k≤2jB(xi,riτ) the Hausdorff dimension of Dα is less than or equal to α. We apply Theorem 22 and we find it is exactly α.


### 4.4.  Proof of Theorem8

Propositions17, 18, 19, and 20 achieve the proof of Theorem 8.


## 4.1. Case of Dyadic Points

We will prove the following proposition.Proposition 17.
Ifx is a dyadic point, and X=(x,F(x)) then
(67)EΩcw(X)=1α-1,EΩw(X)=0,EΩcs(X)=1α-1,EΩs(X)=0,ufpX=1p1α-1.Ifx is a dyadic point, that is, x=2-NK with K∈2N+1, we consider its binary expansion in which iN(x)=1 and in(x)=0 for n>N. For n>N the number x-2-n is dyadic. Since 2-n=∑j=n+1∞‍2-j then x-2-n=(∑j=1N-1‍ij2-j)+2-(N+1)+2-(N+2)+⋯+2-n. On the other hand x+2-n has the simple expansion (∑j=1N(ijx/2j))+(1/2n).Remark thatFN-1(x)=Fn-1(x)=F(x) for n>N and Fn-1(x+2-n)=F(x+2-n).Any pointy in the interval ]x,x+2-n[ satisfies the expansion iN(y)=iN+1(y)=⋯=in-1(y)=in(y)=0.It follows that(68)Fx+2-n-Fx=Fn-1x+2-n-Fn-1x=2-nCn-1(y)
with y any of the points of the interval ]x,x+2-n[.Following Lemma15 and Case 1 there exist two constants A>0 and B>0 and J0≥N (which depend only on the given dyadic point x) such that
(69)∀n≥J0A2-αn≤F(x+2-n)-F(x)≤B2-αn.
Thus we have F(x+2-n)-F(x)≥A.2-αn.On the other hand, following remarks of Section3.2, for any y∈]x-2-n,x[ we have iN(y)=1=⋯=in(y). Thus
(70)Fx-2-n-Fx=Fn-1x-2-n-Fn-1x=-Cn-1(y)2-n.Whence, following Lemma15 and Case 1 we have for n≥J0(71)∀n≥J0A2-αn≤-F(x-2-n)+F(x)≤B2-αn.Letρ>0 and J≥J0 such that 2-J-1≤ρ≤2-J.SinceF≥FJ, then Ωj⊂Ω, where ΩJ is the domain below the graph of FJ. So
(72)meas⁡(B(X,ρ)∩Ωc)≤meas⁡(B(X,ρ)∩ΩJc).Butmeas⁡(B(X,ρ)∩ΩJc) is smaller than the area hb/2 of a triangle with altitude h issued from X and a corresponding hypotenuse b (see Figures 1 and 2).Figure 1
Overview of the function.Figure 2
Zoom at the pointx=1/22.Clearly, we can takeh~2-J. On the other hand, if we write 2-(j+1)<b/2<2-j with j≥J, then, using properties (69) and (71) (in which we replace nα by J), we get b/2~2-J/α. Since α<1, (69) and (71) are valid with n=J/α≥J0.Whence(73)meas⁡(B(X,ρ)∩Ωc)≤Cρ1+(1/α).
We conclude that
(74)EΩcw(X)≥1α-1.SinceEΩcw(X)≤EΩcs(X)≤(1/α)-1 this yields
(75)EΩcs(X)=1α-1.Sincef=1Ω then
(76)ufpX≥1p1α-1.Sincemeas⁡(B(X,r))=meas⁡(B(X,r)⋂Ω)+meas⁡(B(X,r)⋂Ωc) we get EΩs(X)=0; hence EΩw(X)=0.Whence Proposition17 holds.


## 4.2. Case of a Local Maximum ofF

We will prove the following proposition.Proposition 18.
LetX=(x,F(x)).
Ifx is a local maximum,
(77)EΩw(X)=1α-1,EΩcw(X)=0,(78)EΩs(X)=1α-1,EΩcs(X)=0,(79)ufp(X)=1p1α-1.Letx be a local maximum of F. There is an interval I containing x such that for all x′∈I, F(x)≥F(x′). Let N be such that the dyadic interval kN/2N,(kN+1)/2N which contains x is contained in I.Following Proposition14, we know that x has the binary expansion (51); that is, x=(kN0/2N0)+∑l=0∞(1/22l+1+N0).As a consequence of Lemma15, and following Case (2), one can find J0 and two constants A and B such that for n≥J0  (47) holds. Remark that it implies clearly that for n≥J0in(x)=1 if n is odd, and in(x)=0 if n is even.Our goal now is to evaluateF(x)-F(x′) with x′ in the interval kn/2n,(kn+1)/2n⊂kN/2N,kN+1/2N and x′≠x. If x′ is a dyadic then we take its expansion of type ij(x′)=0 for j to be large enough.Letm≥n be the smallest integer such that im(x)=im(x′) and im+1(x)≠im+1(x′). To fix the ideas, suppose that im+1(x)=1 and im+1(x′)=0. Thus
(80)132m-1≥x-x′≥12m+1+12m+3-∑j=m+2∞‍12j≥12m+3.Sinceim(x)=0 we have Cm-1(x)>0.Thus(81)Fx-Fx′=Cm-1xx-x′+∑k=m+∞‍2-kαΛτkx-Λτkx′.We have(82)A2m1-α2-m-3=C12-mα≤Cm-1xx-x′≤B2m1-α2-m-13=C22-mα.Since fork≥m-1 we have Λ(τkx)=1/3, this yields
(83)0≤∑k=m+∞‍2-kαΛτkx-Λτkx′≤C32-mα.Thus we have(84)C12-mα≤F(x)-Fx′≤C3+C22-mα.Let us compute the weak and strong exponents atx.Letρ and J≥J0 such that 2-J-1≤ρ≤2-J. Thus obviously
(85)meas⁡(B(X,ρ)⋂‍Ω)≤meas⁡(B(X,2-j)⋂‍Ω).Remark first that if(x′,y′)∈B(X,2-J)⋂Ω then |x-x′|<2-J, |y′-F(x)|<2-J, and y′≤F(x′). Since x is a local maximum on the interval kJ/2J,(kJ+1)/2J, then y′≤F(x′)≤F(x) and so 0≤F(x)-F(x′)<2-J. Hence (x′,F(x′))∈B(X,2-J)⋂Ω.Furthermore sinceF(x′) satisfies 0≤F(x)-F(x′)≤2-J, and following (84)  x′ belongs to [x-C2-J/α,x+C2-J/α] with C depending only on C3+C2.ThusB(X,2-J)⋂Ω is contained in a rectangle of length 2-J and width C2-J/α.This yields(86)meas⁡(B(X,ρ)⋂‍Ω)≤C2-J(1+(1/α))≤C′ρ1+(1/α).
We can conclude that
(87)EΩw(X)≥1α-1.
Since EΩw(X)≤EΩs(X)≤(1/α)-1 this yields
(88)EΩw(X)=EΩs(X)=1α-1.Sincemeas⁡(B(X,ρ))=meas⁡(B(X,ρ)⋂Ω)+meas⁡(B(X,ρ)⋂Ωc) we get
(89)EΩcs(X)=EΩcw(X)=0.And finally(90)ufp(X)=1p1α-1.


## 4.3. Case ofx∉D⋃S

Ifx∉D⋃S then we will compute separately the weak and strong exponents. We will first prove that for any point x in [0,1] which is not a maximum or a minimum of F the two weak exponents vanish.Proposition 19.
Letx∉D⋃S and X=(x,f(x)). Then EΩw(X)=EΩcw(X)=0.Proof.
We will prove first that we always haveEΩcw(X)=0 but will separate the proofs in cases r(x)>1 and r(x)=1. Then we will prove that EΩw(X)=0 and prove it separately for s(x)>1 and s(x)=1.
(i) Case
r
(
x
)
>
1
. We follow the notations of Case  3 of Lemma 15; that is, one can find two subsequences Jn and Jn′ such that Jn′/Jn>1 and iJn(x)=iJn′+1(x), ij(x)+iJn(x)=1 for Jn<j<Jn′+1. Suppose without lose of generality that iJn+1(x)=0. Let x~n=KJn/2Jn=∑j=1Jnij(x)/2j. Thus we have
(91)2-Jn′-1≤x-x~n≤2-Jn′.
Since Case  3 of Lemma 15 holds, we get
(92)A21-αJn′2-Jn′-1≤FJn′-1x-FJn′-1x~n≤B21-αJn′2-Jn′,A′2-αJn′≤FJn′-1(x)-FJn′-1(x~n)≤B′2-αJn′.
We have F(x)=FJn′-1(x)+∑k=Jn′+∞2-kαΛ(τkx) and FJn′-1(x~n)=F(x~n). Clearly ∑k=Jn′+∞2-kαΛ(τkx)≥0. Thus following (92) we have
(93)A21-αJn′2-Jn′-1≤Fx-Fx~n≤B21-αJn′2-Jn′+2-Jn′α∑k=0+∞‍2-kαΛτk+Jn′xA′2-αJn′≤F(x)-F(x~n)≤B′2-Jn′+2-αJn′F(τJn′x)A′2-αJn′≤F(x)-F(x~n)≤B′2-Jn′+2-αJn′FτJn′13A′2-αJn′≤F(x)-F(x~n)≤C2-αJn′
because the global maxima of F are reached at points 1/3 and 2/3.
We can now apply the mean value theorem and get that for eachn≥J0 we can find yn∈]x-2-Jn′,x+2-Jn′[ such that F(x)-F(yn)=A′·2-αn/2.
Thus using Lemma10 we can conclude that EΩcw(X)=0.
(ii) Case
r
(
x
)
=
1
. Let Jn be defined just as in Lemma 16, that is, that one can find δ>0, δ′>0, and Jn such that (54) is satisfied.
Following the definition ofr(x), for all γ>0, there exists n0 such that for all j≥Jn0Kj2-j-x>2-j(1+γ). Thus in particular for all n≥n0 we have
(94)2-Jn>KJn2-Jn-x>2-Jn(1+γ).
Suppose on one hand CJn-1(x)≥0. Then choose x~n=KJn2-Jn if x∈]KJn2-Jn,2-Jn+KJn2-Jn[(resp.x~n=KJn2-Jn-2-Jn if x∈]KJn2-Jn-2-Jn,KJn2-Jn[).
We have obviously(95)FJn-1(x)-FJn-1(x~n)=CJn-1(x)(x-x~n)≥0.
If we suppose on the other hand CJn-1(x)≤0, then we can choose in the same way a dyadic number x~n=k/2Jn such that
(96)FJn-1(x)-FJn-1(x~n)=CJn-1(x)(x-x~n)≥0.
Together with (54) this yields in any of these cases that
(97)δ′2-αJn≥FJn-1x-FJn-1x~n≥δ2-Jn1+γ21-αJn,δ′2-αJn≥FJn-1(x)-FJn-1(x~n)≥δ2-Jn(1+γ)2(1-α)Jn.
Since ∑k=Jn′+∞2-kαΛ(τkx)≥0 and FJn-1(x~n)=F(x~n) we get
(98)δ′2-αJn+2-αJnF13≥Fx-Fx~n≥FJn-1x-FJn-1x~n≥δ2-Jn1+γ21-αJn,C2-αJn≥F(x)-F(x~n)≥δ2-Jn(γ+α).
To get EΩcw(X) we only have to adapt the proof of Lemma 10 to the case rn=2-Jn. Suppose without loss of generality that x<x~n (the other case can be treated in a similar way) and let rn=2-Jn for n≥n0.
Indeed, since forγ being small enough and for n being large enough 2-Jn is negligible in front of 2-(α+γ)Jn (what we denote 2-(α+γ)Jn≫2-Jn), following the mean value theorem we can find bn∈]min⁡(x,x~n),max⁡(x,x~n)[ such that bn=sup⁡{un∈]x,x~n[,f(un)=f(x)-rn}. For all t∈]bn,x~n[, we have f(t)<f(x)-rn. Thus following the same method as in Lemma 10 we can find C>0 such that
(99)C′rn2≥meas⁡B(X,rn)⋂‍Ωc≥Crn(1+(γ/α))+1.
This yields
(100)2≤liminf⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωclog⁡⁡rn≤limsup⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωclog⁡⁡rn≤2+γα.
Since γ>0 is arbitrary and rn is independent of γ, we have the result and EΩcw(X)=0.
(iii) Case
s
(
x
)
>
1
. Following Case 4, then one can find two subsequences Jn and Jn′ with Jn′/Jn>1 for all n, such that ij(x)+ij+1(x)=1 for Jn<j<Jn′ and iJn′(x)=iJn′+1(x).
Suppose without losing generality thatiJn′(x)=0.
LetX~n such that X~n=(kJn/2Jn)+(2/32Jn)=∑j=1Jn(ijx/2j)+(2/3(2Jn)). We have clearly
(101)2-Jn′-1≤-x+X~n≤2-Jn′.
Following the same sketch as in the proof with r(x)>1 we can say that using Case  4 of Lemma 15(102)A′2-αJn′≤FJn′-1(X~n)-FJn′-1(x)≤B′2-αJn′
and since 2-αJn′F(τJn′1/3)≥∑k=Jn′+∞2-kαΛ(τkX~n)-∑k=Jn′+∞2-kαΛ(τkx)≥0 we have indeed
(103)A′2-αJn′≤F(X~n)-F(x)≤C2-αJn′.
Thus using the mean value theorem and Lemma 11 as in the previous case we conclude that EΩw(X)=0.
(iv) Case
s
(
x
)
=
1
. Let Jn be defined just as in Lemma 16, that is, that one can find δ>0, δ′>0, and Jn such that (54) is satisfied as well as Point 2 of Lemma 16.
Following the definition ofs(x), for all γ>0 there exists n0 such that for all j≥Jn0mj-x>2-j(1+γ). Thus in particular for all n≥n0 we have
(104)2-Jn>mJn-x>2-Jn(1+γ).
Suppose on one hand that CJn-1(x)>0. Then take X~n=(KJn/2Jn)+(2/3(2Jn)). Since iJn+1(x)=0 we have FJn-1(X~n)-FJn-1(x)>0.
Together with (54) this yields that
(105)δ′2-αJn≥FJn-1(x)-FJn-1(X~n)≥δ2-Jn(1+γ)2(1-α)Jnδ′2-αJn≥-FJn-1(x)+FJn-1(X~n)≥δ2-Jn(1+γ)2(1-α)Jn.
The same computation used previously yields
(106)δ′2-αJn+2-αJnF13≥-Fx+FX~n≥-FJn-1x+FJn-1X~n≥δ2-Jn1+γ21-αJn,C2-αJn≥-F(x)+F(X~n)≥δ2-Jn(γ+α).
To get EΩw(X) we only have to adapt the proof of Lemma 11 in the same way we adapt the one of Lemma 10 in the case r(x)=1.
Thus takingrn=2-Jn and following the same method used previously we can find C>0 such that
(107)C′rn2≥meas⁡B(X,rn)⋂‍Ω≥C2-Jn(1+(γ/α))2-Jn.
This yields
(108)2≤liminf⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωlog⁡⁡rn≤limsup⁡n→+∞log⁡meas⁡BX,rn⋂‍Ωlog⁡⁡rn≤2+γα.
Since γ>0 is arbitrary, we have the result and EΩw(X)=0.
Hence Proposition19 holds.For what concerns the strong accessibility exponent we have the following result.Proposition 20.
Supposex∉D⋃‍S and let X0=(x,F(x)). Then we have the following. (1)
Ifr(x)>1/α then EΩcs(X0)=(1/α)-1.(2)
Ifs(x)>1/α then EΩs(X0)=(1/α)-1.(3)
LetDα the set of x∉D⋃S such that r(x)>1/α and s(x)>1/α. Then the Hausdorff dimension of Dα is α.Proof.
(1) Let us prove Point 1. Sincer(x)>1/α and following Point 3 of Lemma 15, for δ>0 such that r(x)-δ>1/α, we can find Jn′ and Jn such that (i)
xn=KJn2-Jn and |x-xn|≤2-Jn′;(ii)
|F(x)-F(xn)|≤2-αJn′.
Since we can chooseJn′ such that 2-Jn′≤C2(r(x)-δ)Jn (see the proof of Point 3 of Lemma 15), then |x-xn| is negligible in front of 2-Jn (what we denote |x-xn|≪2-Jn) and |F(x)-F(xn)|≤2-αJn′≪2-Jn.
Thus we can choose a constantC such that with ρn=C2-Jn and B(X0,ρn/2)⊂B((xn,F(xn)),ρn). Following the proof of Proposition 17 and more precisely (73) we have
(109)meas⁡BX,ρn2⋂‍Ωc≤meas⁡BKJn2-Jn,FKJn2-Jn,ρn⋂‍Ωc≤Cρn1+1/α.
This yields EΩcs(X)≥(1/α)-1.
SinceEΩcs(X)≤(1/α)-1, we get EΩcs(X)=(1/α)-1.
(2) We follow exactly the same proof as previously replacingxn by X~n the sequence of local maxima defined in the proof of Point 4 of Lemma 15.
(3) We follow here the results proved by [18] and summarized in [19] for our special case. Indeed recall the definition given in [18] of an ubiquitous system in a real interval of R.Definition 21.
LetU be a real open interval. Let (xi)i≥1 be points in U and let (ri)i≥1 be a sequence of positive real numbers such that lim⁡i→∞⁡ri=0. The family (xi,ri)i≥1 is a homogeneous ubiquitous system in U if the set limsup⁡iB(xi,ri) is of full Lebesgue measure in U.Theorem D of [19] proved in [18] yields the following result.Theorem 22.
Letτ be a real number with τ≥1. With the above notations if the families (xi,ri)i≥1 and (xi′,ri′)i≥1 are two homogeneous ubiquitous systems in U, then the Hausdorff dimension of the set limsup⁡B(xi,riτ)⋂limsup⁡B(x~j,r~jτ) is at least equal to 1/τ.LetU=]0,1[ and consider K1={(k/2j,2-j),k∈N,0<k<2j,j≥1}. It is a countable set and can be written as K1={(xi,ri),i≥1} with xi a dyadic number for all i≥1. Let K2={(x,r),x∈S,r=2-j/3forj≥1}. It is again a countable set and we can rewrite it as K2={(x~i,r~i),i≥1} with x~i∈S for all i≥1.It is clear thatlimsup⁡B(xi,ri) and limsup⁡B(x~i,r~i) are of full Lebesgue measure.Remark then thatDα=limsup⁡B(xi,riτ)⋂limsup⁡B(x~i,r~iτ) with τ=1/α. Since Dα⊂⋃j≥J,0≤k≤2jB(xi,riτ) the Hausdorff dimension of Dα is less than or equal to α. We apply Theorem 22 and we find it is exactly α.


## 4.4.  Proof of Theorem8

Propositions17, 18, 19, and 20 achieve the proof of Theorem 8.



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*Source: 587347-2015-01-22.xml*