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# The multifractal nature of a parametrized family of von Koch functions

Zoltán Buczolich, Yann Demichel, Stéphane Seuret

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Zoltán Buczolich, Yann Demichel, Stéphane Seuret. The multifractal nature of a parametrized family of von Koch functions. 2025. hal-04987686

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THE MULTIFRACTAL NATURE OF A PARAMETRIZED FAMILY
OF VON KOCH FUNCTIONS

ZOLTÁN BUCZOLICH¹, YANN DEMICHEL² AND STÉPHANE SEURET³

ABSTRACT. In a famous paper published in 1904, Helge von Koch introduced the curve that still serves nowadays as an iconic representation of fractal shapes. In fact, von Koch's main goal was the construction of a continuous but nowhere differentiable function, very similar to the snowflake, using elementary geometric procedures, and not analytical formulae. We prove that a parametrized family of functions (including and) generalizing von Koch's example enjoys a rich multifractal behavior, thus enriching the class of historical mathematical objects having surprising regularity properties. The analysis relies on the study of the orbits of an underlying dynamical system and on the introduction of self-similar measures and non-trivial iterated functions systems adapted to the problem.

# 1. INTRODUCTION

In 1872, Weierstrass presented to the Royal Prussian Academy of Sciences the first explicit example of a family of real functions that are continuous but nowhere differentiable, see [Wei72], definitively refuting the common belief that a continuous curve must admit well-defined tangents at each of its points. From this date, several examples and parametrized families of such functions modeled by lacunary Fourier-like series have been successively proposed by Darboux [Dar75], Dini [Din77], Cellérier [Cel90], or Takagi [Tak03]. Let us also mention the famous Riemann-Fourier series $\sum_{n \ge 1} \frac{\sin(n^2 \pi x)}{n^2}$ which continues to attract the attention of mathematicians from different fields, but which is known to be differentiable at some rational points; see [Ger69]. Very disappointed by the fact that the construction of all these functions, as well as the proofs of their pathological irregularity, were based on analytical formulas rather than geometric and intuitive arguments, the Swedish mathematician Helge von Koch looked for different examples. In 1904 he succeeded in constructing a continuous but nowhere differentiable Jordan curve using a basic geometric method; see [vK04, vK06]. The curve he exhibited gave birth to the so-called snowflake curve and became one of the most iconic curves of the 20th century. In its original form, this curve does not represent a function, but with a slight modification of the construction, von Koch finally obtained a function with the desired properties.

**The generalized von Koch function** $F^\lambda$. Roughly speaking, the graph of the von Koch function is obtained in the same way as the snowflake curve, but the triangles added at each stage of the procedure are straightened up vertically. Precisely, fix a parameter $\lambda > 0$ and consider in the plane a line segment $AB$ that forms an arbitrary angle with the $x$-axis. Divide $AB$ into three equal parts $AC, CE, EB$, and erect a triangle $CDE$ with base $CE$ and median $MD$ parallel to the $y$-axis, directed towards the positive $y$ direction and with length $MD = \lambda AB$,

<footnote n="2020">2020 Mathematics Subject Classification. Primary: 28A78; Secondary: 28A80, 37A05, 37B20, 37A25, 37D25.
Key words and phrases. Fractals, multifractals, dynamical systems, invariant measures, Hausdorff dimension.</footnote>
see Figure 1. Following the footsteps of von Koch, we call $\Omega^\lambda$ the operation that transforms the line segment AB into the polygonal line ACDEB.

<figure><img src="image_2.png" /><figcaption>Figure 1. The basic geometric operation $\Omega^\lambda$ transforms a given line segment AB into the polygonal line ACDEB made up of 4 consecutive line segments.</figcaption></figure>

We call $F_0^\lambda$ the zero function on $[0, 1]$ and $I$ the segment that represents it. Then define $F_1^\lambda$ as the piecewise affine function whose graph is the polygonal line $\Omega^\lambda(I)$, i.e.

$$F_1^\lambda(x) = \begin{cases} 0 & \text{if } x \in [0, \frac{1}{3}], \\ 6\lambda(x - \frac{1}{3}) & \text{if } x \in [\frac{1}{3}, \frac{1}{2}], \\ 6\lambda(\frac{2}{3} - x) & \text{if } x \in [\frac{1}{2}, \frac{2}{3}], \\ 0 & \text{if } x \in [\frac{2}{3}, 1]. \end{cases}$$

Applying $\Omega^\lambda$ to each line segment corresponding to the graph of $F_1^\lambda$ on intervals $[0, \frac{1}{3}]$, $[\frac{1}{3}, \frac{1}{2}]$, $[\frac{1}{2}, \frac{2}{3}]$ and $[\frac{2}{3}, 1]$ respectively, we obtain the graph of the function $F_2^\lambda$. The construction can be continued infinitely and produces a sequence of piecewise affine functions $F_n^\lambda$, $n \ge 0$ (note that $F_n^\lambda$ is affine over exactly $4^n$ intervals). Since the sequence $(F_n^\lambda)_{n \ge 0}$ is non-decreasing, its limit function, denoted by $F^\lambda$ and named the generalized von Koch function with parameter $\lambda$, exists (with eventually some infinite values); see Figure 2 for some examples. The function originally considered by von Koch corresponds to the choice $\lambda = \frac{\sqrt{3}}{6}$, the thick red curve in the figure. In this case, $\Omega^\frac{\sqrt{3}}{6}$ transforms horizontal segments by replacing the median interval by the two sides of an equilateral triangle as for the classical snowflake curve.

In [vK04, vK06], von Koch proved by elementary but delicate arguments that $F^\frac{\sqrt{3}}{6}$ is nowhere differentiable on $[0, 1]$. In fact, he only stated that this function has nowhere a finite derivative, leaving open the question of whether there might be points at which the derivative is infinite. He also did not go further in studying the pointwise regularity of $F^\frac{\sqrt{3}}{6}$, and it seems that this question has never been asked. Even with the rise of multifractal analysis, while most famous pathological functions constructed in the last century have been rediscovered and their multifractal spectrum determined, see e.g. [Jaf97, JM18], von Koch's example has not been considered. Curiously, von Koch's function seems to have been forgotten by mathematicians, while the snowflake has become one of the most iconic mathematical objects. Perhaps it was thought to have the same self-similarity property and monofractal behavior as the snowflake.

**Main results.** The present paper offers a thorough analysis of the local and multiscale behavior of the generalized von Koch function $F^\lambda$ when $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6})$, especially its multifractal properties.
<figure><img src="image_3.png" /><figcaption>Figure 2. The generalized von Koch function $F^{\lambda}$ for various parameters $\lambda$. The original function by von Koch corresponding to the choice $\lambda = \frac{\sqrt{3}}{6}$ is shown by the thick red curve.</figcaption></figure>

In general, the multifractal analysis of a mathematical object $X$ consists in describing its local regularity properties together with its multiscale properties, and connecting them with each other. On one hand, one computes the multifractal spectrum of $X$, i.e. the mapping $d_X$ which associates with every $h \ge 0$ the Hausdorff dimension of the set of points having a pointwise regularity equal to $h$. On the other hand, one describes the global statistical properties of $X$ via the study of its $L^q$-scaling function $\tau_X$ (see Definitions 1.6 and Section 4.1 for more details on the definitions of pointwise regularity for functions and measures, respectively). Multifractals originate from the study of turbulence [FP85] and thermodynamics [HJK+86], and have proved insightful in many mathematical domains, from real and harmonic analysis [Jaf00, BS23] to probability theory [Jaf99, VL12] and dynamical systems [Pes97, Shm19], with deep connection to geometric measure theory and metric number theory [JM18]. Recently, multifractality has been proven for solutions to PDEs connected to the binormal flow [BV22], and investigating the multifractal nature of mathematical objects is a very active research area.

We perform the multifractal analysis of $F^{\lambda}$ in a certain range of $\lambda$, proving that $F^{\lambda}$ enjoys an extremely rich multifractal behavior, which probably von Koch did not suspect. Surprisingly, the multifractal study of $F^{\lambda}$ is deeply dependent on different subranges of $\lambda$. A key fact, as well as a challenging issue, is that, unlike classical functions such as the Weierstrass series, there is no closed analytic representation for $F^{\lambda}$. In the same vein, despite the resemblance of its graph to the so-called “alternating Takagi function”, see e.g. [AK12], the von Koch function is not a self-similar de Rham curve (see Theorem 1.4 below), and cannot be directly studied by using functional equations or as the attractor of an Iterated Function System (IFS). The analysis of $F^{\lambda}$ requires the study of a dynamical system on $[0, 1]$, the multifractal properties of auxiliary self-similar probability measures, and the construction of several IFSs tailored to the problem.

In this paper, we start with the range of parameters $(\frac{1}{6}, \frac{5}{6})$. In terms of the geometric construction of $F_n^{\lambda}$, $\frac{1}{6}$ corresponds to the critical value under which it may occur that after iterating
the construction scheme $\Omega^\lambda$ on a segment with positive (resp. negative) slope, the four new segments all have a positive (resp. negative) slope (see Figure 3 below). In the range $(\frac{1}{6}, \frac{5}{6})$, such a phenomenon does not occur, and the functions $F^\lambda$ follow a construction scheme similar to that of von Koch's original example. Also, due to this geometrical phenomenon, the strategy we develop, based on the study of the dynamical system $([0, 1], \mathcal{B}([0, 1]), T)$ introduced below, is not adapted and the whole problem should be approached differently.

<figure><img src="image_4.png" /><figcaption>Figure 3. When $\lambda \in (0, \frac{1}{6})$, the operation $\Omega^\lambda$ may lead to new slopes all with the same sign, hence a monotonous part in the graph of $F_n^\lambda$.</figcaption></figure>

**Theorem 1.1.** When $\lambda \in (\frac{1}{6}, \frac{5}{6})$, the sequence $(F_n^\lambda)_{n \ge 1}$ converges uniformly to $F^\lambda$, which is continuous but nowhere differentiable. In addition, the set $\mathcal{V}$ of those points $x$ where the derivative of $F^\lambda$ is infinite has Hausdorff dimension $s$, where $s$ is the solution to

$$2 \cdot 3^{-s} + 6^{-s} = 1. \qquad (1.1)$$

When $\lambda \ge \frac{5}{6}$, there are points $x \in [0, 1]$ at which $F^\lambda(x) = \lim_{n \to \infty} F_n^\lambda(x) = +\infty$.

In particular, we answer the question of the existence of infinite derivatives of $F^\lambda$ at some points $x$ which was left open by von Koch.

**Remark 1.2.** Observe that for every $\lambda > 0$ one has $F^\lambda(x) \ge 0$ for every $x \in [0, 1]$ and $F^\lambda(x) = 0$ if and only if $x$ belongs to the classical triadic Cantor set.

**Remark 1.3.** Clearly if $\lambda \le \lambda'$ then $F^\lambda(x) \le F^\lambda'(x)$ for every $x \in [0, 1]$. Therefore it follows from Theorem 1.1 that the sequence $(F_n^\lambda)_{n \ge 1}$ actually converges for all $\lambda \in (0, \frac{5}{6})$. Intuitively, the larger the parameter $\lambda$, the more irregular $F^\lambda$ should be, see Figure 2.

It follows from its construction that $F^\lambda$ satisfies some de Rham's type equations:

$$\forall x \in [0, 1], \quad \begin{cases} F^\lambda(1-x) = F^\lambda(x), \\ F^\lambda(\frac{x}{3}) = \frac{1}{3}F^\lambda(x). \end{cases} \qquad (1.2)$$

However, in contrast to the snowflake, the graph of $F^\lambda$ is not a self-similar set.

**Theorem 1.4.** When $\lambda \in (\frac{1}{6}, \frac{5}{6})$, the graph $\text{Gr}(F^\lambda) := \{(x, F^\lambda(x)) : x \in [0, 1]\}$ of $F^\lambda$ is not a self-similar compact set of the plane, i.e. there is no finite set of contracting similarities $\{\Phi_i : \mathbb{R}^2 \to \mathbb{R}^2\}_{i=1,...,N}$ such that $\text{Gr}(F^\lambda) = \bigcup_{i=1}^N \Phi_i(\text{Gr}(F^\lambda))$.
Let us now turn to the pointwise regularity of $F^\lambda$. Recall how the pointwise behavior of a locally bounded function is quantified.

**Definition 1.5.** Let $f \in L^\infty([0, 1])$ and $x_0 \in [0, 1]$. For $\alpha \ge 0$, $f$ is said to belong to $C^\alpha(x_0)$ if there are a polynomial $P$ of degree less than $\lfloor \alpha \rfloor$ and two constants $C, \delta > 0$ such that

$$ \forall x \in [0, 1], \quad |x - x_0| < \delta \implies |f(x) - P(x - x_0)| \le C|x - x_0|^\alpha. \quad (1.3) $$

The pointwise Hölder exponent of $f$ at $x_0$ is

$$ h_f(x_0) = \sup\{\alpha \ge 0 : f \in C^\alpha(x_0)\}. \quad (1.4) $$

More specifically, the quantity of interest is the multifractal spectrum of $f$.

**Definition 1.6.** Let $f \in L^\infty([0, 1])$. Consider the level sets $E_f(\alpha)$ for $h_f$:

$$ E_f(\alpha) = \{x \in [0, 1] : h_f(x) = \alpha\}. \quad (1.5) $$

The multifractal spectrum of $f$ is the mapping $d_f : \mathbb{R}^+ \to [0, 1] \cup \{-\infty\}$ defined by

$$ d_f(\alpha) = \dim_H(E_f(\alpha)) \quad (1.6) $$

where $\dim_H$ stands for the Hausdorff dimension, with the convention $\dim_H(\emptyset) = -\infty$. The function $f$ is said to be multifractal if the support $\{\alpha : d_f(\alpha) \ne -\infty\}$ of $d_f$ is larger than a single point.

In Theorem 1.9 we determine the multifractal spectrum $d_{F^\lambda}(\alpha)$ of $F^\lambda$ when $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6})$. This parameter range includes the classical von Koch function corresponding to $\lambda = \frac{\sqrt{3}}{6}$. To state this result, we need to introduce a dynamical system that plays a key role in the analysis of $F^\lambda$, as well as in the definition of the self-similar measure $\mu_\lambda$ that is intrinsically related to $F^\lambda$ (these relationships will be explored in the following sections).

Consider the interval $[0, 1]$, $\mathcal{B}([0, 1])$ the Borel $\sigma$-algebra over $[0, 1]$, and the transformation $T : [0, 1] \to [0, 1]$ defined as the piecewise affine map defined by (1.7) and displayed in Figure 4:

$$ T(x) = \begin{cases} 3x & \text{if } 0 \le x < \frac{1}{3}, \\ 6x - 2 & \text{if } \frac{1}{3} \le x < \frac{1}{2}, \\ 4 - 6x & \text{if } \frac{1}{2} \le x < \frac{2}{3}, \\ 3x - 2 & \text{if } \frac{2}{3} \le x \le 1. \end{cases} \quad (1.7) $$

Given $x \in [0, 1]$, the pointwise regularity of $F^\lambda$ at $x$ will be determined by the orbit of $x$ under the action of $T$, see Sections 2 and 3.

Next, we build the IFS corresponding to the inverse branches of $T$. Consider the four mappings $S_i : [0, 1] \to [0, 1]$, $i = 0, \dots, 3$, defined by

$$ S_0(x) = \frac{x}{3}, \quad S_1(x) = \frac{x}{6} + \frac{1}{3}, \quad S_2(x) = -\frac{x}{6} + \frac{2}{3} \quad \text{and} \quad S_3(x) = \frac{x}{3} + \frac{2}{3}. \quad (1.8) $$

These mappings are contracting similarities with ratios $r_i \in (0, 1)$ respectively given by

$$ r_0 = r_3 = \frac{1}{3}, \quad r_1 = r_2 = \frac{1}{6}. \quad (1.9) $$

There exists a unique non-empty compact set $K$ satisfying $K = \bigcup_{i=0}^3 S_i(K)$. This set, called the attractor of the IFS, is clearly the interval $[0, 1]$. Moreover, one easily checks that this IFS
<figure><img src="image_5.png" /><figcaption>Figure 4. Illustration for the dynamics of T associated with the construction of the function $F^{\lambda}$.</figcaption></figure>

satisfies the Open Set Condition (OSC, see [BMP92, LN99] and [Hoc14, BF21, Fen23] for recent developments), i.e. $S_i((0, 1)) \cap S_j((0, 1)) = \emptyset$ as soon as $i \neq j$.

Let us now introduce the particular self-similar measure $\mu_{\lambda}$ associated with this IFS and supported on its attractor $[0, 1]$.

**Definition 1.7.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$ and let $\gamma_{\lambda} \ge 1$ be the unique real number such that

$$ \frac{1}{3^{\gamma_{\lambda}}} + \frac{6\lambda + 1}{6^{\gamma_{\lambda}}} + \frac{6\lambda - 1}{6^{\gamma_{\lambda}}} + \frac{1}{3^{\gamma_{\lambda}}} = 1. \qquad (1.10) $$

Consider the probability vector $(p_{\lambda,i})_{i=0,...,3}$ where

$$ p_{\lambda,0} = p_{\lambda,3} = \frac{1}{3^{\gamma_{\lambda}}}, p_{\lambda,1} = \frac{6\lambda + 1}{6^{\gamma_{\lambda}}} \text{ and } p_{\lambda,2} = \frac{6\lambda - 1}{6^{\gamma_{\lambda}}}. \qquad (1.11) $$

Then $\mu_{\lambda}$ is the unique probability measure satisfying

$$ \mu_{\lambda} = \sum_{i=0}^{3} p_{\lambda,i} \mu_{\lambda} \circ S_{i}^{-1}. \qquad (1.12) $$

The multifractal analysis of $\mu_{\lambda}$ is a classical issue and is based on its $L^q$-spectrum $\tau_{\mu_{\lambda}}$ and its Legendre transform $\tau_{\mu_{\lambda}}^*$. At this point, we only need their definitions; see Section 4.1 for comments and recalls on their properties.

**Definition 1.8.** The $L^q$-spectrum of $\mu_{\lambda}$ is the mapping $\tau_{\mu_{\lambda}}: q \in \mathbb{R} \to \mathbb{R} \cup \{-\infty, +\infty\}$ such that the value $\tau_{\mu_{\lambda}}(q)$ is the unique solution to the equation

$$ \sum_{i=0}^{3} p_{\lambda,i}^{q} r_{i}^{-\tau_{\mu_{\lambda}}(q)} = 1. \qquad (1.13) $$

Moreover, the Legendre transform of $\tau_{\mu_{\lambda}}$ is the mapping $\tau_{\mu_{\lambda}}^*: \alpha \in [0, +\infty) \to \mathbb{R}$ defined by

$$ \tau_{\mu_{\lambda}}^{*}(\alpha) = \inf_{q \in \mathbb{R}} (\alpha q - \tau_{\mu_{\lambda}}(q)). \qquad (1.14) $$
We are now ready to state our main result which asserts that the multifractal spectrum of $F^\lambda$ is that of $\mu_\lambda$ up to a translation and possibly a truncation.

**Theorem 1.9.** Let $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6})$ and consider $\mu_\lambda$, $\tau_{\mu_\lambda}$ and $\tau_{\mu_\lambda}^*$.

(i) The support of $d_{F^\lambda}(\alpha)$ is $[\alpha_{\lambda,\min}, 1] := [1 - \frac{\log(6\lambda + 1)}{\log 6}, 1]$.

(ii) For every $\alpha \in [\alpha_{\lambda,\min}, 1]$,

$$d_{F^\lambda}(\alpha) = \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1). \qquad (1.15)$$

In particular, $d_{F^\lambda}$ is strictly concave on its support, the maximum of $d_{F^\lambda}$ is 1 and is reached at the exponent

$$\alpha_{\lambda, \mathcal{L}} := 1 - \frac{\log(36\lambda^2 - 1)}{4 \log 3 + 2 \log 6}. \qquad (1.16)$$

(iii) Denote by $\mathrm{Gr}(d_{F^\lambda})$ the graph of $d_{F^\lambda}$, and $\mathcal{K}(\mathbb{R}^2)$ the set of non-empty compact sets of $\mathbb{R}^2$ endowed with the Hausdorff distance. The mapping $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6}) \mapsto \mathrm{Gr}(d_{F^\lambda}) \in \mathcal{K}(\mathbb{R}^2)$ is continuous except at $\lambda = \frac{1}{3}$.

Figure 5 displays the multifractal spectrum $d_{F^\lambda}$ of $F^\lambda$ for various parameter values $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6})$.

<figure><img src="image_6.png" /><figcaption>Figure 5. The multifractal spectrum $d_{F^\lambda}$ of $F^\lambda$ for different values of $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6})$. The spectrum of the original von Koch function corresponding to the choice $\lambda = \frac{\sqrt{3}}{6}$ is shown by the thick red curve.</figcaption></figure>

Although the statement of Theorem 1.9 is short, it appears that there are significant differences between two parts of the parameter range. Let us gather some remarks on this.

**Comments on Theorem 1.9.**

(a) From the statement, the multifractal spectrum of $F^\lambda$ is increasing on $[\alpha_{\lambda,\min}, \alpha_{\lambda,\mathcal{L}}]$ and decreasing on $[\alpha_{\lambda,\mathcal{L}}, 1]$. The value $d_{F^\lambda}(1)$ is always strictly positive, while $d_{F^\lambda}(\alpha_{\lambda,\min})$ is always 0. In addition, Lebesgue-almost every $x \in [0, 1]$ is such that $h_{F^\lambda}(x) = \alpha_{\lambda,\mathcal{L}}$. Finally, there is no point $x$ with a pointwise exponent strictly greater than 1, reinforcing the initial result by von Koch that $F^\lambda$ is nowhere differentiable.
(b) When $\lambda$ increases to $\frac{5}{6}$, the abscissa of the leftmost point of the spectrum tends to 0. This is consistent with Proposition 1.1, since $F^\lambda$ does not converge when $\lambda = \frac{5}{6}$. Also, the ordinate of the rightmost point is non-decreasing with $\lambda$ and converges to 1 when $\lambda \searrow \frac{\sqrt{2}}{6}$.

(c) The two mappings $\lambda \mapsto \alpha_{\lambda, \min}$ and $\lambda \mapsto \alpha_{\lambda, \mathcal{L}}$ are decreasing with $\lambda$. This is natural since, as said before, the larger $\lambda$ is, the more irregular $F^\lambda$ is, i.e. the lowest and the Lebesgue almost sure Hölder exponents both decrease as $\lambda$ grows.

(d) Surprisingly, the analysis reveals that there is a discontinuity in the mapping $\lambda \mapsto \mathrm{Gr}(d_{F^\lambda})$ at $\lambda = \frac{1}{3}$. When $\lambda > \frac{1}{3}$, $d_{F^\lambda}(1)$ is constant, it equals $\frac{\log 2}{\log 3}$, and a phase transition occurs at $\lambda = \frac{1}{3}$, where $d_{F^\lambda}(1)$ jumps to $s$, to the unique solution to

$$2 \cdot 3^{-s} + 6^{-s} = 1. \qquad (1.17)$$

Observe that (1.17) coincides with (1.1), but the two corresponding sets are a bit different (this will follow from the proof). Let us give the reason for this phenomenon. When $\lambda > \frac{1}{3}$, the largest local dimension of $\mu_\lambda$ equals $\frac{-\log p_{\lambda,0}}{\log 3}$. After some computations, see Section 4, we can show that the maximum of the support of $d_{F^\lambda}(\alpha) = \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$ equals 1, and its value at 1 is constant, it equals $\frac{\log 2}{\log 3}$. Intuitively, this follows from the fact the points of the triadic Cantor set (recall Remark 1.3) all satisfy $h_{F^\lambda}(x) = 1$. Then, precisely at $\lambda = \frac{1}{3}$, $\frac{-\log p_{\lambda,0}}{\log 3} = \frac{-\log p_{\lambda,2}}{\log 6}$, and this coincidence implies that $\mu_\lambda$ has the same local dimension not only on the triadic Cantor set, but on the larger inhomogeneous Cantor set obtained by iteratively keeping, from any interval $[a, b]$, three out of the four subintervals $[a, a + \frac{1}{3}(b-a)]$, $[a + \frac{1}{3}(b-a), a + \frac{1}{2}(b-a)]$, $[a + \frac{1}{2}(b-a), a + \frac{2}{3}(b-a)]$, and $[a + \frac{2}{3}(b-a), b]$, and removing either $(a + \frac{1}{3}(b-a), a + \frac{1}{2}(b-a))$ or $(a + \frac{1}{2}(b-a), a + \frac{2}{3}(b-a))$ (both of length $\frac{1}{6}(b-a)$), depending on the monotonicity of the approximating function $F_n^\lambda$ on $[a, b]$. This set has dimension given by the solution $s$ to (1.17), and now the set of points $x$ satisfying $h_{F^\lambda}(x) = 1$ has Hausdorff dimension $s \approx 0.8533 > \frac{\log 2}{\log 3}$. This justifies the “jump” of the graph of the spectrum $\mathrm{Gr}(d_{F^\lambda})$ at $\lambda = \frac{1}{3}$ and item (iii) of Theorem 1.9.

(e) Let us emphasize that although the statement is the same for every $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6})$, the analysis is much more involved in the range $(\frac{\sqrt{2}}{6}, \frac{1}{3}]$ (which includes the classical von Koch function) than in the range $(\frac{1}{3}, \frac{5}{6})$. This is due to the fact that when $\lambda > \frac{1}{3}$, the slopes of the functions $F_n^\lambda$ approximating $F^\lambda$ after $n$ steps of the construction tend to infinity (almost) everywhere, while this is not the case when $\lambda \le \frac{1}{3}$. This part deserves a specific treatment in Section 6.

Let us finish this introduction by some remarks on the other range of parameters $\lambda \in (0, \frac{\sqrt{2}}{6}]$. Precisely at $\lambda = \frac{\sqrt{2}}{6}$, the Lebesgue-almost sure exponent $\alpha_{\lambda, \mathcal{L}}$ equals 1, so when $\lambda$ goes below this value, it is expected that the decreasing part disappears. In [BDS25], we investigate the range $\lambda \in (\frac{1}{6}, \frac{\sqrt{2}}{6})$ and we prove that the equality (1.15) does not hold when $\alpha$ is close to 1, and that the right hand-side term is only a lower bound for $d_{F^\lambda}(\alpha)$ when $\alpha \in [\alpha_{\lambda, \min}, 1]$ (we also prove that $d_{F^\lambda}$ is continuous at $\alpha = 1$). The exact spectrum $d_{F^\lambda}$ for these $\lambda$s remains an open question.

When $\lambda$ becomes smaller than $\frac{1}{6}$, our approach is no longer relevant, since the weight $p_{\lambda,2}$ in the construction of $\mu_\lambda$ would become negative. As mentioned above, it corresponds to the critical value when it may happen that the situation of Figure 3 occurs. Then the dynamical
system ([0, 1], $\mathcal{B}([0, 1])$, $T$) is not relevant any more to describe the pointwise behavior of $F^\lambda$ any more, and alternative methods must then be developed.

We emphasize that the limit case $\lambda = \frac{1}{6}$ is of particular interest, if only for historical reasons. In 1944, G. Bouligand returned to the subject of continuous but nowhere differentiable functions in a book he wrote on the role of intuition in mathematics. He gives an example of such a function constructed in exactly the same way as von Koch's original one, but with right triangles added at each stage, so that the curve representing this function admits “corners” everywhere; see [Bou44, pp. 222-223]. In fact, his function coincides with $F^{\frac{1}{6}}$. Curiously, Bouligand gives no proof of his result, and no references to von Koch's work.

**Open questions.** Let us indicate some open questions:

(1) What is the value of the multifractal spectrum $d_{F^\lambda}$ when $\lambda \in (0, \frac{\sqrt{2}}{6}]$? We conjecture that the concavity of $d_{F^\lambda}$ is preserved at all values of $\lambda$.

(2) Does $F^\lambda$ satisfy a multifractal formalism?

(3) What is the Hausdorff dimension of the graph of $F^\lambda$?

(4) What can we say about the (points of) convergence of $F^\lambda$ when $\lambda \ge \frac{5}{6}$?

**Outline of the paper.** The paper is organized as follows. In Section 2, we study the dynamical system ([0, 1], $\mathcal{B}([0, 1])$, $T$) and obtain a key decomposition of real numbers $x \in [0, 1]$ useful for the analysis of $F^\lambda$. Then, in Section 3 we study the slopes of the intermediary steps of the construction of $F^\lambda$, resulting in estimates for the pointwise Hölder exponents. Theorems 1.1 and 1.4 are proved in this section. In Section 4 we collect and prove various regularity properties of $\mu_\lambda$. The connection between the local dimensions of $\mu_\lambda$ and the pointwise Hölder exponents of $F^\lambda$ is determined in Section 5. From this a lower estimate is deduced for $d_{F^\lambda}$, which turns out to be sharp when $\lambda \in (\frac{\sqrt{2}}{6}, \frac{5}{6})$. This sharpness is obtained in very different ways depending on the value of $\lambda$ and on whether the increasing or the decreasing part of the spectrum is considered. In particular, the decreasing part is treated in Section 5, while the increasing part when $\lambda \in (\frac{\sqrt{2}}{6}, \frac{1}{3}]$ is treated separately in Section 6. It requires an intricate proof based on the introduction of iterated function systems adapted to the problem.

## 2. THE DYNAMICAL SYSTEM AND THE ASSOCIATED DECOMPOSITION OF $x \in [0, 1]$

In this section, we study the dynamical system ([0, 1], $\mathcal{B}([0, 1])$, $T$) associated with the recursive construction of $F^\lambda$, where the transformation $T$ was given in (1.7).

Let

$$ U(x) = \begin{cases} 0 & \text{if } 0 \le x < \frac{1}{3}, \\ 1 & \text{if } \frac{1}{3} \le x < \frac{1}{2}, \\ 2 & \text{if } \frac{1}{2} \le x < \frac{2}{3}, \\ 3 & \text{if } \frac{2}{3} \le x \le 1, \end{cases} \quad \text{and} \quad \tilde{U}(x) = \begin{cases} 0 & \text{if } 0 \le x < \frac{1}{3}, \\ \frac{1}{3} & \text{if } \frac{1}{3} \le x < \frac{1}{2}, \\ \frac{2}{3} & \text{if } \frac{1}{2} \le x < \frac{2}{3}, \\ \frac{2}{3} & \text{if } \frac{2}{3} \le x \le 1. \end{cases} \quad (2.1) $$

For every $x \in [0, 1]$, consider its orbit $(T^n(x))_{n \ge 0}$ under the action of $T$ (with $T^0(x) = x$ by definition). Then, associate with $x$ the two sequences $(u_n(x))_{n \ge 0} \in \Omega = \{0, 1, 2, 3\}^N$ and $(\tilde{u}_n(x))_{n \ge 0} \in \{0, \frac{1}{3}, \frac{2}{3}\}^N$ defined as follows:

$$ u_n(x) = U(T^n(x)) \quad \text{and} \quad \tilde{u}_n(x) = \tilde{U}(T^n(x)). \quad (2.2) $$
The sequence $(u_n(x))_{n \ge 0}$ can be thought of as the “digits” of $x$ associated with $T$. However, since $T$ is not defined over intervals of the same length, and has a part with a negative slope, we need the sequence $(\tilde{u}_n(x))_{n \ge 0}$ to obtain a canonical decomposition of $x$ with respect to $T$ using the lower or upper endpoints of the division intervals according to the monotonicity of $T^n$.

**Definition 2.1.** For $x \in [0, 1]$, the number of occurrences of the digits of $x$ are given by

$$ \beta_i(x, n) = \#\{k < n : u_k(x) = i\} \text{ and } \beta_{i,j}(x, n) = \#\{k < n : u_k(x) = i \text{ or } u_k(x) = j\}, \quad (2.3) $$

where $i, j \in \{0, 1, 2, 3\}$, $i \neq j$, and $n \ge 0$. Notice that $\beta_{i,j}(x, n) = \beta_i(x, n) + \beta_j(x, n)$ and $\beta_{0,3}(x, n) + \beta_{1,2}(x, n) = n$. In particular, $\beta_i(x, 0) = \beta_{i,j}(x, 0) = 0$.

Next, we introduce the sequences $(\varepsilon_n(x))_{n \ge 0}$, $(\ell_n(x))_{n \ge 0}$ and $(a_n(x))_{n \ge 0}$ defined as follows:

$$ \varepsilon_n(x) = (-1)^{\beta_2(x,n)}, \quad \ell_n(x) = 3^{-\beta_{0,3}(x,n)} 6^{-\beta_{1,2}(x,n)} \quad (2.4) $$

and for $n \ge 1$, $a_n(x) = \sum_{k=0}^{n-1} \tilde{u}_k(x)\varepsilon_k(x)\ell_k(x)$.

Then $\varepsilon_0(x) = 1$, $\ell_0(x) = 1$ and by definition $a_0(x) = 0$. Moreover, one can easily see that $\varepsilon_{n+1}(x) = \varepsilon_1(T^n(x))\varepsilon_n(x)$.

According to their digits, two subsets of points $x \in [0, 1]$ are of particular interest for us.

**Definition 2.2.**

(i) The countable set $\mathcal{E}$ consists of those points with digits all equal to 0 or all equal to 3 ultimately:

$$ \mathcal{E} = \{x \in [0, 1] : u_n(x) \text{ is eventually constant 0, or constant 3}\}. \quad (2.5) $$

(ii) The set $\tilde{\mathcal{E}}$ consists of those points with a finite number of 1s and 2s in their decomposition:

$$ \tilde{\mathcal{E}} = \{x \in [0, 1] : \text{the sequence } (\beta_{1,2}(x, n))_{n \ge 0} \text{ is eventually constant}\}. \quad (2.6) $$

Notice that $\mathcal{E} \subset \tilde{\mathcal{E}}$, and that $\mathcal{E}$ and $\tilde{\mathcal{E}}$ are (strongly) $T$ invariant, i.e. $T(\mathcal{E}) = \mathcal{E}$ and $T(\tilde{\mathcal{E}}) = \tilde{\mathcal{E}}$.

We denote by $I = (a, b)$ the open interval consisting of $x$ between $a$ and $b$, even when $b < a$. The corresponding closed interval $[a, b]$ will be denoted by $\bar{I}$, and $|I|$ will stand for the common length of $I$ and $\bar{I}$.

**Proposition 2.3.** Let $x \in [0, 1] \setminus \mathcal{E}$. Then, for every $n \ge 0$,

$$ x \in I_n(x) := (a_n(x), b_n(x)) \text{ where } b_n(x) = a_n(x) + \varepsilon_n(x)\ell_n(x). \quad (2.7) $$

The sign of the slope of $T^n$ on $I_n(x)$ is $\varepsilon_n(x)$, $T^n(a_n(x)) = 0$, and $T^n$ maps $I_n(x)$ onto $(0, 1)$.

*Proof.* We proceed by induction on $n \ge 0$. Since $I_0(x) = (0, 1)$ and $T^0(x) = x$ the result is true when $n = 0$. Assume that the statements of Proposition 2.3 hold for some fixed $n \ge 0$.

First, suppose that $\varepsilon_n(x) = 1$. Then $T^n$ maps monotone increasingly $I_n(x) = (a_n(x), a_n(x) + \ell_n(x))$ onto $(0, 1)$, and $T^n(a_n(x)) = 0$. The definition of $T$ and (2.4) imply that $T^{n+1}(a_{n+1}(x)) = 0$ and $\varepsilon_{n+1}(x) = \varepsilon_1(T^n(x))\varepsilon_n(x) = \varepsilon_1(T^n(x))$ is the sign of the slope of $T^{n+1}$ on $I_{n+1}(x)$. Using the definition (2.4) of $a_n(x)$, we get $a_{n+1}(x) = a_n(x) + \tilde{u}_n(x)\varepsilon_n(x)\ell_n(x)$. In addition to the definition of $I_n(x)$ in (2.7), we obtain

$$ I_{n+1}(x) = (a_n(x) + \tilde{u}_n(x)\varepsilon_n(x)\ell_n(x), a_n(x) + \tilde{u}_n(x)\varepsilon_n(x)\ell_n(x) + \varepsilon_{n+1}(x)\ell_{n+1}(x)), \quad (2.8) $$
which coincides with $(a_{n+1}(x), b_{n+1}(x))$. Thus, our construction and $x \in I_n(x)$ imply $x \in I_{n+1}(x)$, and formula (2.7) is proved.

Suppose now that $\varepsilon_n(x) = -1$. Then $T^n$ maps monotone decreasingly $(a_n(x) - \ell_n(x), a_n(x))$ onto $(0, 1)$, and $T^n(a_n(x)) = 0$. Again, recalling (2.4), we see that $T^{n+1}(a_{n+1}(x)) = 0$ and $\varepsilon_{n+1}(x) = \varepsilon_1(T^n(x))\varepsilon_n(x) = -\varepsilon_1(T^n(x))$ is the sign of the slope of $T^{n+1}$ on $I_{n+1}(x)$. The proof of (2.7) is analogous to the case $\varepsilon_n(x) = 1$. $\square$

Let us emphasize that when $x \in \mathcal{E}$, $I_n(x)$ is well-defined and we check that $x \in \overline{I_n(x)} := \overline{I_n(x)}$. In addition, observe that $\mathcal{E}$ coincides with the points of non-differentiability $F_n^\lambda$ which are the points $a_n(x)$ and $b_n(x)$ for every $x \in [0, 1]$ and $n \ge 0$.

Also, we see that $\tilde{\mathcal{E}}$ is a countable union of affine copies of the triadic Cantor set and in particular $\dim_H(\tilde{\mathcal{E}}) = \frac{\log 2}{\log 3}$.

For further ease of notation, we introduce the (finite) set $\mathcal{I}_n$ of all open intervals of generation $n \ge 0$, namely

$$ \mathcal{I}_n = \{I_n(x) : x \in [0, 1] \setminus \mathcal{E}\}. \qquad (2.9) $$

**Corollary 2.4.** Every $x \in [0, 1]$ is uniquely written

$$ x = \sum_{k=0}^{\infty} \tilde{u}_k(x) \varepsilon_k(x) \ell_k(x). \qquad (2.10) $$

*Proof.* When $x \notin \mathcal{E}$, using that $\ell_n(x) \to 0$ it is a direct consequence of (2.7). The fact that (2.10) also holds when $x \in \mathcal{E}$ is left to the reader, and it will not be used in the sequel. $\square$

### 3. ESTIMATES ON THE POINTWISE REGULARITY OF $F^\lambda$ WHEN $\lambda \in (\frac{1}{6}, \frac{5}{6})$

#### 3.1. The approximating functions $F_n^\lambda$.

This subsection contains technical lemmas that are unavoidable in proceeding to the analysis of $F^\lambda$. Our first key lemma is an analytic transcription of the operation $\Omega^\lambda$, explaining how $F_{n+1}^\lambda$ is locally obtained from $F_n^\lambda$ (recall Figure 1 that explains the splitting process).

**Lemma 3.1.** Let $\lambda \ge \frac{1}{6}$ and $n \ge 0$. Consider an interval $[a, b]$ on which the function $F_n^\lambda$ is affine with slope $m > 0$. Set $\ell = b - a > 0$. Then, the function $F_{n+1}^\lambda$ is affine on the four consecutive subintervals

$$ I_0 = (a, a + \frac{\ell}{3}), \quad I_1 = (a + \frac{\ell}{3}, a + \frac{\ell}{2}), \quad I_2 = (a + \frac{\ell}{2}, a + \frac{2\ell}{3}) \quad \text{and} \quad I_3 = (a + \frac{2\ell}{3}, a + \ell), \qquad (3.1) $$

with slopes

$$ m, m + 6\lambda\sqrt{1 + m^2} > 0, m - 6\lambda\sqrt{1 + m^2} < 0 \text{ and } m \text{ respectively.} \qquad (3.2) $$

When $m < 0$ an analogous statement is valid and (3.2) should be replaced by

$$ m, m - 6\lambda\sqrt{1 + m^2} > 0, m + 6\lambda\sqrt{1 + m^2} < 0 \text{ and } m \text{ respectively.} $$

*Proof.* We only deal with the case $m \ge 0$, the other case is left to the reader. Identifying the line segment AB with the graph of $F_n^\lambda$ we can make use of Figure 1 to make some easy calculations. Since we work with slopes after rescaling and translating we can assume moreover that $(a, b) = (0, 1)$ (so $\ell = 1$) and that the projection of AB onto the x-axis is $[0, 1]$ with A being the origin. In addition, by the definition of $\Omega^\lambda$, it follows that $A = (a, 0) = (0, 0)$, $B = (b, m) = (1, m)$, $C = (\frac{1}{3}, \frac{m}{3})$, $D = (\frac{1}{2}, \frac{m}{2} + \lambda\sqrt{1 + m^2})$, $E = (\frac{2}{3}, \frac{2m}{3})$ and $M = (\frac{1}{2}, \frac{m}{2})$.
On $I_0 = (0, \frac{1}{3})$ the slope of AC equals $m$. On $I_3 = (\frac{2}{3}, 1)$ the slope of EB also equals $m$. On $I_1 = (\frac{1}{3}, \frac{1}{2})$ the slope of CD is given by

$$ \frac{\left(\frac{m}{2} + \lambda\sqrt{1+m^2}\right) - \frac{m}{3}}{\frac{1}{2} - \frac{1}{3}} = m + 6\lambda\sqrt{1+m^2}. $$

On $I_2 = (\frac{1}{2}, \frac{2}{3})$ the slope of DE is given by

$$ \frac{\frac{2m}{3} - \left(\frac{m}{2} + \lambda\sqrt{1+m^2}\right)}{\frac{2}{3} - \frac{1}{2}} = m - 6\lambda\sqrt{1+m^2}. $$

Finally, for all $\lambda \ge \frac{1}{6}$ and all $m \in \mathbb{R}$:

$$ \begin{cases} m + 6\lambda\sqrt{1+m^2} \ge m + \sqrt{1+m^2} > m + |m| \ge 0, \\ m - 6\lambda\sqrt{1+m^2} \le m - \sqrt{1+m^2} < m - |m| \le 0. \end{cases} $$

This concludes the proof.

□

Notice that a slope $m > 0$ (resp. $m < 0$) gives birth to four consecutive new slopes with the sequence of signs $+, +, -, +$ (resp. $-, +, -, -$) whatever $\lambda \ge \frac{1}{6}$ is. But when $\lambda < \frac{1}{6}$, the sign of the third (resp. the second) slope may change according to the value of the initial slope $m$. Precisely, when $m > 0$ is large enough, then the four new slopes are all positive, so the new piecewise part in the iterative construction of $F^\lambda$ is monotonous increasing. This is consistent with the fact that the function $F^\lambda$ is more regular for smaller $\lambda$.

Returning to $\lambda \in (\frac{1}{6}, \frac{5}{6})$, we focus on the pointwise behavior of $F_n^\lambda$ at points $x \in [0, 1] \setminus \mathcal{E}$. Denote by $m_n(x)$ the slope of $F_n^\lambda$ in $x$. One has $m_n(x) = 0$ if and only if $\beta_{1,2}(x, n) = 0$, and in this case $\varepsilon_n(x) = 1$. In particular, this is true for $n = 0$. The next lemma connects the dynamical system $([0, 1], \mathcal{B}([0, 1]), T)$ to the construction of the function $F^\lambda$. The sign of $x$ is denoted by sgn($x$): sgn($x$) = 1 if $x > 0$, sgn($x$) = -1 if $x < 0$, and sgn($x$) = 0 if $x = 0$.

**Lemma 3.2.** Let $x \in [0, 1] \setminus \mathcal{E}$. For every $n \ge 0$, $I_n(x)$ is exactly the open interval of maximal length containing $x$ and on which the function $F_n^\lambda$ is affine. In particular,

$$ m_n(x) = \frac{F_n^\lambda(x) - F_n^\lambda(a_n(x))}{x - a_n(x)} \qquad (3.3) $$

and sgn($m_n(x)$) = sgn($\beta_{1,2}(x, n)$) $\varepsilon_n(x)$.

We can rephrase the lemma by saying that the intervals $I \in \mathcal{I}_n$ of generation $n$ are exactly those on which $F_n^\lambda$ is affine.

*Proof.* We proceed by induction on $n \ge 0$. Since $I_0(x) = (0, 1)$ and $F_0^\lambda(x) = 0$ the result is true when $n = 0$. Suppose now that the statement of Lemma 3.2 holds for some fixed $n \ge 0$. Our strategy is to use Lemma 3.1 with $(a, b) = I_n(x)$ and (2.7) to identify the interval $I_{n+1}(x)$ with a certain $I_k$ from (3.1). Using (2.7), we consider the two cases $\varepsilon_n(x) = 1$ and $\varepsilon_n(x) = -1$, and then distinguish four cases according to the value of $u_n(x) \in \{0, 1, 2, 3\}$. Straightforward but useful calculations following from the definitions (2.3) and (2.4) are collected in the table below. Case 1: Suppose that $\varepsilon_n(x) = 1$. We apply Lemma 3.1 with $a = a_n(x)$ and $b = a_n(x) + \ell_n(x)$.
<table><thead><tr><th>u<sub>n</sub>(x)</th><th>ũ<sub>n</sub>(x)</th><th>β<sub>2</sub>(x, n + 1)</th><th>β<sub>0,3</sub>(x, n + 1)</th><th>β<sub>1,2</sub>(x, n + 1)</th><th>ε<sub>n+1</sub>(x)</th><th>ℓ<sub>n+1</sub>(x)</th></tr></thead><tbody><tr><td>0</td><td>0</td><td>β<sub>2</sub>(x, n)</td><td>β<sub>0,3</sub>(x, n) + 1</td><td>β<sub>1,2</sub>(x, n)</td><td>ε<sub>n</sub>(x)</td><td><sup>1</sup>&frasl;<sub>3</sub>ℓ<sub>n</sub>(x)</td></tr><tr><td>1</td><td><sup>1</sup>&frasl;<sub>3</sub></td><td>β<sub>2</sub>(x, n)</td><td>β<sub>0,3</sub>(x, n)</td><td>β<sub>1,2</sub>(x, n) + 1</td><td>ε<sub>n</sub>(x)</td><td><sup>1</sup>&frasl;<sub>6</sub>ℓ<sub>n</sub>(x)</td></tr><tr><td>2</td><td><sup>2</sup>&frasl;<sub>3</sub></td><td>β<sub>2</sub>(x, n) + 1</td><td>β<sub>0,3</sub>(x, n)</td><td>β<sub>1,2</sub>(x, n) + 1</td><td>-ε<sub>n</sub>(x)</td><td><sup>1</sup>&frasl;<sub>6</sub>ℓ<sub>n</sub>(x)</td></tr><tr><td>3</td><td><sup>2</sup>&frasl;<sub>3</sub></td><td>β<sub>2</sub>(x, n)</td><td>β<sub>0,3</sub>(x, n) + 1</td><td>β<sub>1,2</sub>(x, n)</td><td>ε<sub>n</sub>(x)</td><td><sup>1</sup>&frasl;<sub>3</sub>ℓ<sub>n</sub>(x)</td></tr></tbody></table>

* Suppose that $u_n(x) = 0$. Using (2.8) with the second line of the table above and using notation $I_j, j = 0, \dots, 3$ from (3.1) we get

$$I_{n+1}(x) = (a_n(x), a_n(x) + \frac{1}{3}\ell_n(x)) = I_0.$$

In accordance with Lemma 3.1, this latter interval is a maximal open interval on which $F_{n+1}^{\lambda}$ is affine, contains $x$, and the slope of $F_{n+1}^{\lambda}$ is $m_{n+1}(x) = m_n(x)$. Thus,

$$\operatorname{sgn}(m_{n+1}(x)) = \operatorname{sgn}(m_n(x)) = \operatorname{sgn}(\beta_{1,2}(x, n))\varepsilon_n(x) = \operatorname{sgn}(\beta_{1,2}(x, n+1))\varepsilon_{n+1}(x).$$

* Suppose that $u_n(x) = 1$. Using (2.8) with the third line of the table above we obtain

$$\begin{aligned} I_{n+1}(x) &= (a_n(x) + \frac{1}{3}\ell_n(x), a_n(x) + \frac{1}{3}\ell_n(x) + \frac{1}{6}\ell_n(x)) \\ &= (a_n(x) + \frac{1}{3}\ell_n(x), a_n(x) + \frac{1}{2}\ell_n(x)) = I_1. \end{aligned}$$

By Lemma 3.1, $I_1$ is a maximal open interval on which $F_{n+1}^{\lambda}$ is affine, contains $x$, and the slope of $F_{n+1}^{\lambda}$ is $m_{n+1}(x) > 0$. Hence,

$$\operatorname{sgn}(m_{n+1}(x)) = 1 = \operatorname{sgn}(\beta_{1,2}(x, n) + 1)\varepsilon_n(x) = \operatorname{sgn}(\beta_{1,2}(x, n+1))\varepsilon_{n+1}(x).$$

* Suppose that $u_n(x) = 2$. Using (2.8) with the fourth line of the table above we obtain

$$\begin{aligned} I_{n+1}(x) &= (a_n(x) + \frac{2}{3}\ell_n(x), a_n(x) + \frac{2}{3}\ell_n(x) - \frac{1}{6}\ell_n(x)) \\ &= (a_n(x) + \frac{2}{3}\ell_n(x), a_n(x) + \frac{1}{2}\ell_n(x)) \\ &= (a_n(x) + \frac{1}{2}\ell_n(x), a_n(x) + \frac{2}{3}\ell_n(x)) = I_2. \end{aligned}$$

In accordance with Lemma 3.1, $I_2$ is a maximal open interval on which $F_{n+1}^{\lambda}$ is affine, contains $x$, and the slope of $F_{n+1}^{\lambda}$ is $m_{n+1}(x) < 0$. Observe that in this case there is an inversion in the relative position of the extremities of $I_2$ due to the negativeness of the slope. Hence,

$$\operatorname{sgn}(m_{n+1}(x)) = -1 = \operatorname{sgn}(\beta_{1,2}(x, n) + 1)(-\varepsilon_n(x)) = \operatorname{sgn}(\beta_{1,2}(x, n+1))\varepsilon_{n+1}(x).$$

* Finally, suppose that $u_n(x) = 3$. Using (2.8) with the fifth line of the table above we get

$$\begin{aligned} I_{n+1}(x) &= (a_n(x) + \frac{2}{3}\ell_n(x), a_n(x) + \frac{2}{3}\ell_n(x) + \frac{1}{3}\ell_n(x)) \\ &= (a_n(x) + \frac{2}{3}\ell_n(x), a_n(x) + \ell_n(x)) = I_3. \end{aligned}$$

By Lemma 3.1, $I_3$ is a maximal open interval on which $F_{n+1}^{\lambda}$ is affine, contains $x$, and the slope of $F_{n+1}^{\lambda}$ is $m_{n+1}(x) = m_n(x)$. Thus,

$$\operatorname{sgn}(m_{n+1}(x)) = \operatorname{sgn}(m_n(x)) = \operatorname{sgn}(\beta_{1,2}(x, n))\varepsilon_n(x) = \operatorname{sgn}(\beta_{1,2}(x, n+1))\varepsilon_{n+1}(x).$$

*Case 2:* The case $\varepsilon_n(x) = -1$ is treated similarly by applying Lemma 3.1 with $a = a_n(x) - \ell_n(x)$ and $b = a_n(x)$, and is left to the reader. □
Next lemma provides a recursive relation for the sequences of slopes $(m_n(x))_{n \ge 0}$ when $x \notin \mathcal{E}$.

**Lemma 3.3.** Let $x \in [0, 1] \setminus \mathcal{E}$. Suppose that there exists a finite $p \ge 0$ such that $\beta_{1,2}(x, p) = 0$ and $\beta_{1,2}(x, p+1) = 1$. Then, $m_0(x) = \dots = m_p(x) = 0$,

$$m_{p+1}(x) = \begin{cases} 6\lambda & \text{if } u_p(x) = 1, \\ -6\lambda & \text{if } u_p(x) = 2, \end{cases} \qquad (3.4)$$

and, for all $n \ge p+1$, the slopes $m_n(x)$ are recursively given by

$$m_{n+1}(x) = \begin{cases} m_n(x) & \text{if } u_n(x) \in \{0, 3\}, \\ \operatorname{sgn}(m_n(x))(|m_n(x)| + 6\lambda\sqrt{1+m_n^2(x)}) & \text{if } u_n(x) = 1, \\ \operatorname{sgn}(m_n(x))(|m_n(x)| - 6\lambda\sqrt{1+m_n^2(x)}) & \text{if } u_n(x) = 2. \end{cases} \qquad (3.5)$$

*Proof.* The proof is analogous to that of Lemma 3.2. If $u_n(x) \in \{0, 3\}$, $m_{n+1}(x) = m_n(x)$ for both cases $\varepsilon_n(x) = 1$, and $\varepsilon_n(x) = -1$. In particular, since $F_0^\lambda(x) = 0$ and $m_0(x) = 0$, one has $m_{n+1}(x) = 0$ as long as $u_n(x) \in \{0, 3\}$, that is, as long as $\beta_{1,2}(x, n+1) = 0$. Hence, the first nonzero slope appears in the first step $p \ge 0$ such that $u_p(x) \in \{1, 2\}$: for this $p$, $\beta_{1,2}(x, n) = 0$ for all $n \le p$ but $\beta_{1,2}(x, p+1) = 1$. Then, the value of $m_{p+1}(x)$ is computed using Lemma 3.1 with $(a, b) = I_p(x)$ and $m = m_p(x) = 0$. The result (3.4) comes from the fact that in this case $x \in I_{u_p(x)}(x)$. Finally, let us prove (3.5). For all $n \ge p+1$, $m_n(x) \ne 0$, $\beta_{1,2}(x, n) > 0$, and from Lemma 3.2, $\operatorname{sgn}(m_n(x)) = \varepsilon_n(x)$. We distinguish two cases:

* If $u_n(x) = 1$ then

$$\begin{aligned} m_{n+1}(x) &= \begin{cases} m_n(x) + 6\lambda\sqrt{1+m_n^2(x)} & \text{if } \varepsilon_n(x) = 1, \\ m_n(x) - 6\lambda\sqrt{1+m_n^2(x)} & \text{if } \varepsilon_n(x) = -1 \end{cases} \\ &= \operatorname{sgn}(m_n(x))|m_n(x)| + \varepsilon_n(x)(6\lambda\sqrt{1+m_n^2(x)}) \\ &= \operatorname{sgn}(m_n(x))(|m_n(x)| + 6\lambda\sqrt{1+m_n^2(x)}). \end{aligned}$$

* If $u_n(x) = 2$ then

$$\begin{aligned} m_{n+1}(x) &= \begin{cases} m_n(x) - 6\lambda\sqrt{1+m_n^2(x)} & \text{if } \varepsilon_n(x) = 1, \\ m_n(x) + 6\lambda\sqrt{1+m_n^2(x)} & \text{if } \varepsilon_n(x) = -1 \end{cases} \\ &= \operatorname{sgn}(m_n(x))|m_n(x)| - \varepsilon_n(x)(6\lambda\sqrt{1+m_n^2(x)}) \\ &= \operatorname{sgn}(m_n(x))(|m_n(x)| - 6\lambda\sqrt{1+m_n^2(x)}). \end{aligned}$$

This concludes the proof. $\square$

**Remark 3.4.** We deduce from Lemma 3.3 that for every $x \in [0, 1] \setminus \mathcal{E}$, the following upper bound holds for the successive slopes: $|m_{n+1}(x)| \le (1 + 6\lambda)|m_n(x)| + 6\lambda$.

Given a very large slope $|m_n(x)|$ above a point $x$ at step $n$, the next slope $|m_{n+1}(x)|$ can be much smaller and even close to 0 when $\lambda$ gets close to $\frac{1}{6}$. When $\lambda$ is large enough, this phenomenon does not occur and it may be possible to determine asymptotics for $|m_n(x)|$, at least when $x \notin \tilde{\mathcal{E}}$, which is enough for us. The next lemma makes these observations precise.

**Lemma 3.5.** Let $x \in [0, 1] \setminus \mathcal{E}$. If $m_n(x) \ne 0$, then $|m_n(x)| \in [m_n^-, m_n^+]$ with $m_n^- = \sqrt{36\lambda^2 - 1}$ and $m_n^+ \asymp (6\lambda + 1)^n$. Moreover, when $\lambda > \frac{1}{3}$, for every $x \notin \tilde{\mathcal{E}}$, $\lim_{n \to \infty} |m_n(x)| = +\infty$.
*Proof.* We first focus on the lower bound $m_n^-$ for any $|m_n(x)| \neq 0$. Applying Lemma 3.3, one sees that in step $n \geq 1$, the minimal absolute value of the nonzero slope of $F_n^\lambda$ is greater than $6\lambda\sqrt{1+m^2} - |m|$ where $m$ is one of the $4^{n-1}$ slopes of $F_{n-1}^\lambda$. Straightforward calculations show that the function $t \in \mathbb{R} \mapsto 6\lambda\sqrt{1+t^2} - |t|$ is minimal in $\pm 1/\sqrt{36\lambda^2-1}$ with minimal value $\sqrt{36\lambda^2-1}$, hence the value for $m_n^-$.

Let us now estimate the upper bound $m_n^+$. We call $\bar{x}$ the real number whose digits are all equal to 1, i.e. $u_n(\bar{x}) = 1$ for every $n$, and consider the function $f(t) = t + 6\lambda\sqrt{1+t^2}$. Applying Lemma 3.3, it is clear from the construction that $m_n(\bar{x})$ is always the maximal absolute slope of $F_n^\lambda$. Hence, $m_1^+ = m_1(\bar{x}) = 6\lambda$ and $m_n^+ = m_n(\bar{x}) = f^{n-1}(6\lambda)$. We have

$$0 < (1 + 6\lambda)m \leq f(m) = m + 6\lambda\sqrt{1 + m^2} \leq (1 + 6\lambda)m + 6\lambda,$$

and, since $f$ is increasing, by iterating this inequality we get for any $n \geq 1$ that

$$0 < (1 + 6\lambda)^n m \leq f^n(m) \leq (1 + 6\lambda)^n m + 6\lambda \left( \frac{1 - (1 + 6\lambda)^n}{1 - (1 + 6\lambda)} \right).$$

Therefore, putting $m = 6\lambda$ to get the maximum of $f^{n-1}(m)$, we conclude that

$$0 < 6\lambda(1 + 6\lambda)^{n-1} \leq m_n^+ \leq (1 + 6\lambda)^n - 1,$$

which proves the claim about $m_n^+$.

Finally, assume that $\lambda > \frac{1}{3}$ and fix $\varepsilon > 0$ such that $\lambda > \frac{2+\varepsilon}{6} > \frac{1}{3}$. Let $x \notin \tilde{\mathcal{E}}$, which implies that $\beta_{1,2}(x, n) \to \infty$. Proceeding as for $m_n^-$, we have, as soon as $m_n(x) \neq 0$ and $u_n(x) = 1$ or $u_n(x) = 2$,

$$|m_{n+1}(x)| - (1 + \varepsilon)|m_n(x)| \geq 6\lambda\sqrt{1 + m_n^2(x)} - (2 + \varepsilon)|m_n(x)| \geq (6\lambda - (2 + \varepsilon))|m_n(x)| > 0,$$

hence

$$|m_{n+1}(x)| > (1 + \varepsilon)|m_n(x)| > 0.$$

Each time $u_n(x) = 1$ or $u_n(x) = 2$, the above inequality can be applied and when $u_n(x) = 0$ or $u_n(x) = 3$, $m_{n+1}(x) = m_n(x)$. Since $x \notin \tilde{\mathcal{E}}$, the sequence of its digits $u_n(x)$ has to take at least one of the two values 1 or 2 for infinitely many $n$, hence $|m_n(x)| \to \infty$. $\square$

We end this section by stating an upper bound for the difference $F_{n+1}^\lambda(z) - F_n^\lambda(z)$, $z \in [0, 1]$, which will be used, in particular, in the proof of Theorem 1.1.

**Lemma 3.6.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$. Then, for all $n \geq 0$, it holds that

$$\sup_{z \in [0,1]} |F_{n+1}^{\lambda}(z) - F_{n}^{\lambda}(z)| \leq \begin{cases} \lambda \left(\frac{1}{3} + 2\lambda\right)^n & \text{if } \lambda \in (\frac{1}{6}, \frac{1}{3}), \\ \lambda \left(\frac{1}{6} + \lambda\right)^n & \text{if } \lambda \in [\frac{1}{3}, \frac{5}{6}). \end{cases} \quad (3.6)$$

The key point is that for all $\lambda \in (\frac{1}{6}, \frac{5}{6})$, (3.6) gives an exponentially decreasing upper bound.

*Proof.* Consider an interval $I_n(x) = (a_n(x), b_n(x))$ of generation $n$ for some $x \in [0, 1] \setminus \mathcal{E}$, on which $F_n^\lambda$ is affine with slope $m_n(x)$. Assume that $m_n(x) \geq 0$ (the case $m_n(x) \leq 0$ is treated in a similar way). It follows from the geometric construction of $F_n^\lambda$ that

$$S_n(x) := \sup_{z \in I_n(x)} |F_{n+1}^{\lambda}(z) - F_n^{\lambda}(z)| = \lambda L_n(x), \quad (3.7)$$

where $L_n(x) = \ell_n(x)\sqrt{1+m_n^2(x)}$ is the length of the line segment $A_n B_n$, denoted by $|A_n B_n|$ see Figure 6 below.
We claim that, for all $k, n \ge 0$,

$$L_{k+n}(x) \le \begin{cases} \left(\frac{1}{3} + 2\lambda\right)^k L_n(x) & \text{if } \lambda \in (\frac{1}{6}, \frac{1}{3}), \\ \left(\frac{1}{6} + \lambda\right)^k L_n(x) & \text{if } \lambda \in [\frac{1}{3}, \frac{5}{6}). \end{cases} \quad (3.8)$$

First, using (3.5) we obtain

$$\begin{aligned} L_{n+1}^2(x) &= \ell_{n+1}^2(x)(1 + m_{n+1}^2(x)) \\ &\le \left(\frac{1}{3}\ell_n(x)\right)^2 \left(1 + (m_n(x) + 6\lambda\sqrt{1+m_n^2(x)})^2\right) \\ &= \frac{1}{9}\ell_n^2(x)(1 + m_n^2(x) + 12\lambda m_n(x)\sqrt{1+m_n^2(x)} + 36\lambda^2(1+m_n^2(x))) \\ &\le \frac{1}{9}\ell_n^2(x)(1 + 12\lambda + 36\lambda^2)(1 + m_n^2(x)) = \left(\frac{1+6\lambda}{3}\right)^2 L_n^2(x), \end{aligned}$$

where $L_{n+1}(x)$ is defined as $L_n(x)$ by using $I_{n+1}(x)$ and $F_{n+1}^\lambda$. Hence, $L_{n+1}(x) \le (\frac{1}{3}+2\lambda)L_n(x)$, which gives the first bound in (3.8) when $k=1$. An induction combined with (3.7) yields the first part of (3.6).

Let us assume now that $\lambda \ge \frac{1}{3}$. Looking at Figure 6 below, we claim that

$$L_{n+1}(x) \le L'_{n+1}(x) \le (\frac{1}{6} + \lambda)L_n(x), \quad (3.9)$$

where $L'_{n+1}(x)$ is the length of the line segment $A'_{n+1}B'_{n+1}$ achieving the maximal (absolute) slope of $F_n^\lambda$ over $I_n(x)$.

<figure><img src="image_7.png" /><figcaption>Figure 6. An upper bound for $|F_{n+1}^\lambda(z) - F_n^\lambda(z)|$ over $I_n(x)$.</figcaption></figure>

Indeed, using (3.5) to set $m'_{n+1}(x) = m_n(x) + 6\lambda\sqrt{1+m_n^2(x)}$, we obtain

$$\begin{aligned} (L'_{n+1}(x))^2 &= (b'_{n+1}(x) - a'_{n+1}(x))^2 (1 + (m'_{n+1}(x))^2) \\ &= \left(\frac{1}{6}\ell_n(x)\right)^2 \left(1 + (m_n(x) + 6\lambda\sqrt{1+m_n^2(x)})^2\right) \\ &= \frac{1}{36}\ell_n^2(x)(1 + m_n^2(x) + 12\lambda m_n(x)\sqrt{1+m_n^2(x)} + 36\lambda^2(1+m_n^2(x))). \end{aligned} \quad (3.10)$$
It follows that

$$ (L'_{n+1}(x))^2 \geq \lambda^2 \ell_n^2(x)(1+m_n^2(x)) \geq (\frac{1}{3}\ell_n(x))^2(1+m_n^2(x)) = |A_n A'_{n+1}|^2 = |C_n B_n|^2. \quad (3.11) $$

Obviously, $L'_{n+1}(x) \geq |B'_{n+1}C_n|$. Thus, independent of whether $I_{n+1}(x)$ equals $(a_n(x), a'_{n+1}(x))$, $(a'_{n+1}(x), b'_{n+1}(x))$, $(b'_{n+1}(x), c_n(x))$, or $(c_n(x), b_n(x))$, we have

$$ L'_{n+1}(x) \geq L_{n+1}(x). \quad (3.12) $$

Next, for the upper bound, one has from (3.10)

$$ \begin{aligned} (L'_{n+1}(x))^2 &= \frac{1}{36} \ell_n^2(x) (1 + m_n^2(x) + 12\lambda m_n(x) \sqrt{1 + m_n^2(x)} + 36\lambda^2(1 + m_n^2(x))) \\ &\leq \frac{1}{36} \ell_n^2(x) (1 + 12\lambda + 36\lambda^2)(1 + m_n^2(x)) = \left(\frac{1+6\lambda}{6}\right)^2 L_n^2(x). \end{aligned} \quad (3.13) $$

Combining (3.12) and (3.13) we obtain (3.9), and the second bound in (3.8).

Finally, since $S_n(x) \leq \lambda L_n(x)$ and the interval $[0, 1]$ may be decomposed into a finite number of intervals of the form $I_n(x)$, we obtain (3.6) using (3.7) with $k=0$ and $L_0(x) = 1$. $\square$

## 3.2. Proof of the convergence part of Theorem 1.1 and non-differentiability of $F^\lambda$.

For every $x \in [0, 1]$, it follows from (3.6) that the series $\sum |F_{n+1}^\lambda(x) - F_n^\lambda(x)|$ is convergent for all values of $\lambda \in (\frac{1}{6}, \frac{5}{6})$. Hence, the existence of the finite value $F^\lambda(x)$.

Next, let us prove that $F^\lambda$ is nowhere differentiable. Let $x_0 \in [0, 1]$.

First, assume that $x_0 \in \mathcal{E}$. Then, there exists a minimal integer $n_0 \geq 0$ such that $x_0$ is a breakpoint of $F_n^\lambda$ for all $n \geq n_0$. Assume $x_0 \neq 1$ and denote by $(x_0, b_n)$ the interval of generation $n \geq n_0$ with extremity $x_0$ on which $F_n^\lambda$ is affine with slope $m_n$. Set $b'_{n+1} = x_0 + \frac{1}{2}(b_n - x_0)$. By construction of the sequence $(F_n^\lambda)_{n \geq n_0}$, we have $m_n = m_{n_0}$ for all $n \geq n_0$, see Figure 7.

<figure><img src="image_8.png" /><figcaption>Figure 7. Non-differentiability of $F^\lambda$ at a point $x_0 \in \mathcal{E}$.</figcaption></figure>

Assume that $m_{n_0} \geq 0$ (the case $m_{n_0} \leq 0$ can be treated similarly). Then,

$$ \forall n \geq n_0, \quad \begin{cases} F^\lambda(b_n) - F^\lambda(x_0) = F_{n+1}^\lambda(b_n) - F_{n+1}^\lambda(x_0) = m_{n_0}(b_n - x_0), \\ F^\lambda(b'_{n+1}) - F^\lambda(x_0) = F_{n+1}^\lambda(b'_{n+1}) - F_{n+1}^\lambda(x_0) = (m_{n_0} + 2\lambda\sqrt{1+m_{n_0}^2})(b'_{n+1} - x_0). \end{cases} $$

Since $b_n, b'_{n+1} \to x_0$ and $m_{n_0} \neq m_{n_0} + 2\lambda\sqrt{1+m_{n_0}^2}$, the function $F^\lambda$ cannot be differentiable at $x_0$. Finally, the case $x_0 = 1$ is a consequence of the relation $F^\lambda(1-x) = F^\lambda(x)$ (see (1.2)).
Assume now that $x_0 \notin \mathcal{E}$. Assuming $F^\lambda$ is differentiable at $x_0$, the mapping

$$D : (h, h') \in (0, \infty)^2 \rightarrow \frac{F^\lambda(x_0 + h) - F^\lambda(x_0 - h')}{h + h'}$$

admits a finite limit at $(0,0)$ equal to $(F^\lambda)'(x_0)$. Consider the interval $I_n(x_0) = (a_n(x_0), b_n(x_0))$ of generation $n$ that contains $x_0$ on which $F_n^\lambda$ is affine with slope $m_n(x_0)$. Suppose, for example, that $m_n(x_0) \ge 0$ (the case $m_n(x_0) \le 0$ may be treated similarly). Set $m = m_n(x_0)$, $a_0 = a_n(x_0)$, $a_1 = a_0 + \frac{1}{3}\ell_n(x_0)$, $a_2 = a_0 + \frac{1}{2}\ell_n(x_0)$, $a_3 = a_0 + \frac{2}{3}\ell_n(x_0)$ and $a_4 = a_0 + \ell_n(x_0) = b_n(x_0)$, see Figure 8.

<figure><img src="image_9.png" /><figcaption>Figure 8. Non-differentiability of $F^\lambda$ at a point $x_0 \notin \mathcal{E}$.</figcaption></figure>

We claim that we can always select points $a_{i_1}, a_{j_1} \in [a_0, x_0)$ and $a_{i_2}, a_{j_2} \in (x_0, a_4]$ such that

$$\left| \frac{F_{n+1}^{\lambda}(a_{j_2}) - F_{n+1}^{\lambda}(a_{j_1})}{a_{j_2} - a_{j_1}} - \frac{F_{n+1}^{\lambda}(a_{i_2}) - F_{n+1}^{\lambda}(a_{i_1})}{a_{i_2} - a_{i_1}} \right| \ge 2\lambda > 0. \quad (3.14)$$

We indicate in the table below how to select $a_{i_1}, a_{i_2}$ and $a_{j_1}, a_{j_2}$ according to the position of $x_0$ inside $I_n(x_0)$, and the values for the corresponding increments $\rho_{i_1,i_2}$ and $\rho_{j_1,j_2}$ of $F_{n+1}^\lambda$ needed in (3.14).

<table><thead><tr><th>x<sub>0</sub> &isin;</th><th>a<sub>i<sub>1</sub></sub></th><th>a<sub>i<sub>2</sub></sub></th><th>&rho;<sub>i<sub>1</sub>,i<sub>2</sub></sub></th><th>a<sub>j<sub>1</sub></sub></th><th>a<sub>j<sub>2</sub></sub></th><th>&rho;<sub>j<sub>1</sub>,j<sub>2</sub></sub></th><th>|&rho;<sub>j<sub>1</sub>,j<sub>2</sub></sub> &minus; &rho;<sub>i<sub>1</sub>,i<sub>2</sub></sub>|</th></tr></thead><tbody><tr><td>(a<sub>0</sub>, a<sub>1</sub>)</td><td>a<sub>0</sub></td><td>a<sub>1</sub></td><td>m</td><td>a<sub>0</sub></td><td>a<sub>2</sub></td><td>m + 2&lambda;&radic;1 + m<sup>2</sup></td><td>2&lambda;&radic;1 + m<sup>2</sup></td></tr><tr><td>(a<sub>1</sub>, a<sub>2</sub>)</td><td>a<sub>1</sub></td><td>a<sub>2</sub></td><td>m + 6&lambda;&radic;1 + m<sup>2</sup></td><td>a<sub>1</sub></td><td>a<sub>3</sub></td><td>m</td><td>6&lambda;&radic;1 + m<sup>2</sup></td></tr><tr><td>(a<sub>2</sub>, a<sub>3</sub>)</td><td>a<sub>1</sub></td><td>a<sub>3</sub></td><td>m</td><td>a<sub>2</sub></td><td>a<sub>3</sub></td><td>m &minus; 6&lambda;&radic;1 + m<sup>2</sup></td><td>6&lambda;&radic;1 + m<sup>2</sup></td></tr><tr><td>(a<sub>3</sub>, a<sub>4</sub>)</td><td>a<sub>2</sub></td><td>a<sub>4</sub></td><td>m &minus; 2&lambda;&radic;1 + m<sup>2</sup></td><td>a<sub>3</sub></td><td>a<sub>4</sub></td><td>m</td><td>2&lambda;&radic;1 + m<sup>2</sup></td></tr></tbody></table>

The lower bound in (3.14) then holds, since $6\lambda\sqrt{1+m^2} \ge 2\lambda\sqrt{1+m^2} \ge 2\lambda$.
Since $F_{n+1}^{\lambda}$ and $F^{\lambda}$ coincides at points $a_k$, we can rewrite (3.14) as

$$|D(a_{j2} - x_0, x_0 - a_{j1}) - D(a_{i2} - x_0, x_0 - a_{i1})| \ge 2\lambda.$$

Since $a_k \to x_0$, the function $D$ does not satisfy Cauchy's condition at $(0,0)$ and then cannot have a limit at this point. Hence, a contradiction.

Finally, let $\lambda \ge \frac{5}{6}$, and consider again the special point $\bar{x} \in [0,1]$ whose digits are all $u_n(x) = 1$. We call $\bar{x}_n = 0.11\cdots11$ its approximation with $n$ digits equal to 1 and $y_n = F^{\lambda}(\bar{x}_n) = F_n^{\lambda}(\bar{x}_n)$. By construction, for all $n \ge 1$, one has $\ell_n(\bar{x}) = \frac{1}{6^n}$, and $\bar{x}_n = \frac{1}{3}(1 + \frac{1}{6} + \cdots + \frac{1}{6^n})$. The slopes $m_n(\bar{x})$ satisfy $m_{n+1}(\bar{x}) = m_n(\bar{x}) + 6\lambda\sqrt{1 + m_n^2(\bar{x})}$. From $m_1(\bar{x}) = 6\lambda \ge 5$, we deduce that $m_n(\bar{x})$ tends to $\infty$, and in particular

$$m_{n+1}(\bar{x}) \ge (1 + 6\lambda)m_n(\bar{x}) \ge 6m_n(\bar{x}).$$

Again, since $m_1(\bar{x}) \ge 5$, we conclude that for some positive constant $C > 0$, $m_n(\bar{x}) \ge C \cdot 6^n$. Finally, observe that $y_{n+1} = y_n + (\bar{x}_{n+1} - \bar{x}_n)m_{n+1}(\bar{x})$. Combining this, the fact that $\bar{x}_{n+1} - \bar{x}_n = \frac{1}{3 \cdot 6^{n+1}}$ with the previous inequality on $m_n(\bar{x})$, yields $y_{n+1} - y_n \ge C'$ for some positive constant $C' > 0$. In particular, $\lim_{n \to \infty} y_n = \infty$, and $F^{\lambda}(x) = \infty$. This concludes the proof.

**Remark 3.7.** Observe that the last part of the proof holds true for any $x \in [0,1]$ whose digits are all ultimately 1, and actually for many more points. It would be interesting to characterize the set of convergence of $F^{\lambda}$ in this range of parameters.

### 3.3. The increments of $F^{\lambda}$.

We begin this section with a lemma which will be useful in the sequel. It aims to compare $I_n(x)$ and $I_n(y)$ when $x$ and $y$ are close enough.

**Lemma 3.8.** Let $x, y \in [0,1] \setminus \mathcal{E}$ such that $|x - y| \le \frac{1}{2}\ell_n(x)$. Then, either $I_n(x) = I_n(y)$ or $I_n(x)$ and $I_n(y)$ are contiguous. Moreover, $\frac{1}{2}\ell_n(y) \le \ell_n(x) \le 2\ell_n(y)$ and

$$|m_n(y)| \le (1 + 6\lambda)|m_n(x)| + 6\lambda \le 6(|m_n(x)| + 1). \quad (3.15)$$

*Proof.* If $y \in I_n(x)$ then $I_n(y) = I_n(x)$, hence $\ell_n(y) = \ell_n(x)$ and $m_n(y) = m_n(x)$ so the result is obvious. Assume that $y \notin I_n(x)$. Without limiting generality, we can assume that $x < y$. Consider the unique integer $0 \le p \le n - 1$ such that $I_p(x) = I_p(y)$ but $I_{p+1}(x) \ne I_{p+1}(y)$. The construction process implies that

$$\frac{1}{2}\ell_{p+1}(y) \le \ell_{p+1}(x) \le 2\ell_{p+1}(y). \quad (3.16)$$

Obviously $\ell_n(x) \le \ell_{p+1}(x) \le \ell_p(x)/3$ and any interval of generation $p+1$ included in $I_p(x)$ is of length at least $\ell_p(x)/6$. Since by assumption $|x - y| \le \frac{1}{2}\ell_n(x)$, $x$ and $y$ must belong to consecutive intervals of generation $p+1$. Any interval $I \in \mathcal{I}_n$ contained in $I_{p+1}(x)$ is of length at most $\ell_{p+1}(x)/3^{n-p-1}$ and the ones at the extremities of $I_{p+1}(x)$ are exactly of this length. So, if there is a point $y \notin I_p(x)$ at distance $|x - y| < \ell_n(x)/2 < \ell_n(x)$, then $I_n(x)$ must be the rightmost subinterval of $I_{p+1}(x)$ of generation $n$, and $\ell_n(x) = \ell_{p+1}(x)/3^{n-p-1}$. Similarly, we see that $I_n(y)$ should be the leftmost subinterval in $\mathcal{I}_n$ of the generation $n$ included in $I_{p+1}(y)$ of length $\ell_{p+1}(y)/3^{n-p-1}$. Then it follows from (3.16) that $\frac{1}{2}\ell_n(y) \le \ell_n(x) \le 2\ell_n(y)$. Next, given the relative position of $I_n(x)$ and $I_n(y)$, the construction process implies that

$$m_k(y) = m_k(x) \text{ for all } k \in \{0, \dots, p\} \text{ but } m_{p+1}(y) \ne m_{p+1}(x). \quad (3.17)$$
It remains to prove (3.15) for $n = p + 1$. Noting that $m_p(y) = m_p(x)$, it is a direct consequence of Lemma 3.3 (see also Remark 3.4). Since $x$ and $y$ are close to the intersection point of $I_k(x)$ and $I_k(y)$ for $k = p + 1, \dots, n$, one concludes that $m_k(x) = m_{p+1}(x)$ and $m_k(y) = m_{p+1}(y)$ for $k = p + 1, \dots, n$. $\square$

Next, we compute $h_{F^\lambda}(x_0)$ for $x_0 \notin \mathcal{E}$. To this end, first we give an upper bound of the local increments of $F^\lambda$ using the slopes of $F_n^\lambda$.

**Proposition 3.9.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$ and $n \ge 0$. Let $x, y \in [0, 1]$ such that

$$ \frac{1}{12}\ell_n(x) \le |x - y| \le \frac{1}{2}\ell_n(x). \qquad (3.18) $$

Assume that $x \notin \mathcal{E}$. Then we have

$$ |F^{\lambda}(x) - F^{\lambda}(y)| \le C(|m_n(x)| + 1)|x - y|, \qquad (3.19) $$

where the constant $C > 0$ does not depend on $n$, nor on the points $x, y$.

*Proof.* First, we deal with the case $y \in I_n(x)$. Since $F_n^\lambda$ is affine on $I_n(x)$ with slope $m_n(x)$,

$$ \begin{aligned} |F^{\lambda}(x) - F^{\lambda}(y)| &\le |F_n^{\lambda}(x) - F_n^{\lambda}(y)| + |(F^{\lambda} - F_n^{\lambda})(x) - (F^{\lambda} - F_n^{\lambda})(y)| \\ &\le |m_n(x)||x - y| + 2 \sup_{z \in I_n(x)} |F^{\lambda}(z) - F_n^{\lambda}(z)|. \end{aligned} \qquad (3.20) $$

Writing $F^{\lambda}(z) - F_n^{\lambda}(z) = \sum_{k=n}^{\infty} F_{k+1}^{\lambda}(z) - F_k^{\lambda}(z)$, it follows from (3.7) and (3.8) that

$$ \sup_{z \in I_n(x)} |F^{\lambda}(z) - F_n^{\lambda}(z)| \le \lambda \sum_{k=n}^{\infty} L_k(x) \le \lambda \sum_{k=0}^{\infty} L_{k+n}(x) \le C_1 L_n(x) \qquad (3.21) $$

with $C_1 > 0$. Moreover,

$$ L_n(x) = \ell_n(x) \sqrt{1 + m_n^2(x)} \le \ell_n(x)(1 + |m_n(x)|) \le 12|x - y|(1 + |m_n(x)|) \qquad (3.22) $$

which yields (3.19) by combining (3.18), (3.20), (3.21) and (3.22).

Suppose now that $y \notin I_n(x)$. From Lemma 3.8, the interval $I_n(y)$ is contiguous to $I_n(x)$. We assume that $I_n(y)$ precedes $I_n(x)$ with $m_n(x) \ge 0$ (the other cases are treated similarly). We proceed as for the first case but using the intermediate point $a_n(x)$ by writing

$$ |F^{\lambda}(x) - F^{\lambda}(y)| \le |F^{\lambda}(x) - F^{\lambda}(a_n(x))| + |F^{\lambda}(y) - F^{\lambda}(a_n(x))|. \qquad (3.23) $$

Noticing that $F^{\lambda}(a_n(x)) = F_n^{\lambda}(a_n(x))$, we obtain, with (3.18), (3.21) and (3.22):

$$ \begin{aligned} |F^{\lambda}(x) - F^{\lambda}(a_n(x))| &\le |F_n^{\lambda}(x) - F_n^{\lambda}(a_n(x))| + |F^{\lambda}(x) - F_n^{\lambda}(x)| \\ &\le |m_n(x)||x - a_n(x)| + C_1 L_n(x) \\ &\le |m_n(x)|\ell_n(x) + C_1 \ell_n(x)(1 + |m_n(x)|) \\ &\le C'_1(1 + |m_n(x)|)|x - y|, \end{aligned} \qquad (3.24) $$

and

$$ \begin{aligned} |F^{\lambda}(y) - F^{\lambda}(a_n(x))| &\le |F_n^{\lambda}(y) - F_n^{\lambda}(a_n(x))| + |F^{\lambda}(y) - F_n^{\lambda}(y)| \\ &\le |m_n(y)||y - a_n(x)| + C_1 L_n(y) \\ &\le |m_n(y)|\ell_n(y) + C_1 \ell_n(y)(1 + |m_n(y)|). \end{aligned} \qquad (3.25) $$
Using Lemma 3.8 to compare $\ell_n(y)$ with $\ell_n(x)$, and $m_n(y)$ with $m_n(x)$, (3.25) implies that

$$|F^{\lambda}(y) - F^{\lambda}(a_n(x))| \le (1 + 6C_1)\ell_n(x)(1 + |m_n(x)|) \le C_1''|x - y|(1 + |m_n(x)|). \quad (3.26)$$

Finally, by combining (3.23), (3.24) and (3.26) we obtain the desired inequality (3.19). □

**Proposition 3.10.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$ and $n \ge 0$. Then, for all $x \in [0, 1] \setminus \mathcal{E}$,

$$|F^{\lambda}(x) - F^{\lambda}(a_n(x))| \ge |m_n(x)||x - a_n(x)|. \quad (3.27)$$

*Proof.* Recall the definition of $\mathcal{E}$ given in Definition 2.1. The result is obvious when $m_n(x) = 0$. Suppose that $m_n(x) \ne 0$. Since $I_n(x) = (a_n(x), a_n(x) + \varepsilon_n(x)\ell_n(x))$, it follows from Lemma 3.2 that sgn($m_n(x)$) = sgn($x - a_n(x)$). Thus, $F_n^{\lambda}(x) - F_n^{\lambda}(a_n(x)) > 0$ and then $F^{\lambda}(x) \ge F_n^{\lambda}(x) > F_n^{\lambda}(a_n(x)) = F^{\lambda}(a_n(x))$. Therefore, we obtain successively

$$\begin{aligned} |F^{\lambda}(x) - F^{\lambda}(a_n(x))| &= F^{\lambda}(x) - F^{\lambda}(a_n(x)) = F^{\lambda}(x) - F_n^{\lambda}(a_n(x)) \\ &= (F^{\lambda}(x) - F_n^{\lambda}(x)) + (F_n^{\lambda}(x) - F_n^{\lambda}(a_n(x))) \\ &\ge F_n^{\lambda}(x) - F_n^{\lambda}(a_n(x)) = m_n(x)(x - a_n(x)) = |m_n(x)||x - a_n(x)|. \end{aligned}$$

That concludes the proof. □

### 3.4. Estimates on the pointwise Hölder exponents of $F^{\lambda}$.

The first proposition concerns the pointwise Hölder exponent of $F^{\lambda}$.

**Proposition 3.11.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$. For every $x_0 \in [0, 1]$ we have $h_{F^{\lambda}}(x_0) \le 1$, and $h_{F^{\lambda}}(x_0) = 1$ if $x \in \tilde{\mathcal{E}}$.

*Proof.* By Theorem 1.1 the function $F^{\lambda}$ is nowhere differentiable, hence $h_{F^{\lambda}}(x_0) \le 1$ for all $x_0 \in [0, 1]$.

Let $x_0 \in \tilde{\mathcal{E}}$.

First, assume that $x_0 \in \mathcal{E}$, so $x_0$ is a breakpoint of $F_n^{\lambda}$ for some $n \ge 1$. We call $[b', x_0]$, $[x_0, b] \in \mathcal{I}_n$ the two consecutive intervals of generation $n$ with extremity $x_0$, on which $F_n^{\lambda}$ is affine. Let us deal with the interval $[x_0, b]$, the case $[b', x_0]$ is similar. Observe that from its construction, $F^{\lambda}$ enjoys a sort of “limited self-similarity” property (see (1.2)):

$$F^{\lambda}(x_0 + 3(x - x_0)) - F^{\lambda}(x_0) = 3(F^{\lambda}(x) - F^{\lambda}(x_0)) \quad (3.28)$$

holds for all $x \in (x_0, x_0 + \frac{b-x_0}{3}]$. Denoting by $M > 0$ the maximum of $F^{\lambda}(x)$ on $[0, 1]$, we obtain

$$\left| \frac{F^{\lambda}(x) - F^{\lambda}(x_0)}{x - x_0} \right| \le \frac{3 \cdot 2M}{b - x_0} \quad (3.29)$$

for all $x \in (x_0 + \frac{b-x_0}{3}, b]$. Using repeatedly (3.28), for any $n \ge 1$, if $x \in (x_0 + \frac{b-x_0}{3n}, x_0 + \frac{b-x_0}{3n-1}]$ then (3.29) holds. This implies that the absolute values of the right Dini derivatives of $F^{\lambda}$ at $x_0$ are bounded. A similar argument shows the same result for the left Dini derivatives of $F^{\lambda}$. Thus $h_{F^{\lambda}}(x_0) \ge 1$, hence $h_{F^{\lambda}}(x_0) = 1$.

Assume now that $x_0 \in \tilde{\mathcal{E}} \setminus \mathcal{E}$. By construction, there exists $n_{x_0} \ge 1$ such that for every $n \ge n_{x_0}$, $m_n(x_0) = m \in \mathbb{R}$. For every $y$ such that $|y - x_0| \le \ell_{n_{x_0}}(x_0)/12$, there exists (at least) one integer $n$ such that (3.18) holds true (this follows from the fact that $\ell_n(x)/6 \le \ell_{n+1}(x) \le \ell_n(x)/3$). Then Proposition 3.9 applies, and $|F^{\lambda}(x_0) - F^{\lambda}(y)| \le C(|m_n(x_0)| + 1)|x_0 - y| = C(m + 1)|x_0 - y|$. This gives $h_{F^{\lambda}}(x_0) \ge 1$, which allows us to conclude. □
In particular, it follows that the pointwise Hölder exponent of $F^\lambda$ at every $x \in [0, 1]$ is conveniently given by the formula

$$h_{F^\lambda}(x) = \liminf_{y \to x} \frac{\log |F^\lambda(y) - F^\lambda(x)|}{\log |y - x|}. \qquad (3.30)$$

Now, with Propositions 3.9, 3.10 and 3.11 we are able to obtain a first expression for $h_{F^\lambda}(x)$.

**Proposition 3.12.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$. For every $x \in [0, 1] \setminus \mathcal{E}$, we have

$$h_{F^\lambda}(x) = 1 - \limsup_{n \to \infty} \frac{\log |m_n(x)|}{|\log \ell_n(x)|}. \qquad (3.31)$$

*Proof.* Fix such an $x$. Since $x \notin \mathcal{E}$ we have $m_n(x) = 0$ only for finitely many $n$. We set $\alpha_n(x) = \log(|m_n(x)|\ell_n(x))/\log \ell_n(x)$ and $\alpha(x) = \liminf_{n \to \infty} \alpha_n(x)$. We have to prove that $h_{F^\lambda}(x) = \alpha(x)$, which is equivalent to (3.31).

We first show that $h_{F^\lambda}(x) \ge \alpha(x)$. Let $\varepsilon > 0$. By the definition of $\alpha(x)$, there exists an integer $N \ge 1$ such that, for all $n \ge N$, it holds that $\alpha_n(x) > \alpha(x) - \varepsilon$. This provides $|m_n(x)| \le \ell_n(x)^{\alpha(x)-\varepsilon-1}$. Let $\delta = \frac{1}{2}\ell_N(x) > 0$. Let $y \in [0, 1] \setminus \mathcal{E}$ such that $|x - y| < \delta$. We can find $n \ge N$ such that $\frac{1}{12}\ell_n(x) \le |x - y| \le \frac{1}{2}\ell_n(x)$. Hence, by using Proposition 3.9, we obtain, with absolute constants $C, C' > 0$,

$$|F^\lambda(x) - F^\lambda(y)| \le C|m_n(x)||x - y| \le C'|x - y|^{\alpha(x)-\varepsilon}.$$

Therefore, $h_{F^\lambda}(x) \ge \alpha(x) - \varepsilon$. Since this is true for every $\varepsilon > 0$ we get $h_{F^\lambda}(x) \ge \alpha(x)$.

Next, we prove that $h_{F^\lambda}(x) \le \alpha(x)$. Let $\varepsilon > 0$ and $\delta > 0$. Let $N \ge 1$ be an integer such that $\ell_N(x) \le \delta$. By the definition of $\alpha(x)$, there exists an integer $n \ge N$ such that $\alpha_n(x) < \alpha(x) + \varepsilon$. This provides $|m_n(x)| \ge \ell_n(x)^{\alpha(x)+\varepsilon-1}$. Hence, by using Proposition 3.10, we obtain the existence of a point $y = a_n(x)$ satisfying $|x - y| \le \ell_n(x) < \delta$ and such that

$$|F^\lambda(x) - F^\lambda(y)| \ge |m_n(x)||x - y| \ge |x - y|^{\alpha(x)+\varepsilon}.$$

Therefore, $h_{F^\lambda}(x) \le \alpha(x) + \varepsilon$. Since this is true for every $\varepsilon > 0$ we get $h_{F^\lambda}(x) \le \alpha(x)$. $\square$

The next step is to express $h_{F^\lambda}(x)$ in terms of the frequencies of the digits of $x$ through successive slopes $m_n(x)$. Since a precise result will be obtained when $\lim_{n \to \infty} |m_n(x)| = \infty$, we introduce the following particular set of points.

**Definition 3.13.** We denote by $\mathcal{I}$ the set of those $x \in [0, 1] \setminus \mathcal{E}$ for which

$$\lim_{n \to \infty} |m_n(x)| = +\infty.$$

**Proposition 3.14.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$. For every $x \in [0, 1] \setminus \mathcal{E}$, we have

$$h_{F^\lambda}(x) \le 1 - \max \left( \limsup_{n \to \infty} \frac{\beta_1(x, n) \log(6\lambda + 1) + \beta_2(x, n) \log(6\lambda - 1)}{\beta_{0,3}(x, n) \log 3 + \beta_{1,2}(x, n) \log 6}, 0 \right), \qquad (3.32)$$

and equality holds if $x \in \mathcal{I}$.

*Proof.* Fix $x \in [0, 1] \setminus \mathcal{E}$. If the $\limsup$ in (3.32) is negative or zero, then by Lemma 3.11 $h_{F^\lambda}(x) \le 1$, and the result is obvious. Assume that the $\limsup$ is positive. By Proposition 3.12, we have to prove that

$$\limsup_{n \to \infty} \frac{\log |m_n(x)|}{|\log \ell_n(x)|} \ge \limsup_{n \to \infty} \frac{\beta_1(x, n) \log(6\lambda + 1) + \beta_2(x, n) \log(6\lambda - 1)}{\beta_{0,3}(x, n) \log 3 + \beta_{1,2}(x, n) \log 6}, \qquad (3.33)$$
and that the equality holds when $x \in \mathcal{I}$.

Let us introduce the following functions $(f_i)_{i=0,...,3}$ and $(g_i)_{i=0,...,3}$:

$$f_i(t) = \begin{cases} t & \text{if } i \in \{0, 3\}, \\ (6\lambda + 1)t & \text{if } i = 1, \\ (6\lambda - 1)t & \text{if } i = 2, \end{cases} \quad \text{and} \quad g_i(t) = \begin{cases} 1 & \text{if } i \in \{0, 3\}, \\ \frac{6\lambda\sqrt{1 + 1/t^2} + 1}{6\lambda + 1} & \text{if } i = 1, \\ \frac{6\lambda\sqrt{1 + 1/t^2} - 1}{6\lambda - 1} & \text{if } i = 2. \end{cases} \quad (3.34)$$

We also set $g_i(0) = 1$ for all $i$. Observe that $g_i(t) \to 1$ when $t \to +\infty$. Since $x \notin \mathcal{E}$ and the lim sup in (3.32) is positive, we may assume without loss of generality that the integer $p$ associated with $x$ in Lemma 3.3 is equal to $p = 0$ and $\beta_1(x, 1) = 1$, that is $u_1(x) = 1$. Then $m_0(x) = 0$ and $|m_1(x)| = 6\lambda$. For all $n \ge 1$,

$$|m_{n+1}(x)| = f_{u_n(x)}(|m_n(x)|)g_{u_n(x)}(|m_n(x)|).$$

Due to the functions $g_i$, the sequence of slopes $(m_n(x))_{n \ge 0}$ is not a multiplicative process. Our strategy is to compare it with the “idealized” sequence $(\tilde{m}_n(x))_{n \ge 0}$ we would obtain if there were no $g_i$, and show that they have the same asymptotics when $x \in \mathcal{I}$. Thus, let us define the sequence $(\tilde{m}_n(x))_{n \ge 0}$ by $\tilde{m}_0(x) = 0$, $\tilde{m}_1(x) = 6\lambda$ and for all $n \ge 1$,

$$\tilde{m}_{n+1}(x) = f_{u_n(x)}(\tilde{m}_n(x)).$$

Elementary computation gives that for all $n \ge 1$,

$$\tilde{m}_n(x) = 6\lambda(6\lambda + 1)^{\beta_1(x,n)-1}(6\lambda - 1)^{\beta_2(x,n)} > 0. \quad (3.35)$$

Recalling $\ell_n(x) = 3^{-\beta_{0,3}(x,n)}6^{-\beta_{1,2}(x,n)}$, see (2.4), we obtain

$$\limsup_{n \to \infty} \frac{\log \tilde{m}_n(x)}{|\log \ell_n(x)|} = \limsup_{n \to \infty} \frac{\beta_1(x, n) \log(6\lambda + 1) + \beta_2(x, n) \log(6\lambda - 1)}{\beta_{0,3}(x, n) \log 3 + \beta_{1,2}(x, n) \log 6}. \quad (3.36)$$

Therefore, (3.33) will be a consequence of

$$\limsup_{n \to \infty} \frac{\log |m_n(x)|}{|\log \ell_n(x)|} \ge \limsup_{n \to \infty} \frac{\log \tilde{m}_n(x)}{|\log \ell_n(x)|}. \quad (3.37)$$

We easily prove by induction that, for all $n \ge 2$,

$$|m_n(x)| = \tilde{m}_n(x) \prod_{k=1}^{n-1} g_{u_k(x)}(|m_k(x)|). \quad (3.38)$$

It follows that

$$\frac{\log |m_n(x)|}{|\log \ell_n(x)|} = \frac{\log \tilde{m}_n(x)}{|\log \ell_n(x)|} + \left[ \frac{1}{|\log \ell_n(x)|} \sum_{k=1}^{n-1} \log g_{u_k(x)}(|m_k(x)|) \right]. \quad (3.39)$$

Straightforward calculations yield $g_2(x) \ge g_1(x) \ge g_0(x) = g_3(x) = 1$. It follows that the term inside the brackets above is non-negative, which in turn yields (3.37).

Next, using (2.4) and the observation after (2.3), we get

$$|\log \ell_n(x)| = \beta_{0,3}(x, n) \log 3 + \beta_{1,2}(x, n) \log 6 \ge (\beta_{0,3}(x, n) + \beta_{1,2}(x, n)) \log 3 = n \log 3.$$
Therefore, the following upper bound holds:

$$0 \le \frac{1}{|\log \ell_n(x)|} \sum_{k=1}^{n-1} \log g_{u_k(x)}(|m_k(x)|) \le \frac{1}{\log 3} \left[ \frac{1}{n} \sum_{k=1}^{n-1} \log \left( \frac{6\lambda\sqrt{1+1/m_k^2(x)}-1}{6\lambda-1} \right) \right].$$

Assuming that $x \in \mathcal{I}$, i.e. $|m_n(x)| \to +\infty$, we can apply Cesàro's convergence lemma to show that the term inside brackets tends to 0. With (3.39), this implies that (3.37) is an equality and concludes the proof. □

Observe that the lim sup in (3.32) always belongs to $[0, 1]$ when $\lambda \in [\frac{1}{3}, \frac{5}{6}]$ since

$$0 \le \beta_1(x, n) \log(6\lambda + 1) + \beta_2(x, n) \log(6\lambda - 1) \le \beta_{1,2}(x, n) \log 6.$$

We end this section by identifying the points $x$ at which $F^\lambda$ has an infinite derivative, concluding the proof of Theorem 1.1.

**Proposition 3.15.** For every $x \in [0, 1]$, we call $\bar{\beta}_i(x) = \lim_{n \to \infty} \beta_i(x, n) \in [0, +\infty]$, and set

$$\mathcal{V} = \{x \in [0, 1] : \bar{\beta}_1(x) = +\infty \text{ and } \bar{\beta}_2(x) < +\infty\}.$$

Then, $F^\lambda$ has a positive (resp. negative) infinite derivative at every $x \in \mathcal{V}$ such that $\bar{\beta}_2(x)$ is even (resp. odd), and $\dim_{\mathcal{H}}(\mathcal{V}) = s$ where $s$ is the solution to (1.17).

*Proof.* This follows from what precedes. Indeed, when $\bar{\beta}_2(x)$ is even, $\bar{\beta}_1(x) = +\infty$ implies that $\lim_{n \to \infty} m_n(x) = +\infty$, and then (3.19) and (3.27) show that $F^\lambda$ has a positive infinite derivative at $x$. When $\bar{\beta}_2(x)$ is odd, $\lim_{n \to \infty} m_n(x) = -\infty$ and the same argument proves that $F^\lambda$ has a negative infinite derivative at $x$. This shows that $F^\lambda$ has an infinite derivative at every $x \in \mathcal{V}$. The value of the dimension $\dim_{\mathcal{H}}(\mathcal{V}) = s$ has already been discussed in the introduction.

Finally, if $\bar{\beta}_2(x) = +\infty$, then the sign of the slope of $F_n^\lambda$ at $x$ changes infinitely many times with $n$. In particular, noting that the value of $F^\lambda$ and $F_n^\lambda$ coincide at every point of non-differentiability of $F_n^\lambda$, there are two sequences of points $(x_m)_{m \ge 1}$ and $(y_m)_{m \ge 1}$ both converging to $x$, such that for every $m$, one has $F^\lambda(x_m) - F^\lambda(x) < 0 < F^\lambda(y_m) - F^\lambda(x)$. This means that the only possibility for $F^\lambda$ to be differentiable at $x$ would be that $F^\lambda$ has derivative 0 at $x$, which contradicts Theorem 1.1 and the nowhere differentiability of $F^\lambda$.

As a conclusion, the only points where $F^\lambda$ has an infinite derivative are those of $\mathcal{V}$. □

### 3.5. Proof of Theorem 1.4: non-self-similarity of $F^\lambda$.

The existence of points with infinite derivative is probably not what von Koch really hoped for, since a “true” nowhere differentiable function should not have an infinite derivative either. It turns out that these points are useful to prove that the graph $\mathrm{Gr}(F^\lambda)$ of $F^\lambda$ is not a self-similar set. For simplification, let us set $\mathcal{G}_\lambda := \mathrm{Gr}(F^\lambda)$.

Since we work with row vectors, the $N \ge 2$ similarities $(\Phi_i)_{i=1,...,N}$ we consider can be written as

$$\Phi_i(x, y) = \rho_i \cdot (x, y) \begin{pmatrix} \cos \alpha_i & \sin \alpha_i \\ -\sin \alpha_i & \cos \alpha_i \end{pmatrix} + \tau_i \quad \text{with } \rho_i \ge 0, \alpha_i \in [0, 2\pi) \text{ and } \tau_i \in \mathbb{R}^2.$$

Since $\mathcal{G}_\lambda$ is symmetric about the line $x = \frac{1}{2}$, and hence its mirror image can be obtained by a rotation and translation, we can suppose that $\rho_i > 0$ and $\Phi_i$ is orientation preserving.

Let $x_\infty \in \mathcal{V} \cap (0, 1)$ and $P_\infty = (x_\infty, F^\lambda(x_\infty))$ be a point where $F^\lambda$ has derivative $+\infty$, i.e. $\mathcal{G}_\lambda$ has a vertical tangent at $P_\infty$. Observe that for every $i$, $\Phi_i(\mathcal{G}_\lambda) \subset \mathcal{G}_\lambda$ implies that $\Phi_i(\mathcal{G}_\lambda)$ is a connected component of $\mathcal{G}_\lambda$. Hence $\mathcal{G}_\lambda$ has a tangent at the point $\Phi_i(P_\infty)$. Since $F^\lambda$ has no
finite derivative, this is possible only if this tangent is vertical. Hence the rotation angle of $\Phi_i$ is either $\alpha_i = 0$ or $\alpha_i = \pi$.

Set $A = (0,0)$, $B = (1,0)$, $C = (\frac{1}{3},0)$, $E = (\frac{2}{3},0)$ and, for a small $\epsilon > 0$, $C_\epsilon = (\frac{1}{3}+\epsilon, F^\lambda(\frac{1}{3}+\epsilon))$. Denote also by $C_3$ the triadic Cantor set in the interval $AB$ and recall Remark 1.2.

Since $\{\Phi_i(\mathcal{G}_\lambda)\}_{i=1,...,N}$ is a finite covering of $\mathcal{G}_\lambda$ with connected sets, there is an index $j$ and a small $\epsilon > 0$ such that $C, C_\epsilon \in \{(x, F^\lambda(x)): x \in [\frac{1}{3}, \frac{1}{3}+\epsilon]\} \subset \Phi_j(\mathcal{G}_\lambda)$.

First suppose that $\alpha_j = 0$. Since $F^\lambda \ge 0$ and $C \in \Phi_j(\mathcal{G}_\lambda)$, $\Phi_j(A)$ and $\Phi_j(B)$ should both lie on the x-axis: $\Phi_j(A) = (a,0)$ and $\Phi_j(B) = (b,0)$, where $a \le \frac{1}{3} < b$. Actually, since $\Phi_j(B) \in \mathcal{G}_\lambda$ and $F^\lambda$ is strictly positive on $(\frac{1}{3}, \frac{2}{3})$, one must have $b \ge \frac{2}{3}$.

The set $\Phi_j(C_3)$ is a subset of $\mathcal{G}_\lambda$ and lies on the x-axis. Since $\Phi_j(C_3)$ contains all the points of $\Phi_j(\mathcal{G}_\lambda) \cap AB$, one deduces that $\Phi_j(C_3) \subset C_3$. Moreover, since $\Phi_j(\mathcal{G}_\lambda)$ is a connected set and $C, E$ are intermediate points between $A$ and $B$, one necessarily has $C, E \in \Phi_j(C_3)$. This is only possible if $\rho_j = 1$ and $\Phi_j(\mathcal{G}_\lambda) = \mathcal{G}_\lambda$, which is not allowed, since we work with contractions.

Next suppose that $\alpha_j = \pi$. In this situation, one must have $\Phi_j(A) = (a,y)$ and $\Phi_j(B) = (b,y)$ with $0 < b < a$ and $y > 0$. Then $\Phi_j(\mathcal{G}_\lambda) = \{(x, F^\lambda(x)): x \in [b,a]\} \subset \mathcal{G}_\lambda$ and it is an upside-down similar copy of $\mathcal{G}_\lambda$ containing the point $C$.

Denote by $\mathcal{M} = \{(x, F^\lambda(x)): \forall t \in [0,1], F^\lambda(x) \ge F^\lambda(t)\}$ the set of maximum points of $F^\lambda$. Clearly, $\Phi_j(\mathcal{M})$ equals exactly $C_3 \cap \{(x,0): x \in [b,a]\}$. Since $F^\lambda$ has a minimum at $\frac{1}{3}$, $\alpha_j = \pi$ implies that $C \in \Phi_j(\mathcal{M})$. In particular, $0 < b < \frac{1}{3} < a$, and we can find $x_C \in (0,1)$ such that $(x_C, F^\lambda(x_C)) \in \mathcal{M}$ and $\Phi_j((x_C, F^\lambda(x_C))) = C$.

Let us show that this leads to a contradiction. Indeed, note that by using the de Rham's type equations (1.2) and the definition of $F^\lambda$, one has

$$ \text{if } (x, F^{\lambda}(x)) \in \mathcal{M}, \text{ then } \frac{1}{3} + \frac{1}{3 \cdot 6} \le x \le \frac{2}{3} - \frac{1}{3 \cdot 6}. \quad (3.40) $$

Hence, setting $x_M = \max\{x \in [0,1]: (x, F^\lambda(x)) \in \mathcal{M}\}$, it holds that $x_M - x < \frac{1}{3} < 1 - x_M$ for every $(x, F^\lambda(x)) \in \mathcal{M}$.

Now set $x_j = \min\{x \in [0,1]: (x, 0) \in \Phi_j(\mathcal{M})\}$. Since $\Phi_j$ is a similarity with $\alpha_j = \pi$, one has $(x_j, 0) = \Phi_j((x_M, F^\lambda(x_M))) \in C_3$ and $x_j$ is necessarily a left-endpoint of one of the components of $C_3$. Moreover, since $b \le x_j$, the interval $[b, x_j]$ should be fully included in the complementary set of $C_3$, hence in the left gap immediately preceding $x_j$, see Figure 9 below.

<figure><img src="image_10.png" /><figcaption>Figure 9. The relative positions of points $A$, $(b,0)$, $(x_j,0)$ and $C$ when $\alpha_j = \pi$.</figcaption></figure>

It follows in particular that $x_j - b \le \frac{1}{3} - x_j$. But applying (3.40) to $x = x_C$ and then applying $\Phi_j$ yields $x_j - b > \frac{1}{3} - x_j$. This means that the gap in the Cantor set before $x_j \in C_3$ is larger than the distance between $x_j$ and $\frac{1}{3}$, which is impossible for a point in the middle third Cantor set. Hence a contradiction.
## 4. THE MEASURE $\mu_\lambda$ AND ITS MULTIFRACTAL SPECTRUM

In this section, we investigate the multifractal spectrum of $\mu_\lambda$, which paves the way for the proof of Theorem 1.9.

### 4.1. Recalls on the multifractal properties of $\mu_\lambda$.

Let us begin by recalling key notions about multifractal analysis of measures. The lower and upper local dimensions of a probability measure $\mu$ defined on $\mathbb{R}$ are

$$ \underline{\dim}_{\text{loc}}(\mu, x) = \liminf_{r \to 0} \frac{\log \mu(B(x, r))}{\log r} \quad \text{and} \quad \overline{\dim}_{\text{loc}}(\mu, x) = \limsup_{r \to 0} \frac{\log \mu(B(x, r))}{\log r}, \qquad (4.1) $$

and, when these quantities agree, their common value $\dim_{\text{loc}}(\mu, x)$ is called the local dimension of $\mu$ at the point $x$. In addition, the level sets associated with these local dimensions are

$$ \underline{E}_{\mu}(\alpha) = \{x \in \mathbb{R} : \underline{\dim}_{\text{loc}}(\mu, x) = \alpha\}, \quad \overline{E}_{\mu}(\alpha) = \{x \in \mathbb{R} : \overline{\dim}_{\text{loc}}(\mu, x) = \alpha\} $$

and

$$ E_{\mu}(\alpha) = \{x \in \mathbb{R} : \dim_{\text{loc}}(\mu, x) = \alpha\}. $$

We are especially interested in the multifractal spectrum of $\mu$, that is in the mapping $d_\mu : \mathbb{R}^+ \to [0, 1] \cup \{-\infty\}$ defined by

$$ d_{\mu}(\alpha) = \dim_{\mathbb{H}}(\underline{E}_{\mu}(\alpha)) = \dim_{\mathbb{H}}(\{x \in \mathbb{R} : \underline{\dim}_{\text{loc}}(\mu, x) = \alpha\}). \qquad (4.2) $$

Determination of $d_\mu(\alpha)$ requires to consider the following set

$$ \underline{E}_{\mu}^{\ge}(\alpha) = \{x \in \mathbb{R} : \underline{\dim}_{\text{loc}}(\mu, x) \ge \alpha\} \supset \underline{E}_{\mu}(\alpha). \qquad (4.3) $$

Let us return to the measure $\mu_\lambda$ from Definition 1.7. Since it is associated with a hyperbolic IFS satisfying the Open Set Condition, $\mu_\lambda$ is known to enjoy many multifractal properties that will be of great importance in the following and are listed at the end of the section. Before, let us observe that this measure can be alternatively constructed by using the Mass Distribution Principle and considering the successive intervals of generation $n$ from $\mathcal{I}_n$. Precisely, starting with $\mu_\lambda((0, 1)) = 1$ then set $\mu_\lambda(S_i((0, 1))) = p_{\lambda,i}$ for $i = 0, \dots, 3$. In general, assuming that $\mu_\lambda(I_n(x_0))$ is already defined for an interval $I_n(x_0) = (a_n(x_0), a_n(x_0) + \varepsilon_n(x_0)\ell_n(x_0))$ with $x_0 \in [0, 1] \setminus \mathcal{E}$, then set

$$ \left\{ \begin{array}{l} \mu_{\lambda}((a_n(x_0), a_n(x_0) + \frac{1}{3}\varepsilon_n(x_0)\ell_n(x_0))) = p_{\lambda,0}\mu_{\lambda}(I_n(x_0)), \\ \mu_{\lambda}((a_n(x_0) + \frac{1}{3}\varepsilon_n(x_0)\ell_n(x_0), a_n(x_0) + \frac{1}{2}\varepsilon_n(x_0)\ell_n(x_0))) = p_{\lambda,1}\mu_{\lambda}(I_n(x_0)), \\ \mu_{\lambda}((a_n(x_0) + \frac{1}{2}\varepsilon_n(x_0)\ell_n(x_0), a_n(x_0) + \frac{2}{3}\varepsilon_n(x_0)\ell_n(x_0))) = p_{\lambda,2}\mu_{\lambda}(I_n(x_0)), \\ \mu_{\lambda}((a_n(x_0) + \frac{2}{3}\varepsilon_n(x_0)\ell_n(x_0), a_n(x_0) + \varepsilon_n(x_0)\ell_n(x_0))) = p_{\lambda,3}\mu_{\lambda}(I_n(x_0)). \end{array} \right. \qquad (4.4) $$

Since the family $\{\mathcal{I}_n : n \ge 0\}$ generates the Borel $\sigma$-algebra in $[0, 1]$, the iterative construction above defines a Borel measure supported in $[0, 1]$ satisfying (1.12). In particular, it follows from this construction that, for every $x \in [0, 1] \setminus \mathcal{E}$,

$$ \mu_{\lambda}(I_{n}(x)) = \prod_{i=0}^{3} p_{\lambda,i}^{\beta_{i}(x,n)}. \qquad (4.5) $$

Therefore, it will be convenient to deal with quantities $\mu_\lambda(I_n(x))$ instead of $\mu_\lambda(B(x, r))$ to compute the local dimensions in (4.1). The following property of $\mu_\lambda$ makes this possible.
**Lemma 4.1.** Let $\lambda \in (\frac{1}{6}, +\infty)$. Then $\mu_\lambda$ is a doubling measure, i.e. there exists a constant $C_\lambda \ge 1$ such that, for every $x \in [0, 1]$ and every $r > 0$, $\mu_\lambda(B(x, 2r)) \le C_\lambda \mu_\lambda(B(x, r))$.

*Proof.* This will follow from $p_{\lambda,0} = p_{\lambda,3}$ and from Lemma 3.8. Without limiting generality suppose that $0 < 2r < \frac{1}{6}$ and select $n \ge 1$ such that

$$ \ell_n(x) \le 2r < \ell_{n-1}(x). \qquad (4.6) $$

We divide the interval $I_n(x) = (a_n(x), b_n(x))$ into two subintervals $I$ and $I'$ of equal length $\frac{1}{2}|I_n(x)| = \frac{1}{2}\ell_n(x)$. From the definition of $\mu_\lambda$, $\min(\mu_\lambda(I), \mu_\lambda(I')) \ge (\min_{0 \le i \le 3} p_{\lambda,i})\mu_\lambda(I_n(x))$. Since $B(x, r)$ necessarily contains one of the two subintervals $I$ or $I'$, we deduce that $\mu_\lambda(B(x, r)) \ge C\mu_\lambda(I_n(x))$ with some positive constant $C > 0$.

Suppose that $x \le \frac{1}{2}(a_{n-1}(x) + b_{n-1}(x))$, the other case is similar. Choose $y$ such that $I_{n-1}(y)$ and $I_{n-1}(x)$ are contiguous and $I_{n-1}(y)$ is to the left of $I_{n-1}(x)$. By Lemma 3.8 and (4.6), we have $B(x, 2r) \subset I_{n-1}(x) \cup I_{n-1}(y)$. From the definition of $\mu_\lambda$ it follows that $\mu_\lambda(I_{n-1}(x)) \le C'\mu_\lambda(I_n(x))$, where $C' = \max_{0 \le i \le 3} p_{\lambda,i}^{-1} > 0$.

Let $m$ be the largest integer such that $I_{n-1}(x)$ and $I_{n-1}(y)$ are included in the same interval of generation $m \ge 0$. Observe that $m \le n - 2$. This implies that $u_p(x) = u_p(y)$ for every $1 \le p \le m$. The intervals $I_{n-1}(x)$ and $I_{n-1}(y)$ being contiguous, the only possibility for the values of $u_p(x)$ and $u_p(y)$ are $u_{m+1}(x) = u_{m+1}(y) + 1$, and $u_p(x) = 0$ and $u_p(y) = 3$ for every $p \in \{m + 2, \dots, n - 1\}$. This shows that $\mu_\lambda(I_{n-1}(x)) = \mu_\lambda(I_m(x))p_{\lambda, u_{m+1}(x)}^{n-m+2}$ and $\mu_\lambda(I_{n-1}(y)) = \mu_\lambda(I_m(y))p_{\lambda, u_{m+1}(y)}^{n-m+2}$. Since $p_{\lambda,0} = p_{\lambda,3}$ and $I_m(x) = I_m(y)$ we obtain $\mu_\lambda(I_{n-1}(y)) \le C''\mu_\lambda(I_{n-1}(x))$ with a suitable constant $C'' > 0$. Therefore,

$$ \begin{aligned} \mu_\lambda(B(x, 2r)) \le \mu_\lambda(I_{n-1}(x)) + \mu_\lambda(I_{n-1}(y)) &\le (1 + C'')\mu_\lambda(I_{n-1}(x)) \\ &\le C'(1 + C'')\mu_\lambda(I_n(x)) \qquad (4.7) \\ &\le C'(1 + C'')C^{-1}\mu_\lambda(B(x, r)), \end{aligned} $$

which proves the claim by taking $C_\lambda = C'(1 + C'')C^{-1}$. $\square$

From Lemma 4.1, especially from (4.7), it follows that for every $x \in [0, 1]$, in order to compute the local dimensions, one can replace the centered ball $B(x, r)$ by $I_n(x)$, i.e.

$$ \underline{\dim}_{\text{loc}}(\mu_{\lambda}, x) = \liminf_{n \to \infty} \frac{\log \mu_{\lambda}(I_{n}(x))}{\log |I_{n}(x)|}, \qquad (4.8) $$

and similarly for $\overline{\dim}_{\text{loc}}(\mu_\lambda, x)$ and $\dim_{\text{loc}}(\mu_\lambda, x)$. This will simplify many computations in the following.

**Proposition 4.2.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$. For every $x \in [0, 1]$, we have

$$ \underline{\dim}_{\text{loc}}(\mu_{\lambda}, x) = \gamma_{\lambda} - \limsup_{n \to \infty} \frac{\beta_{1}(x, n) \log(6\lambda + 1) + \beta_{2}(x, n) \log(6\lambda - 1)}{\beta_{0,3}(x, n) \log 3 + \beta_{1,2}(x, n) \log 6}. \qquad (4.9) $$
Proof. Using (2.7) and (2.4), we obtain $|I_n(x)| = \ell_n(x) = 3^{-\beta_{0,3}(x,n)}6^{-\beta_{1,2}(x,n)}$. In addition by (1.11) and (4.5), we obtain

$$ 
\begin{aligned}
\frac{\log \mu_{\lambda}(I_n(x))}{\log |I_n(x)|} &= \frac{\beta_0(x, n) \log p_{\lambda,0} + \beta_1(x, n) \log p_{\lambda,1} + \beta_2(x, n) \log p_{\lambda,2} + \beta_3(x, n) \log p_{\lambda,3}}{-\beta_{0,3}(x, n) \log 3 - \beta_{1,2}(x, n) \log 6} \\
&= \frac{-\beta_{0,3}(x, n)\gamma_{\lambda} \log 3 + \beta_1(x, n) \log(6\lambda + 1) + \beta_2(x, n) \log(6\lambda - 1) - \beta_{1,2}(x, n)\gamma_{\lambda} \log 6}{-\beta_{0,3}(x, n) \log 3 - \beta_{1,2}(x, n) \log 6} \\
&= \gamma_{\lambda} - \frac{\beta_1(x, n) \log(6\lambda + 1) + \beta_2(x, n) \log(6\lambda - 1)}{\beta_{0,3}(x, n) \log 3 + \beta_{1,2}(x, n) \log 6}.
\end{aligned}
$$

The result is then a consequence of (4.8). □

Looking at (4.9), we introduce the notion of digit frequencies of $x$.

**Definition 4.3.** A point $x \in [0, 1]$ is called regular if, for every $i = 0, \dots, 3$, the frequency $\beta_i(x)$ of the digit $i$ exists, i.e.

$$ \lim_{n \to \infty} \frac{\beta_i(x, n)}{n} = \beta_i(x). \qquad (4.10) $$

We denote by $\mathcal{R}$ the set of regular points. In particular, if $x \in \mathcal{R}$, then necessarily $(\beta_i(x))_{i=0,\dots,3}$ is a probability vector. Observe that any $x \in \mathcal{E}$ is regular.

Let us now turn to the $L^q$-spectrum $\tau_{\mu_{\lambda}}$ of the measure $\mu_{\lambda}$. We have chosen with Definition 1.8 an implicit but tractable definition of $\tau_{\mu_{\lambda}}(q)$. Let us mention here that a “direct” definition of $\tau_{\mu_{\lambda}}$ in terms of the $q$-moments of $\mu_{\lambda}$ is the following, see e.g. [BMP92]:

$$ \tau_{\mu} : q \in \mathbb{R} \mapsto \liminf_{j \to \infty} \frac{-1}{\log 3^j} \log \sum_{k=0}^{3^j-1} (\mu([k3^{-j}, (k+1)3^{-j}]))^q \in \mathbb{R} \cup \{-\infty, +\infty\}, \quad (4.11) $$

where the summation runs only through indices such that $\mu([k3^{-j}, (k+1)3^{-j}]) \neq 0$. The equality between (4.11) and the solution to (1.13) in Definition 1.8 is proved for self-similar measures that satisfy the Open Set Condition in Corollary C of [LN99].

The properties of $\mu_{\lambda}$ are gathered in the two following propositions. They follow from Theorem 1 of [BMP92] for the upper bounds for all dimensions, Theorems 2 and 3 of [BMP92] for the lower bounds since $\mu_{\lambda}$ is doubling, or Theorems A and B, Corollary C of [LN99] which treats more general cases. The reader may also check [Ols95] for other notions of dimensions and formalisms, and [FL09, SS16, BF21, Shm19, Fen23] for recent spectacular advances on the multifractal formalism for self-similar measures with overlaps (our contractions do not overlap).

**Proposition 4.4.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$ and consider $\mu_{\lambda}$, $\tau_{\mu_{\lambda}}$ and $\tau_{\mu_{\lambda}}^*$. Denote by $[\tilde{\alpha}_{\lambda,\min}, \tilde{\alpha}_{\lambda,\max}]$ the support of $\tau_{\mu_{\lambda}}^*$. Then, the mapping $\alpha \mapsto \tau_{\mu_{\lambda}}^*(\alpha)$ is real analytic concave on the interior of its

support of $\tau_{\mu_{\lambda}}^*$. Then, the mapping $\alpha \mapsto \tau_{\mu_{\lambda}}^*(\alpha)$ is real analytic concave on the interior of its
support, and we have

$$ \tilde{\alpha}_{\lambda, \min} = \lim_{q \to +\infty} \tau'_{\mu_{\lambda}}(q) = \min_{0 \le i \le 3} \frac{\log p_{\lambda, i}}{\log r_i} = \gamma_{\lambda} - \frac{\log(6\lambda + 1)}{\log 6} $$

$$ \tilde{\alpha}_{\lambda, \max} = \lim_{q \to -\infty} \tau'_{\mu_{\lambda}}(q) = \max_{0 \le i \le 3} \frac{\log p_{\lambda, i}}{\log r_i} = \max \left( \gamma_{\lambda}, \gamma_{\lambda} - \frac{\log(6\lambda - 1)}{\log 6} \right). $$

Moreover, $\tau^*_{\mu_{\lambda}}$ reaches its maximum at point $\tilde{\alpha}_{\lambda, \mathcal{L}}$, with $\tau^*_{\mu_{\lambda}}(\tilde{\alpha}_{\lambda, \mathcal{L}}) = 1$, where

$$ \tilde{\alpha}_{\lambda, \mathcal{L}} = \tau'_{\mu_{\lambda}}(0) = \gamma_{\lambda} - \frac{\log(36\lambda^2 - 1)}{4 \log 3 + 2 \log 6}. \quad (4.12) $$

Observe that as long as $\lambda > \frac{1}{3}$, then $\tilde{\alpha}_{\lambda, \max} = \gamma_{\lambda}$, and hence it depends only on $\lambda$ through the constant $\gamma_{\lambda}$. When $\lambda \le \frac{1}{3}$, $\tilde{\alpha}_{\lambda, \max} = \gamma_{\lambda} - \frac{\log(6\lambda - 1)}{\log 6}$ so that the dependence on $\lambda$ is more intricate. We will come back to this fact in Section 4.2 below where in point (7) we also comment on (4.12).

**Proposition 4.5.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$ and consider $\mu_{\lambda}$, $\tau_{\mu_{\lambda}}$ and $\tau^*_{\mu_{\lambda}}$.

(i) For every $\alpha \in [\tilde{\alpha}_{\lambda, \min}, \tilde{\alpha}_{\lambda, \max}]$,

$$ d_{\mu_{\lambda}}(\alpha) = \tau^*_{\mu_{\lambda}}(\alpha) = \dim_{\mathbb{H}}(E_{\mu_{\lambda}}(\alpha)). \quad (4.13) $$

In particular, $\mu_{\lambda}$ satisfies the multifractal formalism.

(ii) For every $\alpha \in [\tilde{\alpha}_{\lambda, \mathcal{L}}, \tilde{\alpha}_{\lambda, \max}]$, i.e. in the decreasing part of the spectrum,

$$ d_{\mu_{\lambda}}(\alpha) = \tau^*_{\mu_{\lambda}}(\alpha) = \dim_{\mathbb{H}}(E^{\ge}_{\mu_{\lambda}}(\alpha)). $$

Let us make some remarks about these results.

**Comments on Proposition 4.5.**

(a) The main reason for the equality between $d\mu_{\lambda}(\alpha) = \dim E_{\mu_{\lambda}}(\alpha)$ and the Hausdorff dimension of $E_{\mu_{\lambda}}(\alpha)$ (last equality of item (i)) is the application of the law of large numbers and Birkhoff's ergodic theorem.

(b) Given $\alpha \ge \tilde{\alpha}_{\lambda, \mathcal{L}}$, item (ii) follows from the large deviations theory (and in particular, Theorem 1 of [BMP92]), and states that the majority of the points that have a lower local dimension greater than $\alpha$ are actually those with the limit local dimension equal to $\alpha$. This is not true when $\alpha < \tilde{\alpha}_{\lambda, \mathcal{L}}$, in this case one has the opposite property that $\dim_{\mathbb{H}}(\{x \in [0, 1] : h_{\mu_{\lambda}}(x) \le \alpha\}) = \tau^*_{\mu_{\lambda}}(\alpha)$, but we will not use this later on.

(c) Writing that $\{x \in [0, 1] : \dim_{\mathbb{H}}(\mu_{\lambda}, x) \ne \tilde{\alpha}_{\lambda, \mathcal{L}}\} = \bigcup_{n \ge 1} \{x \in [0, 1] : \dim_{\mathbb{H}}(\mu_{\lambda}, x) \le \tilde{\alpha}_{\lambda, \mathcal{L}} - 1/n\} \cup \{x \in [0, 1] : \dim_{\mathbb{H}}(\mu_{\lambda}, x) \ge \tilde{\alpha}_{\lambda, \mathcal{L}} + 1/n\}$, comment (b) combined with Proposition 4.4 shows that every set appearing on the right hand-side has a Hausdorff dimension strictly less than 1, hence a Lebesgue measure 0. Hence, the set $\{x \in [0, 1] : \dim_{\mathbb{H}}(\mu_{\lambda}, x) = \tilde{\alpha}_{\lambda, \mathcal{L}}\}$ has full Lebesgue measure in $[0, 1]$.

The next result asserts that one can actually specify that those most important points in a given $E_{\mu_{\lambda}}(\alpha)$ are regular, i.e. they have limit digit frequencies in their decomposition (2.10) and (2.2).

**Lemma 4.6.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$ and consider $\mu_{\lambda}$, $\tau_{\mu_{\lambda}}$ and $\tau^*_{\mu_{\lambda}}$. For any $(\beta_0, \beta_1, \beta_2, \beta_3) \in [0, 1]^4$ set

$$ F_{\beta_0, \beta_1, \beta_2, \beta_3} = \{x \in [0, 1] : x \text{ is regular and } \beta_i(x) = \beta_i \text{ for } i = 0, \dots, 3\}. \quad (4.14) $$
Then, for every $\alpha \in [\tilde{\alpha}_{\lambda,\min}, \tilde{\alpha}_{\lambda,\max}]$, there exists a probability vector $(\beta_0, \beta_1, \beta_2, \beta_3) \in [0, 1]^4$ such that $F_{\beta_0,\beta_1,\beta_2,\beta_3} \subset E_{\mu_\lambda}(\alpha)$ and

$$ \dim_H(F_{\beta_0,\beta_1,\beta_2,\beta_3}) = d_{\mu_\lambda}(\alpha). $$

*Proof.* In Lemmas 2.4 and 5.1 of [GP97] it is proved that given a probability vector $(\beta_i)_{i=0,...,3}$, the set $F_{\beta_0,\beta_1,\beta_2,\beta_3}$ has Hausdorff dimension equal to

$$ \dim_H(F_{\beta_0,\beta_1,\beta_2,\beta_3}) = \frac{\sum_{i=0}^{3} \beta_i \log \beta_i}{\sum_{i=0}^{3} \beta_i \log r_i}. \qquad (4.15) $$

Next, given a possible local dimension $\alpha$ for $\mu_\lambda$, we call $q_\alpha \in \mathbb{R}$ the unique real number such that $\alpha = \tau'_{\mu_\lambda}(q_\alpha)$ (which exists since $\tau_{\mu_\lambda}$ is analytic concave). An immediate computation using the implicit formula (1.13) gives

$$ \alpha = \frac{\sum_{i=0}^{3} p_{\lambda,i}^{q_{\alpha}} r_{i}^{-\tau_{\mu_{\lambda}}(q_{\alpha})} \log p_{\lambda,i}}{\sum_{i=0}^{3} p_{\lambda,i}^{q_{\alpha}} r_{i}^{-\tau_{\mu_{\lambda}}(q_{\alpha})} \log r_{i}}. $$

The multifractal formalism ensures that $\tau_{\mu_\lambda}^*(\alpha) = \alpha q_\alpha - \tau_{\mu_\lambda}(q_\alpha)$. Then, setting $\beta_i = p_{\lambda,i}^{q_\alpha} r_i^{-\tau_{\mu_\lambda}(q_\alpha)}$, (1.13) shows that $(\beta_i)_{i=0,...,3}$ is a probability vector, and then (4.15) yields

$$ \dim_H(F_{\beta_0,\beta_1,\beta_2,\beta_3}) = \frac{\sum_{i=0}^{3} \beta_i \log \beta_i}{\sum_{i=0}^{3} \beta_i \log r_i} = \frac{\sum_{i=0}^{3} \beta_i (q_\alpha \log p_{\lambda,i} - \tau_{\mu_\lambda}(q_\alpha) \log r_i)}{\sum_{i=0}^{3} \beta_i \log r_i} \\ = \alpha q_\alpha - \tau_{\mu_\lambda}(q_\alpha) = \tau_{\mu_\lambda}^*(\alpha), $$

hence the result by (4.13). □

## 4.2. Behavior of $d_{\mu_\lambda}$ in terms of $\lambda$.

This subsection gathers additional remarks, complementary to the previous ones but enlightening the dependence of $d_{\mu_\lambda}$ on $\lambda$.

(1) The minimal and maximal exponents $\tilde{\alpha}_{\lambda,\min}$ and $\tilde{\alpha}_{\lambda,\max}$ are analytic functions of $\lambda$, since the dependence on $\gamma_\lambda$ in $\lambda$ given by (1.10) is analytic. Also, in the range of parameters $\lambda$ of interest, $d_{\mu_\lambda}(\tilde{\alpha}_{\lambda,\min}) = \tau_{\mu_\lambda}^*(\tau_{\mu_\lambda}(+\infty)) = 0$.

(2) When $\lambda \in (\frac{\sqrt{2}}{6}, \frac{1}{3})$, we have

$$ \tilde{\alpha}_{\lambda,\min} = \gamma_{\lambda} - \frac{\log(6\lambda + 1)}{\log 6} = \frac{\log p_{\lambda,1}}{\log r_1} < \frac{\log p_{\lambda,0}}{\log r_0} < \frac{\log p_{\lambda,2}}{\log r_2} = \gamma_{\lambda} - \frac{\log(6\lambda - 1)}{\log 6} = \tilde{\alpha}_{\lambda,\max}. $$

The level set $E_{\mu_\lambda}(\tilde{\alpha}_{\lambda,\max})$ consists of the points for which $\lim_{n \to \infty} \frac{\beta_2(x,n)}{n} = 1$. Indeed, recalling (4.4) which explains how $\mu_\lambda$ spreads the mass of an interval of generation $n$ on its four subintervals, the only possibility for a point $x$ to have a local dimension for $\mu_\lambda$ equal to $\frac{\log p_{\lambda,2}}{\log r_2}$ is that most of the digits of $x$ equal 2, i.e. $\lim_{n \to \infty} \frac{\beta_2(x,n)}{n} = 1$. Then, applying again the multifractal formalism satisfied by $\mu_\lambda$, this set has a Hausdorff dimension equal to $d_{\mu_\lambda}(\tilde{\alpha}_{\lambda,\max}) = \tau_{\mu_\lambda}^*(-\infty) = 0$.

(3) The mapping $(\lambda, \alpha) \mapsto \tau_{\mu_\lambda}^*(\alpha)$ is continuous on $\{(\lambda, \alpha) : \lambda \in (\frac{1}{3}, \frac{5}{6}), \alpha \in [\tilde{\alpha}_{\lambda,\min}, 1]\}$, by properties of the Legendre transform. Hence, applying (4.13), the mapping

$$ \lambda \in (\frac{\sqrt{2}}{6}, \frac{1}{3}) \mapsto \mathrm{Gr}(d_{\mu_{\lambda}}) \in \mathcal{K}(\mathbb{R}^2) \qquad (4.16) $$
is continuous, where we recall that Gr(d<sub>μ<sub>λ</sub></sub>) denotes the (compact) graph of d<sub>μ<sub>λ</sub></sub> and K(R<sup>2</sup>) the space of compact subsets in R<sup>2</sup> equipped with the Hausdorff metric. Propositions 5.4, 5.5 and 6.1 will imply that for these parameter values the spectrum of F<sup>λ</sup> is a translation of that of μ<sub>λ</sub>, but truncated. Figure 10 displays d<sub>F<sup>λ</sup></sub>(α) and d<sub>μ<sub>λ</sub></sub>(α + γ<sub>λ</sub> - 1) for the classical von Koch function, i.e. when λ = 3/6. On [α<sub>λ,min</sub>, 1] these two curves coincide, but the domain of d<sub>μ<sub>λ</sub></sub>(α + γ<sub>λ</sub> - 1) is larger and as Remark (2) implies it reaches the x axis at ˜α<sub>λ,max</sub> - γ<sub>λ</sub> + 1.

(4) Precisely at $\lambda = \frac{1}{3}$,

$$ \frac{\log p_{\lambda,2}}{\log r_2} = \frac{\log p_{\lambda,0}}{\log r_0}, \qquad (4.17) $$

and now the level set E<sub>μ<sub>λ</sub></sub>(˜α<sub>λ,max</sub>) contains the points for which lim<sub>n→∞</sub> β<sub>1</sub>(x,n)/n = 0. In particular, it contains the set {x ∈ [0, 1] : β<sub>1</sub>(x, n) = 0 for every integer n}, which is the inhomogeneous Cantor set described in the introduction, with Hausdorff and box dimensions both equal to the solution to (1.17). It is not difficult to see that d<sub>μ<sub>1</sub> 3</sub>(˜α<sub>1 3,max</sub>) also equals s, again using the multifractal formalism and the coincidence (4.17). So, even if the multifractal spectrum α ↦→ d<sub>μ<sub>1</sub> 3</sub>(α) is still continuous, there is a jump discontinuity at λ = 1/3 from both sides since d<sub>μ<sub>1</sub> 3</sub>(˜α<sub>1 3,max</sub>) > log 2 / log 3 = d<sub>μ<sub>λ</sub></sub>(˜α<sub>λ,max</sub>) for λ ∈ (1/3, 5/6), and d<sub>μ<sub>λ</sub></sub>(˜α<sub>λ,max</sub>) = 0 for λ ∈ (1/6, 1/3), as we observed in (2).

(5) When $\lambda \in (\frac{1}{3}, \frac{5}{6})$, we have

$$ \tilde{\alpha}_{\lambda, \min} = \gamma_{\lambda} - \frac{\log(6\lambda + 1)}{\log 6} = \frac{\log p_{\lambda,1}}{\log r_1} < \frac{\log p_{\lambda,2}}{\log r_2} < \frac{\log p_{\lambda,0}}{\log r_0} = \gamma_{\lambda} = \tilde{\alpha}_{\lambda, \max}. $$

In particular, the level set E<sub>μ<sub>λ</sub></sub>(˜α<sub>λ,max</sub>) contains the points for which lim<sub>n→∞</sub> β<sub>0,3</sub>(x,n)/n = 1. This set has Hausdorff dimension τ<sup>*</sup><sub>μ<sub>λ</sub></sub>(α<sub>λ,max</sub>) = lim<sub>q→-∞</sub> τ'<sub>μ<sub>λ</sub></sub>(q) = log 2 / log 3, and is independent of λ in this range (it contains and has the same dimension as the triadic Cantor set).

(6) When $\lambda \in (\frac{1}{3}, \frac{5}{6}]$, the mapping (4.16) is again continuous, still by (4.13).

(7) When $x$ is chosen randomly Lebesgue uniformly in $[0, 1]$, by the ergodic theorem applied to $T$, for Lebesgue almost every $x$ we have

$$ \frac{\beta_{0,3}(x, n)}{n} \to \frac{2}{3}, \quad \frac{\beta_{1,2}(x, n)}{n} \to \frac{2}{6}, \quad \frac{\beta_{1}(x, n)}{n} \to \frac{1}{6} \quad \text{and} \quad \frac{\beta_{2}(x, n)}{n} \to \frac{1}{6}. \qquad (4.18) $$

Thus, we get heuristically $|I_n(x)| \approx 3^{-2n/3} 6^{-n/3}$ and, using (4.5),

$$ \mu_{\lambda}(I_{n}(x)) \approx p_{\lambda,0}^{n/3} p_{\lambda,1}^{n/6} p_{\lambda,2}^{n/6} p_{\lambda,3}^{n/3}. $$

Considering log μ<sub>λ</sub>(I<sub>n</sub>(x))/log|I<sub>n</sub>(x)|, simplifying and taking its limit when n tends to infinity explains the Lebesgue-almost sure value ˜α<sub>λ,L</sub> given in (4.12).

## 5. THE LOWER BOUND FOR d<sub>F<sup>λ</sup></sub> WHEN λ ∈ (1/6, 5/6) AND THE DECREASING PART OF d<sub>F<sup>λ</sup></sub>

The lower bound we are going to prove for d<sub>F<sup>λ</sup></sub> holds for every λ ∈ (1/6, 5/6) but it turns out that the analysis in the previous section allows us to deal with the subcase when λ ∈ (1/3, 5/6) in an easier way. Indeed, on one hand, Proposition 4.2 gives the explicit value of dim<sub>loc</sub>(μ<sub>λ</sub>, x), and on the other hand, Proposition 3.14 provides an upper bound for the pointwise exponent
of $F^\lambda$. Recalling Definition 3.13 of the set $\mathcal{I}$, the combination of these facts yields the following key result.

**Corollary 5.1.** For every $x \in [0, 1]$, $h_{F^\lambda}(x) \le 1 - \gamma_\lambda + \dim_{\text{loc}}(\mu_\lambda, x)$, with equality when $x \in \mathcal{I}$ and the right hand-side belongs to $[0, 1]$.

Observe also that $1 - \gamma_\lambda + \alpha_{\lambda, \min} \to 0$ when $\lambda \nearrow \frac{5}{6}$, which confirms that our analysis is not applicable when $\lambda > \frac{5}{6}$. Proposition 5.2 below gives the multifractal spectrum of $F^\lambda$, i.e. Theorem 1.9, when $\lambda \in (\frac{1}{3}, \frac{5}{6})$.

**Proposition 5.2.** Let $\lambda \in (\frac{1}{3}, \frac{5}{6})$. Then $d_{F^\lambda}(\alpha) = \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$ for every $\alpha \in [\alpha_{\lambda, \min}, 1]$.

*Proof.* In this range of parameters, by Definition 3.13 and Lemma 3.5 we have $\mathcal{I} = [0, 1] \setminus \tilde{\mathcal{E}}$. When $x \in \tilde{\mathcal{E}}$, $h_{F^\lambda}(x) = 1$, thus

$$d_{F^\lambda}(1) \ge \dim_H(\tilde{\mathcal{E}}) = \frac{\log 2}{\log 3} = d_{\mu_\lambda}(\gamma_\lambda) = d_{\mu_\lambda}(1 + \gamma_\lambda - 1).$$

When $x \notin \tilde{\mathcal{E}}$, then by Lemma 3.5 $x \in \mathcal{I}$, and by Corollary 5.1 $h_{F^\lambda}(x) = \dim_{\text{loc}}(\mu_\lambda, x) + 1 - \gamma_\lambda$. This yields the result, since for every $\alpha \ge 0$, $E_{F^\lambda}(\alpha) = \underline{E}_{\mu_\lambda}(\alpha + \gamma_\lambda - 1)$. $\square$

Now we look for an extension of Proposition 5.2 to the subcase when $\lambda \in (\frac{1}{6}, \frac{1}{3}]$.

**Lemma 5.3.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$. Let $(\beta_i)_{i=0,\dots,3}$ be a probability vector and $x \in F_{\beta_0, \beta_1, \beta_2, \beta_3}$ (recall (4.14)).

(i) If $\beta_1 \log(6\lambda + 1) + \beta_2 \log(6\lambda - 1) > 0$, then $x \in \mathcal{I}$.

(ii) If $\beta_1 \log(6\lambda + 1) + \beta_2 \log(6\lambda - 1) \le 0$, then $h_{F^\lambda}(x) = 1$.

*Proof.* Assume that $\lambda \in (\frac{1}{6}, \frac{1}{3})$ so that $0 < 6\lambda - 1 < 1$. Since $x \in F_{\beta_0, \beta_1, \beta_2, \beta_3}$, there exists $N_\varepsilon \ge 1$ such that

$$\left| \frac{\beta_i(x, n)}{n} - \beta_i \right| < \varepsilon \text{ for all } i = 0, 1, 2, 3 \text{ and } n \ge N_\varepsilon. \qquad (5.1)$$

Let us prove (i). From the assumption in (i), $x \in F_{\beta_0, \beta_1, \beta_2, \beta_3}$ and (2.5) it also follows that $x \notin \mathcal{E}$. Therefore, going back to Lemma 3.3, for all $n \ge 0$,

$$|m_{n+1}(x)| = \begin{cases} |m_n(x)| & \text{if } u_n(x) \in \{0, 3\}, \\ 6\lambda\sqrt{1+m_n^2(x)} + |m_n(x)| & \text{if } u_n(x) = 1, \\ 6\lambda\sqrt{1+m_n^2(x)} - |m_n(x)| & \text{if } u_n(x) = 2. \end{cases} \qquad (5.2)$$

We may assume without loss of generality that $m_1(x) = 6\lambda$, hence

$$|m_n(x)| \ge 6\lambda(6\lambda + 1)^{\beta_1(x,n)-1}(6\lambda - 1)^{\beta_2(x,n)}. \qquad (5.3)$$

By assumption we can select $\varepsilon > 0$ small enough such that

$$(\beta_1 - \varepsilon) \log(6\lambda + 1) + (\beta_2 + \varepsilon) \log(6\lambda - 1) > 0. \qquad (5.4)$$

Recalling that $\log(6\lambda - 1) < 0$, it follows that, for all $n \ge N_\varepsilon$,

$$\log |m_n(x)| \ge \log(6\lambda) + (\beta_1(x, n) - 1) \log(6\lambda + 1) + \beta_2(x, n) \log(6\lambda - 1) \\ \ge \log(6\lambda) + (n-1)((\beta_1 - \varepsilon) \log(6\lambda + 1) + (\beta_2 + \varepsilon) \log(6\lambda - 1)). \qquad (5.5)$$

Hence, $\log |m_n(x)| \to +\infty$, i.e. $x \in \mathcal{I}$.
Let us prove (ii). Suppose that $x \in F_{\beta_0, \beta_1, \beta_2, \beta_3}$ is fixed and assume without loss of generality that $m_1(x) > 0$. By Proposition 3.12 we only have to prove that $h_{F^\lambda}(x) \ge 1$. Given $\varepsilon_1 > 0$, choose $\varepsilon > 0$ small enough such that

$$ \max(\beta_1 \log(6\lambda + 1 + \varepsilon) + \beta_2 \log(6\lambda - 1 + \varepsilon), 2\varepsilon|\log(6\lambda + 1 + \varepsilon)|, 2\varepsilon|\log(6\lambda - 1 + \varepsilon)|) \le \frac{\varepsilon_1}{3} \quad (5.6) $$

and $0 < 6\lambda - 1 + \varepsilon < 1$. Fix $N_\varepsilon \ge 1$ such that (5.1) holds for $i = 0, \dots, 3$, and $M_\lambda > 0$ so large that for all $m \ge M_\lambda$,

$$ \left|1 + 6\lambda\sqrt{1 + \frac{1}{m^2}} - (6\lambda + 1)\right| < \varepsilon \quad \text{and} \quad \left|-1 + 6\lambda\sqrt{1 + \frac{1}{m^2}} - (6\lambda - 1)\right| < \varepsilon. \quad (5.7) $$

**Case 1:** The sequence $(|m_n(x)|)_{n \ge 1}$ is bounded by some $M > 0$.

Then, by Lemma 3.5, $0 < m_n^- = \sqrt{36\lambda^2 - 1} \le |m_n(x)| \le M$, and since $\ell_n(x) \le 3^{-n}$, one has $\lim_{n \to \infty} \frac{\log |m_n(x)|}{|\log \ell_n(x)|} = 0$. Then, Proposition 3.12 yields $h_{F^\lambda}(x) = 1$.

**Case 2:** $\lim_{n \to \infty} |m_n(x)| = +\infty$.

Let $N \ge N_\varepsilon$ be so large that for every $n \ge N$, $|m_n(x)| \ge M_\lambda$. By construction, it follows from (5.7) and Lemma 3.3 that, for all $n \ge N$,

$$ |m_{n+1}(x)| \le \begin{cases} |m_n(x)| & \text{if } u_n(x) \in \{0, 3\}, \\ (6\lambda + 1 + \varepsilon)|m_n(x)| & \text{if } u_n(x) = 1, \\ (6\lambda - 1 + \varepsilon)|m_n(x)| & \text{if } u_n(x) = 2. \end{cases} $$

This implies that for every $n \ge N$,

$$ |m_n(x)| \le |m_N(x)| (6\lambda + 1 + \varepsilon)^{\beta_1(x,n) - \beta_1(x,N)} (6\lambda - 1 + \varepsilon)^{\beta_2(x,n) - \beta_2(x,N)}. $$

By the choice of $N_\varepsilon$ we can use (5.1) to obtain

$$ \begin{aligned} |m_n(x)| & \le |m_N(x)| (6\lambda + 1 + \varepsilon)^{(\beta_1+2\varepsilon)(n-N)} (6\lambda - 1 + \varepsilon)^{(\beta_2-2\varepsilon)(n-N)} \\ & \le C (6\lambda + 1 + \varepsilon)^{(\beta_1+2\varepsilon)n} (6\lambda - 1 + \varepsilon)^{(\beta_2-2\varepsilon)n}, \end{aligned} \quad (5.8) $$

where $C > 0$ depends on $x$ and $N$ only. Taking logarithm and using (5.6), we infer

$$ \log(\sqrt{36\lambda^2 - 1}) \le \log|m_n(x)| \le \log C + n\varepsilon_1. \quad (5.9) $$

The parameter $\varepsilon_1$ being arbitrarily small, one deduces that $\lim_{n \to \infty} \frac{\log |m_n(x)|}{|\log \ell_n(x)|} = 0$, and again Proposition 3.12 gives $h_{F^\lambda}(x) = 1$.

**Case 3:** $\liminf_{n \to \infty} |m_n(x)| < \limsup_{n \to \infty} |m_n(x)| = +\infty$.

There exist $M \ge M_\lambda$ and a unique strictly increasing sequence $(n_k)_{k \ge 1}$ such that $n_1 \ge N_\varepsilon$, $|m_{n_1}(x)| \le M$ and, for every $k \ge 1$,

$$ \left\{ \begin{array}{ll} |m_n(x)| > M & \text{when } n_{2k-1} < n \le n_{2k}, \\ \sqrt{36\lambda^2 - 1} \le |m_n(x)| \le M & \text{when } n_{2k} < n \le n_{2k+1}. \end{array} \right. $$

Let $k \ge 1$. When $n \ge n_{2k-1} \ge N_\varepsilon$, one has by (5.1)

$$ |\beta_i(x, n) - \beta_i(x, n_{2k-1})| \le (n - n_{2k-1})\beta_i + 2n\varepsilon, \text{ for } i = 0, \dots, 3. $$
Then by (5.6), and recalling that $6\lambda - 1 + \varepsilon < 1$,

$$ egin{aligned} |m_n(x)| &\le |m_{n_{2k-1}}(x)|(6\lambda + 1 + \varepsilon)^{\beta_1(x,n)-\beta_1(x,n_{2k-1})}(6\lambda - 1 + \varepsilon)^{\beta_2(x,n)-\beta_2(x,n_{2k-1})} \\ &\le M(6\lambda + 1 + \varepsilon)^{(n-n_{2k-1})\beta_1+2n\varepsilon}(6\lambda - 1 + \varepsilon)^{(n-n_{2k-1})\beta_2-2n\varepsilon} \le MC_{\lambda,\varepsilon}^{n\varepsilon_1} \quad (5.10) \end{aligned} $$

for some constant $C_{\lambda,\varepsilon} \ge 1$ that depends on $\lambda$ and $\varepsilon$ only. Therefore, (5.9) holds for $n_{2k_1} < n \le n_{2k}$ with $C_{\lambda,\varepsilon}$ instead of $C$. When $n_{2k} < n \le n_{2k+1}$, $|m_n(x)| \le M$, so (5.10) also holds in this case. We conclude that since $\varepsilon_1$ is arbitrarily small, $\lim_{n\to\infty} \frac{\log|m_n(x)|}{\log\ell_n(x)} = 0$, and Proposition 3.12 gives $h_{F^\lambda}(x) = 1$.

Finally, the case $\lambda = \frac{1}{3}$ is treated similarly, except that in (5.5), (5.8) and (5.10), $-2n\varepsilon$ is replaced by $+2n\varepsilon$, since $6\lambda - 1 = 1$ and this time we cannot assume that $6\lambda - 1 + \varepsilon < 1$. $\square$

**Proposition 5.4.** Let $\lambda \in (\frac{1}{6}, \frac{5}{6})$. Then, for every $\alpha \in [\alpha_{\lambda,\min}, 1]$, $d_{F^\lambda}(\alpha) \ge \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$.

*Proof.* Let $\alpha \in [\alpha_{\lambda,\min}, 1]$ and set $\tilde{\alpha} = \alpha + \gamma_\lambda - 1$. By Lemma 4.6, let us choose parameters $(\beta_i)_{i=0,...,3}$ such that $F_{\beta_0,\beta_1,\beta_2,\beta_3} \subset E_{\mu_\lambda}(\tilde{\alpha})$ and

$$ d_{\mu_{\lambda}}(\tilde{\alpha}) = \tau_{\mu_{\lambda}}^{*}(\tilde{\alpha}) = \dim_{H}(F_{\beta_{0},\beta_{1},\beta_{2},\beta_{3}}). $$

Next, we show that $F_{\beta_0,\beta_1,\beta_2,\beta_3} \subset E_{F^\lambda}(\alpha)$. Suppose $x \in F_{\beta_0,\beta_1,\beta_2,\beta_3}$. Then its digit frequencies (4.10) exist and equal $\beta_i(x) = \beta_i$ for $i = 0, \dots, 3$, and then the local dimension of $\mu_\lambda$ in formula (4.9) is given by

$$ \tilde{\alpha} = \dim_{\text{loc}}(\mu_{\lambda}, x) = \gamma_{\lambda} - \frac{\beta_{1} \log(6\lambda + 1) + \beta_{2} \log(6\lambda - 1)}{(\beta_{0} + \beta_{3}) \log 3 + (\beta_{1} + \beta_{2}) \log 6}. $$

If $\alpha < 1$ then $\tilde{\alpha} < \gamma_\lambda$, which implies that $\beta_1 \log(6\lambda + 1) + \beta_2 \log(6\lambda - 1) > 0$. Hence, by (i) of Lemma 5.3, we have $x \in \mathcal{I}$. Thus, by Corollary 5.1, we have $h_{F^\lambda}(x) = 1 - \gamma_\lambda + \tilde{\alpha} = \alpha$ and hence $F_{\beta_0,\beta_1,\beta_2,\beta_3} \subset E_{F^\lambda}(\alpha)$. Therefore, $d_{F^\lambda}(\alpha) \ge \dim_H(F_{\beta_0,\beta_1,\beta_2,\beta_3}) = \tau_{\mu_\lambda}^*(\tilde{\alpha}) = \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$.

If $\alpha = 1$ then $\tilde{\alpha} = \gamma_\lambda$, which implies that $\beta_1 \log(6\lambda + 1) + \beta_2 \log(6\lambda - 1) = 0$. Hence, by (ii) of Lemma 5.3, we have $h_{F^\lambda}(x) = \alpha = 1$ and hence $F_{\beta_0,\beta_1,\beta_2,\beta_3} \subset E_{F^\lambda}(1)$. Therefore, $d_{F^\lambda}(\alpha) = d_{F^\lambda}(1) \ge \dim_H(F_{\beta_0,\beta_1,\beta_2,\beta_3}) = \tau_{\mu_\lambda}^*(\tilde{\alpha}) = \tau_{\mu_\lambda}^*(\gamma_\lambda) = \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$. $\square$

Finally, we conclude on the decreasing part of the multifractal spectrum $d_{F^\lambda}$ when $\lambda \in (\frac{\sqrt{2}}{6}, \frac{1}{3}]$.

**Proposition 5.5.** Let $\lambda \in (\frac{\sqrt{2}}{6}, \frac{1}{3}]$. Then, for every $\alpha \in [\alpha_{\lambda,\mathcal{L}}, 1]$, $d_{F^\lambda}(\alpha) = \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$.

*Proof.* Let $\alpha \in [\alpha_{\lambda,\mathcal{L}}, 1]$. By Proposition 5.4, it is enough to prove the upper bound $d_{F^\lambda}(\alpha) \le \tau_{\mu_\lambda}^*(\alpha+\gamma_\lambda-1)$. By Propositions 3.14 and 4.2, if $x \in E_{F^\lambda}(\alpha)$, then necessarily $x \in \underline{E}_{\mu_\lambda}^>(\alpha+\gamma_\lambda-1)$. Thus item (ii) of Proposition 4.5 implies that $d_{F^\lambda}(\alpha) \le \dim_H(\underline{E}_{\mu_\lambda}^>(\alpha+\gamma_\lambda-1)) = \tau_{\mu_\lambda}^*(\alpha+\gamma_\lambda-1)$, hence the result. $\square$

## 6. THE UPPER BOUND FOR THE INCREASING PART OF $d_{F^\lambda}$ WHEN $\lambda \in [\frac{\sqrt{2}}{6}, \frac{1}{3})$

It is interesting that the above arguments are not sufficient to prove that $d_{F^\lambda}(\alpha) = \tau_{\mu_\lambda}^*(\alpha + 1 - \gamma_\lambda)$ in the increasing part of the spectrum, since item (ii) of Proposition 4.5 does not apply any more. We have to develop a much more involved argument to prove the upper bound. Proposition 5.4 and Proposition 6.1 below will imply (1.15), i.e. item (ii) of Theorem 1.9.

**Proposition 6.1.** Let $\lambda \in [\frac{\sqrt{2}}{6}, \frac{1}{3}]$. Then, for every $\alpha \in [\alpha_{\lambda,\min}, \alpha_{\lambda,\mathcal{L}}]$, $d_{F^\lambda}(\alpha) \le \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$.
The rest of this section is devoted to the proof of Proposition 6.1.

*Proof ideas.* To make the reading of the proof easier, we present here the strategy and the different steps of the proof.

• Observe that the case $\lambda = \frac{\sqrt{2}}{6}$ and $\alpha_0 = 1$ is obvious since in this case $d_{F^\lambda}(\alpha_0) = 1 = \tau_{\mu_\lambda}^*(\alpha_0 + \gamma_\lambda - 1)$. So in the rest of our argument for the case $\lambda = \frac{\sqrt{2}}{6}$ we suppose $\alpha_0 < \alpha_{\lambda,\mathcal{L}} = 1$.

• Suppose that for some $\alpha_0 \in [\alpha_{\lambda,\min}, \alpha_{\lambda,\mathcal{L}})$, points not in $\mathcal{I}$ can modify the spectrum, that is, $\dim_H(\{x \notin \mathcal{I} : h_{F^\lambda}(x) = \alpha_0\}) > \dim_H(\{x \in \mathcal{I} : h_{F^\lambda}(x) = \alpha_0\})$. By the previous results, we know that $\dim_H(\{x \in \mathcal{I} : h_{F^\lambda}(x) = \alpha_0\}) = \tau_{\mu_\lambda}^*(\alpha_0 + \gamma_\lambda - 1)$. When $x \notin \mathcal{I}$,

$$ \liminf_{n \to \infty} |m_n(x)| < +\infty. \qquad (6.1) $$

Since $|m_n(x)|$ is small infinitely often, for a sufficiently large $M$ select a sequence $n_j(x) \to \infty$ such that for every $j$,

$$ \frac{M}{2(6\lambda + 1)} < |m_{n_j}(x)| < M. \qquad (6.2) $$

Our goal is to build a large set of regular points $x$ where $h_{F^\lambda}(x)$ is small. In order to obtain such a small $h_{F^\lambda}(x)$, it will be necessary that sometimes $|m_n(x)|$ is “rapidly increasing” for $n_j(x) \le n \le n'_j(x)$. Using these rapidly increasing stretches, one can build a self-similar set $\mathbf{F}$, attractor of an IFS satisfying the Open Set Condition, which contains a subset of points in $\mathcal{I}$ with local $\mu_\lambda$-dimension equal to some $\alpha_{\mathbf{F}}$ with Hausdorff dimension larger than $\tau_{\mu_\lambda}^*(\alpha_{\mathbf{F}})$, leading to a contradiction.

• Therefore, proceeding towards a contradiction, suppose that for $\alpha_0 \in (\alpha_{\lambda,\min}, \alpha_{\lambda,\mathcal{L}})$

$$ 1 = \dim_H(E_{F^\lambda}(\alpha_{\lambda,\mathcal{L}})) > \dim_H(E_{F^\lambda}(\alpha_0)) := d_0 > \tau_{\mu_\lambda}^*(\alpha_0 + \gamma_\lambda - 1) := d_0^*. \qquad (6.3) $$

We choose $d_1, \dots, d_4$ such that

$$ d_0^* < d_2 < d_3 < d_1 < d_0 < d_4 < 1, \qquad (6.4) $$

and by continuity and monotonicity of $\tau_{\mu_\lambda}^*$, we select $\varepsilon_0 > 0$ such that $\alpha_{\lambda,\min} < \alpha_0 + \varepsilon_0 < \alpha_{\lambda,\mathcal{L}}$ and

$$ d_2 > \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1) \text{ for } \alpha \in (\alpha_{\lambda,\min}, \alpha_0 + \varepsilon_0), \qquad (6.5) $$

see Figure 10 below. As later arguments show, $d_4$ can be chosen as close as we wish to 1, therefore we select $d_2$ and $d_3$ before fixing $d_4$, this way we can ensure

$$ d_3 + d_4 - d_2 > 1. \qquad (6.6) $$

## 6.1. Construction of the IFS.

In this section, we build an IFS $\{\tilde{\Phi}_k\}_{k=1,...,K}$ satisfying the Open Set Condition. Each map $\tilde{\Phi}_k$ will be a bijective affine map from $[0,1]$ onto some interval $\mathbf{J}_k$ (the monotonicity of $\tilde{\Phi}_k$ will be explicit). All intervals $\mathbf{J}_k$ are of the form $\overline{I}_n(x)$ for some $x$ and $n$ (recall our notation $\overline{I}_n(x)$ for the closed interval $I_n(x)$, the closure of the open interval $I_n(x)$). The delicate part is to identify these intervals $\mathbf{J}_k$, since we want to control very precisely the behavior of $\mu_\lambda$ and $F^\lambda$ on them. To do so, we will proceed in two steps, first building some intervals on which we can ensure that $|m_n(x)|$ is quite large, and then inside each of them we will find smaller intervals where the slopes $m_n(x)$ are precisely controlled.
<figure><img src="image_11.png" /><figcaption>Figure 10. Illustration to the proof of Proposition 6.1. The thick red continuous curve is $d_{F^\lambda}(\alpha)$ when $\lambda = \frac{\sqrt{3}}{6}$, while the thick dotted line shows $d_{\mu_\lambda}(\alpha + \gamma_\lambda - 1)$.</figcaption></figure>

### 6.1.1. The family of the intervals $J'_j$.

Recall (6.3). Since $\dim_H(E_{F^\lambda}(\alpha_0)) = d_0$ and $d_1 < d_0$, there exists a compact set $E \subset E_{F^\lambda}(\alpha_0) \setminus (\mathcal{E} \cup \mathcal{I})$ such that $\mathcal{H}^{d_1}(E) = +\infty$. Since $E \cap \mathcal{I} = \emptyset$, one has

$$E = \bigcup_{M \ge 1} \{x \in E : |m_n(x)| < M \text{ for infinitely many integers } n\}.$$

Recall that by Corollary 5.1 and (6.3), $\dim_H(\{x \in \mathcal{I} : h_{F^\lambda}(x) = \alpha_0\}) = \tau_{\mu_\lambda}^*(\alpha_0 + \gamma_\lambda - 1) = d_0^* < d_1$. Since $\mathcal{H}^{d_1}(E) = +\infty$, we can suppose, up to a small modification of $d_1$, that the compact set $E$ was chosen so that there exists $M > 100$ such that for all $x \in E$, $|m_{n_j(x)}(x)| < M$ for infinitely many integers $n_j(x)$, and $\mathcal{H}^{d_1}(E) = +\infty$.

On the other hand, let $\varepsilon_1 > 0$ be such that $\alpha_0 + \varepsilon_1 < 1$. The fact that $h_{F^\lambda}(x) = \alpha_0$ for $x \in E$ implies, for instance by (3.31), that for infinitely many integers $n'_j(x)$ one has $|m_{n'_j(x)}(x)| |I_{n'_j(x)}(x)| \ge |I_{n'_j(x)}(x)|^{\alpha_0 + \varepsilon_1}$, i.e.

$$\log |m_{n'_j(x)}(x)| \ge (1 - \alpha_0 - \varepsilon_1) |\log \ell_{n'_j(x)}(x)|. \qquad (6.7)$$

Hence, one can choose a suitably small $\varepsilon_2 > 0$ and construct iteratively two sequences $(n_j(x))_{j \ge 0}$ and $(n'_j(x))_{j \ge 0}$ as follows:

(a) $n_0(x) = 0$ and $n'_0(x)$ is so large that

$$\varepsilon_2(1 - \alpha_0 - \varepsilon_1) |\log \ell_{n'_0(x)}(x)| > \log M. \qquad (6.8)$$

(b) For $j \ge 1$, let $\tilde{n}_j(x)$ be the smallest integer strictly larger than $n'_{j-1}(x)$ such that

$$|m_{\tilde{n}_j(x)}(x)| < M. \qquad (6.9)$$
(c) Then let $n'_j(x)$ be the smallest integer strictly larger than $\tilde{n}_j(x)$ satisfying (6.7). Since $n'_j(x)$ is larger than $n'_0(x)$ we obtain $|\log \ell_{n'_j(x)}(x)| > |\log \ell_{n'_0(x)}(x)|$. Therefore, by (6.8) and (6.9)

$$ \log |m_{\tilde{n}_j(x)}(x)| < \varepsilon_2(1 - \alpha_0 - \varepsilon_1) |\log \ell_{n'_j(x)}(x)|. \quad (6.10) $$

(d) Finally, choose $n_j(x)$ as the largest integer which is less than $n'_j(x)$ but the largest integer $n \ge \tilde{n}_j(x)$ for which (6.10) holds with $\tilde{n}_j(x)$ replaced by $n = n_j(x)$.

Items (b) and (c) are made possible by the remarks preceding the construction. Thus we have:

* For every $j \ge 1$, $n_j(x) < n'_j(x) < n_{j+1}(x)$, and $\lim_{j \to \infty} n_j(x) = \lim_{j \to \infty} n'_j(x) = \infty$.

* For $n_j(x) < n < n'_j(x)$, $|\bar{I}_n(x)| > \ell_{n'_j(x)}(x) = |\bar{I}_{n'_j(x)}(x)|$, hence

$$ \log |m_n(x)| < (1 - \alpha_0 - \varepsilon_1) |\log |\bar{I}_n(x)|| < (1 - \alpha_0 - \varepsilon_1) |\log \ell_{n'_j(x)}(x)| \quad (6.11) $$

and by (d) we know that (6.10) does not hold for $n \in (n_j(x), n'_j(x))$ and hence

$$ \log |m_n(x)| \ge \varepsilon_2(1 - \alpha_0 - \varepsilon_1) |\log \ell_{n'_j(x)}(x)| \ge \varepsilon_2(1 - \alpha_0 - \varepsilon_1) n'_j(x) \log 3. \quad (6.12) $$

Intuitively, $|m_{n_j(x)}(x)|$ is quite small, while $|m_{n'_j(x)}(x)|$ is quite large, and $|m_n(x)|$ remains controlled between the two integers $n_j(x) < n < n'_j(x)$.

**Remark 6.2.** At the very end of the proof, $\varepsilon_1 > 0$ and $\varepsilon_2 > 0$ will be supposed to be so small that some conditions are satisfied. It is important to note that they do not depend on $\varepsilon_0$. We do not write the conditions here because they will naturally appear along the construction. We will refer to this remark at the appropriate time.

For all integers $j, k \ge 0$, set

$$ X_{j,k} = \{x \in E : 6^{-(k+1)} \le |\bar{I}_{n_j(x)}(x)| < 6^{-k}\}, \quad (6.13) $$

and select a maximal integer $K_j$ such that, for all $k < K_j$, $X_{j,k} = \emptyset$. It is easy to see from its definition that $K_j \to \infty$ as $j \to \infty$. Using the hierarchical structure of the intervals from $\mathcal{I}_n$, one can extract from the family $\{\bar{I}_{n_j(x)}(x)\}_{x \in X_{j,k}}$ a finite number say $L_{j,k}$ many non-overlapping intervals $\{\mathbf{J}_{j,k,l} := \bar{I}_{n_j(x_l)}(x_l)\}_{l \in \{1, \dots, L_{j,k}\}}$, such that

$$ \bigcup_{x \in X_{j,k}} \bar{I}_{n_j(x)}(x) = \bigcup_{l=1}^{L_{j,k}} \mathbf{J}_{j,k,l}. \quad (6.14) $$

By non-overlapping intervals, we mean here and in the rest of the paper that the interiors of the intervals have empty intersection. Observe that by (6.13), $L_{j,k} < 6 \cdot 6^k$ and the intervals $\mathbf{J}_{j,k,l}$ are approximately of the same length.

Now one can extract in a same way from the family $\{\bar{I}_{n'_j(x)}(x)\}_{x \in X_{j,k}}$ a finite number, say $L'_{j,k}$ many intervals $\{\bar{I}_{n'_j(x'_l)}(x'_l)\}_{l' \in \{1, \dots, L'_{j,k}\}}$ such that

$$ \bigcup_{x \in X_{j,k}} \bar{I}_{n'_j(x)}(x) = \bigcup_{l'=1}^{L'_{j,k}} \bar{I}_{n'_j(x'_l)}(x'_l). \quad (6.15) $$

We reorganize them by grouping those that belong to the same $\mathbf{J}_{j,k,l}$. Let

$$ \mathcal{L}_{j,k,l} = \{l' \ge 1 : \bar{I}_{n'_j(x'_l)}(x'_l) \subset \mathbf{J}_{j,k,l}\}, $$
and write $\mathbf{J}'_{j,k,l,l'} := \bar{I}_{n'_j(x'_l)}(x'_l)$ for those intervals $\bar{I}_{n'_j(x'_l)}(x'_l)$ that are included in $\mathbf{J}_{j,k,l}$, i.e. the set in (6.15) is rewritten as

$$ \bigcup_{l=1}^{L_{j,k}} \mathbf{J}'_{j,k,l} \text{ with } \mathbf{J}'_{j,k,l} = \bigcup_{l' \in \mathcal{L}_{j,k,l}} \mathbf{J}'_{j,k,l,l'} \subset \mathbf{J}_{j,k,l}. \qquad (6.16) $$

Observe that the definition makes sense since each $\mathbf{J}'_{j,k,l,l'}$ is a subset of exactly one $\mathbf{J}_{j,k,l}$, and these intervals are non-overlapping and organized in a hierarchical manner.

**Lemma 6.3.** Let $d_3 > 0$ be such that $d_2 < d_3 < d_1$. Then, there are infinitely many integers $j$ with the following property: there exist integers $k_j$ and $l_j \le L_{j,k_j}$ such that

$$ \sum_{l' \in \mathcal{L}_{j,k_j,l_j}} |\mathbf{J}'_{j,k_j,l_j,l'}|^{d_3} > 100 \cdot 6^{-k_j}. \qquad (6.17) $$

*Proof.* Suppose that for $j > j_0$, for every $k$ and $l \le L_{j,k_j}$,

$$ \sum_{l' \in \mathcal{L}_{j,k,l}} |\mathbf{J}'_{j,k,l,l'}|^{d_3} \le 100 \cdot 6^{-k}. \qquad (6.18) $$

Consider the cover of $E$ given by $\bigcup_{k=K_j}^{\infty} \bigcup_{l=1}^{L_{j,k}} \bigcup_{l' \in \mathcal{L}_{j,k,l}} \mathbf{J}'_{j,k,l,l'}$. Then, using (6.13), (6.14), $L_{j,k} \le 6 \cdot 6^k$, $d_1 > d_3$ and later (6.18), we obtain

$$ \begin{aligned} \sum_{k=K_j}^{\infty} \sum_{l=1}^{L_{j,k}} \sum_{l' \in \mathcal{L}_{j,k,l}} |\mathbf{J}'_{j,k,l,l'}|^{d_1} &= \sum_{k=K_j}^{\infty} \sum_{l=1}^{L_{j,k}} \sum_{l' \in \mathcal{L}_{j,k,l}} |\mathbf{J}'_{j,k,l,l'}|^{d_1-d_3} |\mathbf{J}'_{j,k,l,l'}|^{d_3} \\ &\le \sum_{k=K_j}^{\infty} \sum_{l=1}^{L_{j,k}} \sum_{l' \in \mathcal{L}_{j,k,l}} 6^{-(d_1-d_3)k} |\mathbf{J}'_{j,k,l,l'}|^{d_3} \\ &\le \sum_{k=K_j}^{\infty} 6^{-(d_1-d_3)k} \sum_{l=1}^{L_{j,k}} \sum_{l' \in \mathcal{L}_{j,k,l}} |\mathbf{J}'_{j,k,l,l'}|^{d_3} \\ &\le \sum_{k=K_j}^{\infty} 6^{-(d_1-d_3)k} L_{j,k} \cdot 100 \cdot 6^{-k} \le 600 \sum_{k=K_j}^{\infty} 6^{-(d_1-d_3)k} < \infty. \end{aligned} $$

This would imply $d_1 \ge \dim_H(E) > d_1$, a contradiction. $\square$

**Definition 6.4.** The family $\mathbf{J}'_j$ consisting of finitely many closed intervals is defined for those $j \ge 1$ for which $l_j$ exists in Lemma 6.3. We put $\bar{\mathbf{J}}_j = \mathbf{J}_{j,k_j,l_j}$, $\mathcal{L}_j = \mathcal{L}_{j,k_j,l_j}$, and $\mathbf{J}'_{j,k_j,l_j,l'} = \mathbf{J}'_{j,l'}$, and the family $\mathbf{J}'_j$ is defined by

$$ \mathbf{J}'_j = \bigcup_{l' \in \mathcal{L}_j} \mathbf{J}'_{j,l'} \subset \bar{\mathbf{J}}_j. \qquad (6.19) $$
### 6.1.2. The family of the intervals $\tilde{\mathbf{J}}_{l,\rho_j}$, and the construction of $\tilde{\mathbf{F}}$.

By the multifractal formalism and Comment (c) after Proposition 4.5, the set $E_{F^\lambda}(\alpha_{\lambda,\mathcal{L}})$ is of full Lebesgue measure in $[0, 1]$. For this exponent $\alpha_{\lambda,\mathcal{L}}$, by the Lemma 5.3, the set $E_{F^\lambda}(\alpha_{\lambda,\mathcal{L}}) \cap \mathcal{I}$ still has full Lebesgue measure, and the elements $x$ of this subset satisfy $\lim_{n \to \infty} |m_n(x)| = +\infty$. Up to extraction, there exist an integer $N_\mathcal{L}$ and another, smaller, subset of $E_{F^\lambda}(\alpha_{\lambda,\mathcal{L}}) \cap \mathcal{I}$ with positive Lebesgue measure of points $x$ such that, for all $n \ge N_\mathcal{L}$,

$$|\tilde{m}_n(x)| > 100. \qquad (6.20)$$

Consider then $E_\mathcal{L} \subset E_{F^\lambda}(\alpha_{\lambda,\mathcal{L}}) \cap \mathcal{I}$, a compact subset such that $\dim_H(E_\mathcal{L}) = 1$ and, for every $x \in E_\mathcal{L}$ and $n \ge N_\mathcal{L}$, (6.20) is true.

Given $\rho \in (0, 1)$, for every $x \in E_\mathcal{L}$, choose $n(x, \rho) \ge 1$ such that $\ell_{n(x,\rho)}(x) < \rho \le \ell_{n(x,\rho)-1}(x)$. We can extract from the family $\{\bar{I}_{n(x,\rho)}(x)\}_{x \in E_\mathcal{L}}$ a finite number say $\tilde{L}_\rho$ many non-overlapping intervals $\{\bar{I}_{n(x_l,\rho)}(x_l) := \tilde{\mathbf{J}}_{l,\rho}\}_{l=1,\dots,\tilde{L}_\rho}$, such that

$$\bigcup_{x \in E_\mathcal{L}} \bar{I}_{n(x, \rho)}(x) = \bigcup_{l=1}^{\tilde{L}_\rho} \tilde{\mathbf{J}}_{l, \rho}. \qquad (6.21)$$

By construction, for any $l \ge 1$ one has

$$\frac{\rho}{6} \le |\tilde{\mathbf{J}}_{l,\rho}| < \rho. \qquad (6.22)$$

Let us now choose $d_4 < 1$. We suppose that it is very close to 1, such that (6.6) holds. Since $\dim_H(E_\mathcal{L}) = 1 > d_4$, there is $\rho_\mathcal{L} > 0$ such that if $\rho < \rho_\mathcal{L}$, then $\sum_{l=1}^{\tilde{L}_\rho} |\tilde{\mathbf{J}}_{l,\rho}|^{d_4} > 100$, which by (6.22) implies

$$\tilde{L}_{\rho} > \frac{100}{\rho^{d_4}}. \qquad (6.23)$$

**Definition 6.5.** Fix large integers $j$ and $k_j > N_\mathcal{L}$ such that (6.17) is satisfied. Set $\rho_j = 6^{-k_j}$ and $\tilde{L}_j = \tilde{L}_{\rho_j}$. The family of intervals $\tilde{\mathbf{J}}_{l,\rho_j}$ is used to define

$$\tilde{\mathbf{F}} = \bigcup_{l=1}^{\tilde{L}_j} \tilde{\mathbf{J}}_{l,\rho_j} \text{ with } \tilde{L}_j > 100 \cdot 6^{k_j d_4}, \qquad (6.24)$$

where the lower bound for $\tilde{L}_j$ can be obtained from (6.23).

By (6.22) used with $\rho_j$, one has

$$6^{-(k_j+1)} < |\tilde{\mathbf{J}}_{l,\rho_j}| \le 6^{-k_j}. \qquad (6.25)$$

Recalling that the intervals $\tilde{\mathbf{J}}_{l,\rho_j}$ are of the form $\bar{I}_{n(x_l,\rho_j)}(x_l)$ with $x_l \in E_\mathcal{L}$, let us introduce the simple notation

$$\tilde{\mathbf{n}}(x_l) = n(x_l, \rho_j), \text{ that is } x_l \in \bar{I}_{\tilde{\mathbf{n}}(x_l)}(x_l) = \tilde{\mathbf{J}}_{l, \rho_j} \subset \tilde{\mathbf{F}}. \qquad (6.26)$$

The set $\tilde{\mathbf{F}}$ consists of intervals that satisfy (6.20) and (6.23) with $\rho = \rho_j$. Intuitively, the increment of the “idealized” slope of $F_n^\lambda$ as $n$ increases on these intervals is uniformly bounded below with a reasonably large constant and by (6.24) we have “many” of them.
### 6.1.3. The construction of $\mathbf{F}$ and its dimension.

The integer $j$ is now supposed to be fixed and satisfies the previous constraints so that the family $\mathbf{J}'_j$ and $\tilde{\mathbf{F}}$ are well defined.

To build $\mathbf{F}$, we start with the intervals $\tilde{\mathbf{J}}_{l,\rho_j}$ of $\tilde{\mathbf{F}}$, where by (6.20) the intermediary functions $F_n^\lambda$ have uniformly sufficiently large slopes on them. Then, inside each interval $\tilde{\mathbf{J}}_{l,\rho_j}$ of $\tilde{\mathbf{F}}$ we are going to build subintervals using affine images, $\Psi_l(\mathbf{J}'_j)$ of intervals from families of type $\mathbf{J}'_j$ constructed in Subsection 6.1.1. On the intervals $\Psi_l(\mathbf{J}'_j)$ we control the growth of the slope of the intermediate steps $F_n^\lambda$ which lead to $F^\lambda$. Our objective is then to use the subintervals from the families $\Psi_l(\mathbf{J}'_j)$, denoted by $\Psi_l(\mathbf{J}_{j,l'})$ to define affine maps from $[0, 1]$ onto these intervals that will form a finite IFS. The intuition is that since the value of the slopes $|m_n(x)|$ of $F^\lambda$ is controlled at every scale between 1 and the length of these subintervals, by self-similarity, the growth of $|m_n(x)|$ will be controlled on the attractor of the IFS, ensuring that points belonging to this attractor will satisfy $\lim_{n\to\infty} |m_n(x)| = +\infty$ and thus will belong to $\mathcal{I}$. Finally, we will prove that the attractor has a too large Hausdorff dimension, contradicting our assumptions (6.3) and (6.5).

The set $\tilde{\mathbf{F}}$ being fixed, we now use the family of intervals $\mathbf{J}'_j \subset \bar{\mathbf{J}}_j$ from Definition 6.4. For any $l \in \{1, \dots, \tilde{L}_j\}$,

$$ \text{we select an affine bijection } \Psi_l : \bar{\mathbf{J}}_j \to \tilde{\mathbf{J}}_{l,\rho_j}. \qquad (6.27) $$

This choice is meant to ensure that the construction will keep memory only of what happens between $\bar{\mathbf{J}}_j$ and the intervals $\mathbf{J}'_{j,l'}$, where the variation of the slopes is controlled by (6.11) and (6.12). There are two such bijections, and to determine the monotonicity of $\Psi_l$, we look at its domain and its range. More precisely, let $x_0 \in \tilde{\mathbf{J}}_{l,\rho_j} \setminus \mathcal{E}$ and $x'_0 \in \mathbf{J}_j \setminus \mathcal{E}$. By definition $\bar{I}_{\tilde{\mathbf{n}}(x_0)}(x_0) = \tilde{\mathbf{J}}_{l,\rho_j}$, and one can select an integer $n'_l(x'_0)$ such that $\bar{I}_{n'_l(x'_0)}(x'_0) = \bar{\mathbf{J}}_j$. Now $\Psi_l$ is chosen monotone increasing (resp. monotone decreasing) if $\varepsilon_{\tilde{\mathbf{n}}(x_0)}(x_0)\varepsilon_{n'_l(x'_0)}(x'_0) = 1$ (resp. $\varepsilon_{\tilde{\mathbf{n}}(x_0)}(x_0)\varepsilon_{n'_l(x'_0)}(x'_0) = -1$). This choice is natural since it reflects the possible change in monotonicity between the functions $F_{\tilde{\mathbf{n}}(x_0)}^\lambda$ and $F_{n'_l(x'_0)}^\lambda$ on the corresponding intervals. Observe that the family of intervals $(\Psi_l(\mathbf{J}'_{j,l'}))_{l=1,\dots,\tilde{L}_j, l' \in \mathcal{L}_j}$ consists of non-overlapping intervals.

Consider now the affine contractive bijections $\Phi_{l,l'} : [0, 1] \to \Psi_l(\mathbf{J}'_{j,l'})$, $l = 1, \dots, \tilde{L}_j$, $l' \in \mathcal{L}_j$. As above, the monotonicity of $\Phi_{l,l'}$ must be compatible with our construction. For this, pick $x_0 \in \Psi_l(\mathbf{J}'_{j,l'}) \setminus \mathcal{E}$ and choose $n_{l,l'}(x_0)$ in such a way that $\bar{I}_{n_{l,l'}(x_0)}(x_0) = \Psi_l(\mathbf{J}'_{j,l'})$. Then $\Phi_{l,l'}$ is monotone increasing (resp. monotone decreasing) if $\varepsilon_{n_{l,l'}(x_0)}(x_0) = 1$ (resp. $\varepsilon_{n_{l,l'}(x_0)}(x_0) = -1$).

**Definition 6.6.** The first generation, $\tilde{\mathbf{G}}$, of the compact attractor $\mathbf{F}$ is defined by

$$ \tilde{\mathbf{G}} = \bigcup_{l=1}^{\tilde{L}_j} \Psi_l(\mathbf{J}'_j) = \bigcup_{l=1}^{\tilde{L}_j} \bigcup_{l' \in \mathcal{L}_j} \Psi_l(\mathbf{J}'_{j,l'}) \subset \bigcup_{l=1}^{\tilde{L}_j} \tilde{\mathbf{J}}_{l,\rho_j}. \qquad (6.28) $$

By (6.13) and (6.22),

$$ 6^{-(k_j+1)} \le |\tilde{\mathbf{J}}_{l,\rho_j}| < \rho_j = 6^{-k_j} \text{ and } 6^{-(k_j+1)} \le |\mathbf{J}_j| < 6^{-k_j}, \qquad (6.29) $$

hence

$$ \text{the absolute value of the slope of } \Psi_l \text{ is between } \frac{1}{6} \text{ and } 6. \qquad (6.30) $$
In particular, denoting by $r_{l,l'} = |\Psi_l(\mathbf{J}'_{j,l'})|$ the contraction ratio of $\Phi_{l,l'}$, one has

$$ \frac{1}{6} |\mathbf{J}'_{j,l'}| \le r_{l,l'} \le 6 |\mathbf{J}'_{j,l'}|. \qquad (6.31) $$

By (6.17) and (6.31), for a fixed $l$,

$$ \sum_{l' \in \mathcal{L}_j} |r_{l,l'}|^{d_3} > \frac{100}{6^{d_3}} \cdot 6^{-k_j} > 10 \cdot 6^{-k_j}. \qquad (6.32) $$

By (6.23) we have $\tilde{L}_j > 100 \rho_j^{-d_4} = 100 \cdot 6^{k_j d_4}$. Therefore, using (6.31), (6.32) and the fact that $d_2 < d_3$, and finally using (6.6), we obtain

$$ \begin{aligned} \sum_{l=1}^{\tilde{L}_j} \sum_{l' \in \mathcal{L}_j} |r_{l,l'}|^{d_2} &= \sum_{l=1}^{\tilde{L}_j} \sum_{l' \in \mathcal{L}_j} |r_{l,l'}|^{d_3} |r_{l,l'}|^{d_2-d_3} \ge \sum_{l=1}^{\tilde{L}_j} \sum_{l' \in \mathcal{L}_j} |r_{l,l'}|^{d_3} 6^{-k_j(d_2-d_3)} \qquad (6.33) \\ &\ge \sum_{l=1}^{\tilde{L}_j} 10 \cdot 6^{-k_j} 6^{-k_j(d_2-d_3)} \ge 100 \cdot 6^{k_j(d_4+d_3-d_2)} \cdot 10 \cdot 6^{-k_j} > 1000. \end{aligned} $$

We can now give the precise definition of the set $\mathbf{F}$.

**Definition 6.7.** We call $\mathbf{F}$ the attractor of the IFS $(\Phi_{l,l'})_{l \in \{1, \dots, \tilde{L}_j\}, l' \in \mathcal{L}_j}$.

For ease of notation, to avoid double indexing, we re-index the functions $\Phi_{l,l'}$ to obtain the bijections $\tilde{\Phi}_k : [0, 1] \to \mathbf{J}_k$, $k = 1, \dots, K$, with $K = \tilde{L}_j \cdot \#(\mathcal{L}_j)$. Note that

$$ \text{the intervals } \mathbf{J}_k \text{ are of the form } \Psi_l(\mathbf{J}'_{j,l'}) \text{ with suitable values of } l \text{ and } l'. \qquad (6.34) $$

The contraction ratio of $\tilde{\Phi}_k$ is denoted by $r_k$.

**Lemma 6.8.** Let $d_5$ be the unique solution to

$$ \sum_{l=1}^{\tilde{L}_j} \sum_{l' \in \mathcal{L}_j} |r_{l,l'}|^{d_5} = 1. \qquad (6.35) $$

Then $\dim_{\mathbf{H}}(\mathbf{F}) = d_5$, and for every $\alpha \in (\alpha_{\min}, \alpha_0 + \varepsilon_0)$ one has $\dim_{\mathbf{H}}(\mathbf{F}) > d_2 > \tau_{\mu_{\lambda}}^*(\alpha + \gamma_{\lambda} - 1)$.

*Proof.* Let us introduce the symbolic dynamical system associated with the IFS $(\tilde{\Phi})_{k=1, \dots, K}$. Take the symbolic space $\tilde{\Omega} = \{\omega = (\omega_1 \omega_2 \cdots) : \omega_i \in \{1, \dots, K\}\}$, with the one-sided shift $\sigma$. The distance of the points in $\tilde{\Omega}$ is given by $d_{\tilde{\Omega}}(\omega, \omega') = \sum_{i \ge 1} |\omega_i - \omega_i'| 2^{-i}$. One can define the projection $\pi$ of $\tilde{\Omega}$ onto $\mathbf{F}$ by

$$ \pi(\omega) = \bigcap_{p=1}^{\infty} (\tilde{\Phi}_{\omega_1} \circ \cdots \circ \tilde{\Phi}_{\omega_p})([0, 1]). $$

Since the intervals $\mathbf{J}_k$ are non-overlapping there is a countable set $\mathcal{E}' \subset \mathbf{F}$ such that if $x \in \mathbf{F} \setminus \mathcal{E}'$, then there is a unique $\tilde{\omega}(x) = (\tilde{\omega}_p(x))_{p \ge 1} \in \tilde{\Omega}$ such that

$$ \{x\} = \bigcap_{p=1}^{\infty} (\tilde{\Phi}_{\tilde{\omega}_1(x)} \circ \cdots \circ \tilde{\Phi}_{\tilde{\omega}_p(x)})([0, 1]). $$
One easily sees that $\mathcal{E}' \subset \mathcal{E}$. Since we work with non-atomic measures, the countable exceptional set will be of measure zero and has no influence on the dimensions considered hereafter. On $\mathbf{F} \setminus \mathcal{E}'$ one can define the “inverse dynamical system” to the IFS. Indeed, for $x \in \mathbf{F} \setminus \mathcal{E}'$ there is a unique interval $\mathbf{J}_k$ containing $x$ and we can define $S(x) = \tilde{\Phi}_k^{-1}(x)$. For this $x$ we have $\tilde{\omega}(S(x)) = \sigma\tilde{\omega}(x)$ and in general $S^p(x) \in \mathbf{J}_{\tilde{\omega}_{p+1}(x)}$ for $p \ge 0$. Given a finite word $\overline{\omega}_1 \cdots \overline{\omega}_j$, we introduce the cylinder sets $[\overline{\omega}_1 \cdots \overline{\omega}_j] = \{\tilde{\omega} \in \tilde{\Omega} : \tilde{\omega}_i = \overline{\omega}_i \text{ for } i = 1, \dots, j\}$. Clearly $\pi([k]) = \mathbf{J}_k$.

By construction, the IFS $(\tilde{\Phi}_k)_{k=1,\dots,K}$ satisfies the Open Set Condition, so, for example, by Theorem 9.3 in [Fal97] it follows from (6.35) that $\dim_{\mathbf{H}}(\mathbf{F}) = d_5$. In particular, recalling (6.5), and (6.33) $\dim_{\mathbf{H}}(\mathbf{F}) = d_5 > d_2$. By construction one has $d_2 > \tau_{\mu_\lambda}^*(\alpha + \gamma_\lambda - 1)$ for $\alpha \in (\alpha_{\min}, \alpha_0 + \varepsilon_0)$. $\square$

## 6.2. Regularity of $\mu_\lambda$ and $F^\lambda$ at points in $\mathbf{F}$.

We start by proving that $\mathbf{F}$ contains essentially regular points, i.e. points whose digit frequencies in (4.10) all converge.

**Lemma 6.9.** $\mathcal{H}^{d_5}$-almost every point of $\mathbf{F}$ is regular.

*Proof.* The idea here is to introduce a Borel probability measure $\mu_{\mathbf{F}}$ living naturally on $\mathbf{F}$. We use the dynamical system introduced in the proof of Lemma 6.8. For each integer $k$, let $\mathbf{n}_k$ be the integer such that $\overline{I}_{\mathbf{n}_k}(x) = \mathbf{J}_k$ for $x \in \mathbf{J}_k \setminus \mathcal{E}'$. Observe that for $i = 0, \dots, 3$ the quantity

$$ \beta_{i,k} := \beta(x, \mathbf{n}_k) \text{ does not depend on which } x \in \mathbf{J}_k \setminus \mathcal{E}' \text{ is chosen.} \quad (6.36) $$

Let us define $\xi_1(x) = x$ and, for $p \ge 2$,

$$ \xi_p(x) = \tilde{\Phi}_{\tilde{\omega}_{p-1}(x)}^{-1} \circ \dots \circ \tilde{\Phi}_{\tilde{\omega}_1(x)}^{-1}(x) = \pi(\sigma^{p-1}(\pi^{-1}(x))) = S^{p-1}(x). $$

By definition, for every $p$, $\xi_p(x) \in \mathbf{J}_{\tilde{\omega}_p(x)}$. In addition, for every $x \in \mathbf{J}_k$, set

$$ \tilde{I}_1(x) = \mathbf{J}_k \text{ and } \mathbf{n}(x) = \mathbf{n}_k, \quad (6.37) $$

that is $\mathbf{J}_k = \tilde{I}_1(x) = \overline{I}_{\mathbf{n}(x)}(x) = \overline{I}_{\mathbf{n}(\xi_1(x))}(\xi_1(x))$. Next, set

$$ \tilde{I}_p(x) = \overline{I}_{\mathbf{n}_p(x)}(x) \text{ where } \mathbf{n}_p(x) = \mathbf{n}(\xi_1(x)) + \dots + \mathbf{n}(\xi_p(x)). $$

Since our IFS satisfies the Open Set Condition, it holds $0 < \mathcal{H}^{d_5}(\mathbf{F}) < +\infty$. Then, for every Borel set $A \subset \mathbb{R}$, define

$$ \mu_{\mathbf{F}}(A) = \frac{\mathcal{H}^{d_5}(A \cap \mathbf{F})}{\mathcal{H}^{d_5}(\mathbf{F})}. $$

In particular, $\mu_{\mathbf{F}}(\mathbf{J}_k) = \mu_{\mathbf{F}}(\pi[k]) = r_k^{d_5}$.

It is then natural to consider the measure $\nu$ on $\tilde{\Omega}$ defined by $\nu(A) = \mu_{\mathbf{F}}(\pi(A))$ for Borel sets $A \subset \tilde{\Omega}$. By construction, it satisfies $\nu([\overline{\omega}_1 \cdots \overline{\omega}_j]) = r_1^{d_5} \cdots r_j^{d_5}$. Hence, it is a Bernoulli measure on $\tilde{\Omega}$, and so a fortiori an ergodic measure on $(\tilde{\Omega}, \sigma)$, likewise $\mu_{\mathbf{F}}$ is ergodic for the dynamical system $(\mathbf{F} \setminus \mathcal{E}', S)$. Applying the ergodic theorem, we obtain that for $\mu_{\mathbf{F}}$-almost every $x \in \mathbf{F}$ and every $l = 1, \dots, K$, the limit

$$ \tilde{\beta}_l := \lim_{p \to \infty} \frac{1}{p} \#\{j \le p : \tilde{\omega}_j(x) = l\} $$

exists and $\sum_{l=1}^K \tilde{\beta}_l = 1$.
Recalling (6.37), the limit $\kappa = \lim_{p \to \infty} \frac{\mathbf{n}_p(x)}{p} = \sum_{l=1}^{K} \tilde{\beta}_l \mathbf{n}_l$ also exists. This implies that, for $i = 0, \dots, 3$,

$$ \frac{\beta_i(x, \mathbf{n}_p(x))}{\mathbf{n}_p(x)} = \frac{1}{\mathbf{n}_p(x)} \sum_{s=1}^{p} \beta_i(\xi_s(x), \mathbf{n}(\xi_s(x))) = \frac{p}{\mathbf{n}_p(x)} \frac{1}{p} \sum_{s=1}^{p} \beta_i(\xi_s(x), \mathbf{n}(\xi_s(x))) \\ \xrightarrow{p \to \infty} \frac{1}{\kappa} \sum_{l=1}^{K} \tilde{\beta}_l \beta_{i,l} =: \beta_i, \qquad (6.38) $$

where $\beta_{i,l}$ was given at (6.36). Hence $\mu_F$-almost every point of $F$ is regular. $\square$

In order to compute the pointwise Hölder exponents of $F^\lambda$ at the typical points in $F$, we first prove that these points $x$ belong to $\mathcal{I}$.

**Lemma 6.10.** For $\mu_F$-almost every $x \in F$, one has $\lim_{n \to \infty} |m_n(x)| = \lim_{n \to \infty} \tilde{m}_n(x) = +\infty$.

*Proof.* To prove that $x \in \mathcal{I}$, we are actually going to prove that the “idealized” slopes $\tilde{m}_n(x)$ defined by (3.4) tend to infinity. Since $|m_n(x)| \ge \tilde{m}_n(x)$, the result follows.

Recall that $\mathbf{J}_{\tilde{\omega}_s(x)} = \bar{I}_{\mathbf{n}(\xi_s(x))}(\xi_s(x))$, (6.26) and the definition (6.28) of $G$. When $x \in G$ one has $x \in \mathbf{J}_{l, \rho_j}$ with a suitable $l$ and $\bar{I}_{\tilde{\mathbf{n}}(x)}(x) = \mathbf{J}_{l, \rho_j}$. Set $\tilde{x} = \Psi_l^{-1}(x)$.

The monotonicity of the affine maps $\Psi_l$ defined in (6.27) is compatible with our construction, that is we copied the construction at $\tilde{x} \in \mathbf{J}_j$ by $\Psi_l$ to the one at $x \in \mathbf{J}_{l, \rho_j}$, recall also (6.19). In other words, $\tilde{m}_n(x)$ changes for $n$ between $\tilde{\mathbf{n}}(x)$ and $\mathbf{n}(x)$ the same way as $\tilde{m}_n(\tilde{x})$ changes when $n$ varies between $n_j(\tilde{x})$ and $n'_j(\tilde{x})$, see (6.11) and (6.12) applied to $\tilde{x}$. By (6.37)

$$ \tilde{m}_{\mathbf{n}(x)}(x) = \tilde{m}_{\tilde{\mathbf{n}}(x)}(x) \frac{\tilde{m}_{\mathbf{n}(x)}(x)}{\tilde{m}_{\tilde{\mathbf{n}}(x)}(x)} = \tilde{m}_{\tilde{\mathbf{n}}(x)}(x) \frac{\tilde{m}_{\tilde{\mathbf{n}}(x)+n'_j(\tilde{x})-n_j(\tilde{x})}(x)}{\tilde{m}_{\tilde{\mathbf{n}}(x)}(x)} = \tilde{m}_{\tilde{\mathbf{n}}(x)}(x) \frac{\tilde{m}_{n'_j(\tilde{x})}(\tilde{x})}{\tilde{m}_{n_j(\tilde{x})}(\tilde{x})}. \quad (6.39) $$

By (6.20), $\tilde{m}_{\tilde{\mathbf{n}}(x)}(x) > 100$. Thus

$$ \log \tilde{m}_{\mathbf{n}(x)}(x) > \log 100 + \log \tilde{m}_{n'_j(\tilde{x})}(\tilde{x}) - \log \tilde{m}_{n_j(\tilde{x})}(\tilde{x}). \quad (6.40) $$

Also, by (6.7) and (6.10), one knows that

$$ \log |m_{n'_j}(\tilde{x})| - \log |m_{n_j}(\tilde{x})| \ge (1 - \varepsilon_2)(1 - \alpha_0 - \varepsilon_1) |\log \ell_{n'_j}(\tilde{x})|. \quad (6.41) $$

This provides us with information about the growth of $\log |m_{n'_j}(\tilde{x})| - \log |m_{n_j}(\tilde{x})|$, but additional asymptotic information is needed about $\log \tilde{m}_{n'_j}(\tilde{x}) - \log \tilde{m}_{n_j}(\tilde{x})$. To obtain this, recall (3.38) and (3.34). With a suitable constant $C > 0$, for all $|t| > 1$ and for any $i \in \{0, \dots, 3\}$,

$$ 0 \le \log g_i(t) \le \frac{C}{t^2}. \quad (6.42) $$

Hence, with another suitable constant $C' > 0$, using (6.12) and recalling the definition of the digits $u_k(\tilde{x})$ of $\tilde{x}$ in (2.2),

$$ \begin{aligned} & \left| \log |m_{n'_j}(\tilde{x})| - \log |m_{n_j}(\tilde{x})| - (\log \tilde{m}_{n'_j}(\tilde{x}) - \log \tilde{m}_{n_j}(\tilde{x})) \right| \\ & = \sum_{k=n_j(\tilde{x})}^{n'_j(\tilde{x})-1} \log g_{u_k(\tilde{x})}(|m_k(\tilde{x})|) \le C'(n'_j(\tilde{x}) - n_j(\tilde{x}))3^{-2\varepsilon_2(1-\alpha_0-\varepsilon_1)n'_j(\tilde{x})} \quad (6.43) \\ & \le C'n'_j(\tilde{x})3^{-2\varepsilon_2(1-\alpha_0-\varepsilon_1)n'_j(\tilde{x})} \le 1 \le \frac{\varepsilon_2}{6}(1-\alpha_0-\varepsilon_1)|\log \ell_{n'_j}(\tilde{x})|. \end{aligned} $$
if $n_j(\tilde{x})$ (and then $n'_j(\tilde{x})$ since $n'_j(\tilde{x}) \ge n_j(\tilde{x})$) is sufficiently large.

By (6.40), (6.43) and (6.39),

$$ \log \tilde{m}_{\mathbf{n}(x)}(x) > (1 - 2\varepsilon_2)(1 - \alpha_0 - \varepsilon_1) |\log \ell_{n'_j(\tilde{x})}(\tilde{x})|. \quad (6.44) $$

Observe that by (6.30) and (6.34), for every $x \in \tilde{\mathbf{G}}$,

$$ \frac{1}{6} < \frac{|\mathbf{J}'_{j,l'}|}{|\mathbf{J}_k|} = \frac{\ell_{n'_j(\tilde{x})}(\tilde{x})}{|\overline{\mathcal{I}}_{\mathbf{n}(x)}(x)|} < 6 \quad \text{and} \quad \ell_{n'_j(\tilde{x})}(\tilde{x}) \le 3^{-n'_j(\tilde{x})}. $$

This implies

$$ |\log |\overline{\mathcal{I}}_{\mathbf{n}(x)}(x)|| \le |\log \ell_{n'_j(\tilde{x})}(\tilde{x})| - \log 6 \le (1 - \varepsilon_2) |\log \ell_{n'_j(\tilde{x})}(\tilde{x})|, \quad (6.45) $$

if $\log 6 \le \varepsilon_2 n'_j(\tilde{x}) \log 3 \le \varepsilon_2 |\log \ell_{n'_j(\tilde{x})}(\tilde{x})|$, which can be achieved by choosing a sufficiently large $n'_j(\tilde{x})$.

Hence for $x \in \tilde{\mathbf{G}}$ we have (6.39), (6.40), (6.41), (6.45) and therefore

$$ \frac{\log \tilde{m}_{\mathbf{n}(x)}(x)}{|\log |\overline{\mathcal{I}}_{\mathbf{n}(x)}(x)||} \ge (1 - \varepsilon_2)(1 - 2\varepsilon_2)(1 - \alpha_0 - \varepsilon_1) > (1 - 3\varepsilon_2)(1 - \alpha_0 - \varepsilon_1). \quad (6.46) $$

In addition, using notation from the proof of Lemma 6.9, we have

$$ \ell_{\mathbf{n}_p(x)}(x) = \prod_{s=1}^{p} |\mathbf{J}_{\tilde{\omega}_s(x)}| \quad \text{and} \quad \tilde{m}_{\mathbf{n}_p(x)}(x) = \prod_{s=1}^{p} \tilde{m}_{\mathbf{n}(\xi_s(x))}(\xi_s(x)). \quad (6.47) $$

Therefore, by using (6.46) we obtain

$$ \log \tilde{m}_{\mathbf{n}_p(x)}(x) = \sum_{s=1}^{p} \log \tilde{m}_{\mathbf{n}(\xi_s(x))}(\xi_s(x)) \ge (1 - 3\varepsilon_2)(1 - \alpha_0 - \varepsilon_1) \sum_{s=1}^{p} |\log |I_{\mathbf{n}(\xi_s(x))}(\xi_s(x))|| \\ = (1 - 3\varepsilon_2)(1 - \alpha_0 - \varepsilon_1) \sum_{s=1}^{p} |\log |\mathbf{J}_{\tilde{\omega}_s(x)}||. $$

Hence, combining (6.47) with the fact that $\mathbf{n}_{p+1}(x) - \mathbf{n}_p(x)$ is uniformly bounded in $p$ and $x$, one deduces that, for $\mu_{\mathbf{F}}$-almost every $x \in \mathbf{F}$,

$$ \lim_{n \to \infty} \frac{\log \tilde{m}_n(x)}{|\log \ell_n(x)|} = \lim_{p \to \infty} \frac{\log \tilde{m}_{n_p(x)}(x)}{|\log \ell_{n_p(x)}(x)|} \ge (1 - 3\varepsilon_2)(1 - \alpha_0 - \varepsilon_1) > 0. $$

In particular, $\lim_{n \to \infty} \tilde{m}_n(x) = +\infty$, hence the conclusion. $\square$

We now show that most points of $\mathbf{F}$ share the same pointwise Hölder exponent for $F^\lambda$.

**Lemma 6.11.** $\mathcal{H}^{d_5}$-almost all points $x \in \mathbf{F}$ have the same pointwise Hölder exponent for $F^\lambda$, and this exponent $h_{F^\lambda}(x)$ satisfies $h_{F^\lambda}(x) = \dim_{\text{loc}}(\mu_\lambda, x) + 1 - \gamma_\lambda \le \alpha_0 + \varepsilon_0$.

*Proof.* Using Lemma 6.9, for $\mathcal{H}^{d_5}$-almost every $x \in \mathbf{F}$, we derive from (4.9) that

$$ \underline{\dim}_{\text{loc}}(\mu_\lambda, x) = \gamma_\lambda - \frac{\beta_1 \log(6\lambda + 1) + \beta_2 \log(6\lambda - 1)}{(\beta_0 + \beta_3) \log 3 + (\beta_1 + \beta_2) \log 6} \quad (6.48) $$

where $\beta_i$ is given by (6.38). By Lemma 6.10, $\mu_{\mathbf{F}}$-almost every $x$ belongs to $\mathcal{I}$, so by (3.33)

$$ \lim_{n \to \infty} \frac{\log |m_n(x)|}{|\log \ell_n(x)|} = \frac{\beta_1 \log(6\lambda + 1) + \beta_2 \log(6\lambda - 1)}{(\beta_0 + \beta_3) \log 3 + (\beta_1 + \beta_2) \log 6} = \lim_{n \to \infty} \frac{\log \tilde{m}_n(x)}{|\log \ell_n(x)|}. \quad (6.49) $$
Now (6.48) and (6.49) together imply that for $\mu_F$-almost every $x$,

$$h_{F^{\lambda}}(x) = \dim_{\text{loc}}(\mu_{\lambda}, x) + 1 - \gamma_{\lambda} \leq (\gamma_{\lambda} - (1 - 3\varepsilon_2)(1 - \alpha_0 - \varepsilon_1)) + 1 - \gamma_{\lambda} \leq \alpha_0 + \varepsilon_0$$

if $\varepsilon_1 > 0$ and $\varepsilon_2 > 0$ are chosen sufficiently small from the beginning (recall Remark 6.2). Hence, the result. $\square$

We are now able to conclude on Proposition 6.1.

*Proof of Proposition 6.1.* Lemma 6.11 states that $h_{F^{\lambda}}(x)$ is $\mu_F$-almost everywhere constant given by (6.48). Calling $\alpha_F$ this exponent, one has by (6.5),

$$d_5 = \dim_H(\mathbf{F}) \leq \tau_{\mu_{\lambda}}^*(\alpha_{\mathbf{F}}) < d_2, \qquad (6.50)$$

contradicting the result in Lemma 6.8. This concludes the proof of Proposition 6.1. $\square$

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<footnote n="1">¹ DEPARTMENT OF ANALYSIS, ELTE EÖTVÖS LORÁND, UNIVERSITY, PÁZMÁNY PÉTER SÉTÁNY 1/C, 1117 BUDAPEST, HUNGARY</footnote>

<footnote>Email address: zoltan.buczolich@ttk.elte.hu</footnote>

<footnote>URL: http://buczo.web.elte.hu, ORCID Id: 0000-0001-5481-8797</footnote>

<footnote n="2">² LABORATOIRE MODAL'X, UMR CNRS 9023, UPL, UNIVERSITÉ PARIS NANTERRE, 200 AVENUE DE LA RÉPUBLIQUE, 92001 NANTERRE, FRANCE.</footnote>

<footnote>Email address: ydemichel@parisnanterre.fr</footnote>

<footnote>URL: https://www.parisnanterre.fr/m-yann-demichel, ORCID Id: 0009-0008-4306-9469</footnote>

<footnote n="3">³ STÉPHANE SEURET, UNIV PARIS EST CRETEIL, UNIV GUSTAVE EIFFEL, CNRS, LAMA UMR8050, F-94010 CRETEIL, FRANCE</footnote>

<footnote>Email address: seuret@u-pec.fr</footnote>

<footnote>URL: https://sites.math.u-pem.fr/sseuret/</footnote>