Hölder spectrum of typical monotone continuous functions

Zoltán Buczolich^{∗}, Department of Analysis, Eötvös Loránd University, Rákóczi út 5, 1088 Budapest, Hungary

and

Judit Nagy, Department of Analysis, Eötvös Loránd University, Rákóczi út 5, 1088 Budapest, Hungary

June 19, 1999

Abstract

We compute the Hölder spectrum and the moment sums of the typical (in the sense of Baire category) monotone continuous function, f. We show that these functions are of multifractal nature, for example α equals the Hausdorff dimension of those points where such a function, f is of exact Hölder class α ∈ [0, 1]. We also prove that f “grows” only at those points where it has bad Hölder properties, that is, at those points where it is of Hölder class 0. The upper moment sum \overline{τ}_{f} (q) = ∞ if q < 0, \overline{τ}_{f} (q) = 1 − q if 0 ≤ q ≤ 1, and \overline{τ}_{f} (q) = 0 if 1 < q. The lower moment sum \underline{τ}_{f} (q) = 1 − q if q < 0, \underline{τ}_{f} (q) = 0 if 0 ≤ q ≤ 1, and \underline{τ}_{f} (q) = 1 − q if 1 < q.

1 Introduction

In this paper we investigate the multifractal (Hölder) properties of the typical (in the sense of Baire category) monotone continuous function defined on [0, 1]. We denote the space of monotone increasing continuous functions equipped with the sup norm by C_{↗}[0, 1]. It is a separable, complete metric space, being a closed subspace of C[0, 1]. If a property holds for a residual subset of C_{↗}[0, 1] then we say that the typical function f has this property.

In [11, 12] the differentiability properties of typical monotone continuous functions were studied and it was shown that the typical monotone continuous function, f is a strictly monotone increasing singular function and its derivative equals 0 wherever it exists. Of course, f′ exists almost everywhere in [0, 1].

∗This author was supported by Grants FKFP 0192/1999 and Hungarian National Foundation for Scientific Research Grant No. T 016094

1991 Mathematics Subject Classification: Primary 28A80, 26A16 Secondary: 58F12, 26A48.

Keywords: typical monotone continuous function, moment sum, Hölder spectrum.

Jaffard [2, 3, 4, 5] has studied the multifractal properties of several specific continuous functions. For these functions many interesting methods, for example, wavelets, Diophantine approximation etc. were used.

It is not difficult to verify that the typical continuous function on [0, 1] is quite uninteresting from our point of view, it does not have multifractal Hölder properties, in fact, it is Hölder class 0 everywhere in [0, 1]. Hence it is much more interesting to look at the class of typical monotone continuous functions which are of multifractal nature. By studying typical monotone continuous functions we study typical continuous measures on [0, 1]. We point out that typical measures on arbitrary separable metric spaces were considered by [10, 9], for example a typical measure typically has no local dimension.

We denote by E_{α}^{f} those points of [0, 1] where f ∈ C_{↗}[0, 1] is of exact Hölder class α and by E_{≤α}^{f} those points where f is of Hölder class β for any β < α and is not of Hölder class β for any β > α (the detailed definitions will be given in the preliminaries section).

The main results of this paper are given in Theorem 6. In assertions (i-ii) of Theorem 6 we show that dim_{H}E_{α}^{f} = dim_{H}E_{≤α}^{f} = α holds for α ∈ [0, 1], while E_{≤α}^{f} = ∅ for α > 1. This implies that for any α ∈ [0, 1] there are “many points” where the typical f ∈ C_{↗}[0, 1] function is of exact Hölder class α. In assertion (vi) we will show that if α ≠ 0 then f is not “growing” on the sets Ê_{≤α}^{f} ⊃ E_{α}^{f}. In fact, all the growth of f takes place on E_{≤0}^{f}, by this we mean that μ_{f}(E_{≤0}^{f}) = μ_{f}([0, 1]) = f(1)−f(0), where μ_{f} is the Borel measure corresponding to f. Thus μ_{f}(∪_{α≠0}Ê_{α}^{f}) = 0.

Studying multifractal properties quite often the moment sums, τ_{f}(q) (for the definition see (1) below), are used [1, 6, 7] and in “lucky situations” the Legendre transform of τ_{f}(q) provides the spectrum of f. As one can expect the typical f ∈ C_{↗}[0, 1] is not the “lucky case”. In assertions (iii-v) of Theorem 6 we compute these moment sums. It turns out that for the typical f ∈ C_{↗}[0, 1] τ_{f}(0) = 1, τ_{f}(1) = 0 and τ_{f}(q) does not exist for other vaules of q. If the upper moment sums are considered we have \overline{τ}_{f}(q) = ∞ if q < 0, \overline{τ}_{f}(q) = 1 − q if 0 ≤ q ≤ 1 and \overline{τ}_{f}(q) = 0 if 1 < q. For the lower moment sums, \underline{τ}_{f}(q), we show that \underline{τ}_{f}(q) = 1 − q if q < 0, \underline{τ}_{f}(q) = 0 if 0 ≤ q ≤ 1 and \underline{τ}_{f}(q) = 1 − q if 1 < q. It might be worth to mention that an elementary computation shows that for f(x) = x we have τ_{f}(q) = 1 − q for all q ∈ ℝ.

In Section 4 we present the main construction of this paper. This construction will provide the dense G_{δ} sets in C_{↗}[0, 1] showing that certain properties are typical. The proof of Theorem 6 will be split into parts according to the different assertions made.

2 Preliminaries

The integer part of the real number x, that is, the largest integer not exceeding x, will be denoted by [x].

We use the notation λ(A) for the Lebesgue measure of the measurable set A ⊂ [0, 1] while H^{s}(A) denotes its s-dimensional Hausdorff measure [6, 7].

The space of continuous and continuously differentiable functions on [0, 1] is denoted by C[0, 1] and by C^{1}[0, 1], respectively.

Definition 1. Given α ≥ 0 and a point x_{0} ∈ [0, 1] we say that the function f is C^{α}(x_{0}) if there exists a polynomial P of degree at most [α] and a constant C > 0 such that |f(x) − P (x − x_{0})| ≤ C|x − x_{0}|^{α} for all x ∈ [0, 1]. The Hölder exponent of f at x_{0} is defined by

h_{f}(x_{0}) = sup{α : f is C^{α}(x_{0})}.

Set

E_{≤α}^{f} = {x_{0} : h_{f}(x_{0}) ≤ α} and E_{α}^{f} = {x_{0} : h_{f}(x_{0}) = α, and f is C^{α}(x_{0})}.

The set E_{α}^{f} denotes the set of those points where f is of “exact” Hölder class α. It is clear that E_{α}^{f} ⊂ E_{≤α}^{f} \ ∪_{β<α}E_{≤β}^{f} = Ê_{α}^{f}. However this inclusion is proper,

for example if we take g(x) = − log(x/10) · \sqrt{x/10} for x ∈ (0, 1] and g(0) =^{def}0 then g ∈ C_{↗}[0, 1], h_{g}(0) = 1/2, but g is not C^{1/2}(0).

To compute Hausdorff dimensions we will use Proposition 2.2 of [7] (see also Proposition 4.9 of [6]):

Proposition 1. Let E ⊂ ℝ^{n} be a Borel set, let μ be a finite Borel measure on ℝ^{n} and 0 < c < ∞.

(a) If lim sup_{r→0+} μ(B(x, r))/r^{s} ≤ c for all x ∈ E then H^{s}(E) ≥ μ(E)/c.

(b) If lim sup_{r→0+} μ(B(x, r))/r^{s} ≥ c for all x ∈ E then H^{s}(E) ≤ 2^{s}μ(E)/c.

3 Elementary lemmas for the moment sums

Given a monotone increasing continuous function g : [0, 1] → ℝ we denote by μ_{g} the Borel measure corresponding to g, that is for which μ_{g}((a, b)) = g(b) − g(a) for each interval (a, b).

Set

τ_{g,j}(q) = \frac{1}{j log 2} log ∑_{k=0}^{2^{j}−1} (μ_{g} ([\frac{k}{2^{j}} , \frac{k + 1}{2^{j}}]))^{q} and τ_{g}(q) = lim_{j→∞} τ_{g,j}(q). (1)

If the latter limit does not exist we use \overline{τ}_{g}(q) = lim sup_{j→∞} τ_{g,j}(q) and \underline{τ}_{g}(q) = lim inf_{j→∞} τ_{g,j}(q).

It is not difficult to see that τ_{cg}(q) = τ_{g}(q) holds for any c > 0, so there is no problem if the finite measure μ_{g} is not a probability measure as long as g is a nonconstant increasing function, actually, when we compute moment sums we can assume that μ_{g} is a probability measure.

Lemma 2. If g ∈ C^{1}[0, 1] and g′ > 0 then for every ϵ > 0 there exists K > 0 such that τ_{g,j}(q) ≥ 1 − q − ϵ for all j ≥ K, q ∈ [0, 1].

Proof. Using the continuity of g′ > 0 we can choose ϱ_{1} such that g′ > ϱ_{1} > 0. Using the mean value theorem for g from the above estimates we infer

μ_{g} ([\frac{k}{2^{j}} , \frac{k + 1}{2^{j}}]) ≥ ϱ_{1} \frac{1}{2^{j}} .

Then

τ_{g,j}(q) ≥ \frac{1}{j log 2} log (2^{j} (\frac{ϱ_{1}}{2^{j}} )^{q}) = 1 − q + \frac{q log ϱ_{1}}{j log 2} .

This proves the lemma.

Lemma 3. For any q ∈ [0, 1] and g ∈ C_{↗}[0, 1] we have \overline{τ}_{g}(q) ≤ 1 − q and \underline{τ}_{g}(q) ≥ 0.

Proof. Set μ_{k,j} = μ_{g}([\frac{k}{2^{j}} , \frac{k+1}{2^{j}}]). If q = 0 then ∑_{k=0}^{2^{j}−1} μ_{k,j}^{0} = 2^{j} and hence

\overline{τ}_{g}(0) = τ_{g}(0) = 1. In the sequel, we assume q ∈ (0, 1]. Obviously, ∑_{k=0}^{2^{j}−1} μ_{k,j} = g(1) − g(0) =^{def}G_{0} > 0. Since μ_{k,j} ≥ 0 we can use some well-known inequalities between means of nonegative numbers and obtain

(\frac{∑_{k=0}^{2^{j}−1} μ_{k,j}^{q}}{2^{j}} )^{\frac{1}{q}} ≤ \frac{∑_{k=0}^{2^{j}−1} μ_{k,j}}{2^{j}} = \frac{G_{0}}{2^{j}} .  (2)

This implies ∑_{k=0}^{2^{j}−1} μ_{k,j}^{q} ≤ G_{0}^{q}2^{j(1−q)} and an easy calculation implies \overline{τ}_{g}(q) ≤ 1 − q.

Using that τ_{cg} = τ_{g} for any c > 0 we can assume μ_{g}([0, 1]) = G_{0} = 1. Then μ_{k,j}^{q} ≥ μ_{k,j} and hence ∑_{k=0}^{2^{j}−1} μ_{k,j}^{q} ≥ ∑_{k=0}^{2^{j}−1} μ_{k,j} = G_{0} = 1. Thus τ_{g,j}(q) ≥ 0 and hence \underline{τ}_{g}(q) ≥ 0 holds.

The next two lemmas are the q > 1 versions of the previous two ones. Their proofs are analogous so we omit some details.

Lemma 4. If g ∈ C^{1}[0, 1] and g′ ≥ 0 then for every ϵ > 0 there exists K > 0 such that τ_{g,j}(q) ≤ 1 − q + ϵ for all j ≥ K, q ∈ [1, 1/ϵ].

Proof. Using the continuity of g′ > 0 we can choose ϱ_{2} such that ϱ_{2} > g′. By the mean value theorem we infer

μ_{g} ([\frac{k}{2^{j}} , \frac{k + 1}{2^{j}}]) ≤ ϱ_{2} \frac{1}{2^{j}} .

Then

τ_{g,j}(q) ≤ \frac{1}{j log 2} log (2^{j} (\frac{ϱ_{2}}{2^{j}} )^{q}) = 1 − q + \frac{q log ϱ_{2}}{j log 2} .

Lemma 5. For any q > 1 and g ∈ C_{↗}[0, 1] we have \underline{τ}_{g}(q) ≥ 1−q and \overline{τ}_{g}(q) ≤ 0.

Proof. Define μ_{k,j} and G_{0} as in the proof of Lemma 2. From q > 1 it follows that we have the reversed inequality in (2), that is,

(\frac{∑_{k=0}^{2^{j}−1} μ_{k,j}^{q}}{2^{j}} )^{\frac{1}{q}} ≥ \frac{G_{0}}{2^{j}} .

This implies ∑_{k=0}^{2^{j}−1} μ_{k,j}^{q} ≥ G_{0}^{q}2^{j(1−q)} and hence \underline{τ}_{g}(q) ≥ 1 − q.

Like in Lemma 2 we can assume μ_{g}([0, 1]) = 1 and using μ_{k,j}^{q} ≤ μ_{k,j} we can

obtain ∑_{k=0}^{2^{j}−1} μ_{k,j}^{q} ≤ 1 which implies τ_{g,j}(q) ≤ 0 and \overline{τ}_{g}(q) ≤ 0.

4 The main construction

Assume m ∈ ℕ is given. We choose a countable dense subset, {f_{n}}, in C_{↗}[0, 1], such that each f_{n} is continuously differentiable and f′_{n} > 0 on [0, 1]. Using Lemmas 2 and 4 with ϵ = 1/n choose K_{n} > n such that for all q ∈ [0, 1] and j ≥ K_{n} we have

τ_{f_{n},j}(q) ≥ 1 − q − \frac{1}{n};  (3)

and for all q ∈ [1, n] and j ≥ K_{n} we have

τ_{f_{n},j}(q) ≤ 1 − q + \frac{1}{n}  (4)

Using the uniform continuity of f_{n} choose and fix N_{n} ≥ K_{n} such that

f_{n} (\frac{k + 1}{2^{N_{n}}}) − f_{n} (\frac{k}{2^{N_{n}}} ) = μ_{f_{n}} ([\frac{k}{2^{N_{n}}} , \frac{k + 1}{2^{N_{n}}}]) < \frac{1}{2n}  (5)

holds for k = 0, ..., 2^{N_{n}} − 1.

We will choose a large integer ν_{n,m} > max{N_{n}^{2}, m} to be determined later. Set

h_{n,m}^{∗} = \frac{1}{4^{n}2^{mν_{n,m}}} < \frac{1}{4^{n}} ,  h_{n,m} = \frac{h_{n,m}^{∗}}{2^{ν_{n,m}}} ,  (6)
δ_{n,m} = (h_{n,m})^{2m} \frac{1}{4} < \frac{1}{2} \frac{1}{2^{ν_{n,m}}} ,  P (k, n, m) = \frac{k}{2^{ν_{n,m}}} + δ_{n,m}, (7)

where k ∈ {0, ..., 2^{ν_{n,m}} − 1}. In some applications of our construction we will use different values for h_{n,m}^{∗}, however for all these values we will have 0 < h_{n,m}^{∗} ≤ 1/4^{n} ≤ 1/4n. Observe that h_{n,m}^{∗}, h_{n,m}, δ_{n,m} and P (k, n, m) are all dyadic rationals.

Set

W_{n,m} = ⋃_{k=0}^{2^{νn,m} −1} (\frac{k}{2^{ν_{n,m}}} , P (k, n, m)) .

[figure: \frac{k}{2^{νn,m}}; \frac{k+1}{2^{νn,m}}; 0; 1; P (k, n, m); δ_{n,m}; h_{n,m}; h_{n,m}^{∗}; g_{n,m}^{∗∗}; f_{n}; g_{n}^{∗}]

Next we define the auxiliary function g_{n}^{∗}(x) = f_{n}((k′ + 1)/2^{N_{n}}) if x ∈ (k′/2^{N_{n}}, (k′ + 1)/2^{N_{n}}] \ W_{n,m}, (k′ = 0, ..., 2^{N_{n}} − 1) and g_{n}^{∗}(0) = f_{n}(0). On W_{n,m} we define g_{n}^{∗} so that it is continuous on [0, 1] and is linear on each interval (k/2^{ν_{n,m}}, P (k, n, m)).

Set g_{n,m}^{∗∗}(x) = (k + 1)h_{n,m} if x ∈ (k/2^{ν_{n,m}}, (k + 1)/2^{ν_{n,m}}] \ W_{n,m}, (k = 0, ..., 2^{ν_{n,m}} −1) and g_{n,m}^{∗∗}(0) = 0. On W_{n,m} we define g_{n,m}^{∗∗} so that it is continuous on [0, 1] and is linear on each interval (k/2^{ν_{n,m}}, P (k, n, m)).

Finally we put g_{n,m}(x) = g_{n}^{∗}(x) + g_{n,m}^{∗∗}(x).

[figure: h_{n,m}^{∗}; g_{n,m}; f_{n}; g_{n,m} + ϱ_{n,m}; g_{n,m} − ϱ_{n,m}]

By (5)

|f_{n}(x) − g_{n}^{∗}(x)| < \frac{1}{2n} and 0 ≤ g_{n,m}^{∗∗}(x) ≤ g_{n,m}^{∗∗}(1) = h_{n,m}^{∗}.  (8)

We also have

|f_{n}(x) − g_{n,m}(x)| ≤ |f_{n}(x) − g_{n}^{∗}(x)| + h_{n,m}^{∗} ≤ \frac{3}{4n}  (9)

Using

ϱ_{n,m} = \frac{(h_{n,m})^{m}}{8} < \frac{h_{n,m}}{8}  (10)

we set G_{n,m} = B(g_{n,m}, ϱ_{n,m}) and H_{m} = ∪_{n=m}^{∞}G_{n,m}. It is clear that H_{m} is open in C_{↗}[0, 1] and using the density of the functions {f_{n}} and (9) it is easy to see that H_{m} is dense. To prove the main result of the paper we will use the residual set F^{∗} = ∩_{m=1}^{∞}H_{m}.

5 Main results

The purpose of this paper is to prove the following theorem:

Theorem 6. There is a residual subset F ⊂ C_{↗}[0, 1] such that for any f ∈ F we have

(i) Ê_{α}^{f} = ∅ for α > 1;

(ii) dim_{H}E_{α}^{f} = dim_{H} Ê_{α}^{f} = dim_{H}E_{≤α}^{f} = α for all α ∈ [0, 1];

(iii) \overline{τ}_{f}(q) = 1−q and \underline{τ}_{f}(q) = 0 for 0 < q ≤ 1, and \overline{τ}_{f}(0) = \underline{τ}_{f}(0) = τ_{f}(0) = 1;

(iv) \overline{τ}_{f}(q) = 0 and \underline{τ}_{f}(q) = 1 − q for 1 < q;

(v) \overline{τ}_{f}(q) = ∞ and \underline{τ}_{f}(q) = 1 − q for q < 0;

(vi) μ_{f}(E_{≤0}^{f}) = f(1) − f(0), that is μ_{f}([0, 1] \ E_{≤0}^{f}) = 0.

We will split the proof of the theorem according to the assertions (i-vi). It is also clear that if for each assertion we can find a suitable dense G_{δ} set in C_{↗}[0, 1] then the intersection of these sets will provide F.

5.1 Assertion (i)

Proof. Recall that a monotone function is differentiable almost everywhere and by [11, 12] there is a residual, dense G_{δ} subset, F_{0} of C_{↗}[0, 1] such that f′(x) = 0 at all points where f′ exists. We use a slightly modified version of the main construction, in equation (6) we set h_{n,m}^{∗} = \frac{1}{4^{n}}. With these values of h_{n,m}^{∗} the sets H_{m} will still remain dense and open in C_{↗}[0, 1]. By choosing ν_{n,m} ≥ m^{2} sufficiently large we can assume that

h_{n,m} = \frac{h_{n,m}^{∗}}{2^{ν_{n,m}}} = \frac{1}{4^{n}2^{ν_{n,m}}} > 2 (\frac{1}{2^{ν_{n,m}}} )^{1+ \frac{1}{m}} .  (11)

Denote by F_{1} = ∩_{m=1}^{∞}H_{m} the set obtained by the main construction and sup-

pose that f ∈ F_{(i)} =^{def}F_{1} ∩ F_{0}.

Proceeding towards a contradiction assume 1 < α < 2, x_{0} ∈ [0, 1] and f is C^{α}(x_{0}). Then

|f(x) − (a_{0} + a_{1}(x − x_{0}))| ≤ C|x − x_{0}|^{α}

and, clearly, a_{0} = f(x_{0}), f′(x_{0}) exists and a_{1} = f′(x_{0}) = 0 where we also used that f ∈ F_{0}. Hence

|f(x) − f(x_{0})| ≤ C|x − x_{0}|^{α}.  (12)

Fix m > 2 such that α > 1 + \frac{2}{m} and C < m. Choose δ > 0 such that m|r|^{1/m} < 1 holds for 0 < |r| < δ; hence

C|r|^{α} < m|r|^{α} < m|r|^{1+ \frac{2}{m}} < |r|^{1+ \frac{1}{m}}

is also true. From f ∈ F_{1} it follows that there exists n ≥ m such that f ∈ G_{n,m}. We can also suppose that ν_{n,m} is so large that 2^{−ν_{n,m}} < δ. Set r = 1/2^{ν_{n,m}} if x_{0} + r ∈ [0, 1], otherwise set r = −1/2^{ν_{n,m}}. Clearly,

|g_{n,m}^{∗∗}(x_{0} + r) − g_{n,m}^{∗∗}(x_{0})| = h_{n,m} and g_{n}^{∗}, g_{n,m}^{∗∗} are increasing;

hence

|g_{n,m}(x_{0} + r) − g_{n,m}(x_{0})| ≥ h_{n,m}.

Using f ∈ B(g_{n,m}, ϱ_{n,m}) and ϱ_{n,m} < h_{n,m}/8 we obtain

|f(x_{0} + r) − f(x_{0})| > |g_{n,m}(x_{0} + r) − g_{n,m}(x_{0})| − 2ϱ_{n,m} > \frac{3}{4}h_{n,m}.

Therefore by recalling (11) and (12)

\frac{3}{4}h_{n,m} < |f(x_{0} + r) − f(x_{0})| ≤ C|r|^{α} < |r|^{1+ \frac{1}{m}} = (\frac{1}{2^{ν_{n,m}}} )^{1+ \frac{1}{m}} < \frac{h_{n,m}}{2} ,

a contradiction.

We showed that for 1 < α < 2 there is no x_{0} ∈ [0, 1] for which f is C^{α}(x_{0}). If f is C^{β}(x_{0}) and 0 < α < β then f is C^{α}(x_{0}) as well. Thus we showed that h_{f}(x_{0}) ≤ 1 holds for any f ∈ F_{(i)} and x_{0} ∈ [0, 1].

5.2 Assertion (ii)

Before turning to this assertion we state and prove a lemma.

Lemma 7. Assume g ∈ C_{↗}[a, b] and

U_{g} = {x ∈ [a, b] : ∃y ∈ [a, b] such that |g(y) − g(x)| ≥ |y − x|}

then λ(U_{g}) ≤ 2(g(b) − g(a)).

Proof. Set

U_{1} = {x ∈ [a, b] : ∃y ∈ (x, b] such that g(y) > g(x) + (y − x)}

and

U_{2} = {x ∈ [a, b] : ∃y ∈ [a, x) such that g(y) < g(x) + (y − x)}.

Clearly, U_{g} = U_{1} ∪ U_{2}. We will show that λ(U_{1}) ≤ g(b) − g(a), the proof for U_{2} is similar.

Letting g_{0}(x) = g(x) − x we have

U_{1} = {x ∈ [a, b] : ∃y ∈ (x, b] such that g_{0}(y) > g_{0}(x)}.

Set g_{1}(x) = max{g_{0}(t) : t ∈ [x, b]}. Then g_{1}(b) = g_{0}(b), g_{1}(a) ≥ g_{0}(a) and g′_{1}(b) ≥ −1 where this derivative exists.

It is easy to see that g_{1}(x) is monotone decreasing and hence (by using Theorem 2.9.19 of [8])

∫_{[a,b]} g′_{1} ≤ g_{1}(b) − g_{1}(a) ≤ g_{0}(b) − g_{0}(a) = g(b) − g(a) + (a − b).

On the other hand g′_{1} = 0 almost everywhere on U_{1} and hence

∫_{[a,b]} g′_{1} ≥ ∫_{[a,b]\U_{1}}(−1) = (a − b) + λ(U_{1}).

From the above two displayed inequalities it follows that λ(U_{1}) ≤ g(b)−g(a).

Now we verify assertion (ii).

Proof. Any f ∈ F_{1} is almost everywhere differentiable, and hence it is C^{1}(x_{0}) at almost every x_{0} ∈ [0, 1]. This and assertion (i) imply dim_{H}(E_{1}^{f}) = 1.

Assume 0 < α < 1 is fixed and 1 > β > α. Next we show that dim_{H}E_{≤α}^{f} ≤ α. Actually, this is valid for any f ∈ C_{↗}[0, 1] and follows easily from Proposition 1. If x_{0} ∈ E_{≤α}^{f} then f is not C^{β}(x_{0}) and hence there exists x_{n} → x_{0} such that

|f(x_{n}) − f(x_{0})| ≥ |x_{n} − x_{0}|^{β}.

Letting r_{n} = |x_{n} − x_{0}| we have

\frac{μ_{f}(B(x_{0}, r_{n}))}{r_{n}^{β}} ≥ 1.

This implies lim sup_{r→0+} μ_{f}(B(x_{0}, r))/r^{β} ≥ 1. From Proposition 1 it follows that H^{β}(E_{≤α}^{f}) ≤ 2^{β}μ_{f}(E_{≤α}^{f}) < ∞ and hence dim_{H}E_{≤α}^{f} ≤ β. Since this holds for any β > α we obtain dim_{H}E_{≤α}^{f} ≤ α.

We turn to the much more difficult task to verify that dim_{H}E_{α}^{f} ≥ α, where 0 < α < 1 is fixed. Since E_{α}^{f} ⊂ E_{≤α}^{f} this will imply dim_{H}E_{α}^{f} = dim_{H}E_{≤α}^{f} = α. We will use the main construction again with h_{n,m}^{∗} = 1/4^{n} and assume that f ∈ F_{1} = ∩_{m=1}^{∞}H_{m}. This means that for each m there exists n(m) ≥ m such that f ∈ ∩_{m=1}^{∞}G_{n(m),m}. We will choose a suitable subsequence m_{ℓ} ≥ ℓ. For ease of notation we let G_{ℓ} = G_{n(m_{ℓ}),m_{ℓ}}, N_{ℓ} = N_{n(m_{ℓ})}, ν_{ℓ} = ν_{n(m_{ℓ}),m_{ℓ}}, h_{ℓ} = h_{n(m_{ℓ}),m_{ℓ}}, δ_{ℓ} = δ_{n(m_{ℓ}),m_{ℓ}}, ϱ_{ℓ} = ϱ_{n(m_{ℓ}),m_{ℓ}}, and P_{ℓ}(k) = \frac{k}{2^{N_{ℓ}}} + δ_{ℓ} for k = 0, ..., 2^{N_{ℓ}} − 1. Observe that each of the above numbers is a dyadic rational.

We choose h′_{ℓ} such that it is a dyadic rational and

2^{ℓ−1}h_{ℓ}^{1/α} ≤ h′_{ℓ} ≤ 2^{ℓ}h_{ℓ}^{1/α}.  (13)

Set h′′_{ℓ} = h′_{ℓ}/2^{ℓ}. Thus

\frac{h_{ℓ}^{1/α}}{2} ≤ h′′_{ℓ} ≤ h_{ℓ}^{1/α}.  (14)

We assume that the subsequence m_{ℓ} is sufficiently rapidly increasing, namely, 2^{m_{ℓ+1}} exceeds the denominator of the above numbers at “level ℓ”. This assumption will be convenient, for example, this way the intervals used in the definitions of g_{ℓ+1}^{∗} = g_{n(m_{ℓ+1})}^{∗} and g_{ℓ+1}^{∗∗} = g_{n(m_{ℓ+1}),m_{ℓ+1}}^{∗}, will belong to a subdivision of the intervals occuring at “level ℓ”.

Next we define closed sets F′_{ℓ} and F_{ℓ} and using their intersection we define some Cantor type sets. Using Proposition 1 we will estimate the Hausdorff dimension of these sets, furthermore we will also obtain information about the Hölder properties of f on these sets. Put

F′_{ℓ} = ⋃_{k=0}^{2^{ν_{ℓ}}}[P_{ℓ}(k) + h′′_{ℓ}, P_{ℓ}(k) + h′_{ℓ}] and F_{ℓ} = ⋂_{ℓ′≤ℓ} F′_{ℓ}.

We also let F_{∞} = ∩_{ℓ=1}^{∞}F_{ℓ}.

It is clear that if we choose the sequence m_{ℓ} to be sufficiently rapidly increasing then each F_{ℓ} consists of κ(ℓ) > 2^{ℓ} many intervals of the form J_{ℓ,k} = [P_{ℓ}(k)+h′′_{ℓ}, P_{ℓ}(k)+h′_{ℓ}]. By repeated subdivision ([7] pp. 9-10) we define a Borel probability measure, μ_{α} with support on F_{∞} by setting μ_{α}(J_{ℓ,k}) = 1/κ(ℓ), that is, we assign equal weight to each interval J_{ℓ,k}.

Assume ℓ ≥ 2. Choose a component of F_{ℓ−1}. This component is a closed interval of length h′′′_{ℓ−1} =^{def}h′_{ℓ−1} −h′′_{ℓ−1} and F′_{ℓ} has one subinterval in each interval [k/2^{ν_{ℓ}}, (k + 1)/2^{ν_{ℓ}}] (we assumed that m(ℓ) is increasing sufficiently fast; hence from h′′′_{ℓ−1} being at “level ℓ−1” it follows that it is an integer multiple of 1/2^{ν_{ℓ}}). Then κ(ℓ) ≥ h′′′_{ℓ−1}2^{ν_{ℓ}} and hence

μ_{α}(J_{ℓ,k}) ≤ \frac{1}{h′′′_{ℓ−1}2^{ν_{ℓ}}} .  (15)

Our next purpose is to show that for all x ∈ F_{∞} we have

lim sup_{r→0+} \frac{μ_{α}(B(x, r))}{r^{α−ϵ}} < 1  (16)

for all ϵ ∈ (0, α). We consider r ∈ (1/2^{ν_{ℓ+1}}, 1/2^{ν_{ℓ}}] for each ℓ = 2, 3, ... . We split each such case into two subcases

I. h′_{ℓ} < r ≤ 1/2^{ν_{ℓ}};

II. 1/2^{ν_{ℓ+1}} < r ≤ h′_{ℓ}.

Assuming ϵ ∈ (0, α) is fixed first we consider subcase I. In this case B(x, r) can intersect at most three different J_{ℓ,k} and using (13) and (6)

\frac{μ_{α}(B(x, r))}{r^{α−ϵ}} ≤ \frac{\frac{3}{h′′′_{ℓ−1}2^{ν_{ℓ}}}}{(h′_{ℓ})^{α−ϵ}} ≤ \frac{\frac{3}{h′′′_{ℓ−1}2^{ν_{ℓ}}}}{2^{(ℓ−1)(α−ϵ)}h_{ℓ}^{(α−ϵ)/α}} =

= \frac{3 · 2^{(ϵ−α)(ℓ−1)}}{h′′′_{ℓ−1}(h_{ℓ}^{∗})^{(α−ϵ)/α}} 2^{ν_{ℓ}( \frac{α−ϵ}{α} −1)} ≤ \frac{3 · 4^{m_{ℓ}}}{h′′′_{ℓ−1}} 2^{ν_{ℓ}( \frac{α−ϵ}{α} −1)} < \frac{3 · 4^{m_{ℓ}}}{h′′′_{ℓ−1}} 2^{m_{ℓ}^{2}( \frac{α−ϵ}{α} −1)} < \frac{1}{4}

where the last inequality holds when m_{ℓ} is chosen to be sufficiently large.

Subcase II. In this case B(x, r) can intersect less than (2r/2^{−ν_{ℓ+1}}) + 2 ≤ 4r2^{ν_{ℓ+1}} many components of F_{ℓ+1}, each of μ_{α} measure μ_{α}(J_{ℓ,k})2^{−ν_{ℓ+1}}/h′′′_{ℓ}. By (15)

μ_{α}(B(x, r)) ≤ \frac{4r}{h′′′_{ℓ}} μ_{α}(J_{ℓ,k}) ≤ 4\frac{r}{h′_{ℓ}} \frac{1}{h′′′_{ℓ−1}2^{ν_{ℓ}}} .

Thus

\frac{μ_{α}(B(x, r))}{r^{α−ϵ}} ≤ \frac{4\frac{r}{h′_{ℓ}} \frac{1}{h′′′_{ℓ−1}2^{ν_{ℓ}}}}{(\frac{r}{h′_{ℓ}} h′_{ℓ})^{α−ϵ}} =

4 (\frac{r}{h′_{ℓ}} )^{1−(α−ϵ)} \frac{\frac{1}{h′′′_{ℓ−1}2^{ν_{ℓ}}}}{(h′_{ℓ})^{α−ϵ}} ≤ 4\frac{\frac{1}{h′′′_{ℓ−1}2^{ν_{ℓ}}}}{(h′_{ℓ})^{α−ϵ}} < 1

where in the last line we used an estimate similar to the one used in subcase I. Hence for all x ∈ F_{∞} we have (16).

Next we investigate the Hölder properties of f at points of F_{∞}.

Since f ∈ B(g_{ℓ}, ϱ_{ℓ}) we have

|f(y) − f(x)| ≥ |g_{ℓ}(y) − g_{ℓ}(x)| − 2ϱ_{ℓ} ≥ |g_{ℓ}(y) − g_{ℓ}(x)| − h_{ℓ}/4.  (17)

Recall that g_{ℓ} = g_{ℓ}^{∗} + g_{ℓ}^{∗∗} and g_{ℓ}^{∗∗} increases by h_{ℓ} on each interval [k/2^{ν_{ℓ}}, P_{ℓ}(k)].

Assume x ∈ F_{∞} ⊂ F_{ℓ} and α + \frac{1}{ℓ} < β < 1. Then using δ_{ℓ} < h′_{ℓ} there exists y ∈ [x − 2h′_{ℓ}, x] such that |g_{ℓ}^{∗∗}(y) − g_{ℓ}^{∗∗}(x)| = h_{ℓ} and hence using (17) and (13) we obtain

|f(y) − f(x)| ≥ \frac{3}{4}h_{ℓ} ≥ \frac{3}{4} (\frac{h′_{ℓ}}{2^{ℓ}} )^{α} ≥ \frac{3}{8} \frac{1}{(h′_{ℓ})^{1/ℓ}2^{ℓα}} (2h′_{ℓ})^{α+ \frac{1}{ℓ}} ≥

\frac{3}{16} \frac{1}{h_{ℓ}^{1/ℓα}2^{ℓα}} (2h′_{ℓ})^{β} ≥ \frac{3}{16} \frac{1}{h_{ℓ}^{1/ℓα}2^{ℓα}} |x − y|^{β} > |x − y|^{β}

where the last inequality holds if we choose a sufficiently large ℓ and ν_{ℓ}, namely, choosing ν_{ℓ} > m_{ℓ}^{3} ≥ ℓ^{3} we have

\frac{3}{16} \frac{1}{h_{ℓ}^{1/ℓα}2^{ℓα}} = \frac{3}{16} \frac{1}{(h_{ℓ}^{∗})^{1/ℓα}} 2^{\frac{ν_{ℓ}}{ℓα} −ℓα} > \frac{3}{16}2^{\frac{ℓ^{2}}{α} −ℓα} > 1.

Since the above type estimations are valid for all β ∈ (α, 1) when ℓ is sufficiently large we obtain that F_{∞} ⊂ E_{≤α}^{f}. From (16) and Proposition 1 it follows that dim_{H}F_{∞} ≥ α which implies dim_{H}(E_{≤α}^{f}) ≥ α.

Next we refine our argument to obtain dim_{H}E_{α}^{f} ≥ α. To obtain this result we find a subset F_{∞}^{0} ⊂ F_{∞} for which μ_{α}(F_{∞}^{0}) > 0 and F_{∞}^{0} ⊂ E_{α}^{f}. By Proposition 1 this will imply the result about the Hausdorff dimension. To obtain F_{∞}^{0} we delete some components of F_{ℓ} to obtain the closed sets F_{ℓ}^{0}. We will have a lower estimate on the μ_{α} measures of these sets yielding μ_{α}(F_{∞}^{0}) = μ_{α}(∩_{ℓ=1}^{∞}F_{ℓ}^{0}) > 0.

[figure: g_{ℓ} + ϱ_{ℓ}; f; g_{ℓ}; g_{ℓ} − ϱ_{ℓ}; \frac{k}{2^{ν_{ℓ}}}; P_{ℓ}(k); P_{ℓ}(k) + h′_{ℓ}; \frac{k+1}{2^{ν_{ℓ}}}; h′′_{ℓ}; h′′′_{ℓ}]

Assume x ∈ F′_{ℓ}. Then x ∈ [P_{ℓ}(k) + h′′_{ℓ}, P_{ℓ}(k) + h′_{ℓ}] ⊂ [k/2^{ν_{ℓ}}, (k + 1)/2^{ν_{ℓ}}] for a suitable k. If y ∈ [P_{ℓ}(k), (k + 1)/2^{ν_{ℓ}}] then g_{ℓ}^{∗∗}(x) − g_{ℓ}^{∗∗}(y) = 0. If y ∉ [P_{ℓ}(k), (k + 1)/2^{ν_{ℓ}}] then h′′_{ℓ} ≤ |y − x| and one can easily see that

|g_{ℓ}^{∗∗}(y) − g_{ℓ}^{∗∗}(x)| ≤ ([\frac{|y − x|}{2^{−ν_{ℓ}}}] + 1) h_{ℓ} ≤ h_{ℓ} + |y − x|2^{ν_{ℓ}}h_{ℓ} =  (18)
h_{ℓ} + |y − x|h_{ℓ}^{∗} < (2h′′_{ℓ})^{α} + |y − x| ≤ 2^{α}|y − x|^{α} + |y − x| ≤ 3|y − x|^{α}, (19)

where we also used (6) and (14).

In the next argument we assume that ℓ_{0} is sufficiently large and ℓ > ℓ_{0}. We define the sets F_{ℓ}^{0} by induction on ℓ.

We assume that ℓ_{0} is so large that for all ℓ ≥ ℓ_{0} we have

2h_{ℓ}^{ℓ} < h_{ℓ}^{\frac{1}{α} +2}.  (20)

To start our induction we choose a component J_{ℓ_{0}} =^{def}[P_{ℓ_{0}}(k′)+h′′_{ℓ_{0}}, P_{ℓ_{0}}(k′)+h′_{ℓ_{0}}]

of F_{ℓ_{0}} in a way that g_{ℓ_{0}}^{∗} is constant on J′_{ℓ_{0}} =^{def}[k′/2^{ν_{ℓ_{0}}} , (k′ + 1)/2^{ν_{ℓ_{0}}} ] and set F_{ℓ_{0}}^{0} = J_{ℓ_{0}}.

[figure: \frac{k′}{2^{ν_{ℓ−1}}}; P_{ℓ−1}(k′); h′_{ℓ−1}; \frac{k′+1}{2^{ν_{ℓ−1}}}; h′′_{ℓ−1}; J_{ℓ−1}; J′′_{ℓ−1}; J′_{ℓ−1}; h_{ℓ−1}; f; g_{ℓ−1}]

Assume ℓ > ℓ_{0} and J_{ℓ−1} =^{def}[P_{ℓ−1}(k′) + h′′_{ℓ−1}, P_{ℓ−1}(k′) + h′_{ℓ−1}] is a component

of F_{ℓ−1}^{0}. Then J_{ℓ−1} ⊂ J′_{ℓ−1} =^{def}[k′/2^{ν_{ℓ−1}}, (k′+1)/2^{ν_{ℓ−1}}]. We also suppose that g_{ℓ−1}^{∗}

is constant on J′_{ℓ−1}. Set J′′_{ℓ−1} = [P_{ℓ−1}(k′), (k′ + 1)/2^{ν_{ℓ−1}}]. Observe that g_{ℓ−1}^{∗∗} is constant on J′′_{ℓ−1}; hence g_{ℓ−1} is also constant there. From f ∈ B(g_{ℓ−1}, ϱ_{ℓ−1}) we infer

f (\frac{k′ + 1}{2^{ν_{ℓ−1}}}) − f(P_{ℓ−1}(k′)) ≤ 2ϱ_{ℓ−1} = \frac{h_{ℓ−1}^{m_{ℓ−1}}}{4} < h_{ℓ−1}^{ℓ−1} < h_{ℓ−1}^{\frac{1}{α} +2}  (21)

where we used (10), m_{ℓ} ≥ ℓ and (20).

From f ∈ B(g_{ℓ}, ϱ_{ℓ}) and 2ϱ_{ℓ} < h_{ℓ}^{\frac{1}{α} +2} it follows that

|g_{ℓ}(y) − g_{ℓ}(x)| − 2h_{ℓ}^{\frac{1}{α} +2} ≤ |f(y) − f(x)| ≤ |g_{ℓ}(y) − g_{ℓ}(x)| + 2h_{ℓ}^{\frac{1}{α} +2}  (22)

for all x, y.

Assume x ∈ J_{ℓ−1} ⊂ F_{ℓ−1} and y ∈ [k′/2^{ν_{ℓ−1}}, P_{ℓ−1}(k′)]. Then (18), (14), |y − x| ≥ h′′_{ℓ−1} and |f − g_{ℓ−1}| ≤ ϱ_{ℓ−1} < h_{ℓ−1}^{\frac{1}{α} +2} yield

|f(y) − f(x)| ≤ 2h_{ℓ−1}^{\frac{1}{α} +2} + |g_{ℓ−1}(y) − g_{ℓ−1}(x)| = 2h_{ℓ−1}^{\frac{1}{α} +2} +
|g_{ℓ−1}^{∗∗}(y) − g_{ℓ−1}^{∗∗}(x)| ≤ 2hℓ−1^{\frac{1}{α}}h_{ℓ−1}^{2} + 3|y − x|^{α} ≤  (23)
4h′′_{ℓ−1}h_{ℓ−1}^{2} + 3|y − x|^{α} ≤ |y − x| + 3|y − x|^{α} ≤ 4|y − x|^{α}.

Using (21) and (22) we obtain

g_{ℓ} (\frac{k′ + 1}{2^{ν_{ℓ−1}}}) − g_{ℓ}(P_{ℓ−1}(k′)) ≤ h_{ℓ−1}^{\frac{1}{α} +2} + 2h_{ℓ}^{\frac{1}{α} +2} < 2h_{ℓ−1}^{\frac{1}{α} +2}.

Since g_{ℓ} = g_{ℓ}^{∗} + g_{ℓ}^{∗∗} and the functions g_{ℓ}^{∗}, g_{ℓ}^{∗∗} are monotone we also have

g_{ℓ}^{∗} (\frac{k′ + 1}{2^{ν_{ℓ−1}}}) − g_{ℓ}^{∗}(P_{ℓ−1}(k′)) < 2h_{ℓ−1}^{\frac{1}{α} +2} ≤ 4h′′_{ℓ−1}h_{ℓ−1}^{2} < \frac{1}{2}h′′′_{ℓ−1}h_{ℓ−1}  (24)

where we assumed that ℓ_{0} ≤ ℓ is sufficiently large and hence 4h_{ℓ−1} < 1/2. Now Lemma 7 implies that there is an exceptional set Ψ^{∗} such that

λ(Ψ^{∗}) < 2 (g_{ℓ}^{∗} (\frac{k′ + 1}{2^{ν_{ℓ−1}}}) − g_{ℓ}^{∗}(P_{ℓ−1}(k′))) < h′′′_{ℓ−1}h_{ℓ−1}

and from x ∈ J′′_{ℓ−1} \ Ψ^{∗} it follows that |g_{ℓ}^{∗}(y) − g_{ℓ}^{∗}(x)| ≤ |y − x| holds for all y ∈ J′′_{ℓ−1}.

We can assume that ν_{ℓ} is chosen so large that 2/2^{ν_{ℓ}} is less than all the jumps g_{ℓ}^{∗}((k′ + 1)/2^{N_{n}}) − g_{ℓ}^{∗}(k′/2^{N_{n}}) > 0, (k′ ∈ {0, ..., 2^{N_{n}} − 1}). Hence g_{ℓ}^{∗} is either constant on an interval [k/2^{ν_{ℓ}}, (k + 1)/2^{ν_{ℓ}}] or increases by at least 2/2^{ν_{ℓ}}.

This implies that (b − a) > 1/2^{ν_{ℓ}} for each component (a, b) of Ψ^{∗}. Denote by Ψ_{0}^{∗} the union of those intervals of the form [k/2^{ν_{ℓ}}, (k + 1)/2^{ν_{ℓ}}] which intersect Ψ^{∗}. By adding to each component (a, b) of Ψ^{∗} those intervals [k/2^{ν_{ℓ}}, (k +1)/2^{ν_{ℓ}}] which contain a and b we obtain Ψ_{0}^{∗}. Hence λ(Ψ_{0}^{∗}) ≤ 3λ(Ψ^{∗}) < 3h′′′_{ℓ−1}h_{ℓ−1}. Denote by F_{ℓ}^{∗} the union of those components of F_{ℓ} ∩J′′_{ℓ−1} which do not intersect Ψ_{0}^{∗}. We define F_{ℓ}^{0} so that F_{ℓ}^{0} ∩ J′′_{ℓ−1} equals F_{ℓ}^{∗}. Observe that F_{ℓ} ∩ J_{ℓ−1} = F_{ℓ}∩J′_{ℓ−1} consists of h′′′_{ℓ−1}2^{ν_{ℓ}} many components and we need to delete λ(Ψ_{0}^{∗})2^{ν_{ℓ}} < 3h_{ℓ−1}h′′′_{ℓ−1}2^{ν_{ℓ}} many of them to obtain F_{ℓ}^{0} ∩ J′_{ℓ−1} = F_{ℓ}^{0} ∩ J_{ℓ−1}.

Since the μ_{α} measure of each component of F_{ℓ} equals μ_{α}(F_{ℓ}∩J_{ℓ−1})2^{−ν_{ℓ}}/h′′′_{ℓ−1} we obtain

μ_{α}(F_{ℓ}^{0} ∩ J_{ℓ−1}) ≥ (1 − 3h_{ℓ−1})μ_{α}(F_{ℓ} ∩ J_{ℓ−1}) = (1 − 3h_{ℓ−1})μ_{α}(F_{ℓ−1}^{0} ∩ J_{ℓ−1}).

Adding the above inequalities for all J_{ℓ−1} type components of F_{ℓ−1}^{0} we have

μ_{α}(F_{ℓ}^{0}) ≥ (1 − 3h_{ℓ−1})μ_{α}(F_{ℓ−1}^{0}).  (25)

Using ν_{ℓ} ≥ n_{ℓ} ≥ m_{ℓ} ≥ ℓ and (6) we infer

μ_{α}(F_{∞}^{0}) ≥ ∏_{ℓ=ℓ_{0}+1}^{∞}(1 − 3h_{ℓ−1})μ_{α}(F_{ℓ_{0}}^{0}) > 0.  (26)

For each x ∈ F_{ℓ}^{0} and y ∈ J′′_{ℓ−1} we have |g_{ℓ}^{∗}(y) − g_{ℓ}^{∗}(x)| ≤ |y − x|. We also know that x ∈ F_{ℓ}^{0} ⊂ F′_{ℓ} and (18) implies

|g_{ℓ}^{∗∗}(y) − g_{ℓ}^{∗∗}(x)| ≤ 3|y − x|^{α}.

Hence |g_{ℓ}(y) − g_{ℓ}(x)| ≤ 4|y − x|^{α} holds for all x ∈ F_{ℓ}^{0} and y ∈ J′′_{ℓ−1}. By (22)

we obtain |f(y) − f(x)| ≤ 4|y − x|^{α} + 2h_{ℓ}^{\frac{1}{α} +2}. We also note that if J_{ℓ} =^{def}[P_{ℓ}(k) + h′′_{ℓ}, P_{ℓ}(k) + h′_{ℓ}] is a component of F_{ℓ}^{0} then J_{ℓ} ⊂ J′_{ℓ} =^{def}[k/2^{ν_{ℓ}}, (k + 1)/2^{ν_{ℓ}}] and g_{ℓ}^{∗} is constant on J′_{ℓ}, since otherwise |g_{ℓ}^{∗}(y) − g_{ℓ}^{∗}(x)| ≤ |y − x| cannot hold for all y ∈ J′_{ℓ} (recall that we assumed that the jumps of g_{ℓ}^{∗} are at least 2 · 2^{ν_{ℓ}}).

Assume x ∈ F_{ℓ}^{0} and y ∈ J′′_{ℓ−1} \ J′_{ℓ}. Then |y − x| ≥ h′′_{ℓ} ≥ h_{ℓ}^{1/α}/2; hence 2^{α}|y − x|^{α} ≥ h_{ℓ}. Therefore

|f(y) − f(x)| ≤ 4|y − x|^{α} + 2h_{ℓ}^{\frac{1}{α} +2} ≤ 4|y − x|^{α} + 4h_{ℓ}^{2}|y − x| ≤ 5|y − x|^{α}.

If x ∈ F_{ℓ}^{0} ⊂ F_{ℓ−1} and y ∈ [k′/2^{ν_{ℓ−1}}, P_{ℓ−1}(k′)) = J′_{ℓ−1} \ J′′_{ℓ−1} we have already seen in (23) that |f(y)−f(x)| ≤ 4|y −x|^{α}. Hence |f(y)−f(x)| ≤ 5|y −x|^{α} holds for all x ∈ F_{ℓ}^{0} and y ∈ J′_{ℓ−1} \ J′_{ℓ}.

Finally, if x ∈ ∩_{ℓ=ℓ_{0}}^{∞}F_{ℓ}^{0} = F_{∞}^{0} and x ≠ y ∈ J′_{ℓ_{0}} then there exists ℓ ≥ ℓ_{0} + 1 such that y ∈ J′_{ℓ−1} \ J′_{ℓ} and hence |f(y) − f(x)| ≤ 5|y − x|^{α} holds. Therefore F_{∞}^{0} ⊂ E_{α}^{f} and by Proposition 1, (26) and (16) we infer dim_{H}(F_{∞}^{0}) ≥ α.

This shows that for functions in F_{(ii)} =^{def}F_{1} assertion (ii) holds.

5.3 Assertion (iii)

Proof. Assume F_{(iii)} is the dense open set F^{∗} of the main construction (used with the value of h_{n,m}^{∗} given in (6)) and f ∈ F_{(iii)}. Given j_{0} ∈ ℕ set m = j_{0} and choose n ≥ m such that f ∈ G_{n,m}. Then N_{n} ≥ K_{n} > n ≥ m = j_{0}; using (5) and the definition of g_{n}^{∗} we obtain

μ_{f_{n},k′} =^{def}μ_{f_{n}} ([\frac{k′}{2^{N_{n}}} , \frac{k′ + 1}{2^{N_{n}}}]) = g_{n}^{∗} (\frac{k′ + 1}{2^{N_{n}}}) − g_{n}^{∗} (\frac{k′}{2^{N_{n}}} ) .

Since f′_{n} > 0 on [0, 1] we have 0 < μ_{f_{n},min} =^{def} min{μ_{f_{n},k′} : k′ = 0, ..., 2^{N_{n}} − 1} and we can assume that ν_{n,m} was chosen so large that 2^{−ν_{n,m}} < μ_{f_{n},min}/8; hence h_{n,m}^{∗} < 2^{−ν_{n,m}} < μ_{f_{n},min}/8 and ϱ_{n,m} < h_{n,m}/8 < h_{n,m}^{∗}/8 < μ_{f_{n},min}/8.

Recalling (8) we obtain

μ_{f} ([\frac{k′}{2^{N_{n}}} , \frac{k′ + 1}{2^{N_{n}}}]) = |f (\frac{k′ + 1}{2^{N_{n}}}) − f (\frac{k′}{2^{N_{n}}} )| ≥

|g_{n,m} (\frac{k′ + 1}{2^{N_{n}}}) − g_{n,m} (\frac{k′}{2^{N_{n}}} )| − 2ϱ_{n,m} ≥  (27)

|g_{n}^{∗} (\frac{k′ + 1}{2^{N_{n}}}) − g_{n}^{∗} (\frac{k′}{2^{N_{n}}} )| − 2ϱ_{n,m} − 2h_{n,m}^{∗} > \frac{μ_{f_{n},k′}}{2} .

Thus, using (3) we have

τ_{f,N_{n}}(q) ≥ \frac{1}{N_{n} log 2} log ∑_{k′=0}^{2^{Nn} −1}(μ_{f_{n},k′}/2)^{q} = τ_{f_{n},N_{n}}(q) − \frac{q}{N_{n}} ≥ 1 − q − \frac{2}{j_{0}} .

This implies \overline{τ}_{f}(q) = lim sup_{j→∞} τ_{f,j}(q) ≥ 1−q. By Lemma 3 we obtain \overline{τ}_{f}(q) = 1 − q.

Since f is strictly monotone increasing one can easily show that τ_{f}(0) = 1.

Assume 0 < q ≤ 1 is given and fixed. We verify that \underline{τ}_{f}(q) = 0.

Given f ∈ F_{(iii)} and j_{0} > 1/q we choose n, m as above. Observe that out of the 2^{ν_{n,m}} intervals of the form [k/2^{ν_{n,m}}, (k + 1)/2^{ν_{n,m}}], (k = 0, ..., 2^{ν_{n,m}} − 1) there are only 2^{N_{n}} many intervals on which g_{n}^{∗} is nonconstant. We call these intervals irregular and the remaining intervals are called regular. We denote by Γ the set of those k ∈ {0, ..., 2^{ν_{n,m}} − 1} for which [k/2^{ν_{n,m}}, (k + 1)/2^{ν_{n,m}}] is regular. For irregular k’s we use the trivial estimate

(μ_{f} ([\frac{k}{2^{ν_{n,m}}} , \frac{k + 1}{2^{ν_{n,m}}}]))^{q} ≤ μ_{f}([0, 1]) = 1,

where we used that replacing f by cf we can assume that μ_{f} is a probability measure. For k ∈ Γ using g_{n}^{∗}((k + 1)/2^{ν_{n,m}}) − g_{n}^{∗}(k/2^{νn,m}) = 0 we obtain

μ_{f} ([\frac{k}{2^{ν_{n,m}}} , \frac{k + 1}{2^{ν_{n,m}}}]) = f (\frac{k + 1}{2^{ν_{n,m}}}) − f (\frac{k}{2^{ν_{n,m}}} ) ≤

g_{n,m} (\frac{k + 1}{2^{ν_{n,m}}}) − g_{n,m} (\frac{k}{2^{ν_{n,m}}} ) + 2ϱ_{n,m} =

g_{n,m}^{∗∗} (\frac{k + 1}{2^{ν_{n,m}}}) − g_{n,m}^{∗∗} (\frac{k}{2^{ν_{n,m}}} ) + 2ϱ_{n,m} =

h_{n,m} + 2ϱ_{n,m} < 2h_{n,m}.

If j_{0} and hence n ≥ m = j_{0} > 1/q is sufficiently large then we can assume that ν_{n,m} > N_{n}^{2} satisfies 2^{ν_{n,m}} · 2^{q}h_{n,m}^{q} = 2^{ν_{n,m}} · 2^{q}(h_{n,m}^{∗}/2^{ν_{n,m}})^{q} = 2^{ν_{n,m}} · 2^{q}(1/4^{n}2^{(m+1)ν_{n,m}})^{q} ≤ 1 < 2^{N_{n}}. Now using #({0, ..., 2^{ν_{n,m}} − 1} \ Γ) = 2^{N_{n}} #Γ < 2^{ν_{n,m}}, and the above estimates we obtain

τ_{f,ν_{n,m}}(q) ≤ \frac{1}{ν_{n,m} log 2} log(2^{N_{n}} + 2^{ν_{n,m}} · 2^{q}h_{n,m}^{q}) ≤  (28)

\frac{1}{ν_{n,m} log 2} log(2 · 2^{N_{n}}) ≤ \frac{1}{ν_{n,m} log 2} log(2^{\sqrt{ν_{n,m}}}) ≤ \frac{1}{\sqrt{ν_{n,m}}} ≤ \frac{1}{j_{0}} ,

Hence \underline{τ}_{f}(q) ≤ 0 and Lemma 3 completes the proof showing that \underline{τ}_{f}(q) = 0.

5.4 Assertion (iv)

The proof of this assertion is similar at many places to that of Assertion (iii).

Proof. Assume q > 1 is fixed. Let F_{(iv)} = F^{∗} of the main construction and for given j_{0} ∈ ℕ choose m, n and define μ_{f_{n},k′}, μ_{f_{n},min} as in the proof of Assertion (iii). We can again assume that h_{n,m}^{∗} < μ_{f_{n},min}/8 and ϱ_{n,m} < μ_{f_{n} min}/8.

Then similarly to (27) we obtain

μ_{f} ([\frac{k′}{2^{N_{n}}} , \frac{k′ + 1}{2^{N_{n}}}]) ≤ |g_{n}^{∗} (\frac{k′ + 1}{2^{N_{n}}}) − g_{n}^{∗} (\frac{k′}{2^{N_{n}}} )| +

2ϱ_{n,m} + 2h_{n,m}^{∗} < 2μ_{f_{n},k′}.

We can use (4) when n ≥ j_{0} ≥ q and we have

τ_{f,N_{n}}(q) ≤ \frac{1}{N_{n} log 2} log ∑_{k′=0}^{2^{Nn} −1}(2μ_{f_{n},k′})^{q} ≤ 1 − q + \frac{q + 1}{j_{0}}.

This implies \underline{τ}_{f}(q) ≤ 1 − q. By Lemma 5 we obtain \underline{τ}_{f}(q) = 1 − q.

Next we show \overline{τ}_{f}(q) ≥ 0. Observe that from ν_{n,m} > N_{n}^{2} and (7) it follows that if we choose j_{n} such

that 2^{−j_{n}} = δ_{n,m} then \sqrt{j_{n}} > N_{n}. We can also assume that j_{0} ≤ m ≤ n is so large that

g_{n}^{∗}(1) − g_{n}^{∗}(0) ≥ f(1) − f(0) − 2ϱ_{n,m} − h_{n,m}^{∗} ≥ \frac{f(1) − f(0)}{2}.

Observe that if k′′ ∈ {0, ..., 2^{j_{n}} − 1} is chosen so that k′′/2^{j_{n}} = k′/2^{N_{n}} for a suitable k′ ∈ {0, ..., 2^{N_{n}} − 1} then

g_{n}^{∗} (\frac{k′′ + 1}{2^{j_{n}}}) − g_{n}^{∗} (\frac{k′′}{2^{j_{n}}} ) = μ_{f_{n},k′},

we denote the set of these k′′ by Γ′′.

For k′′ ∈ Γ′′ similarly to (27) we obtain

μ_{f} ([\frac{k′′}{2^{j_{n}}} , \frac{k′′ + 1}{2^{j_{n}}}]) ≥ g_{n}^{∗} (\frac{k′ + 1}{2^{N_{n}}}) − g_{n}^{∗} (\frac{k′}{2^{N_{n}}} ) − 2ϱ_{n,m} − 2h_{n,m}^{∗} ≥ \frac{μ_{f_{n},k′}}{2} .

By using an inequality between means of non negative numbers we have

(\frac{∑_{k′=0}^{2^{Nn} −1} μ_{f_{n},k′}^{q}}{2^{N_{n}}} )^{\frac{1}{q}} ≥ \frac{∑_{k′=0}^{2^{Nn} −1} μ_{f_{n},k′}}{2^{N_{n}}} = \frac{g_{n}^{∗}(1) − g_{n}^{∗}(0)}{2^{N_{n}}}.

Hence

∑_{k′=0}^{2^{Nn} −1} μ_{f_{n},k′}^{q} ≥ (g_{n}^{∗}(1) − g_{n}^{∗}(0))^{q}2^{N_{n}(1−q)} ≥ (\frac{f(1) − f(0)}{2})^{q} 2^{N_{n}(1−q)}.

Thus

τ_{f,j_{n}}(q) = \frac{1}{j_{n} log 2} log ∑_{k′′=0}^{2^{jn} −1} (μ_{f} ([\frac{k′′}{2^{j_{n}}} , \frac{k′′ + 1}{2^{j_{n}}}]))^{q} ≥

\frac{1}{j_{n} log 2} log ∑_{k′′∈Γ′′} (μ_{f} ([\frac{k′′}{2^{j_{n}}} , \frac{k′′ + 1}{2^{j_{n}}}]))^{q} ≥

\frac{1}{j_{n} log 2} log ((\frac{f(1) − f(0)}{2})^{q} 2^{N_{n}(1−q)} \frac{1}{2^{q}} ) ≥

\frac{1}{j_{n} log 2} log ((\frac{f(1) − f(0)}{4})^{q} 2^{\sqrt{j_{n}}(1−q)}) ≥ \frac{1 − q}{\sqrt{j_{n}}} + \frac{log ((\frac{f(1)−f(0)}{4})^{q})}{j_{n} log 2}.

This implies \overline{τ}_{f}(q) ≥ 0. By Lemma 5 we obtain \overline{τ}_{f}(q) = 0.

5.5 Assertion (v)

Proof. First we show \underline{τ}_{f}(q) ≥ 1 − q. Assume f ∈ F_{(v)} =^{def}F_{(i)} = F_{1} ∩ F_{0} then f′ = 0 almost everywhere and hence there exists δ > 0 and a measurable set H ⊂ [0, 1] such that λ(H) > 1/2 and for any x ∈ H, y ∈ (x − δ, x + δ) we have

|f(y) − f(x)| < |y − x|.  (29)

Assume j is chosen so that 1/2^{j} < δ. Denote by Γ_{H} the set of those k ∈ {0, ..., 2^{j} − 1} for which H ∩ [k/2^{j}, (k + 1)/2^{j}] ≠ ∅. From λ(H) > 1/2 it follows that #Γ_{H} > 2^{j−1}. From (29) it follows that μ_{f}([k/2^{j}, (k + 1)/2^{j}]) < 2^{−j} holds when k ∈ Γ_{H}. Hence using q < 0 we have

τ_{f,j}(q) ≥ \frac{1}{j log 2} log ∑_{k∈Γ_{H}} (μ_{f} ([\frac{k}{2^{j}} , \frac{k + 1}{2^{j}}]))^{q} >

\frac{1}{j log 2} log(2^{j−1}2^{−jq}) = \frac{j(1 − q) − 1}{j}.

This implies \underline{τ}_{f}(q) ≥ 1 − q.

Next we show \underline{τ}_{f}(q) ≤ 1 − q. For any n ≥ m we have

g_{n,m} (\frac{k + 1}{2^{ν_{n,m}}}) − g_{n,m} (\frac{k}{2^{ν_{n,m}}} ) ≥ h_{n,m}.

Since ϱ_{n,m} < h_{n,m}/8 we have

μ_{f,n,m,k} = f (\frac{k + 1}{2^{ν_{n,m}}}) − f (\frac{k}{2^{ν_{n,m}}} ) >

g_{n,m} (\frac{k + 1}{2^{ν_{n,m}}}) − g_{n,m} (\frac{k}{2^{ν_{n,m}}} ) − 2ϱ_{n,m} >

\frac{3}{4}h_{n,m} = \frac{3}{4^{n+1}2^{ν_{n,m}}} > \frac{1}{4^{2n}2^{ν_{n,m}}} .

Keeping in mind that q < 0 we infer

τ_{f,ν_{n,m}}(q) = \frac{1}{ν_{n,m} log 2} log ∑_{k=0}^{2^{νn,m} −1} μ_{f,n,m,k}^{q} <

\frac{1}{ν_{n,m} log 2} log(2^{ν_{n,m}}4^{−2nq}2^{−qν_{n,m}}) = (1 − q) + \frac{−4nq}{ν_{n,m}} .

From ν_{n,m} > N_{n}^{2} ≥ K_{n}^{2} > n^{2} ≥ m^{2} easily follows that \frac{−4nq}{ν_{n,m}} → 0 as m, and hence n, tends to ∞.

Therefore \underline{τ}_{f}(q) ≤ 1 − q.

Now we verify that \overline{τ}_{f}(q) = ∞.

Observe that by (7), P (k, n, m) < (2k + 1)/2^{ν_{n,m}+1} and hence g_{n,m} is constant on [(2k + 1)/2^{ν_{n,m}+1}, (k + 1)/2^{ν_{n,m}}]. Therefore,

f (\frac{k + 1}{2^{ν_{n,m}}}) − f (\frac{2k + 1}{2^{ν_{n,m}+1}} ) < 2ϱ_{n,m} = \frac{(h_{n,m})^{m}}{4} .

Thus in line (31) below recalling (11) we obtain

τ_{f,ν_{n,m}+1}(q) =

\frac{1}{(ν_{n,m} + 1) log 2} log ∑_{k′=0}^{2^{νn,m+1}−1} (f (\frac{k′ + 1}{2^{ν_{n,m}+1}} ) − f (\frac{k′}{2^{ν_{n,m}+1}} ))^{q} >  (30)

\frac{1}{(ν_{n,m} + 1) log 2} log ∑_{k=0}^{2^{νn,m} −1} (f (\frac{k + 1}{2^{ν_{n,m}}}) − f (\frac{2k + 1}{2^{ν_{n,m}+1}} ))^{q} >

\frac{1}{(ν_{n,m} + 1) log 2} log (2^{ν_{n,m}} \frac{(h_{n,m})^{mq}}{4^{q}} ) =

\frac{1}{(ν_{n,m} + 1) log 2} log(2^{ν_{n,m}}(4^{n}2^{ν_{n,m}})^{−mq}4^{−q}) =  (31)

\frac{ν_{n,m}(1 − mq)}{ν_{n,m} + 1} + \frac{−2mnq log 2 + log(4^{−q})}{(ν_{n,m} + 1) log 2}.

Recalling that q < 0 and ν_{n,m} > N_{n}^{2} ≥ K_{n}^{2} > n^{2} ≥ m^{2} we obtain \overline{τ}_{f}(q) = ∞.

5.6 Assertion (vi)

Assume f ∈ F_{(vi)} =^{def}F_{(ii)}. We will use the notation and choice of h_{n,m}^{∗} introduced in the proof of assertion (ii) and hence choose a suitable subsequence m_{ℓ} ≥ ℓ. We will denote W_{ℓ} = W_{n(m_{ℓ}),m_{ℓ}}. Clearly,

μ_{g_{ℓ}}(W_{ℓ}) = ∑_{k=0}^{2^{ν_{ℓ}} −1} g_{ℓ}(P_{ℓ}(k)) − g_{ℓ} (\frac{k}{2^{ν_{ℓ}}} ) = ∑_{k=0}^{2^{ν_{ℓ}} −1} g_{ℓ} (\frac{k + 1}{2^{ν_{ℓ}}}) − g_{ℓ} (\frac{k}{2^{ν_{ℓ}}} ) =

g_{ℓ}(1) − g_{ℓ}(0) ≥ f(1) − f(0) − 2ϱ_{ℓ}.

Assuming m_{ℓ} ≥ 2, we have by (6) and (10) ϱ_{ℓ} < h_{ℓ}^{2}/2 < h_{ℓ}/2^{ν_{ℓ}+1}. Therefore

μ_{f}(W_{ℓ}) = ∑_{k=0}^{2^{ν_{ℓ}} −1} f(P_{ℓ}(k)) − f (\frac{k}{2^{ν_{ℓ}}} ) ≥

∑_{k=0}^{2^{ν_{ℓ}} −1} (g_{ℓ}(P_{ℓ}(k)) − g_{ℓ} (\frac{k}{2^{ν_{ℓ}}} ) − 2ϱ_{ℓ}) ≥ f(1) − f(0) − 2^{ν_{ℓ}+1}ϱ_{ℓ} >

f(1) − f(0) − h_{ℓ} > f(1) − f(0) − \frac{1}{2^{ℓ}} .

Thus μ_{f}([0, 1] \ W_{ℓ}) < 1/2^{ℓ}. By the Borel-Cantelli lemma μ_{f}-almost every x ∈ [0, 1] belongs to finitely many of the sets [0, 1] \ W_{ℓ}, that is, for μ_{f}- almost every x there exists ℓ_{x} such that for all ℓ ≥ ℓ_{x} we have x ∈ W_{ℓ}.

If x ∈ W_{ℓ}, that is, x ∈ (k/2^{ν_{ℓ}}, P_{ℓ}(k)) for a suitable k then

f(P_{ℓ}(k)) − f (\frac{k}{2^{ν_{ℓ}}} ) ≥ g_{ℓ}(P_{ℓ}(k)) − g_{ℓ} (\frac{k}{2^{ν_{ℓ}}} ) − 2ϱ_{ℓ} ≥

g_{ℓ}^{∗∗}(P_{ℓ}(k)) − g_{ℓ}^{∗∗} (\frac{k}{2^{ν_{ℓ}}} ) − 2ϱ_{ℓ} = h_{ℓ} − 2ϱ_{ℓ} >

\frac{h_{ℓ}}{2} > \frac{δℓ^{\frac{1}{2m_{ℓ}}}}{2} > \frac{δℓ^{\frac{1}{2ℓ}}}{2} .

Taking y = k/2^{ν_{ℓ}} or y = P_{ℓ}(k) we have

|f(y) − f(x)| ≥ \frac{f(P_{ℓ}(k)) − f(k/2^{ν_{ℓ}})}{2} > \frac{δℓ^{\frac{1}{2ℓ}}}{4} .

On the other hand |y − x| < δ_{ℓ} and hence |f(y) − f(x)| > |y − x|^{1/2ℓ}/4. Since we can choose any ℓ ≥ ℓ_{x} we have h_{f}(x) = 0. This shows h_{f} = 0, μ_{f}-almost everywhere in [0, 1], that is, μ_{f}(E_{≤0}^{f}) = f(1) − f(0) = μ_{f}([0, 1]).

References

[1] G. Brown, G. Michon and J. Peyrière, On the multifractal analysis of measures, J. Statist. Phys. 66 no. 3-4 (1992), 775-790.

[2] S. Jaffard, Multifractal formalism for functions Part I: Results valid for all functions, SIAM J. Math. Anal, 28 No. 4 (1997), 944-970.

[3] S. Jaffard, Multifractal formalism for functions Part II: Self-similar functions, SIAM J. Math. Anal, 28 No. 4 (1997), 971-998.

[4] S . Jaffard, The spectrum of singularities of Riemann’s function, Rev. Mat. Iberoamericana, 12 (1996) 441-460.

[5] S. Jaffard, Old friends revisited: the multifractal nature of some classical functions, J. of Fourier Anal. and Appl. 3 No. 1 (1997), 1-22.

[6] K. Falconer, Fractal Geometry, John Wiley & Sons (1990).

[7] K. Falconer, Techniques in Fractal Geometry, John Wiley & Sons (1997).

[8] H. Federer, Geometric Measure Theory, Springer Verlag, New York, 1969.

[9] J. Genyuk, A typical measure typically has no local dimension, Real Anal. Exchange, 23(2), 1997/8, 525-538.

[10] H. Haase, A survey on the dimension of measures, Topology, measures, and fractals (Warnemünde, 1991), 66-75, Math. Res., 66, Akademie-Verlag, Berlin 1992.

[11] T. Zamfirescu, Most monotone functions are singular, Amer. Math. Monthly, 88 (1) (1981), 47–49.

[12] T. Zamfirescu, Typical monotone continuous functions, Arch. Math., 42 (1984), 151–156.