1 Introduction
The Rauzy gasket is a set which lies within the two-dimensional simplex. It is an important subset of parameter space in numerous dynamical or topological problems (see [Reference Gamburd, Magee and Ronan9] and the introductions of [Reference DeLeo and Dynnikov4, Reference Dynnikov, Hubert and Skripchenko5]), having been discovered independently at least three times in different contexts.
It was a conjecture of Arnoux [Reference Arnoux, Starosta, Barral and Seuret1] that the Hausdorff dimension of this set is smaller than two, and this was confirmed by [Reference Avila, Hubert and Skripchenko2]. (It is also a special case of a conjecture of Novikov that this dimension is strictly between one and two.) In this short note, we give a completely elementary proof of this fact.
Our method of proof is fairly flexible, and we illustrate this by applying it to a family of higher-dimensional examples
$\mathcal G_d$
(
$d\geq 3$
) which generalise the Rauzy gasket. That is, the purpose of this note is to give an elementary proof of the following theorem.
Theorem 1.1.
$\dim _H(\mathcal G_d) < d-1$
for any
$d \geq 3$
.
We note that for the
$d=3$
case, this result has been strengthened with explicit (albeit loose) upper bounds [Reference Fougeron8, Reference Pollicott and Sewell11] and complemented with a lower bound [Reference Gutiérrez-Romo and Matheus10].
In §2, we define
$\mathcal G_d$
. In §3, we provide a covering lemma that allows us to reduce to considerations of volumes of certain simplices, and in §4, we give an explicit formula for these volumes. In §5, we use the formula to prove some iterative inequalities, which combine with the renewal theorem to get our main technical result. We then verify the assumptions of this technical result in §6 to complete the proof of Theorem 1.1.
2 Definition of
$\mathcal G_d$
We begin with the general definition of a Rauzy gasket in the standard d-simplex, first defined in [Reference Arnoux, Starosta, Barral and Seuret1]. These were shown to have Hausdorff dimension strictly less than d in [Reference Fougeron8], using estimates of [Reference Baragar3].
Definition 2.1. Fix
$d\geq 3$
as the dimension of the ambient space, and let
$\Delta \subset \mathbb R^d$
be the standard
$d-1$
simplex:
$$ \begin{align*} \Delta = \{x \in[0,1]^d : \|x\| = 1\}, \end{align*} $$
where
$\|x\| = \|x\|_{l^1} = \sum _{j=1}^d v_j$
denotes the usual
$l^1$
norm. For
$j \in \{1,2,\ldots ,d\}$
, define the matrix
$M_j \in \{0,1\}^{d \times d}$
by
$$ \begin{align*} (M_j)_{i,k} = \begin{cases} 1 & \text{if}\ i=j\ \text{or}\ i=k,\\ 0 & \text{otherwise}, \end{cases} \end{align*} $$
that is,
$M_j$
has ones on the main diagonal and the jth row, and zeros elsewhere.
We define
$\mathcal G_{d-1}$
as the attractor of the projectivised maps
$\{T_j\}_{j=1}^d$
,
$$ \begin{align*} T_j:\Delta\to \Delta, \quad T_j (x) = \frac {M_j \cdot x} {\|M_j\cdot x\|}, \end{align*} $$
that is,
$\mathcal G_{d-1}$
is the unique non-empty compact subset of
$\Delta $
such that
$$ \begin{align*} \mathcal G_{d-1} = \bigcup_{j=1}^d T_j(\mathcal G_{d-1}) \end{align*} $$
(see [Reference Falconer6] for equivalent definitions).
Example 2.2. (
$d=3$
)
The original Rauzy gasket, depicted in Figure 1, corresponds to
$d=3$
. The three matrices are
$$ \begin{align*} M_1 = \left( \begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\!, \quad M_2 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right)\!, \quad M_3 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1 \\ \end{array} \right)\!, \end{align*} $$
and the first map is
$$ \begin{align*} T_1(x,y,z) = \left( \frac 1 {2-x}, \frac y {2-x}, \frac z {2-x} \right)\!. \end{align*} $$
Figure 1 The Rauzy gasket,
$\mathcal G_3$
.
We now introduce some notation to be used throughout. We shall also write
$|{\underline i}| = n$
as shorthand for
${\underline i} \in \{1,2,\ldots ,d\}^n$
.
Notation 2.3. Throughout we write, for
${\underline i} = (i_1,\ldots , i_n) \in \{1,2,\ldots , d\}^n$
:
-
•
$M_{\underline i} := M_{i_1}M_{i_2}\cdots M_{i_n}$
; -
•
$T_{\underline i} := T_{i_1}\circ T_{i_2}\circ \cdots \circ T_{i_n}$
; and -
•
$\Delta _{\underline i} := T_{\underline i}(\Delta )$
.
3 Covering lemma
Our first step in the proof of Theorem 1.1 is the following lemma.
Lemma 3.1. Given
$\delta \in (0,1)$
,
$\dim _H(\mathcal G_{d}) \leq d + \delta - 2$
, whenever
$$ \begin{align*} X_n := \sum_{|{\underline i}| = n} \operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^{\delta} \to 0 \end{align*} $$
as
$n \to \infty $
, where
$\operatorname {\mathrm {vol}}_{d-1}(\Delta _{\underline i})$
denotes the
$(d-1)$
-dimensional volume of
$\Delta _{\underline i}$
.
Note that, where appropriate, we write
$f \lesssim g$
to mean that there exists a constant
$C>0$
, depending only on d and
$\delta $
, such that
$f \leq Cg$
.
Proof. From the definition of Hausdorff dimension [Reference Falconer6], to show
$\dim _H(\mathcal G_d) \leq d+\delta - 2$
it suffices to exhibit a family of open covers
$\{\mathcal C_n\}_{n=1}^\infty $
of
$\mathcal G$
such that
$$ \begin{align*} \sum_{S \in \mathcal C_n} \operatorname{\mathrm{diam}}(S)^{d+\delta-2} \to 0 \end{align*} $$
as
$n \to \infty $
(in particular, this implies
$\max _{S\in \mathcal C_n} \operatorname {\mathrm {diam}}(S) \to 0$
). To define these covers, it follows from the definition of
$\mathcal G_d$
, and the fact that
$\mathcal G_d\subset \Delta $
, that
$$ \begin{align*} \mathcal G_d \subset \bigcup_{|{\underline i}|=n} \Delta_{\underline i}. \end{align*} $$
Thereby, providing a cover of each
$\Delta _{\underline i}$
by open balls and taking the union gives an open cover of
$\mathcal G_n$
. For each
$|{\underline i}| = n$
, the construction is as follows.
Figure 2 A collection of balls of diameter
$h_{\underline i}$
covering a copy of
$B_{\underline i}$
. The bases of the balls are denoted by dots. Scaling each ball by a factor of
$\sqrt d$
about its centre gives a cover of
$\Delta _{\underline i}$
.
Since
$T_{\underline i}$
is the projectivization of an injective linear map,
$\Delta _{\underline i}$
is a
$(d-1)$
-simplex. Choose a maximal-volume face
$B_{\underline i}$
of
$\Delta _{\underline i}$
as its base, so that
$h_{\underline i}$
, its height measured from
$B_{\underline i}$
, is the smallest such height. A simple induction shows that
$B_{\underline i}$
is contained in a
${(d-2)}$
-dimensional hypercuboid, whose side lengths are at least
$h_{\underline i}$
and whose volume is at most
$(d-2)\!\operatorname {\mathrm {vol}}_{d-2}(B_{\underline i})$
. It follows that this cuboid and hence
$B_{\underline i}$
can be covered with
$n_{\underline i}$
open balls of radius
$h_{\underline i}$
, where
$$ \begin{align*} n_{\underline i} \lesssim \frac {\operatorname{\mathrm{vol}}_{d-2}(B_{\underline i})} {h_{\underline i}^{d-2}} \lesssim \frac {\operatorname{\mathrm{vol}}_{d-2}(B_{\underline i})^{d-1}} {\operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^{d-2}}. \end{align*} $$
Treating
$\Delta _{\underline i} \subset B_{\underline i} \times [0,h_{\underline i}]$
, taking such a cover of
$B_{\underline i} \times \{h_{\underline i}/2\}$
and enlarging the balls by a factor of
$\sqrt d$
gives a cover of
$\Delta _{\underline i}$
(see Figure 2). Doing this construction simultaneously for all such
$|{\underline i}|=n$
defines the cover
$\mathcal C_n$
. Then, since
$$ \begin{align*} \sum_{S \in \mathcal C_n} \operatorname{\mathrm{diam}}(S)^{d+\delta-1} &\lesssim \sum_{|i|=n} \frac {\operatorname{\mathrm{vol}}_{d-2}(B_{\underline i})^{d-1}} {\operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^{d-2}} \cdot \bigg( \frac {\operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})} {\operatorname{\mathrm{vol}}_{d-2}(B_{\underline i})} \bigg) ^{d+\delta-2}\\ &= \sum_{|i|=n} \operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^{\delta} \operatorname{\mathrm{vol}}_{d-2}(B_{\underline i})^{1-\delta} \end{align*} $$
and
$\operatorname {\mathrm {vol}}_{d-2}(B_{\underline i}) \lesssim 1,$
our assumption gives the required convergence to zero as
$n\to \infty $
.
4 Volume formula
We now show that the volume of
$\Delta _{\underline i}$
can be expressed simply in terms of the entries of
$M_{\underline i}$
.
Lemma 4.1. For any tuple
${\underline i}$
,
$$ \begin{align*} \frac {\operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})} {\operatorname{\mathrm{vol}}_{d-1}(\Delta)} = \nu(M_{\underline i}), \end{align*} $$
where for any
$N \in \mathbb R^{d\times d}$
,
$$ \begin{align*} \nu(N) := \prod_{j=1}^d \bigg( \sum_{k=1}^d N_{k,j} \bigg)^{-1} = \prod_{j=1}^d \|N \cdot e_j\|^{-1} , \end{align*} $$
and where
$e_j = (0,\ldots ,0,1,0,\ldots ,0)$
is the jth standard basis vector in
$\mathbb R^d$
, that is, vertex of
$\Delta $
.
Proof. We define
$\Delta _{\underline i}^\ast := \{\unicode{x3bb} v : \unicode{x3bb} \in [0,1], v\in \Delta _{\underline i}\}$
, that is, the d-simplex with vertices consisting of those of
$\Delta _{\underline i}$
plus the origin, and define
$\Delta ^\ast $
analogously. Integrating appropriately gives
$$ \begin{align} \frac {\operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})} {\operatorname{\mathrm{vol}}_{d-1}(\Delta)} = \frac {\operatorname{\mathrm{vol}}_d(\Delta_{\underline i}^\ast)} {\operatorname{\mathrm{vol}}_d(\Delta^\ast)}. \end{align} $$
The two d-simplices are related by a linear action:
$\Delta _{\underline i}^\ast = M_{\underline i}^\ast \cdot \Delta ^\ast $
, where
$$ \begin{align*} M_{\underline i}^\ast := \big( T_{\underline i}(e_1) \big| T_{\underline i}(e_2) \big| \cdots \big| T_{\underline i}(e_d) \big) = \bigg( \frac {M_{\underline i} \cdot e_1} {\|M_{\underline i}\cdot e_1\|} \;\bigg|\; \frac {M_{\underline i} \cdot e_2} {\|M_{\underline i}\cdot e_2\|} \;\bigg| \cdots \bigg|\; \frac {M_{\underline i} \cdot e_d} {\|M_{\underline i}\cdot e_d\|} \bigg). \end{align*} $$
That is,
$M_{\underline i}^\ast $
is
$M_{\underline i}$
with each jth column multiplied by a factor of
$\|M_{\underline i}\cdot e_j\|^{-1}$
. Therefore, using that
$\det (M_j) = 1$
for each
$j=1,\ldots ,d$
, the right-hand side of equation (1) is none other than
$$ \begin{align*} \det(M_{\underline i}^\ast) = \nu(M_{\underline i}) \det(M_{\underline i}) = \nu(M_{\underline i}), \end{align*} $$
as required.
5 Renewal theorem
We first give a convenient partition of
$\{1,\ldots ,d\}^n$
and decomposition of
$X_n$
. Note that these are incomplete, and this gives rise to the remainder term
$r_n$
in Lemma 5.3.
Definition 5.1. For each
$n> k \geq 1$
, let
$$ \begin{align*} A_{n,k} = \big\{ {\underline i} = (i_1,i_2,\ldots,i_n) \in \{1,2,\ldots, d\}^n \mid i_1 = \cdots = i_k \neq i_{k+1} \big\} \end{align*} $$
and
$$ \begin{align*} X_{n,k} = \sum_{{\underline i} \in A_{n,k}} \operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^{\delta}. \end{align*} $$
The following lemma allows us to focus on the convergence of
$X_{n,1}$
in place of
$X_n$
.
Lemma 5.2.
$X_{n} \lesssim X_{n+2,1}$
.
Proof. By a change of variables, we simply have
$$ \begin{align*} X_{n+2,1} \geq \sum_{|{\underline i}|=n} \operatorname{\mathrm{vol}}_{d-1}(T_1T_2(\Delta_{\underline i}))^\delta \geq \min_{x \in \Delta} \big(\operatorname{\mathrm{Jac}}_x T_1T_2\big)^\delta \sum_{|{\underline i}|=n} \operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^\delta =: C X_n. \end{align*} $$
This constant C is positive as
$T_1$
and
$T_2$
are injective.
The next important lemma is our main tool to guarantee that
$X_{n,1}$
converges to zero.
Lemma 5.3. For all
$n>k\geq 1$
,
$$ \begin{align*} X_{n+1,k+1} \leq b_k X_{n,k} \quad \text{and} \quad X_{n+1,1} \leq \sum_{j=1}^{n-1} a_k X_{n,j} + r_n, \end{align*} $$
where
$$ \begin{align*} b_k = \max_{v \in R_k\cap\Delta_1} \| M_1\cdot v \|^{-d\delta} , \quad a_k = \max_{v \in R_k\cap\Delta_1} \sum_{j=2}^d \| M_j\cdot v \|^{-d\delta} \end{align*} $$
and
$r_n \lesssim n^{(1-d)\delta }$
, and where
$$ \begin{align*} R_k := \mathrm{cl} \bigg( \bigcup_{j=1}^d T_j^k(\Delta) \setminus T_j^{k+1}(\Delta) \bigg), \quad \text{that is}, R_k\cap\Delta_1 = \mathrm{cl} ( T_1^k(\Delta) \setminus T_1^{k+1}(\Delta) ). \end{align*} $$
Proof. First note that by the definition of
$A_{n,k}$
, if
${\underline i} = (i_1,\ldots ,i_n) \in A_{n,k}$
, then:
-
•
$(i_1;{\underline i}):= (i_1,i_1,i_2,\ldots ,i_n) \in A_{n+1,k+1}$
; and -
•
$(j;{\underline i}) := (j,i_1,i_2,\ldots ,i_n) \in A_{n+1,1}$
for any
$j \neq i_1$
.
A similar statement applies for
${\underline i} \in \{1\}^n \cup \{2\}^n \cup \cdots \cup \{d\}^n$
.
From this,
${\underline i}' \in A_{n+1,k+1}$
if and only if there exists a unique
${\underline i} \in A_{n,k}$
such that
${{\underline i}' = (i_1;{\underline i})}$
. Consequently,
$$ \begin{align} X_{n+1,k+1} = \sum_{{\underline i}' \in A_{n+1,k+1}} \operatorname{\mathrm{vol}}_{d-1}(\Delta_{{\underline i}'})^\delta = \sum_{i \in A_{n,k}} \operatorname{\mathrm{vol}}_{d-1}(T_{i_1}(\Delta_{\underline i}))^\delta. \end{align} $$
Using the formula in Lemma 4.1 and that
$$ \begin{align*} {\underline i} \in A_{n,k}\!\implies\!\Delta_{\underline i}\in R_k := \mathrm{cl} \bigg( \bigcup_{j=1}^d T_j^k(\Delta) \setminus T_j^{k+1}(\Delta) \bigg) \end{align*} $$
(that is,
$T_{\underline i}(e_j) \in R_k \cap \Delta _{{\underline i}_1}$
for each j), we have
$$ \begin{align} \frac{\operatorname{\mathrm{vol}}_{d-1}(T_{i_1}(\Delta_{\underline i}))^\delta}{\operatorname{\mathrm{vol}}_{d-1}(\Delta_{{\underline i}'})^\delta} &=\prod_{j=1}^d\frac{\|M_{i_1}M_{\underline i} \cdot e_j\|^{-\delta}}{\|M_{\underline i} \cdot e_j\|^{-\delta}}=\prod_{j=1}^d \bigg\| M_{i_1} \cdot \frac{M_{\underline i} \cdot e_j}{\|M_{\underline i} \cdot e_j\|} \bigg\|^{-\delta}\nonumber\\ &=\prod_{j=1}^d \|M_{i_1} \cdot T_{\underline i} (e_j)\|^{-\delta}\leq \prod_{j=1}^d\max_{v \in R_k \cap \Delta_{i_1}}\|M_{i_1} \cdot v\|^{-\delta}\nonumber \\ &=\prod_{j=1}^d\max_{v \in R_k \cap \Delta_{1}}\!\|M_{1} \cdot v\|^{-\delta}= \max_{v \in R_k \cap \Delta_{1}}\!\|M_{1} \cdot v\|^{-d\delta} \nonumber\\ &=: b_k, \end{align} $$
using symmetry and linearity. Applying this estimate to equation (2) gives the required inequality:
$$ \begin{align*} X_{n+1,k+1} &\leq \sum_{i \in A_{n,k}} b_k\, \operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^\delta = b_k X_{n,k}. \end{align*} $$
The proof of the second inequality in the lemma is slightly more nuanced. From our first consideration, we have
$$ \begin{align*} X_{n+1,1} = \sum_{|{\underline i}|=n} \sum_{\substack{1\leq \omega\leq d:\\\omega \neq i_1}} \operatorname{\mathrm{vol}}_{d-1}(T_\omega(\Delta_{\underline i}))^\delta, \end{align*} $$
and we bound the internal sum in two cases.
Case 1.
${\underline i} \in A_{n,k}$
for some k. This case is similar to the proof of equation (3), but this time we also apply the AM-GM inequality:
$$ \begin{align} &\frac{\sum_{\omega \neq i_1} \operatorname{\mathrm{vol}}_{d-1}(T_\omega(\Delta_{\underline i}))^\delta} {\operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^\delta}\nonumber\\ &\quad= \sum_{\omega\neq i_1} \prod_{j=1}^d \frac {\|M_{\omega}M_{\underline i} \cdot e_j\|^{-\delta}} {\|M_{\underline i} \cdot e_j\|^{-\delta}} = \sum_{\omega \neq i_1} \prod_{j=1}^d \|M_{\omega} \cdot T_{\underline i} (e_j)\|^{-\delta} \nonumber \\ &\quad\leq \frac 1 d \sum_{\omega\neq i_1} \sum_{j=1}^d \|M_{\omega} \cdot T_{\underline i} (e_j)\|^{-d\delta} = \frac 1 d \sum_{j=1}^d \sum_{\omega\neq i_1} \|M_{\omega} \cdot T_{\underline i} (e_j)\|^{-d\delta} \nonumber \\ &\quad\leq \max_{v \in R_k \cap \Delta_{i_1}} \sum_{\omega \neq i_1}\|M_{\omega} \cdot v\|^{-d\delta} = \max_{v \in R_k\cap\Delta_1} \sum_{j=2}^d \|M_j \cdot v\|^{-d\delta} \nonumber \\ &\quad=: a_k. \end{align} $$
Summing over
${\underline i}\in A_{n,k}$
hence gives
$$ \begin{align*} \sum_{i \in A_{n,k}} \sum_{\omega \neq i_1} \operatorname{\mathrm{vol}}_{d-1}(T_\omega(\Delta_{\underline i}))^\delta \leq \sum_{i \in A_{n,k}} a_k \operatorname{\mathrm{vol}}_{d-1}(\Delta_{\underline i})^\delta = a_k X_{n,k}. \end{align*} $$
Case 2.
${\underline i} \in \{1\}^n \cup \{2\}^n \cup \cdots \cup \{d\}^n$
. This is an explicit calculation, using that all
$\binom d 2$
of the summands for this case are equal:
$$ \begin{align*} r_n &:= \sum_{j=1}^d \sum_{\omega \neq j} \operatorname{\mathrm{vol}}_{d-1}(T_\omega T_j^n (\Delta))^\delta \lesssim \operatorname{\mathrm{vol}}_{d-1}(T_1T_2^n(\Delta))^\delta\\ &= \big( 2^{1-d} (2n+1)^{-1} (n+1)^{2-d} \big)^\delta \lesssim n^{\delta(1-d)}, \end{align*} $$
using the formula in Lemma 4.1 and the explicit form of
$M_1M_2^n$
. This case completes the proof of the second inequality and hence of the lemma.
The only missing piece, before we prove the main technical result of this paper, is to furnish
$a_k$
and
$b_k$
with values. The proof is simple, but for convenience, we defer it to the appendix.
Lemma 5.4. For each
$k\in \mathbb N$
,
$$ \begin{align*} b_k = \bigg( \frac {k+2} {k+3} \bigg)^{d\delta} \quad \text{and} \quad a_k = \bigg( \frac {k+1} {2k+1} \bigg)^{d\delta} + 2^{-d\delta} (d-2). \end{align*} $$
We now use the proof of a particular case of the renewal theorem [Reference Feller7, p. 330] to conclude the following result from the previous.
Theorem 5.5. If
$\delta> (d-1)^{-1}$
and
$ \sum _{k=1}^\infty a_k \prod _{j=1}^{k-1} b_j < 1, $
that is,
$$ \begin{align*} 3^{d\delta} \sum_{k=1}^\infty \bigg( \frac {k+1} {2k+1} \bigg)^{d\delta} (k+2)^{-d\delta} + 2^{-d\delta} (d-2) (k+2)^{-d\delta} < 1, \end{align*} $$
then
$X_{n,1} \to 0$
as
$n \to \infty $
. Consequently,
$\dim _H(\mathcal G_d)\leq d-2+\delta $
.
Proof. Applying the first inequality of Lemma 4.3 to the summands of the second gives
$$ \begin{align*} X_{n+1,1} \leq \sum_{k=1}^{n-1} a_k \prod_{j=1}^{k-1} b_j\, X_{n+1-k,1} + r_n. \end{align*} $$
Writing
$\unicode{x3bb} _k = a_k\prod _{j=1}^{k-1}b_j$
for succinctness, we then have
$$ \begin{align*} \sum_{n=1}^N X_{n+1,1} \leq \sum_{n=1}^N \sum_{k=1}^{n-1} \unicode{x3bb}_k X_{n+1-k,1} + \sum_{n=1}^N r_n &= \sum_{k=1}^{N-1} \unicode{x3bb}_k \sum_{n=k+1}^{N} X_{n+1-k,1} + \sum_{n=1}^N r_n \\ &\leq \sum_{k=1}^{N-1} \unicode{x3bb}_k \sum_{j=1}^{N+1} X_{j,1} + \sum_{n=1}^N r_n, \end{align*} $$
that is,
$$ \begin{align*} \sum_{n=1}^N X_{n+1,1} \leq \frac { x_1\sum_{k=1}^{N-1} \unicode{x3bb}_k + \sum_{n=1}^N r_n} {1- \sum_{k=1}^N \unicode{x3bb}_k}. \end{align*} $$
The right-hand side is bounded in N by our assumptions, and the result follows.
6 Proof of Theorem 1.1
The remainder of the proof of Theorem 1.1 is to show that Theorem 5.5 holds for
$\delta =1$
and any
$d\geq 3$
, since by continuity, it will then apply for any
$\delta <1$
sufficiently close to 1. More explicitly, we wish to show that
$$ \begin{align*} &\sum_{k=1}^\infty \bigg( \frac {3(k+1)} {(2k+1)(k+2)} \bigg)^d + (d-2) \bigg( \frac3{2(k+2)} \bigg)^{d}\\ &\quad= \sum_{k=1}^\infty \bigg( \frac 1{2k+1} + \frac1{k+2} \bigg)^d + (d-2) \bigg( \frac3{2(k+2)} \bigg)^{d} \end{align*} $$
is strictly less than 1. This uses elementary calculus. Since each term on the right-hand side is decreasing in d for
$d\geq 3$
, it suffices to show just the
$d=3$
case, that is, that
$$ \begin{align} \frac{27}8 \sum_{k=1}^\infty \bigg( \frac {2(k+1)} {(2k+1)(k+2)} \bigg)^3 + \bigg( k+2 \bigg)^{-3} < 1. \end{align} $$
This follows by simply bounding the tail of the sum by an integral. For any
$n\in \mathbb N$
, the left-hand side of equation (5) is at most
$$ \begin{align*} &\frac{27}8 \sum_{k=1}^n \bigg( \frac {2(k+1)} {(2k+1)(k+2)} \bigg)^3 + \bigg( k+2 \bigg)^{-3} + \frac{27}8 \int_{n-1}^\infty 8(2x+1)^{-3} + (x+2)^{-3} \;\mathrm dx \\ &\quad= \frac{27}8 \sum_{k=1}^n \bigg( \frac {2(k+1)} {(2k+1)(k+2)} \bigg)^3 + \bigg( k+2 \bigg)^{-3} + \frac {27}{4(2n-1)^{2}} + \frac {27}{16(n+1)^{2}}, \end{align*} $$
and for
$n=3$
, this expression equals
$$ \begin{align*} &\bigg(\frac23\bigg)^3 + \bigg(\frac12\bigg)^3 + \bigg(\frac9{20}\bigg)^3 + \bigg(\frac38\bigg)^3 + \bigg(\frac{12}{35}\bigg)^3 + \bigg(\frac3{10}\bigg)^{3} + \frac{27}{4 \times 5^2} + \frac{27}{16 \times 4^2} \\&\quad= \frac{574,898,507}{592,704,000} <1. \end{align*} $$
Hence, equation (5) holds, completing the proof of the theorem.
Acknowledgements
The first author is partly supported by ERC-Advanced Grant 833802-Resonances and EPSRC grant EP/T001674/1, and the second by the Alfréd Rényi Young Researcher Fund.
A Appendix. Proof of Lemma 5.4
Noting that
$\|M_j \cdot v\| = (2-v_j)$
and
$$ \begin{align*} R_k\cap\Delta_1 := \bigg\{ v \in \Delta \; \bigg| \; \frac{k}{k+1} \leq v_1 \leq \frac{k+1}{k+2} \bigg\}, \end{align*} $$
we have
$$ \begin{align*} b_k = \max \bigg\{ (2-v_1)^{-d\delta} \; \bigg| \; \frac{k}{k+1} \leq v_1 \leq \frac{k+1}{k+2} \bigg\} = \bigg( 2 - \frac{k+1}{k+2} \bigg)^{-d\delta} = \bigg( \frac{k+2}{k+3} \bigg)^{d\delta}. \end{align*} $$
Regarding the value of
$a_k$
, we write
$a_k = \max _{R_k\cap \Delta _1} f$
, where
$$ \begin{align*} f(v) := \sum_{j=2}^d (2-v_j)^{-d\delta}. \end{align*} $$
Here, f is convex (since its summands are convex), so the maximum value it takes on
$T_1^j(\Delta )\supset R_k\cap \Delta _1 $
is obtained at one of its d vertices. By symmetry, this maximum is either
$f(e_1)$
or
$f ((k/({k+1})) e_1 + (1/({k+1})) e_j)$
, and we find it is the latter:
$$ \begin{align*} f(e_1) = 2^{-d\delta}(d-1) \leq \bigg( \frac {k+1} {2k+1} \bigg)^{d\delta} + 2^{-d\delta} (d-2) = f\bigg( \frac k{k+1} e_1 + \frac 1{k+1} e_j \bigg), \end{align*} $$
since
$(k/({k+1})) e_1 + (1/({k+1})) e_2$
lies in
$R_k\cap \Delta _1$
,
$a_k$
takes the claimed value.
