††2010 Mathematics Subject Classification:11J06.††Key words and phrases:
Minkowski question mark function, Continued fractions.

On the derivative of the Minkowski question-mark function
Dmitry Gayfulin111Research is supported by RNF grant No. 19-11-00001.

###### Abstract

The Minkowski question-mark function $?(x)$ is a continuous strictly
increasing function defined on $[0,1]$ interval. It is well known fact that
the derivative of this function, if exists, can take only two values: $0$ and
$+\infty$. It is also known that the value of the derivative $?^{\prime}(x)$
at the point $x=[0;a_{1},a_{2},\ldots,a_{t},\ldots]$ is connected with the
limit behavior of the arithmetic mean $(a_{1}+a_{2}+\ldots+a_{t})/t$.
Particularly, N. Moshchevitin and A. Dushistova showed that if
$a_{1}+a_{2}+\ldots+a_{t}<\kappa_{1}t$, where
$\kappa_{1}=2\log\bigl{(}{\frac{1+\sqrt{5}}{2}}\bigr{)}/\log{2}=1.3884\ldots$,
then $?^{\prime}(x)=+\infty$. They also proved that the constant $\kappa_{1}$
is non-improvable. We consider a dual problem: how small can be the quantity
$a_{1}+a_{2}+\ldots+a_{t}-\kappa_{1}t$ if $?^{\prime}(x)=0$? We obtain the
non-improvable estimates of this quantity.

##  1 Introduction

###  1.1 The Minkowski function $?(x)$.

For an arbitrary $x\in[0,1]$ we consider its continued fraction expansion

$x=[0;a_{1},a_{2},\ldots,a_{n},\ldots]=\cfrac{1}{a_{1}+\cfrac{1}{a_{2}+\cdots}},\,\,\,a_{j}\in\mathbb{Z}_{+}$

with natural partial quotients $a_{t}$. This representation is infinite when
$x\not\in\mathbb{Q}$ and finite for rational $x$. For irrational numbers the
continued fraction representation is unique, however each rational $x$ has two
different representations

$x=[0;a_{1},a_{2},\ldots,a_{n-1},a_{n}]\,\,\,\,\,\text{and}\,\,\,\,\,x=[0;a_{1},a_{2},\ldots,a_{n-1},a_{n}-1,1],\,\,\,\,\text{where}\,\,\,\,a_{n}\geqslant
2.$

By

$\frac{p_{k}}{q_{k}}:=[0;a_{1},\ldots,a_{k}]$

we denote the $k$th convergent fraction to $x$. By $B_{n}$ we denote the $n$th
level of the Stern-Brocot tree, that is

$B_{n}:=\\{x=[0;a_{1},\ldots,a_{k}]:a_{1}+\ldots+a_{k}=n+1\\}.$

In [12] Minkowski introduced the function $?(x)$ which may be defined as the
limit distribution function of sets $B_{n}$. This function was rediscovered
several times and studied by many authors (see [11],[10],[1],[4],[13]). For
irrational $x=[0;a_{1},a_{2},\ldots,a_{n},\ldots]$ the formula

$?(x)=\sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{2^{a_{1}+\ldots+a_{k}-1}}$
(1)

introduced by Denjoy [2, 3] and Salem [15] may be considered as one of the
equivalent definitions of the function $?(x)$. If $x$ is rational, then the
infinite series in (1) is replaced by a finite sum. Note that
$?([0;a_{1},\ldots,a_{t}+1])=?([0;a_{1},\ldots,a_{t},1])$ and hence $?(x)$ is
well-defined for rational numbers too. It is known that $?(x)$ is a continuous
strictly increasing function, also its derivative $?^{\prime}(x)$, if exists,
can take only two values – $0$ and $+\infty$. Almost everywhere in [0; 1] in
the sense of Lebesgue measure the derivative exists and equals $0$. If $x$ is
rational, then $?^{\prime}(x)$ also exists and equals zero.

###  1.2 Notation and parameters

We will denote the sequences by capital letters $A,B,C$ and their elements by
the corresponding small letters $a_{i},b_{j},c_{k}$. All sequences of the
present paper contain positive integers unless otherwise stated. For an
arbitrary finite sequence $B=(b_{1},b_{2},\ldots,b_{n})$ we denote

$\overleftarrow{B}=(b_{n},b_{n-1},\ldots,b_{1}),\quad
S(B)=\sum\limits_{i=1}^{n}b_{i}.$

By $\langle A\rangle$ we denote the continuant of (possibly empty) finite
sequence $A=(a_{1},\ldots,a_{t})$. It is defined as follows: the continuant of
the empty sequence equals $1$, $\langle a_{1}\rangle=a_{1}$, if $t\geqslant 2$
one has

$\langle a_{1},a_{2},\ldots,a_{t}\rangle=a_{t}\langle
a_{1},a_{2},\ldots,a_{t-1}\rangle+\langle a_{1},a_{2},\ldots,a_{t-2}\rangle.$
(2)

Note that the finite continued fraction $[0;a_{1},\ldots,a_{t}]$ can be
expressed using continuants:

$[0;a_{1},\ldots,a_{t}]=\frac{\langle a_{2},\ldots,a_{t}\rangle}{\langle
a_{1},a_{2},\ldots,a_{t}\rangle}.$ (3)

Rule (2) can be generalized as follows:

$\begin{split}\langle
a_{1},a_{2},\ldots,a_{t},a_{t+1},\ldots,a_{s}\rangle=\langle
a_{1},a_{2},\ldots,a_{t}\rangle\langle a_{t+1},\ldots,a_{s}\rangle+\langle
a_{1},a_{2},\ldots,a_{t-1}\rangle\langle a_{t+2},\ldots,a_{s}\rangle=\\\
=\langle a_{1},a_{2},\ldots,a_{t}\rangle\langle
a_{t+1},a_{t+2},\ldots,a_{s}\rangle(1+[0;a_{t-1},a_{t-2},\ldots,a_{1}][0;a_{t+2},a_{t+3},\ldots,a_{s}]).\end{split}$
(4)

One can find more about the properties of continuants in [9].

For an irrational $x=[0;a_{1},a_{2},\ldots,a_{n},\ldots]$ we consider the sum
$S_{x}(t)$ of its partial quotients up to $t-$th:

$S_{x}(t)=a_{1}+a_{2}+\ldots+a_{t}.$

For an arbitrary finite sequence $B$ we will denote by $S(B)$ and $\Pi(B)$ the
sum and the product of its elements respectively.

Throughout the paper we always denote the sequence of partial quotients of $x$
by $a_{1},a_{2},\ldots,a_{t},\ldots$ unless otherwise stated. We will also
denote the sequence of $t$ first elements of this infinite sequence by
$A_{t}$. Thus, $S_{x}(t)=S(A_{t})$.

We also need the following constants

$\Phi=\frac{1+\sqrt{5}}{2},\quad\kappa_{1}=\frac{2\log{\Phi}}{\log{2}}=1.3884\ldots,$
(5) $\lambda_{n}=\frac{n+\sqrt{n^{2}+4}}{2},$ (6)
$\kappa_{2}=\frac{4\log\lambda_{5}-5\log\lambda_{4}}{\log\lambda_{5}-\log\lambda_{4}-\log\sqrt{2}}=4.401\ldots,$
(7) $\kappa_{4}=\sqrt{\frac{\kappa_{1}-1}{\log 2}}=0.7486\ldots.$ (8)

For an arbitrary sequence $A$ of length $t$ we denote $S(A)-\kappa_{1}t$ by
$\varphi^{(1)}(A)$. For $x=[0;a_{1},\ldots,a_{t},\ldots]$ we denote
$\varphi^{(1)}(A_{t})$ by $\varphi^{(1)}_{x}(t)$.

###  1.3 Critical values

In the paper [14] it was shown by J. Paradis, P. Viader, L. Bibiloni that the
value of the derivative of the function $?(x)$ is connected with the limit
behavior of $\frac{S_{x}(t)}{t}$. They showed that if
$\frac{S_{x}(t)}{t}<\kappa_{1}$ and $?^{\prime}(x)$ exists, then
$?^{\prime}(x)=+\infty$. On the other side, denote by $z\approx 5.319$ is the
root of the equation $2\log(1+z)=z\log{2}$. If $\frac{S_{x}(t)}{t}\geqslant z$
and $?^{\prime}(x)$ exists, then $?^{\prime}(x)=0$.

In the paper [4] A. Dushistova and N. Moshchevitin improved the results of
[14] and formulated the following two theorems.
Theorem A.
(i) Let for real irrational $x\in(0,1)$ the inequality $S_{x}(t)<\kappa_{1}t$
hold for all $t$ large enough. Then the derivative $?^{\prime}(x)$ exists and
equals $+\infty$.
(ii) For any positive $\varepsilon$ there exists an irrational number ${\bf
x}\in(0,1)$, such that $?^{\prime}(x)=0$ and the inequality
$S_{x}(t)<(\kappa_{1}+\varepsilon)t$ holds for all $t$ large enough.
Theorem B.
(i) Let for real irrational $x\in(0,1)$ the inequality $S_{x}(t)>\kappa_{2}t$
hold for all $t$ large enough. Then the derivative $?^{\prime}(x)$ exists and
equals $0$.
(ii) For any positive $\varepsilon$ there exists an irrational number
$x\in(0,1)$, such that $?^{\prime}(x)=+\infty$ and the inequality
$S_{x}(t)>(\kappa_{2}-\varepsilon)t$ holds for all $t$ large enough.

One can see that the constants $\kappa_{1}$ and $\kappa_{2}$ in theorems A and
B are non-improvable.

###  1.4 The dual problem

In the paper [5] the dual problem was considered. Suppose that
$?^{\prime}(x)=0$. How small can be the difference
$\varphi^{(1)}_{x}(t)=S_{x}(t)-\kappa_{1}t$? Statement (ii) of Theorem $A$
implies that $\varphi^{(1)}_{x}(t)$ can be less than $\varepsilon t$ for any
positive $\varepsilon$. The first non-trivial estimate of $\varphi^{1}_{x}(t)$
was obtained in [5] by A. Dushistova, I. Kan and N. Moshchevitin.
Theorem C.
(i) Let for irrational $x\in(0,1)$ the derivative $?^{\prime}(x)$ exists and
$?^{\prime}(x)=0$. Then for any $\varepsilon>0$ for all $t$ large enough one
has

$\max_{u\leqslant t}\varphi^{(1)}_{x}(u)=\max_{u\leqslant
t}\left(S_{x}(u)-\kappa_{1}u\right)\geqslant(\kappa_{4}-\varepsilon)\sqrt{t\log
t}.$ (9)

(ii) There exists an irrational ${\bf x}\in(0,1)$, such that $?^{\prime}({\bf
x})=0$ and or any $\varepsilon>0$ for all $t$ large enough one has

$\varphi^{(1)}_{x}(t)=S_{x}(t)-\kappa_{1}t\leqslant(2\sqrt{2+\varepsilon})\kappa_{4}\sqrt{t\log
t}.$ (10)

In the paper [8] a strengthened version of the inequality (9) was obtained
within the same condition

$\max_{u\leqslant
t}\varphi^{(1)}_{x}(u)\geqslant\biggl{(}\frac{2}{\sqrt{3}}+\varepsilon\biggr{)}\kappa_{4}\sqrt{t\log
t}.$ (11)

Of course, one can ask the same question when $?^{\prime}(x)=+\infty$. How
small can be the difference $\varphi^{(2)}_{x}(t):=\kappa_{2}t-S_{x}(t)$? The
first result in this area was also obtained in [5]. It was improved several
times and for now the best known estimates are the following:
Theorem D.[7]
(i) Let for real irrational $x\in(0,1)$ the derivative $?^{\prime}(x)$ exists
and $?^{\prime}(x)=+\infty$. Then for all $t$ large enough one has

$\max_{u\leqslant t}\left(\kappa_{2}u-S_{x}(u)\right)\geqslant
0.06222\sqrt{t}.$ (12)

(ii) There exists an irrational $x\in(0,1)$, such that $?^{\prime}(x)=+\infty$
and for all $t$ large enough one has

$\kappa_{2}t-S_{x}(t)\leqslant 0.26489\sqrt{t}.$ (13)

##  2 Main results

Note that there are different quantities on the left-hand sides of the
inequalities (9) and (10). One can say that the inequality (9) considers the
uniform behavior of $\varphi^{(1)}_{x}(t)$, whereas the inequality (10) deals
with the local behavior of this quantity. In the present paper we consider
upper and lower estimates in both cases. Our first theorem states that
statement (ii) of Theorem C is non-improvable.

######  Theorem 1.

Let for irrational $x\in(0,1)$ the derivative $?^{\prime}(x)$ exists and
$?^{\prime}(x)=0$. Then for any $\varepsilon>0$ for infinitely many $t$ one
has

$\varphi^{(1)}_{x}(t)\geqslant(2\sqrt{2}-\varepsilon)\kappa_{4}\sqrt{t\log
t}.$ (14)

Our second theorem provides optimal estimate of the uniform behavior of
$\varphi^{(1)}_{x}(t)$.

######  Theorem 2.

(i)Let for irrational $x\in(0,1)$ the derivative $?^{\prime}(x)$ exists and
$?^{\prime}(x)=0$. Then for any $\varepsilon>0$ for all $t$ large enough one
has

$\max_{u\leqslant
t}\varphi^{(1)}_{x}(u)\geqslant(\sqrt{2}-\varepsilon)\kappa_{4}\sqrt{t\log
t}.$ (15)

(ii) For any $\varepsilon>0$ there exists an irrational $x\in(0,1)$ such that
$?^{\prime}(x)=0$ and for infinitely many $t$ one has

$\max_{u\leqslant
t}\varphi^{(1)}_{x}(u)\leqslant(\sqrt{2}+\varepsilon)\kappa_{4}\sqrt{t\log
t}.$ (16)

##  3 Auxiliary lemmas

###  3.1 Increment lemmas and their corollaries

The following two lemmas shed light on the connection between the behavior of
the sum of partial quotients of $x$ and the value of the derivative
$?^{\prime}(x)$.

######  Lemma 3.1.

([5], Lemma 1) For an irrational $x=[0;a_{1},a_{2},\ldots,a_{t},\ldots]$ and
for $\delta$ small in absolute value, there exists a natural $t=t(x,\delta)$
such that

$\frac{?(x+\delta)-?(x)}{\delta}\geqslant\frac{\langle A_{t}\rangle\langle
A_{t-1}\rangle}{2^{S_{x}(t)+4}}.$ (17)

######  Lemma 3.2.

([5], Lemma 2) For an irrational $x=[0;a_{1},a_{2},\ldots,a_{t},\ldots]$ and
for $\delta$ small in absolute value, there exists a natural $t=t(x,\delta)$
such that

$\frac{?(x+\delta)-?(x)}{\delta}\leqslant\frac{\langle
A_{t}\rangle^{2}}{2^{S_{x}(t)-2}}.$ (18)

It is not convenient for us that the numerators of the right-hand sides of
(17) and (18) do not coincide. That is why we prove a ’’symmetric’’ corollary
of Lemmas 3.1 and 3.2

######  Lemma 3.3.

For an irrational $x=[0;a_{1},a_{2},\ldots,a_{t},\ldots]$ the derivative
$?^{\prime}(x)$ equals zero if and only if

$\lim\limits_{t\to\infty}\frac{\langle A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}=0.$
(19)

###### Proof.

($\Leftarrow$) Follows immediately from Lemma 3.2.
($\Rightarrow$) It follows from Lemma 3.1 that

$\lim\limits_{t\to\infty}\frac{\langle
A_{t-1}\rangle}{\sqrt{2}^{S_{x}(t)}}=0,$ (20)

Suppose that $\frac{\langle A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}$ does not tend
to zero as $t\to\infty$. That is, there exists a positive constant $c$ such
that

$\frac{\langle A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}>c$ (21)

for infinitely many $t$. Consider $N\in\mathbb{N}$ such that for any
$n\geqslant N$

$\frac{\langle A_{n-1}\rangle}{\sqrt{2}^{S_{x}(n)}}<\frac{c}{100}.$

Consider an arbitrary integer $t>N+a_{N}$ such that the inequality (21) holds.
Then

$\frac{(a_{t}+1)c}{100}>\frac{(a_{t}+1)\langle
A_{t-1}\rangle}{\sqrt{2}^{S_{x}(t)}}>\frac{\langle
A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}>c$

Thus, $a_{t}\geqslant 100$. On the other hand,

$\frac{\sqrt{2}^{a_{t}}c}{(a_{t-1}+1)(a_{t}+1)}<\frac{\sqrt{2}^{a_{t}}}{a_{t-1}+1}\frac{\langle
A_{t-1}\rangle}{\sqrt{2}^{S_{x}(t)}}<\frac{\langle
A_{t-2}\rangle}{\sqrt{2}^{S_{x}(t-1)}}<\frac{c}{100}$

Therefore, we have $100\sqrt{2}^{a_{t}}<(a_{t-1}+1)(a_{t}+1)$. As
$a_{t}\geqslant 100$, one can easily see that $a_{t-1}>a_{t}+1$. From (21) one
can also derive that

$\frac{\langle
A_{t-1}\rangle}{\sqrt{2}^{S_{x}(t-1)}}>\frac{a_{t}+1}{\sqrt{2}^{a_{t}}}\frac{\langle
A_{t-1}\rangle}{\sqrt{2}^{S_{x}(t-1)}}>\frac{\langle
A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}>c.$

Repeating the same argument $t-N$ times we obtain that
$a_{N}>a_{t}+t-N>a_{t}+a_{N}$ and therefore we come to a contradiction. ∎

###  3.2 Continuant lower estimate

The following lemma is a useful tool to estimate the values of continuants,
most of whose elements are equal to $1$. We introduce some notation first.
Having a sequence $A_{t}$, denote by $w(A_{t})$ the number of its elements
greater than $1$. The set of such elements forms the sequence that we denote
by $d_{A_{t}}(1),d_{A_{t}}(2),\ldots,d_{A_{t}}(w(A_{t}))$.

######  Lemma 3.4.

For an arbitrary continuant $\langle A_{t}\rangle=\langle
a_{1},a_{2},\ldots,a_{t}\rangle$ one has

$\langle
a_{1},a_{2},\ldots,a_{t}\rangle\geqslant\frac{1}{2}\Phi^{t}\prod\limits_{i=1}^{w(A_{t})}\biggl{(}\frac{d_{A_{t}}(i)}{4}\biggr{)}.$
(22)

###### Proof.

We prove by induction on $w(A_{t})$. If $w(A_{t})$ equals zero, then $\langle
A_{t}\rangle=F_{t+1}$ – $(t+1)$-th Fibonacci number and one can easily verify
(22) using Binet’s formula.

Now suppose that the inequality (22) holds for $w(A_{t})=n$. Consider an
arbitrary continuant $\langle A_{t}\rangle$ having $w(A_{t})=n+1$. Let us also
consider the continuant $\langle A^{\prime}_{t}\rangle$, which is obtained
from $\langle A_{t}\rangle$ by replacing one of its elements greater than one
by $1$. Of course, $w(A^{\prime}_{t})=n$ and one has

$\langle A_{t}\rangle=\frac{\langle A_{t}\rangle}{\langle
A^{\prime}_{t}\rangle}\langle A^{\prime}_{t}\rangle\geqslant\frac{\langle
A_{t}\rangle}{\langle
A^{\prime}_{t}\rangle}\frac{1}{2}\Phi^{t}\prod\limits_{i=1}^{n}\biggl{(}\frac{d_{A^{\prime}_{t}}(i)}{4}\biggr{)}.$
(23)

Applying (4), one can easily see that for arbitrary finite sequences $A$ and
$B$ one has

$\langle A,x,B\rangle=\langle A,x\rangle\langle
B\rangle(1+[0;x,\overleftarrow{A}][0;B])=x\langle A\rangle\langle
B\rangle(1+[0;\overleftarrow{A}][0;x])(1+[0;x,\overleftarrow{A}][0;B]).$

Hence,

$\frac{\langle A,x,B\rangle}{\langle
A,1,B\rangle}=x\frac{1+[0;x,\overleftarrow{A}][0;B]}{1+[0;1,\overleftarrow{A}][0;B]}\frac{1+[0;\overleftarrow{A}][0;x]}{1+[0;\overleftarrow{A}]}\geqslant
x\frac{1}{2}\cdot\frac{1}{2}=\frac{x}{4}.$ (24)

Applying the lower estimate of $\frac{\langle A_{t}\rangle}{\langle
A^{\prime}_{t}\rangle}$ from (24) to (23), we prove the induction step. The
lemma is proved. ∎

######  Lemma 3.5.

Suppose that $?^{\prime}(x)=0$. Then $\varphi^{(1)}_{x}(t)>0$ for all $t$
large enough.

###### Proof.

From Lemma 3.3 one has

$\frac{\langle A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}>\frac{1}{2}.$

Using an obvious estimate $\langle A_{t}\rangle>\frac{\Phi^{t}}{2}$ from (22),
one has

$\frac{1}{2}<\frac{\langle
A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}<\frac{1}{2}\frac{\Phi^{t}}{\sqrt{2}^{\kappa_{1}t+\varphi^{(1)}_{x}(t)}}.$

As $\sqrt{2}^{\kappa_{1}}=\Phi$, we obtain the statement of the lemma. ∎

######  Lemma 3.6.

Consider two arbitrary real numbers $\beta>\alpha\geqslant 3$ and a real
number $s\geqslant\beta$. Let $R(s,\alpha,\beta)$ be the set of all finite
sequences $R=(r_{1},\ldots,r_{k})$ of real numbers such that $\alpha\leqslant
r_{i}\leqslant\beta$ for all $i\leqslant k$ and $S(R)=s$. Then

$\min\limits_{R\in
R(s,\alpha,\beta)}\Pi(R)\geqslant\beta^{\bigl{[}\frac{s}{\beta}\bigr{]}}$ (25)

###### Proof.

Note that the number $k$ is not fixed in the definition of
$R(s,\alpha,\beta)$. Denote

$f(s,\alpha,\beta)=\min\limits_{R\in R(s,\alpha,\beta)}\Pi(R).$

It is clear that the function $f(s,\alpha,\beta)$ is monotonic in the first
argument. Hence, without loss of generality one can say that
$\frac{s}{\beta}\in\mathbb{Z}$. Consider the sequence
$R_{\beta}=(\underbrace{\beta,\beta,\ldots,\beta}_{s/\beta\text{ times}})$.
Using the compactness argument, one can easily see that there exists a
sequence $R_{0}=(r^{0}_{1},\ldots,r^{0}_{m})\in R(s,\alpha,\beta)$ such that
$\Pi(R_{0})=f(s,\alpha,\beta)$. Suppose that $R_{0}\neq R_{\beta}$. One can
easily show that $R_{0}$ cannot contain more than two elements not equal to
$\alpha$ or $\beta$. Indeed, if $\alpha<r^{0}_{i}\leqslant r^{0}_{j}<\beta$,
then there exists $\delta>0$ such that $r^{0}_{i}-\delta>\alpha$ and
$r^{0}_{j}+\delta<\beta$. As
$(r^{0}_{i}-\delta)(r^{0}_{j}+\delta)<r^{0}_{i}r^{0}_{j}$, one can see that

$\Pi(r^{0}_{1},\ldots,r^{0}_{i-1},r^{0}_{i}-\delta,r^{0}_{i+1},\ldots,r^{0}_{j-1},r^{0}_{j}+\delta,r^{0}_{j+1},\ldots,r^{0}_{m})<\Pi(r^{0}_{1},\ldots,r^{0}_{i-1},r^{0}_{i},r^{0}_{i+1},\ldots,r^{0}_{j-1},r^{0}_{j},r^{0}_{j+1},\ldots,r^{0}_{m})$

and we obtain a contradiction with the definition of $R_{0}$.

On the other hand, it follows from the definition of $R_{0}$ that

$\Pi(R_{0})\leqslant\Pi(R_{\beta}).$ (26)

Without loss of generality one can assume that $R_{0}$ does not contain
elements equal to $\beta$. Indeed, if we remove all such elements from $R_{0}$
and the same number of elements from $R_{\beta}$, the inequality (26) will be
still satisfied. Thus, $R_{0}$ has the form
$R_{0}=(x,\underbrace{\alpha,\alpha,\ldots,\alpha}_{(s-x)/\alpha\text{
times}})$ up to transposition of elements. Here $\alpha\leqslant x<\beta$.

Denote $n=\frac{s-x}{\alpha}$. Inequality (26) can be written as

$\alpha^{n}x\leqslant\beta^{\frac{\alpha n+x}{\beta}}.$ (27)

But in fact, the opposite inequality is true for all $\alpha\leqslant
x\leqslant\beta$. Taking into account the fact that
$\alpha^{\beta}>\beta^{\alpha}$ for $\beta>\alpha\geqslant 3$, one can easily
verify that

$\alpha^{n}x>\beta^{\frac{\alpha n+x}{\beta}}$ (28)

for $x=\alpha$ and $x=\beta$. As $\beta^{\frac{\alpha n+x}{\beta}}$ is a
convex downward function of $x$, one can deduce that it lies below the linear
function $\alpha^{n}x$ for all $\alpha\leqslant x\leqslant\beta$. Thus, we
obtain a contradiction with (26) and the lemma is proved. ∎

###  3.3 Elimination of small elements greater than $1$

######  Lemma 3.7.

Let $A,C$ be arbitrary (possibly empty) sequences of positive integers. Let
$B$ be a symmetric sequence. Consider two arbitrary integers $p\geqslant
m\geqslant 1$. Then

$\langle A,1,B,p+m-1,C\rangle\leqslant\langle A,m,B,p,C\rangle.$ (29)

###### Proof.

Denote $q=\frac{p+m}{2}$. Consider the function

$f(x)=\langle A,q+x,B,q-x,C\rangle,$

where $x$ run through the set of integers if $p+m$ is even and runs through
the set of half-integers otherwise. One can see that $f(x)$ is a quadratic
polynomial with negative leading coefficient. The maximum of $f(x)$ is
attained at the point111see [6], Lemma 5.4 for computational details

$x_{m}=\frac{[0;\overleftarrow{B}]-[0;B]+[0;C]-[0;\overleftarrow{A}]}{2}.$

As $B$ is a symmetric sequence, $[0;\overleftarrow{B}]=[0;B]$ and therefore
$|x_{m}|\leqslant\frac{1}{2}$. Hence, as $p\geqslant m\geqslant 1$, one can
easily see that

$\langle A,1,B,p+m-1,C\rangle=f\biggl{(}-\frac{p+m-2}{2}\biggr{)}\leqslant
f\biggl{(}-\frac{p-m}{2}\biggr{)}=\langle A,m,B,p,C\rangle.$

∎

######  Lemma 3.8.

Suppose that $?^{\prime}(x)=0$. Then there exists an irrational number
$y=[0;b_{1},b_{2},\ldots,b_{t},\ldots]$ such that:

  1. 1.

$?^{\prime}(y)=0$.

  2. 2.

For all $i\in\mathbb{N}$ either $b_{i}=1$ or $b_{i}\geqslant 12$.

  3. 3.

For all $i\in\mathbb{N}$ one has
$\varphi^{(1)}_{y}(i)\leqslant\varphi^{(1)}_{x}(i)$.

###### Proof.

First, let us eliminate all elements equal to $2$ from the sequence
$a_{1},a_{2},\ldots$. Denote by $s_{1}$ the smallest index such that
$a_{s_{1}}=2$. Denote by $t_{1}$ the smallest index greater than $s_{1}$ such
that $a_{t_{1}}>1$. Now the procedure is repeated recursively:

$s_{i}=\min\\{n:n>t_{i-1},a_{n}=2\\},\quad t_{i}=\min\\{n:n>s_{i},a_{n}>1\\}$

Thus, we obtain the two (possibly infinite) growing sequences
$s_{1}<t_{1}<s_{2}<t_{2}<\ldots$. Note that if $s_{j}<n<t_{j}$ for some $j$,
then $a_{n}=1$. Define the irrational number
$x^{\prime}=[0;a^{\prime}_{1},a^{\prime}_{2},\ldots,a^{\prime}_{t},\ldots]$ as
follows:

$a^{\prime}_{n}=\begin{cases}a_{n}-1=1\quad\text{if $n=s_{i}$ for some
$i\in\mathbb{N}$}\\\ a_{n}+1\quad\text{if $n=t_{i}$ for some
$i\in\mathbb{N}$}\\\ a_{n}\quad\text{otherwise}\end{cases}$

One can easily see from the definition of $x^{\prime}$ that
$\varphi^{(1)}_{x^{\prime}}(t)\leqslant\varphi^{(1)}_{x}(t)$ for all
$t\geqslant 1$. Let us now show that $?^{\prime}(x^{\prime})=0$.

It follows from Lemma 3.7 that $\langle
a^{\prime}_{1},a^{\prime}_{2},\ldots,a^{\prime}_{t}\rangle:=\langle
A^{\prime}_{t}\rangle\leqslant\langle A_{t}\rangle$ for all $t\geqslant 1$. On
the other hand $|{S_{x^{\prime}}(t)}-{S_{x}(t)}|\leqslant 1$. Hence, as

$\lim\limits_{t\to\infty}\frac{\langle A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}=0$

by Lemma 3.3, we obtain that

$\lim\limits_{t\to\infty}\frac{\langle
A^{\prime}_{t}\rangle}{\sqrt{2}^{S_{x^{\prime}}(t)}}=0$

and therefore $?^{\prime}(x^{\prime})=0$. Using the same argument, one can
eliminate the elements equal to $3,4,\ldots,12$ from the sequence
$a^{\prime}_{1},a^{\prime}_{2},\ldots$. The lemma is proved. ∎

######  Lemma 3.9.

Suppose that $?^{\prime}(x)=0$ and all partial quotients of $x$ are either
equal to $1$ or greater than $11$. There exists $T\in\mathbb{N}$ such that for
all $t>T$ the inequality $\varphi^{(1)}_{x}(t)>3w(A_{t})$ holds.

###### Proof.

Lemma 3.3 implies that there exists $T$ such that $\forall t>T$ one has

$\frac{\langle A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}<\frac{1}{2}.$

Hence, as $\frac{x}{4}\geqslant 3$ for $x\geqslant 12$ and
$\sqrt{2}^{\kappa_{1}}=\Phi$, one has by Lemma 3.4

$\frac{1}{2}>\frac{\langle
A_{t}\rangle}{\sqrt{2}^{S_{x}(t)}}\geqslant\frac{1}{2}\frac{\Phi^{t}3^{w(A_{t})}}{\sqrt{2}^{\kappa_{1}t+\varphi^{(1)}_{x}(t)}}=\frac{1}{2}\frac{3^{w(A_{t})}}{\sqrt{2}^{\varphi^{(1)}_{x}(t)}}.$
(30)

Statement of the lemma immediately follows from (30). ∎

##  4 Blocks structure

###  4.1 Parameters introduction

Let $x=[0;a_{1},a_{2},\ldots,a_{t},\ldots]$ be an irrational number such that
$?^{\prime}(x)=0$. By Lemma 3.8, without loss of generality one can say that
either $a_{i}=1$ or $a_{i}\geqslant 6$ for all $i\in\mathbb{N}$. Throughout
the remaining part of the paper we consider $\varepsilon$ as a fixed positive
real number from the statements of Theorems 1 and 2. Let
$\lambda=\lambda(\varepsilon)$ be an arbitrary rational number such that
$1>\lambda>1-\varepsilon^{6}.$ Define the following integer constants:

$M=\frac{10\log{\varepsilon}}{\log{\lambda}},\ \
P=\biggl{[}\frac{\log{6}}{\log{(1+\varepsilon^{2})}}\biggr{]}+1,\ \
N=2M(P+2).$ (31)

Now we select an integer parameter $t_{0}$ large enough such that
$(1-\lambda)\lambda^{N}t_{0}>\frac{t_{0}}{\log{t_{0}}}$ and

$\lambda^{N}t_{0}\in\mathbb{Z}.$ (32)

Denote $t_{i}=\lambda^{i}t_{0}$, where $1\leqslant i\leqslant N$. We define
$B_{i}=(a_{t_{i}+1},a_{t_{i}+2},\ldots,a_{t_{i-1}})$ – the $i$-th block, where
$1\leqslant i\leqslant N$. Denote $t_{N+1}=0$ and
$B_{N+1}=(a_{t_{N+1}+1},\ldots,a_{t_{N}})$. Thus, we have

$[0;A_{t_{0}}]=[0;a_{1},\ldots,a_{t_{0}}]=[0;B_{N+1},B_{N},\ldots,B_{1}].$

One can easily see that $S_{x}(t_{0})=\sum\limits_{i=1}^{N+1}S(B_{i})$ and
$\varphi^{(1)}_{x}(t_{0})=\sum\limits_{i=1}^{N+1}\varphi^{(1)}_{x}(B_{i})$.

For each block $B_{i}$ denote its greatest element by $M_{i}$ and the index of
such element by $m_{i}$ (if the greatest element is not unique, we take the
rightmost one). Thus, $a_{m_{i}}=M_{i}$. Denote
$c_{k}=\frac{M_{k}}{\sqrt{t_{k-1}\log{t_{0}}}}$. Let us also consider for each
$1\leqslant i\leqslant N+1$ the short block
$B^{\prime}_{i}=(a_{t_{i}+1},\ldots,a_{m_{i}-1})$. Note that

$\varphi^{(1)}_{x}(m_{i})=\varphi^{(1)}(B_{N+1})+\varphi^{(1)}(B_{N})+\ldots+\varphi^{(1)}(B_{i+1})+\varphi^{(1)}(B^{\prime}_{i})+(M_{i}-\kappa_{1}).$
(33)

For each $1\leqslant i\leqslant N+1$ define the real numbers $f_{i}$ and
$f^{\prime}_{i}$ from the following identities:

$\langle
B_{i}\rangle=\sqrt{2}^{S(B_{i})+f_{i}\sqrt{t_{i-1}\log{t_{0}}}},\quad\langle
B^{\prime}_{i}\rangle=\sqrt{2}^{S(B^{\prime}_{i})+f^{\prime}_{i}\sqrt{t_{i-1}\log{t_{0}}}}.$
(34)

###  4.2 Lower estimate of $\varphi^{(1)}(B_{k})$

######  Lemma 4.1.

Suppose that $?^{\prime}(x)=0$. Then for all $1\leqslant i\leqslant N$ one has

$f^{\prime}_{i}\sqrt{t_{i-1}\log{t_{0}}}+\sum\limits_{k=i+1}^{N+1}f_{k}\sqrt{t_{k-1}\log{t_{0}}}<0.$
(35)

###### Proof.

As $?^{\prime}(x)=0$, by Lemma 3.3, without loss of generality one can say
that for all $1\leqslant i\leqslant N$ the inequality

$\frac{\langle
B_{N+1},B_{N},\ldots,B_{i+1},B^{\prime}_{i}\rangle}{\sqrt{2}^{S(B_{N+1})+S(B_{N})+\ldots+S(B_{i+1})+S(B^{\prime}_{i})}}<1$

is satisfied. Using (4) we obtain

$\frac{\langle B_{N+1}\rangle\langle B_{N}\rangle\ldots\langle
B_{i+1}\rangle\langle
B^{\prime}_{i}\rangle}{\sqrt{2}^{S(B_{N+1})+S(B_{N})+\ldots+S(B_{i+1})+S(B^{\prime}_{i})}}<1.$
(36)

Substituting (34) to (36) and taking logarithm of both parts, we get the
statement of the lemma. ∎

######  Lemma 4.2.

Suppose that $?^{\prime}(x)=0$. If the inequality

$|\varphi^{(1)}(B_{k})|\geqslant\kappa_{1}(t_{k-1}-t_{k})\varepsilon^{5},$
(37)

holds for some $1\leqslant k\leqslant N+1$, then one has:

$\max_{t_{N}\leqslant u\leqslant t_{0}}\varphi^{(1)}_{x}(u)\geqslant
t_{0}^{0.9}.$ (38)

###### Proof.

Suppose that

$\varphi^{(1)}(B_{k})\geqslant\kappa_{1}(t_{k-1}-t_{k})\varepsilon^{5}.$

As,

$\varphi^{(1)}_{x}(t_{k-1})=\varphi^{(1)}_{x}(t_{k})+\varphi^{(1)}(B_{k}),$

using the fact that $\varphi^{(1)}_{x}(t_{k})>0$ by Lemma 3.5, we have

$\max_{t_{N}\leqslant u\leqslant
t_{0}}\varphi^{(1)}_{x}(u)\geqslant\varphi^{(1)}_{x}(t_{k-1})>\varphi^{(1)}(B_{k})\geqslant\kappa_{1}(t_{k-1}-t_{k})\varepsilon^{5}=\kappa_{1}\lambda^{k-1}(1-\lambda)t_{0}>t_{0}^{0.9}.$

On the other hand, if

$\varphi^{(1)}(B_{k})\leqslant-\kappa_{1}(t_{k-1}-t_{k})\varepsilon^{5},$

we again use the fact that $\varphi^{(1)}_{x}(t_{k-1})>0$ and obtain

$\max_{t_{N}\leqslant u\leqslant
t_{0}}\varphi^{(1)}_{x}(u)\geqslant\varphi^{(1)}_{x}(t_{k})=\varphi^{(1)}_{x}(t_{k-1})-\varphi^{(1)}(B_{k})\geqslant\kappa_{1}(t_{k-1}-t_{k})\varepsilon^{5}>t_{0}^{0.9}.$

∎

One can easily deduce from Lemma 4.2 that if the inequality (37) holds, then
the inequalities (14) and (15) are also satisfied. Therefore, throughout the
remaining part of the paper we will assume that

$S(B_{k})=\kappa_{1}(t_{k-1}-t_{k})(1+o(\varepsilon^{4}))$ (39)

for all $1\leqslant k\leqslant N+1$.

######  Lemma 4.3.

Suppose that $?^{\prime}(x)=0$. Then for all $1\leqslant k\leqslant N+1$ one
has

$\varphi^{(1)}(B_{k})\geqslant\frac{(\kappa_{1}-1)(t_{k-1}-t_{k})\sqrt{\log
t_{0}}(1+o(\varepsilon^{4}))}{c_{k}\sqrt{t_{k-1}}\log{2}}-f_{k}\sqrt{t_{k-1}\log{t_{0}}}.$
(40)

###### Proof.

We recall that $w(B_{k})$ is the number of elements of the block $B_{k}$ which
are greater than $1$ and $d_{B_{k}}(1),\ldots,d_{B_{k}}(w(B_{k}))$ is the
sequence of such elements. It follows from Lemma 4.2 and Lemma 3.9 that

$\sum\limits_{i=1}^{w(B_{k})}d_{B_{k}}(i)=(\kappa_{1}-1)(t_{k-1}-t_{k})(1+o(\varepsilon^{4})).$

Therefore, as $d_{B_{k}}(i)\geqslant 12$ for all $i$, one has
$\frac{d_{B_{k}}(i)}{4}\geqslant 3$. Now we can obtain a lower estimate of

$\prod\limits_{i=1}^{w(B_{k})}\biggl{(}\frac{d_{B_{k}}(i)}{4}\biggr{)}$

applying Lemma 3.6 for
$s=\frac{1}{4}(\kappa_{1}-1)(t_{k-1}-t_{k})(1+o(\varepsilon^{4}))$,
$\alpha=3$, and $\beta=c_{k}\sqrt{t_{k-1}\log{t_{0}}}$. We have

$\prod\limits_{i=1}^{w(B_{k})}\biggl{(}\frac{d_{B_{k}}(i)}{4}\biggr{)}\geqslant\biggl{(}\frac{c_{k}\sqrt{t_{k-1}\log{t_{0}}}}{4}\biggr{)}^{\frac{(\kappa_{1}-1)(t_{k-1}-t_{k})(1+o(\varepsilon^{4}))}{c_{k}\sqrt{t_{k-1}\log{t_{0}}}}}.$
(41)

Substituting the estimate (41) to (22) we obtain

$\langle
B_{k}\rangle\geqslant\Phi^{t_{k-1}-t_{k}}\biggl{(}\frac{c_{k}\sqrt{t_{k-1}\log{t_{0}}}}{4}\biggr{)}^{\frac{(\kappa_{1}-1)(t_{k-1}-t_{k})(1+o(\varepsilon^{4}))}{c_{k}\sqrt{t_{k-1}\log{t_{0}}}}}.$
(42)

Taking into account that $\sqrt{2}^{\kappa_{1}}=\Phi$, from (34), (42) and
Lemma 3.3 we get:

$\biggl{(}\frac{9c_{i}\sqrt{t_{k-1}\log{t_{0}}}}{16}\biggr{)}^{\frac{(\kappa_{1}-1)(t_{k-1}-t_{k})(1+o(\varepsilon^{4}))}{c_{i}\sqrt{t_{k-1}\log{t_{0}}}}}\leqslant\sqrt{2}^{\varphi^{(1)}(B_{k})+f_{i}\sqrt{t_{k-1}\log{t_{0}}}}.$
(43)

Taking logarithms of both parts of (43), after some transformations we obtain
the inequality

$\frac{(\kappa_{1}-1)(t_{k-1}-t_{k})(1+o(\varepsilon^{4}))}{c_{i}\sqrt{t_{k-1}\log{t_{0}}}}\frac{\log
t_{0}}{2}\leqslant(\varphi^{(1)}(B_{k})+f_{k}\sqrt{t_{k-1}\log{t_{0}}})\log\sqrt{2}$
(44)

which is equivalent to (40). Lemma is proved. ∎

###  4.3 Main estimate

For $k\leqslant N$ one can write (40) as

$\varphi^{(1)}(B_{k})\geqslant\biggl{(}\frac{(\kappa_{1}-1)(1-\lambda)(1+o(\varepsilon^{4}))}{c_{k}\log{2}}-f_{k}\biggr{)}\sqrt{t_{k-1}\log{t_{0}}}.$
(45)

For $k=N+1$ we have

$\varphi^{(1)}(B_{N+1})\geqslant\biggl{(}\frac{(\kappa_{1}-1)(1+o(\varepsilon^{4}))}{c_{N+1}\log{2}}-f_{N+1}\biggr{)}\sqrt{t_{N}\log{t_{0}}}.$
(46)

Let the greatest element of the short block $B^{\prime}_{k}$ be equal to
$c^{\prime}_{k}\sqrt{t_{k-1}\log{t}}$. Using the argument from Lemma 4.3, one
can deduce the following lower estimate of $\varphi^{(1)}(B^{\prime}_{k})$:

$\varphi^{(1)}(B^{\prime}_{k})\geqslant\frac{S(B^{\prime}_{k})\sqrt{\log
t_{0}}(1+o(\varepsilon^{4}))}{c^{\prime}_{k}\sqrt{t_{k-1}}\log{2}}-f^{\prime}_{k}\sqrt{t_{k-1}\log{t_{0}}}.$
(47)

######  Lemma 4.4.

Suppose that $?^{\prime}(x)=0$. Then for all $1\leqslant i\leqslant N$ one has

$\varphi^{(1)}_{x}(m_{i})\geqslant\Biggl{(}\frac{(\kappa_{1}-1)(1-\lambda)}{\log{2}}\biggl{(}\sum\limits_{k=i+1}^{N}\frac{(\sqrt{\lambda})^{k-i}}{c_{k}}\biggr{)}+c_{i}\Biggr{)}\sqrt{t_{i-1}\log{t_{0}}}\bigl{(}1+o(\varepsilon^{4})\bigr{)}.$
(48)

###### Proof.

Substituting the estimates (40) for $k=N,N-1,\ldots,i+1$, (46), and (47) to
(33) and taking into account the inequality (35) we obtain:

$\begin{split}\varphi^{(1)}_{x}(m_{i})\geqslant(1+o(\varepsilon^{4}))\biggl{(}\frac{\kappa_{1}-1}{\log{2}c_{N+1}}\sqrt{t_{N}\log{t_{0}}}+\frac{S(B_{i})\sqrt{\log
t_{0}}}{\log{2}c^{\prime}_{i}\sqrt{t_{i-1}}}+\\\
\frac{(\kappa_{1}-1)\sqrt{\log
t_{0}}}{\log{2}}\sum\limits_{k=i+1}^{N}\frac{t_{k-1}-t_{k}}{c_{k}\sqrt{t_{k-1}}}\biggr{)}+(c_{i}\sqrt{t_{i-1}\log{t_{0}}}-\kappa_{1}).\end{split}$
(49)

Using the trivial estimates

$\frac{S(B_{i})\sqrt{\log
t_{0}}}{\log{2}c^{\prime}_{i}\sqrt{t_{i-1}}}>0,\quad\frac{\kappa_{1}-1}{\log{2}c_{N+1}}\sqrt{t_{N}\log{t_{0}}}>0$

we obtain

$\begin{split}\varphi^{(1)}_{x}(m_{i})\geqslant(1+o(\varepsilon^{4}))\biggl{(}\frac{(\kappa_{1}-1)\sqrt{\log
t_{0}}}{\log{2}}\sum\limits_{k=i+1}^{N}\frac{t_{k-1}-t_{k}}{c_{k}\sqrt{t_{k-1}}}\biggr{)}+(c_{i}\sqrt{t_{i-1}\log{t_{0}}}-\kappa_{1}).\end{split}$
(50)

Taking into account the fact that $t_{k}=\lambda^{k}t_{0}$, we immediately
obtain (48). The lemma is proved. ∎

Inequality (48) is the key tool that we will use in proofs of Theorem 1 and
the first statement of Theorem 2. Let us simplify it using a new notation.
Denote

$\alpha=\frac{(\kappa_{1}-1)}{\log{2}},\quad\eta=\frac{1}{\alpha}.$ (51)

Then, one can rewrite (48) as follows:

$\begin{split}\varphi^{(1)}_{x}(m_{i})\geqslant\biggl{(}(1-\lambda)\sum\limits_{k=i+1}^{N}\frac{(\sqrt{\lambda})^{k-i}}{c_{k}}+\eta
c_{i}\biggr{)}\alpha\sqrt{t_{i-1}\log{t_{0}}}.\bigl{(}1+o(\varepsilon^{4})\bigr{)}.\end{split}$
(52)

##  5 Key lemmas

The inequality (52) reduces the estimation of $\max\limits_{1\leqslant
i\leqslant N}\varphi^{(1)}_{x}(m_{i})$ to the problem of finding maximum of
the following quantity:

$(1-\lambda)\sum\limits_{k=i+1}^{N}\frac{(\sqrt{\lambda})^{k-i}}{c_{k}}+\eta
c_{i},\quad i=1,2,\ldots,N.$

Note that this problem does not deal with the Minkowski function, continued
fractions etc. It is purely combinatorial. The following lemma allows us to
estimate the desired maximum.

######  Lemma 5.1.

Let $\eta,c_{1},c_{2},\ldots,c_{N}$ be arbitrary positive real numbers. Define
the real numbers $\varphi_{i}$ as follows:

$\varphi_{i}=(1-\lambda)\sum\limits_{k=i+1}^{N}\frac{\sqrt{\lambda}^{k-i}}{c_{k}}+\eta
c_{i}.$ (53)

Then the following inequality holds:

$\max\limits_{1\leqslant i\leqslant
N}\varphi_{i}\geqslant\sqrt{8\eta}(1+o(\varepsilon)).$ (54)

###### Proof.

The proof of the lemma will be splitted into several steps. We also recall
that the constants $M$, $N$, and $P$ used in our argument are defined in (31).

######  Lemma 5.2.

Suppose that the inequality (54) is not satisfied. Then there exists a natural
number $i_{1}\leqslant M$ such that
$c_{i_{1}}\geqslant\frac{1}{2\sqrt{\eta}}$. Moreover, for all $i\leqslant N$
one has

$c_{i}<\frac{3}{\sqrt{\eta}}.$ (55)

###### Proof.

Suppose the contrary. Let us estimate $\varphi_{1}$ from below:

$\begin{split}\varphi_{1}\geqslant(1-\lambda)\sum\limits_{k=1}^{N}\frac{\sqrt{\lambda}^{k}}{c_{k+1}}+\eta
c_{1}\geqslant(1-\lambda)\sum\limits_{k=1}^{M}\frac{\sqrt{\lambda}^{k}}{c_{k+1}}\geqslant
2\sqrt{\eta}(1-\lambda)\sum\limits_{k=1}^{M}\sqrt{\lambda}^{k}=\\\
=2\sqrt{\eta}(1-\lambda)\sqrt{\lambda}\frac{1-\sqrt{\lambda}^{M}}{1-\sqrt{\lambda}}=2\sqrt{\eta}(1+\sqrt{\lambda})(1+o(\varepsilon^{4}))>\sqrt{8\eta}(1+o(\varepsilon^{4})).\end{split}$
(56)

We obtain a contradiction with (54). The estimate (55) comes from the trivial
inequality $\varphi_{i}>\eta c_{i}$. Lemma is proved. ∎

######  Lemma 5.3.

Suppose that the inequality (54) is not satisfied. Then for all $i_{m}<N-M,\
m\geqslant 1$ there exists a number $i_{m}<i_{m+1}<i_{m}+M$ such that
$c_{i_{m+1}}>(1+\varepsilon^{2})c_{i_{m}}.$

###### Proof.

Suppose the contrary. Let the inequality
$c_{i_{m}+j}<c_{i_{m}}(1+\varepsilon^{2})$ be satisfied for all $1\leqslant
j\leqslant M$. Then, using the argument from Lemma 5.2 we obtain

$\begin{split}\varphi_{i_{m}}\geqslant(1-\lambda)\sum\limits_{j=1}^{M}\frac{\sqrt{\lambda}^{j}}{c_{i_{m}+j}}+\eta
c_{i_{m}}\geqslant\frac{1-\lambda}{1+\varepsilon^{2}}\sum\limits_{j=1}^{M}\frac{\sqrt{\lambda}^{j}}{c_{i_{m}}}+\eta
c_{i_{m}}\geqslant\\\ \biggl{(}\frac{2}{(1+\varepsilon^{2})c_{i_{m}}}+\eta
c_{i_{m}}\biggr{)}(1+o(\varepsilon^{4}))\geqslant\sqrt{\frac{8\eta}{1+\varepsilon^{2}}}\bigl{(}1+o(\varepsilon^{4})\bigr{)}.\end{split}$
(57)

In the last ’’$\geqslant$’’ of (57) we use the Cauchy-Schwarz inequality. We
come to a contradiction with (54). Lemma is proved. ∎

Now we are ready to prove Lemma 5.1. Suppose that the inequality (54) is not
satisfied. By Lemma 5.2 there exists $1\leqslant i_{1}<M$ such that
$c_{i}\geqslant\frac{1}{2\sqrt{\eta}}$. Applying Lemma 5.3
$P=\biggl{[}\frac{\log{6}}{\log{(1+\varepsilon^{2})}}\biggr{]}+1$ times, we
obtain the number $c_{i_{P+1}}$ such that
$c_{i_{P+1}}>\frac{1}{2\sqrt{\eta}}(1+\varepsilon^{2})^{P}\eta>\frac{3}{\sqrt{\eta}}$.
We come to a contradiction with (55). Lemma is proved. ∎

One can rewrite (52) as

$\begin{split}\varphi^{(1)}_{x}(m_{i})\geqslant\sqrt{\lambda}^{i}\biggl{(}(1-\lambda)\sum\limits_{k=i+1}^{N}\frac{(\sqrt{\lambda})^{k-i}}{c_{k}}+\eta
c_{i}\biggr{)}\alpha\sqrt{t_{0}\log{t_{0}}}(1+o(\varepsilon^{4})).\end{split}$
(58)

Note that $\sqrt{\lambda}=1+o(\varepsilon^{5})$. As Theorem 2 requires us to
show that $\max\limits_{1\leqslant i\leqslant N}\varphi^{(1)}_{x}(m_{i})$ is
greater than $\sqrt{t_{0}\log{t_{0}}}$ multiplied by some constant factor, we
need to estimate the maximum of the following quantity

$\lambda^{i}\biggl{(}(1-\lambda)\sum\limits_{k=i+1}^{N}\frac{(\sqrt{\lambda})^{k-i}}{c_{k}}+\eta
c_{i}\biggr{)},\quad i=1,2,\ldots,N.$

This estimate is provided by the following lemma.

######  Lemma 5.4.

Let $C=(c_{1},c_{2},\ldots,c_{N})$ be an arbitrary sequence of non-negative
real numbers. Let $\eta$ be an arbitrary positive real number. Define the
numbers $\varphi_{i}$ using (53). Define the numbers $\varphi^{\prime}_{i}$ as
follows:

$\varphi^{\prime}_{i}(C)=\sqrt{\lambda}^{i}\varphi_{i}=(1-\lambda)\sum\limits_{k=i+1}^{N}\frac{\sqrt{\lambda}^{k}}{c_{k}}+\sqrt{\lambda}^{i}\eta
c_{i}.$ (59)

Then one has:

$\max\limits_{1\leqslant i\leqslant
N}\varphi^{\prime}_{i}(C)\geqslant\sqrt{2\eta}(1+o(\varepsilon)).$ (60)

###### Proof.

The proof of the lemma will be also splitted into several steps. First, using
the substitution $d_{j}=\sqrt{\lambda}^{j}c_{j}$, we write (59) as

$\tilde{\varphi}^{\prime}_{i}(D):=\varphi^{\prime}_{i}(C)=(1-\lambda)\sum\limits_{k=i+1}^{N}\frac{\lambda^{k}}{d_{k}}+\eta
d_{i}.$ (61)

Here $D=(d_{1},d_{2},\ldots,d_{N})$. Denote
$\tilde{\varphi}^{\prime}_{max}(D)=\max\limits_{1\leqslant k\leqslant
N}\tilde{\varphi}^{\prime}_{k}(D)$. In order to prove the lemma it is enough
for us to show that

$\min\limits_{D\in\mathbb{R}^{N}_{+}}\tilde{\varphi}_{max}(D)=\sqrt{2\eta}(1+o(\varepsilon)).$
(62)

Denote the minimum of (62) by $y_{min}$.

######  Lemma 5.5.

Suppose that $\tilde{\varphi}^{\prime}_{max}(D)=y_{min}$ for some
$D\in\mathbb{R}^{N}_{+}$. Then for all $1\leqslant k\leqslant N$ one has
$\tilde{\varphi}^{\prime}_{k}(D)=y_{min}$.

###### Proof.

Suppose that $\tilde{\varphi}^{\prime}_{max}(D)=y_{min}$, but for some $i$ one
has $\tilde{\varphi}^{\prime}_{i}(D)<y_{min}$. We call the index $n$
minimizing if $\varphi^{\prime}_{n}(D)=y_{min}$. Let $k$ be the largest non-
minimizing index. Without loss of generality one can say that all indices less
than $k$ are not minimizing too. Indeed, consider the sequence

$D^{\prime}=(d_{1},\ldots,d_{k-1},d_{k}+\delta,d_{k+1},\ldots,d_{N}),$

where $\delta>0$ is some small parameter. One can easily see that
$\tilde{\varphi}^{\prime}_{i}(D^{\prime})<\tilde{\varphi}^{\prime}_{i}(D)$ for
$i<k$,
$\tilde{\varphi}^{\prime}_{i}(D^{\prime})>\tilde{\varphi}^{\prime}_{i}(D)$ for
i$=k$ and
$\tilde{\varphi}^{\prime}_{i}(D^{\prime})=\tilde{\varphi}^{\prime}_{i}(D)$ for
$i>k$. As $k$ is a non-minimizing index, there exists $\delta>0$ such that
$\tilde{\varphi}^{\prime}_{i}(D^{\prime})<y_{min}$ for all $i\leqslant k$.

Thus, $k+1$ is the smallest minimizing index. As all indices less than $k+1$
are not minimizing, there exists $\delta>0$ such that for the sequence

$D^{\prime\prime}=(d_{1},\ldots,d_{k},d_{k+1}-\delta,d_{k+2},\ldots,d_{N})$

one has
$y_{min}>\tilde{\varphi}^{\prime}_{i}(D^{\prime\prime})>\tilde{\varphi}^{\prime}_{i}(D)$
for $i\leqslant k$ and
$\tilde{\varphi}^{\prime}_{i}(D^{\prime\prime})<\tilde{\varphi}^{\prime}_{i}(D)=y_{min}$
for $i=k+1$. Thus, we obtained the sequence whose smallest minimizing index is
at least $k+2$. Repeating this argument we obtain the sequence
$D^{(N)}=(d^{(N)}_{1},d^{(N)}_{2},\ldots,d^{(N)}_{N})$ with the smallest
minimizing index equal to $N$. One can easily see that for the sequence

$D^{\prime(N)}=(d^{(N)}_{1},d^{(N)}_{2},\ldots,d^{(N)}_{N}-\delta)$

for $\delta>0$ small enough one has
$\tilde{\varphi}^{\prime}_{max}(D^{\prime(N)})<\tilde{\varphi}^{\prime}_{max}(D^{(N)})=y_{min}$
and we obtain a contradiction with the definition of $y_{min}$. Lemma is
proved. ∎

######  Lemma 5.6.

There exists a unique sequence $D=(d_{1},d_{2},\ldots,d_{N})$ such that
$\tilde{\varphi}^{\prime}_{max}(D)=y_{min}$. The elements of this sequence
satisfy the recurrent equation

$d_{k+1}=\frac{d_{k}+\sqrt{d_{k}^{2}+\frac{4(1-\lambda)\lambda^{k+1}}{\eta}}}{2}$
(63)

with the initial condition $d_{1}=0$.

###### Proof.

Lemma 5.5 implies that $\tilde{\varphi}^{\prime}_{N}(D)=y_{min}$. This fact
yields a linear equation $\eta d_{N}=y_{min}$ from which $d_{N}$ is uniquely
defined. Then we substitute $d_{N}$ to the equation
$\tilde{\varphi}^{\prime}_{N-1}(D)=y_{min}$ and find $d_{N-1}$ etc. Therefore
the sequence $D$ is uniquely defined.

Suppose that $d_{1}>0$. But if we decrease $d_{1}$, we would also decrease
$\tilde{\varphi}^{\prime}_{1}(D)$, but $\tilde{\varphi}^{\prime}_{max}(D)$
would be still equal to $y_{min}$. We obtain a contradiction with the
uniqueness of $D$. Finally, from the equation
$\tilde{\varphi}^{\prime}_{k+1}(D)=\tilde{\varphi}^{\prime}_{k}(D)$ one can
derive

$\eta(d_{k+1}-d_{k})=(1-\lambda)\frac{\lambda^{k+1}}{d_{k+1}}.$ (64)

Considering (64) as the quadratic equation on $d_{k+1}$ and choosing the
positive root, we obtain (63). Lemma is proved. ∎

Thus, $y_{min}=\frac{d_{N}}{\eta}$ where $d_{N}$ can be evaluated from the
recurrent equations (63). Multiplying both sides of (63) by $\sqrt{\eta}$, we
obtain

$\sqrt{\eta}d_{k+1}=\frac{\sqrt{\eta}d_{k}+\sqrt{(\sqrt{\eta}d_{k})^{2}+4(1-\lambda)\lambda^{k+1}}}{2}.$
(65)

Put $e_{k}=\sqrt{\eta}d_{k}$. Using the introduced notation, one can write
(65) as follows:

$e_{k+1}=\frac{e_{k}+\sqrt{e_{k}^{2}+4(1-\lambda)\lambda^{k+1}}}{2}.$ (66)

Denote222Equation (66) is equivalent to
$\frac{e_{k+1}-e_{k}}{\delta_{k+1}}=\frac{1}{e_{k+1}}$. Thus, (66) might be
considered as numerical integration of the differential equation
$y^{\prime}=\frac{1}{y}$ on the non-uniform grid $X_{i}$. I am thankful to I.
Mitrofanov who drew my attention to this fact.
$\delta_{k}=(1-\lambda)\lambda^{k}$ и $X_{n}=\sum\limits_{k=1}^{n}\delta_{n}$.

######  Lemma 5.7.

For $n\geqslant 1$ one has $e_{n}<\sqrt{2X_{n}}$ and
$e_{n+1}-e_{n}>\sqrt{2X_{n+2}}-\sqrt{2X_{n+1}}$.

###### Proof.

The first statement is proved by induction. For $n=1$ one can easily verify
the inequality. As $\sqrt{x}$ is a convex function, one can easily see that

$e_{n+1}=\frac{e_{n}+\sqrt{e_{n}^{2}+4\delta_{n+1}}}{2}<\frac{\sqrt{2X_{n}}+\sqrt{2X_{n}+4\delta_{n+1}}}{2}<\sqrt{2X_{n}+2\delta_{n+1}}=\sqrt{2X_{n+1}}.$
(67)

Now we prove the second statement. Note that

$e_{n+1}-e_{n}=\frac{\sqrt{e_{n}^{2}+4\delta_{n+1}}-e_{n}}{2}=\frac{2\delta_{n+1}}{e_{n}+\sqrt{e_{n}^{2}+4\delta_{n+1}}}=\frac{\delta_{n+1}}{e_{n+1}}.$
(68)

On the other hand, as we already showed,
$\sqrt{2X_{n+2}}>\sqrt{2X_{n+1}}>e_{n+1}$ and therefore

$\sqrt{2X_{n+2}}-\sqrt{2X_{n+1}}=\frac{2\delta_{n+2}}{\sqrt{2X_{n+2}}+\sqrt{2X_{n+1}}}<\frac{\delta_{n+1}}{e_{n+1}}=e_{n+1}-e_{n}.$
(69)

Lemma is proved. ∎

Now we are ready to prove Lemma 5.4. By Lemma 5.7, $\sqrt{2X_{n+1}}-e_{n}$
forms a decreasing sequence of positive real numbers. Hence

$0<\sqrt{2X_{N}}-e_{N-1}<\sqrt{2X_{2}}-e_{1}<2\sqrt{X_{2}}=\sqrt{2(1-\lambda)(\lambda+\lambda^{2})}<2\sqrt{1-\lambda}=o(\varepsilon^{2}).$

On the other hand,
$2X_{N}=2\sum\limits_{k=1}^{N}(1-\lambda)\lambda^{k}=2(1-\lambda^{N+1})=2+o(\varepsilon^{5})$.
Thus, $e_{N-1}=\sqrt{2}+o(\varepsilon^{2})$. It also follows from (66) that
$e_{N}=\sqrt{2}+o(\varepsilon^{2})$. Hence,

$y_{min}=\eta{d_{N}}=\sqrt{\eta}e_{N}=\sqrt{2\eta}+o(\varepsilon^{2}).$

Lemma is proved. ∎

##  6 Proof of Theorem 1 and the first statement of Theorem 2

Now we are ready to prove Theorem 1.

###### Proof.

From (52) and Lemma 5.1 one can deduce that for all $t_{0}$ large enough there
exist integer numbers $i$ and $m_{i}$ satisfying $1\leqslant i\leqslant N$ and
$\frac{t_{0}}{\log{t_{0}}}<m_{i}\leqslant t_{0}$ such that

$\varphi^{(1)}_{x}(m_{i})\geqslant\sqrt{8\eta}\alpha\sqrt{t_{i-1}\log{t_{0}}}(1+o(\varepsilon))=2\sqrt{2}\kappa_{4}\sqrt{t_{i-1}\log{t_{0}}}(1+o(\varepsilon)).$
(70)

As $t_{i}<m_{i}\leqslant t_{i-1}$ and
$\frac{t_{i}}{t_{i-1}}=\lambda=1+o(\varepsilon^{5})$, from (70) one has

$\varphi^{(1)}_{x}(m_{i})\geqslant
2\sqrt{2}\kappa_{4}\sqrt{m_{i}\log{m_{i}}}(1+o(\varepsilon)).$

Theorem is proved. ∎

Let us now prove the first statement of Theorem 2.

###### Proof.

It follows from (58) and Lemma 5.4 that for all $t_{0}$ large enough there
exist integer numbers $i$ and $m_{i}$ satisfying $1\leqslant i\leqslant N$ and
$\frac{t_{0}}{\log{t_{0}}}<m_{i}\leqslant t_{0}$ such that

$\sqrt{\lambda}^{i}\varphi^{(1)}_{x}(m_{i})\geqslant\sqrt{2\eta}\alpha\sqrt{t_{i}\log{t_{0}}}(1+o(\varepsilon))=\sqrt{2}\kappa_{4}\sqrt{t_{i}\log{t_{0}}}(1+o(\varepsilon)).$

In other words,

$\max_{u\leqslant
t_{0}}\varphi^{(1)}_{x}(u)\geqslant\varphi^{(1)}_{x}(m_{i})\geqslant\sqrt{2}\kappa_{4}\sqrt{t_{0}\log{t_{0}}}(1+o(\varepsilon)).$

The first statement of Theorem 2 is proved. ∎

##  7 Proof of the second statement of Theorem 2

###  7.1 Superblocks definition

###### Proof.

In this chapter we will construct an irrational number
$x=[0;a_{1},\ldots,a_{n},\ldots]$ that will satisfy the conditions of the
second statement of Theorem 2. The continued fraction of $x$ will have the
form

$x=[0;\mathfrak{B}^{(0)},\mathfrak{B}^{(1)},\mathfrak{B}^{(2)},\ldots,\mathfrak{B}^{(n)},\ldots],$
(71)

where the segments $\mathfrak{B}^{(i)}$ will be defined later. We will call
these segments superblocks. Recall that the sequence $d_{k}$ is defined from
the equations (63) with the initial condition $d_{1}=0$. It will be convenient
for us to modify the first element of this sequence. We set
$d_{1}=d_{2}=\sqrt{\frac{(1-\lambda)\lambda}{\eta}}=o(\varepsilon^{5})$.

Let us now define the sequence $T_{i}$ that plays the key role in our
construction. We choose an arbitrary integer $T_{1}$, satisfying
$\lambda^{N}T_{1}>\frac{T_{1}}{\log{T_{1}}}$. Then, if the number $T_{i-1}$ is
defined, we put $T_{i}=\bigl{[}\frac{T_{i-1}}{\lambda^{N}}\bigr{]}$. For each
$i\geqslant 1$ put $t^{(i)}_{0}=T_{i}$.

Now we describe the construction of the superblock $\mathfrak{B}^{(i)}$ for an
arbitrary positive integer $i$. For each $k$ such that $1\leqslant k\leqslant
N$ we select three natural numbers $m^{(i)}_{k},n^{(i)}_{k}$ и $t^{(i)}_{k}$
from the following conditions:

$\begin{split}d_{k}\sqrt{T_{i}\log{T_{i}}}\leqslant m^{(i)}_{k}\leqslant
d_{k}\sqrt{T_{i}\log{T_{i}}}(1+\varepsilon^{4})\\\
\frac{d_{k}}{\kappa_{1}-1}\sqrt{T_{i}\log{T_{i}}}\leqslant
n^{(i)}_{k}\leqslant\frac{d_{k}}{\kappa_{1}-1}\sqrt{T_{i}\log{T_{i}}}(1+\varepsilon^{4})\\\
\frac{\log{T_{i}}}{\log{2}}(1+\frac{\varepsilon}{8}-\varepsilon^{3})\leqslant
m^{(i)}_{k}+n^{(i)}_{k}-\kappa_{1}(n^{(i)}_{k}+1)\leqslant\frac{\log{T_{i}}}{\log{2}}(1+\frac{\varepsilon}{8}+\varepsilon^{3})\\\
t^{(i)}_{k}=\lambda^{k}t^{(i)}_{0}+\theta(n^{(i)}_{k}+1),\ \text{where}\
|\theta|\leqslant\frac{1}{2},\quad(n^{(i)}_{k}+1)\mid(t^{(i)}_{k-1}-t^{(i)}_{k}).\end{split}$
(72)

One can easily see that the numbers $m^{(i)}_{k},n^{(i)}_{k},t^{(i)}_{k}$,
satisfying conditions (72) always exists. Consider the block
$B^{(i)}_{k}=(a_{t^{(i)}_{k}+1},\ldots,a_{t^{(i)}_{k-1}})$ having the
following structure:

$B^{(i)}_{k}=(m^{(i)}_{k},\underbrace{1,1,\ldots,1}_{n^{(i)}_{k}\text{\
numbers}},m^{(i)}_{k},\underbrace{1,1,\ldots,1}_{n^{(i)}_{k}\text{\
numbers}},\ldots,m^{(i)}_{k},\underbrace{1,1,\ldots,1}_{n^{(i)}_{k}\text{\
numbers}}).$ (73)

We denote the sequence of $N$ blocks
$(B^{(i)}_{N},B^{(i)}_{N-1},\ldots,B^{(i)}_{1})$ by $\mathfrak{B}^{(i)}$. For
the initial superblock
$\mathfrak{B}^{(0)}=(a_{1},a_{2},\ldots,a_{t^{(1)}_{N}})$ we set all its
elements to be equal to $1$. Thus, the construction of the continued fraction
(71) is fully described. Note that the initial superblock $\mathfrak{B}^{(0)}$
has fixed length and therefore does not affect neither the value of
$?^{\prime}(x)$ nor the behavior of $\varphi^{(1)}_{x}(t)$ as $t$ grows.

###  7.2 $?^{\prime}(x)=0$

As we already mentioned it is enough to show that for
$x^{\prime}=[0;\mathfrak{B}^{(1)},\mathfrak{B}^{(2)},\ldots,\mathfrak{B}^{(n)},\ldots])$
one has $?^{\prime}(x^{\prime})=0$. Denote the elements of continued fraction
expansion of $x^{\prime}$ by $a^{\prime}_{1},a^{\prime}_{2},\ldots$. By Lemma
3.3 it is enough to show that the function

$f_{x^{\prime}}(t)=\frac{\langle
a^{\prime}_{1},a^{\prime}_{2},\ldots,a^{\prime}_{t}\rangle}{\sqrt{2}^{a^{\prime}_{1}+a^{\prime}_{2}+\ldots+a^{\prime}_{t}}}$
(74)

tends to $0$ as $t\to\infty$. One can easily see that the function
$f_{x^{\prime}}(t)$ grows if $a^{\prime}_{t}=1$ and decreases if
$a^{\prime}_{t}\geqslant 5$. Of course, all $m^{(i)}_{k}$ are greater than
$5$. Thus it is enough to consider only the case when $a^{\prime}_{t}=1$, but
$a^{\prime}_{t+1}\geqslant 5$. In this case the continuant $\langle
a^{\prime}_{1},a^{\prime}_{2},\ldots,a^{\prime}_{t}\rangle$ consists of the
sequences of the form
$(m^{(i)}_{l},\underbrace{1,1,\ldots,1}_{n^{(i)}_{l}\text{\ numbers}})$. From
the formula $\langle A,B\rangle\leqslant 2\langle A\rangle\langle B\rangle$
and Lemma 3.3 one can easily see that if we show that

$\frac{2\langle m^{(i)}_{l},\underbrace{1,1,\ldots,1}_{n^{(i)}_{l}\text{\
numbers}}\rangle}{\sqrt{2}^{m^{(i)}_{l}+n^{(i)}_{l}}}<\frac{1}{2},$ (75)

we will obtain the fact that $?^{\prime}(x)=0$. Note that $4\langle
m^{(i)}_{l},\underbrace{1,1,\ldots,1}_{n^{(i)}_{l}\text{\
numbers}}\rangle<8m^{(i)}_{l}\Phi^{n^{(i)}_{l}}$. From the definition (72) one
can easily see that
$m^{(i)}_{l}+n^{(i)}_{l}=\kappa_{1}n^{(i)}_{l}+\frac{\log{T_{i}}}{\log{2}}(1+\frac{\varepsilon}{8}+o(\varepsilon^{2}))$.
Thus, (75) is equivalent to the following:

$\frac{8m^{(i)}_{l}\Phi^{n^{(i)}_{l}}}{\sqrt{2}^{\kappa_{1}n^{(i)}_{l}+\frac{\log{T_{i}}}{\log{2}}(1+\frac{\varepsilon}{8}+o(\varepsilon^{2}))}}=\frac{8m^{(i)}_{l}}{\sqrt{2}^{\frac{\log{T_{i}}}{\log{2}}(1+\frac{\varepsilon}{8}+o(\varepsilon^{2}))}}<1.$
(76)

Taking logarithm of both sides of (76) and substituting $m^{(i)}_{l}$ from
(72) , we obtain that

$\frac{\log{T_{i}}}{2}\Bigl{(}1+o(\varepsilon^{2})\Bigr{)}<\frac{\log{T_{i}}}{\log{2}}\Bigl{(}1+\frac{\varepsilon}{8}+o(\varepsilon^{2})\Bigr{)}\log{\sqrt{2}}=\frac{\log{T_{i}}}{2}\Bigl{(}1+\frac{\varepsilon}{8}+o(\varepsilon^{2})\Bigr{)}.$
(77)

Therefore, $?^{\prime}(x)=0$.

####  7.2.1 The inequality (16) is satisfied

We will show that for $x$ that we built in Section 7.1 for all $t=T_{i}$,
$i\geqslant 1$ the inequality (16) is satisfied. This inequality is equivalent
to the following:

$\varphi^{(1)}_{x}(\nu)=S_{x}(\nu)-\kappa_{1}\nu\leqslant(\sqrt{2}+\varepsilon)\kappa_{4}\sqrt{t\log
t}\quad\forall\nu\leqslant t.$ (78)

We will prove (78) by induction on $i$. Suppose that $t=T_{1}=t^{(1)}_{0}$. As
the segment $\mathfrak{B}^{(0)}=(a_{1},a_{2},\ldots,a_{t^{(1)}_{N})}$ has
length $\bigl{(}o(\varepsilon^{6})\bigr{)}t$, we will not take it into account
in our further argument.

One can easily see that $\varphi_{1}(\nu)>\varphi_{1}(\nu-1)$ if and only if
$a_{\nu}>1$. Thus, it is enough to verify the inequality (78) only in the case
when $a_{\nu}>1$. Suppose that $t^{(1)}_{i}<\nu\leqslant t^{(1)}_{i-1}$ It
follows from the definition (72) that

$m^{(1)}_{i}+n^{(1)}_{i}-\kappa_{1}(n^{(1)}_{i}+1)>0.$

Hence for any finite sequence $B$ and for any $1\leqslant j\leqslant N$ one
has

$\varphi^{(1)}(B,\underbrace{1,\ldots,1}_{n^{(1)}_{j}\text{\
numbers}},m^{(1)}_{j})>\varphi^{(1)}(B).$

Thus, it is enough to verify (78) only for the largest
$t^{(1)}_{k}<\nu\leqslant t^{(1)}_{k-1}$ such that $a_{\nu}>1$. In this case
we have

$\begin{split}\varphi^{(1)}_{x}(\nu)\leqslant\sum\limits_{i=k}^{N}\varphi^{(1)}(B^{(1)}_{i})+m^{(1)}_{k}=\sum\limits_{i=k}^{N}\biggl{(}\frac{T_{1}(\lambda^{i-1}-\lambda^{i})}{n^{(1)}_{i}+1}\frac{\log{T_{1}}}{\log{2}}\Bigl{(}1+\frac{\varepsilon}{8}\Bigr{)}\biggr{)}\bigl{(}1+o(\varepsilon^{2})\bigr{)}+m^{(1)}_{k}.\end{split}$
(79)

This can be written as follows:

$\varphi^{(1)}_{x}(\nu)\leqslant\Biggl{(}\frac{(1-\lambda)(\kappa_{1}-1)}{\log{2}}\Bigl{(}1+\frac{\varepsilon}{8}\Bigr{)}\sum\limits_{i=k+1}^{N}\frac{\lambda^{i-k}}{d_{i}}+d_{k}\Biggr{)}\sqrt{T_{1}\log{T_{1}}}\bigl{(}1+o(\varepsilon^{2})\bigr{)}.$
(80)

Denote $\alpha=\frac{\kappa_{1}-1}{\log{2}}(1+\frac{\varepsilon}{8})$,
$\eta=\frac{1}{\alpha}$. Using the introduced notation we have

$\varphi^{(1)}_{x}(\nu)\leqslant\Biggl{(}(1-\lambda)\sum\limits_{i=k+1}^{N}\frac{\lambda^{i-k}}{d_{i}}+\eta
d_{k}\Biggr{)}\alpha\sqrt{T_{1}\log{T_{1}}}\bigl{(}1+o(\varepsilon^{2})\bigr{)}.$
(81)

We recall that $d_{1},d_{2},\ldots,d_{N}$ is the minimizing sequence for (61).
Thus from Lemma 5.5 we obtain that

$\varphi^{(1)}_{x}(\nu)\leqslant\alpha\sqrt{2\eta}\sqrt{T_{1}\log{T_{1}}}(1+o(\varepsilon))=\sqrt{2\alpha}\sqrt{T_{1}\log{T_{1}}}(1+o(\varepsilon))\leqslant\biggl{(}\sqrt{2}+\frac{\varepsilon}{2}\biggr{)}\kappa_{4}\sqrt{T_{1}\log{T_{1}}}.$
(82)

Thus the inequality (78) is satisfied for $t=T_{1}$. Using the same argument
one can derive that for all $n\in\mathbb{N}$ one has

$\varphi^{(1)}(\mathfrak{B^{(n)}})\leqslant\biggl{(}\sqrt{2}+\frac{\varepsilon}{2}\biggr{)}\kappa_{4}\sqrt{T_{n}\log{T_{n}}}$
(83)

and for all $1\leqslant k\leqslant N$, $n\in\mathbb{N}$

$\sum\limits_{i=k}^{N}\varphi^{(1)}(B^{(n)}_{i})+m^{(n)}_{k}\leqslant\biggl{(}\sqrt{2}+\frac{\varepsilon}{2}\biggr{)}\kappa_{4}\sqrt{T_{n}\log{T_{n}}}.$
(84)

Now we show that (78) is satisfied for $t=T_{n}$ when $n\geqslant 2$. Using
the same argument, one can deduce

$\varphi^{(1)}_{x}(\nu)\leqslant\sum\limits_{i=1}^{n-1}\varphi_{1}(\mathfrak{B}^{(i)})+\sum\limits_{i=k}^{N}\varphi^{(1)}(B^{(n)}_{i})+m^{(n)}_{k}$
(85)

for some $1\leqslant k\leqslant N$. From (83) it follows that

$\begin{split}\sum\limits_{i=1}^{n-1}\varphi_{1}(\mathfrak{B}^{(i)})\leqslant\biggl{(}\sqrt{2}+\frac{\varepsilon}{2}\biggr{)}\kappa_{4}\sum\limits_{i=1}^{n-1}\sqrt{T_{i}\log{T_{i}}}\leqslant\biggl{(}\sqrt{2}+\frac{\varepsilon}{2}\biggr{)}\kappa_{4}\sqrt{\log{T_{n}}}\sum\limits_{i=1}^{n-1}\sqrt{T_{i}}=\\\
=\biggl{(}\sqrt{2}+\frac{\varepsilon}{2}\biggr{)}\kappa_{4}\sqrt{T_{n}\log{T_{n}}}\sum\limits_{i=1}^{n-1}\sqrt{\lambda}^{Ni}\leqslant\biggl{(}\sqrt{2}+\frac{\varepsilon}{2}\biggr{)}\kappa_{4}\sqrt{T_{n}\log{T_{n}}}\frac{\sqrt{\lambda}^{N}}{1-\sqrt{\lambda}^{N}}=\biggl{(}\sqrt{T_{n}\log{T_{n}}}\biggr{)}o(\varepsilon^{4}).\end{split}$
(86)

Substituting the estimates (84) and (86) to (85) we obtain that

$\varphi^{(1)}_{x}(\nu)\leqslant\biggl{(}\sqrt{2}+\frac{2\varepsilon}{3}\biggr{)}\kappa_{4}\sqrt{T_{n}\log{T_{n}}}$
(87)

and the second statement of Theorem 2 is proved. ∎ Acknowledgements: I would
like to thank Igor Kan for fruitful discussions. Most results of this paper
are based on his ideas.

## References

  * [1] G. Alkauskas, Integral transforms of the Minkowski question mark function, PhD thesis, Nottingham 2008.
  * [2] A. Denjoy, Sur une fonction reelle de Minkowski, C. R. Acad. Sci. Paris 194 (1932), 44-46.
  * [3] A. Denjoy, Sur une fonction de Minkowski, J. Math. Pures Appl. 17 (1938), 105-151.
  * [4] A.A. Dushistova, N.G. Moshchevitin, On the derivative of the Minkowski question mark function ?(x) (Russian), Fundam. Prikl. Mat. 16 (2010), 33-44; translation in J. Math. Sci. (N.Y.) 182 (2012), 463-471.
  * [5] A.A. Dushistova, I.D. Kan, N.G. Moshchevitin, Differentiability of the Minkowski question mark function, J. Math. Anal. Appl. 401 (2) (2013), 774-794.
  * [6] D Gayfulin, Derivatives of two functions belonging to the Denjoy–Tichy–Uits family St. Petersburg Mathematical Journal, 2016, 27:1, 51–85.
  * [7] Д. Р. Гайфулин, И. Д. Кан, Производная функции Минковского., Изв. РАН. Сер. матем., 85:4 (2021) .
  * [8] Д. Р. Гайфулин, И. Д. Кан, Стационарные точки функции Минковского., Матем. сб. (to appear in 2021).
  * [9] I. D. Kan, Methods for estimating continuants, J. Math. Sci. (NY), 182 (4) (2012), 508-517.
  * [10] M. Kesseböhmer, B. O. Stratmann, Fractal analysis for sets of non-differentiability of Minkowski’s question mark function, J. Number Theory 128 (2008), 2663-2686.
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  * [12] H. Minkowski, Zur Geometrie der Zahlen, Verhandlungen des III Internationalen Mathematiker-Kongresous, Heidelberg 1904, 164-173.
  * [13] J. Paradís, P. Viader, L. Bibiloni, A new light on Minkowski’s $?(x)$ function, J. Number Theory 73 (2) (1998), 212-227.
  * [14] J. Paradís, P. Viader, L. Bibiloni, The derivative of Minkowski’s $?(x)$ function. J. Math. Anal. and Appl. 253 (2001), 107 - 125.
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Dmitry Gayfulin,
Steklov Mathematical Institute of Russian Academy of Sciences
ul. Gubkina, 8, Moscow, Russia, 119991
<EMAIL_ADDRESS>


# Dark Matter in CCDs at Modane (DAMIC-M): a silicon detector apparatus
searching for low-energy physics processes

S. J. Lee,11footnotetext: Corresponding author. B. Kilminster  A. Macchiolo

###### Abstract

Dark Matter In CCDs (DAMIC) is a silicon detector apparatus used primarily for
searching for low-mass dark matter using the silicon bulk of Charge-Coupled
Devices (CCDs) as targets. The silicon target within each CCD is
$675\text{\,}\mathrm{\SIUnitSymbolMicro m}$ thick and its top surface is
divided into over 16 million $15\text{\,}\mathrm{\SIUnitSymbolMicro m}$
$\times$ $15\text{\,}\mathrm{\SIUnitSymbolMicro m}$ pixels. The DAMIC
collaboration has installed a number of these CCDs at SNOLAB. As of 2019,
DAMIC at SNOLAB has reached operational conditions with leakage current less
than
$8.2\text{\times}{10}^{-22}\text{\,}\mathrm{A}\text{\,}{\mathrm{cm}}^{-2}$ and
a readout noise of $1.6\text{\,}\mathrm{e}$, achieved with 5 CCDs. A new DAMIC
apparatus will be installed at Laboratoire Souterrain de Modane in a few
years. The DAMIC at Modane (DAMIC-M) collaboration will be using an improved
version of CCDs designed by Lawrence Berkeley National Laboratory with skipper
amplifiers that use non-destructive readout with multiple-sampling, enabling
the CCDs to achieve a readout noise of $0.068\text{\,}\mathrm{e}$. The low
readout noise, in conjunction with low leakage current of these skipper CCDs,
will allow DAMIC-M to observe physics processes with collisions energies as
low as $1\text{\,}\mathrm{eV}$. The DAMIC-M experiment will consist of an
array of 50 large-area skipper CCDs with more than 36 million pixels in each
CCD. The following proceeding will introduce the DAMIC apparatus at SNOLAB and
its results and as well as the capabilities and the status of the new DAMIC-M
experiment.

##  1 Introduction

The DArk Matter In CCDs (DAMIC) experiment originates from Fermi National
Accelerator Laboratory (Fermilab). It was originally assembled using spare
Charge-Coupled Devices (CCDs) used in the Dark Energy Survey (DES)
experiment222The DES experiment can be reached at
https://www.darkenergysurvey.org/. In order to use the high-resolution and
low-noise scientific CCDs to search for dark matter (DM), the DAMIC
collaboration has installed a shield around the devices to reduce the
radioactive backgrounds. The right of Figure 1 illustrates a layout of a DAMIC
experiment installed in the Sudbury Neutrino Observatory Laboratory (SNOLAB)
[1]. The experiment was installed underground to take advantage of
$2\text{\,}\mathrm{km}$ of rock ( $6\text{\,}\mathrm{km}$ water equivalent) to
reduce the cosmogenic particles. This experiment is called DAMIC at SNOLAB
(DAMIC-SNOLAB).

###  1.1 DAMIC at SNOLAB

In addition to the natural shielding provided by the rocks at SNOLAB, DAMIC-
SNOLAB has an additional $42\text{\,}\mathrm{cm}$ of polyethylene and
$21\text{\,}\mathrm{cm}$ of lead surrounding the experiment to shield the CCDs
from secondary particles from cosmogenic sources and surrounding radioactive
background sources. Inside a copper vessel containing the CCDs, a selected
number of CCDs are further surrounded by radio-pure ancient lead to provide
more shielding from local radioactive sources. The left of Figure 1 shows
DAMIC-SNOLAB without all of its shielding. Within the copper box there are
currently 8 specialized CCDs produced by the Lawrence Berkeley National
Laboratory (LBNL) and Teledyne DALSA333Teledyne Dalsa can be reached at
https://www.teledynedalsa.com/en/home/.

Figure 1:  Left shows the DAMIC-SNOLAB experiment without its external shields
and right shows a diagram demonstrating the configuration of external shield
that surrounds the experiment.

###  1.2 CCDs for DAMIC

Since the original DAMIC experiment at Fermilab, the DAMIC-SNOLAB
collaboration has modified several components of the LBNL CCDs. Some of the
modifications include increasing the thickness of the CCD, removal of
reflection layer in the backside of the CCD (which was a requirement to
increase sensitivity to the infrared spectrum for the DES experiment) and
radio-purity and resistance of the bulk. A proper classification of the CCDs
used by the DAMIC collaboration is large area, thick, 3-stage high voltage
compatible, p channel in n bulk, fully-depleted back-illuminated scientific
grade CCDs. Each of the CCDs installed in SNOLAB is
$675\text{\,}\mathrm{\SIUnitSymbolMicro m}$ thick, weighs
$6.0\text{\,}\mathrm{g}$ and is comprised of over 16 million pixels and with
each pixel covering an area of $15\text{\,}\mathrm{\SIUnitSymbolMicro m}$
$\times$ $15\text{\,}\mathrm{\SIUnitSymbolMicro m}$. These CCDs are fully
depleted at $40\text{\,}\mathrm{V}$ and kept at a pressure of
${10}^{-7}\text{\,}\mathrm{mbar}$ and a temperature of
$135\text{\,}\mathrm{K}$ [2].

###  1.3 Scientific CCDs

It is worth noting that these CCDs are scientific-grade CCDs and therefore
have an advantage that the operation and configuration of each exposure or
image taken can be controlled. This includes power variables such as bias
voltages, drain voltage, and reset voltage, as well as the clock voltages used
for holding charges within each pixel during the exposure, and the clock
sequences for reading out the charges in each pixel.

Figure 2 exemplifies the general top surface structure of an individual pixel
in these CCDs generated using Synopsys Sentarus TCAD444More information on
TCAD tools can be found at https://www.synopsys.com/silicon/tcad.html. The top
brown layer is an oxide passivation layer. Some of the brown oxide layer is
thermally grown in between green polysilicon gate structures. The n doped
polysilicon gate structures are in electrical contact with a buried p channel
through a dielectric junction made out of a combination of SiO2 and Si3N4.
Under the buried p channel is the bulk of the CCD. Approaching the backside of
the CCD, the bulk is negatively doped until reaching in-situ doped polysilicon
(ISDP) deposition layer, and passivated with alternating layers of SiO2 and
polysilicon.

Figure 2:  A TCAD-generated illustration of top structure of CCDs used by
DAMIC during charge transfer along its buried p channel.

Figure 2 also demonstrates the charge flow and collection in the top structure
of the CCDs. There are 3 different geometries of polysilicon gate structure.
Starting at the leftmost component of the figure are gate structures 1, 2, 3
and repeating. These are the 3-stage gate structures. Gate structure 1 is
defined as the center of one pixel and each pixel shares the borders where
gate structures 2, and 3 overlap as seen in the black rectangle. In addition,
the figure illustrates in teal a large concentration of charge carriers
(holes) being localized to under a single gate structure (gate 3). This TCAD
simulation demonstrates the charge transfer between each gate structure.
Similar simulations can be used to optimize the charge collection and transfer
efficiency.

##  2 Results from DAMIC at SNOLAB

DAMIC-SNOLAB has been operational for over 6 years searching for DM. During
the search, the collaboration has seen some standard model particle tracks,
and radio-impurities from within the shielding, cables and CCDs themselves.
The collaboration has been able to model the depth of the standard model (SM)
particle tracks, set constraints on some DM candidates, model the radio-
impurity of detector components and, in addition have been able to optimize
the operating parameters of the CCDs. DAMIC-SNOLAB CCDs have been operating
with a leakage current
$2\text{\times}{10}^{-22}\text{\,}\mathrm{A}\text{\,}{\mathrm{cm}}^{-2}$ with
a readout noise equivalent to $1.6\text{\,}\mathrm{e}$555$\text{\,}\mathrm{e}$
denotes electron charges.

###  2.1 Particle tracks in CCDs

While CCDs are 2 dimensional imaging devices, each CCD can be used to
reconstruct 3 dimensional particle tracks. As incident particles ionize and
produce excess charge carriers while the device is fully depleted, the charge
carriers must travel up to the top polysilicon gate structure in order to be
collected. As previously stated the CCDs installed at SNOLAB are
$675\text{\,}\mathrm{\SIUnitSymbolMicro m}$ thick and fully depleted at
$40\text{\,}\mathrm{V}$ and are held at $135\text{\,}\mathrm{K}$. This leaves
some distance and time before charge carriers reach the gate structure,
allowing some of the charge carriers to diffuse into adjacent pixel
structures. Figure 3 illustrates a simplified diagram of charge diffusion to
the right. The left of Figure 3 shows the actual particle tracks left by some
SM particles. As long as the CCD is not overdepleted, the diffusion of charge
carriers can be modelled using the Point Spread Function (PSF) and can be used
to determine the depth of particle tracks within the bulk [3, 4, 6, 5].

Figure 3:  The left shows particle tracks from alpha, muon, electron and
diffusion-limited x-ray tracks with relative scale. The center and right show
a diagram illustrating the diffusion of charge carriers as a function of depth
resulting in diffused particle tracks [1].

###  2.2 Search for dark matter candidates

By modelling PSF and optimizing the operating parameters of the CCDs using SM
particle interactions, DAMIC-SNOLAB has been able to search for some of the DM
candidates. Two primary DM candidates searched for at DAMIC-SNOLAB are low
mass, Weakly Interacting Massive Particles (WIMPs) and hidden photons (often
denoted as $\gamma_{\chi}$). Some of the latest results can be found in Figure
4 and in [8, 7].

Figure 4:  Left shows upper limit on the cross section and mass of Weakly
Interacting Massive Particles (WIMP) bold red line demonstrates the
constraints set by the DAMIC-SNOLAB experiment [8]. Right shows constraints of
the cross section and mass of hidden photon interacting with standard model
electron set by the DAMIC-SNOLAB experiment [7].

###  2.3 Background model

One of the key challenges of searches for dark matter is to eliminate the flux
of SM particles. The SM particles can be introduced into a DM experiment by
minuscule amounts of radioactive contamination. The unit of background in DM
experiments is typically the $\text{\,}\mathrm{dru}$
($1\text{\,}\mathrm{E}\mathrm{v}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{/}\mathrm{k}\mathrm{eV}\mathrm{/}\mathrm{k}\mathrm{g}\mathrm{/}\mathrm{d}$666$\text{\,}\mathrm{kg}$
here should not be confused with $\text{\,}\mathrm{kGy}$ and
$\text{\,}\mathrm{d}$ denotes day ). Obvious large contributors to these
radioactive sources are naturally occuring unstable radon, and radioactive
isotopes in lead such as 210Pb. Radon can be reduced through a radon-free,
overpressure environment, typically using nitrogen gas, and by selecting
ancient lead, with reduced isotope levels. However there are other smaller
background contributors that together make a large impact to a DM experiment.
Almost all of the components that make up the DAMIC-SNOLAB experiment have
been tested for their radioactivity in destructive tests such as Inductively
Coupled Plasma-Mass-Spectrometry (ICP-MS). Using the results of the ICP-MS of
all of the materials used to construct the detector, the radioactivity at the
position of each sensor can be simulated using GEANT 4. Combining the lab
tests results and simulations, DAMIC-SNOLAB currently has achieved a
background of $11.8\text{\,}\mathrm{dru}$. This can further be improved by
increasing or decreasing the detector material, depending on the radio-purity
of the materials used.

##  3 DAMIC at Modane

As stated in Section 2, DAMIC-SNOLAB has been operating for over 6 years. To
achieve better sensitivity to dark matter, the experiment would need to
achieve lower background rates and increase the active detector mass. As of
2018, a new collaboration called DAMIC at Modane (DAMIC-M) has been formed to
install a new DAMIC experiment at Laboratoire Souterrain de Modane (LSM). At
LSM, The DAMIC-M collaboration plans to install new larger CCDs with lower
background rates and improved electronics, along with improvements to the
design of the experiment777The DAMIC-M collaboration can be reached at
https://damic.uchicago.edu/.

###  3.1 Preliminary design of DAMIC-M

One of the limiting factors for the DAMIC-SNOLAB was the active cooling and
heating elements and the vacuum pump, which generated electrical noise. As the
CCDs are operated in vacuum at a pressure of
${10}^{-7}\text{\,}\mathrm{mbar}$, the cooling element must be in thermal and
electrical contact with each device, introducing a challenge for providing a
common device and earth ground. Furthermore, since the experiment is
underground, the definition of common earth ground is not well defined.
Without a proper earth ground, the motors used in vacuum pump and cryocooler
generate minuscule amount of excess charge that produces multiple artifacts in
CCDs. DAMIC-M will be designed to hold vacuum passively using charcoal cryo-
pump and will be using liquid nitrogen to cool its sensors. This will reduce
the possibility of introducing electrical noise into the sensors.

###  3.2 Reducing the background

As mentioned in Section 2.3, DAMIC-SNOLAB had many well-studied small
background sources. Some of these can be reduced by simply reducing the amount
material used and also by a careful production of the detector material.
First, all of the components of DAMIC-M will be tested in laboratory for their
radioimpurity. Second, all of these components will be transported and
produced under some form of controlled shielding from cosmogenic particles.

The most crucial detector elements are obviously the CCDs. The silicon ingot
used for producing the CCDs is purchased from Topsil888Topsil can be reached
at http://www.topsil.com/ located in Scandinavia. This ingot will be processed
into wafers and transported first to the east coast and then the west coast of
North America. Once unshielded, the silicon ingot is expected to be exposed to
cosmogenic particles at an unacceptable level during transportation.
Therefore, a specially designed sea container providing shielding from
cosmogenic particles will be used to transport the silicon components that
will eventually become CCDs. Furthermore, the CCDs will be produced under
shielding further reducing the exposure to cosmogenic particles. In addition,
the copper components of the detector will be made using electroforming
techniques in an underground laboratory and similar care will be taken during
transportation. With such care in production and transportation of the
detector components, DAMIC-M aims to have background levels in the CCDs of
$0.1\text{\,}\mathrm{dru}$.

###  3.3 The skipper CCDs

Since the installation of DAMIC-SNOLAB, LBNL has been able to produce a
skipper amplifier for their CCDs [9]. The CCDs equipped with these amplifiers
are called skipper CCDs. The special feature of the skipper amplifier is that
the output node of the readout structure is floating rather than being in
constant contact with the rest of the readout structure. Once the charge
reaches the output node, it is either collected into the readout electronics,
or discarded through the drain. By being able to break the contact to the
output node, the charge can skip the output node and be sampled multiple times
without being destroyed before being readout [9, 3].

Figure 5:  Left shows a conventional amplifier structure of a 2004 LBNL CCD
seen from the top surface. Right shows an image of skipper amplifier structure
seen from top surface from [9] with permission.

Figure 5 shows the surface structure of both conventional and skipper
amplifiers. Most of the CCD top surface structures have been labelled. The red
circle illustrates the path of the charge. In a conventional amplifier, the
charge can only be sent in one direction, towards the output node. For a CCD
with a skipper amplifier, a single packet of charge (charge collected from a
single pixel) can enter the video output gate, while the next packet is held
behind the summing well. From this point on, a packet of charge can be sampled
multiple times skipping the output node. Once the packet enters the output
node, the charges are finally read or destroyed. By being able to perform non-
destructive sampling of a single charge packet multiple times, readout noise
can be reduced.

The DAMIC-M collaboration has been operating the prototype skipper CCDs on
test stands at above ground laboratories, and has been able to read out
charges from prototype skipper CCDs with readout noise less than 1 electron as
demonstrated in Figure 6. DAMIC-M intends to install 50 skipper CCDs, with
each CCD having over 36 million $15\text{\,}\mathrm{\SIUnitSymbolMicro m}$
$\times$ $15\text{\,}\mathrm{\SIUnitSymbolMicro m}$ pixels.

Figure 6:  Readout performed by DAMIC-M collaboration using a 1k x 6k
prototype skipper CCD to demonstrate sub-electron resolution readout noise.
Here, the readout noise is equivalent to
($7.22\text{\,}$$\pm$$0.05\text{\,}$)${10}^{-2}\text{\,}\mathrm{e}$.

###  3.4 Simulations

Since all of the components of the DAMIC-M detector will be assayed for their
radioactivity, the radioactivity of the detector can be simulated using
GEANT4. As a result, the amount of radiation reaching the CCDs will be
simulated. These types of simulations will be used to determine the optimal
sizes of shielding to maximize the shielding from cosmogenic and other local
radioactive sources but with minimal radioactivity within the shielding
materials.

DAMIC-M will also be using Synopsys Sentarus TCAD to optimize the design of
the new skipper CCDs and the operating parameters. TCAD will also be used to
better model the PSF and improve the accuracy of reconstructed particle tracks
within the CCDs.

##  4 Conclusion

DAMIC-SNOLAB has been operating up to eight $16\text{\,}\mathrm{Mpx}$ CCDs at
$135\text{\,}\mathrm{K}$, in vacuum at a pressure of
${10}^{-7}\text{\,}\mathrm{mbar}$, fully depleted at $40\text{\,}\mathrm{V}$
with leakage current down to
$2\text{\times}{10}^{-22}\text{\,}\mathrm{A}\text{\,}{\mathrm{cm}}^{-2}$,
readout noise equivalent to $1.6\text{\,}\mathrm{e}$ and with radioactive
background of $11.8\text{\,}\mathrm{dru}$.

Since the installation of DAMIC-SNOLAB, technology and techniques for
developing and operating DM detectors have significantly improved, and a new
collaboration called DAMIC-M has been formed to install a new DAMIC detector
setup at Laboratoire Souterrain de Modane. DAMIC-M has since become a CERN
recognized experiment and will also search for displacement damage at atomic
scale and model low energy non-ionizing energy loss in collaboration with the
RD50 collaboration at CERN999The RD50 collboration can be reached at
https://rd50.web.cern.ch/rd50/.

In 2020, a proof-of-concept prototype of DAMIC-M will be installed. The final
detector design of DAMIC-M will have 50 skipper CCDs with readout noise less
than ($7.22\text{\,}$$\pm$$0.05\text{\,}$)${10}^{-2}\text{\,}\mathrm{e}$ in a
charcoal cryo-pump with background noise lower than
$0.1\text{\,}\mathrm{dru}$.

## Acknowledgments

The author would like to acknowledge the Swiss National Science Foundation
(SNSF) and their Funding LArge international REsearch projects (FLARE) grant
for enabling this research.

The author would like to also acknowledge S. E. Holland and C. J. Bebek from
LBNL for their development of their CCDs and correspondence which has been
supporting research for both DAMIC-SNOLAB and DAMIC-Modane.

The author would also like to acknowledge Sub-Electron-Noise SkipperCCD
Experiment Instrument (SENSEI) and Fermilab for their work together with LBNL
developing sub-electron noise skipper CCDs.

The author would also like to acknowledge American Association of Variable
Star Observers (AAVSO) and their CEO Dr. Stella Kafka for writing and
maintaining a wonderful comprehensible reference resource for CCD photometry.

The author would also finally like to acknowledge the extended members of the
DAMIC-SNOLAB and DAMIC-M collaborations.

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