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Derivative of the Minkowski function: optimal estimates D. R. Gayfulin Steklov Mathematical Institute of Russian Academy of Sciences, Moscow, Russia
Russian version:
Matematicheskii Sbornik, 2022, Volume 213, Number 10, DOI: https://doi.org/10.4213/sm9692 
Bibliographic databases:
    § 1. Introduction1.1. The Minkowski function $?(x)$The function traditionally denoted by $?(x)$ was originally considered by Minkowski in 1904. For the first time this notation was used in his work [1]. Minkowski constructed this function as an example of a continuous monotone bijective map of the closed interval $[0,1]$ onto itself which has the following property: the set of algebraic numbers of degree at most $2$ (that is, the rational numbers and quadratic irrationalities) goes to the set of rational numbers of the closed interval. This property of the function $?(x)$ becomes obvious, if we consider an equivalent definition introduced by Salem [2]. Namely, if
$$
\begin{equation}
x= [0; a_1,a_2,\dots,a_t,\dots]= \cfrac{1}{\displaystyle{a_1+\cfrac{1}{\displaystyle{a_2 + \dots+\cfrac{1}{\displaystyle{a_t + \displaystyle{\dotsb} }}}}}}
\end{equation}
\tag{1}
$$
is a finite or infinite expansion of a number $x\in [0,1]$ in a continued fraction with positive integer partial quotients $a_1,a_2,\dots,a_t,\dots$, then
$$
\begin{equation}
?(x) =\frac{1}{2^{a_1-1}} - \frac{1}{2^{a_1+a_2-1}}+ \dots+ \frac{(-1)^{n+1}}{2^{a_1+a_2+\dots+a_n-1}}+ \dotsb.
\end{equation}
\tag{2}
$$
Now the bijection between the sets of algebraic numbers of degree at most 2 and the set of rational numbers follows immediately from the Euler-Lagrange theorem on periodic continued fractions. As is well known, the derivative $?'(x)$ (when it exists at a given point $x$) can only take two values, $0$ or $+\infty$. At the same time, according to Lebesgue’s theorem, the derivative is zero almost everywhere. It is also known that the derivative of $?(x)$ is always zero at rational points. Other curious properties of the Minkowski function can be found, for example, in [2]–[4]. 1.2. Notation and parametersWe denote sequences by uppercase letters, for example, $A$, $B$, $C$, and their elements by lowercase ones, for example, $a_i$, $b_j$, $c_k$. All sequences in this paper consist of positive integers unless otherwise stated. For an arbitrary finite sequence $B=(b_1,b_2,\dots,b_n)$ we set
$$
\begin{equation*}
\overleftarrow{B}=(b_n,b_{n-1},\dots,b_1), \qquad S(B)=\sum_{i=1}^{n}b_i.
\end{equation*}
\notag
$$
We set $4_n=\underbrace{4,\dots,4}_{n \text{ numbers}}$; other sequences of repeating numbers are denoted similarly. We let $\langle A\rangle$ denote the continuant of a (possibly empty) sequence $A\,{=}\,(a_1,\dots,a_t)$. This quantity is defined recursively as follows: the continuant of an empty sequence is equal to $1$. The continuant of a one-element sequence is equal to this element: $\langle a_1\rangle=a_1$. Further, if $t\geqslant 2$, then we set
$$
\begin{equation}
\langle a_1,a_2,\dots,a_t\rangle=a_t\langle a_1,a_2,\dots,a_{t-1}\rangle +\langle a_1,a_2,\dots,a_{t-2}\rangle.
\end{equation}
\tag{3}
$$
It can readily be seen that the finite continued fraction $[0;a_1,\dots,a_t]$ can be expressed in terms of continuants:
$$
\begin{equation}
[0;a_1,\dots,a_t]=\frac{\langle a_2,\dots,a_t\rangle}{\langle a_1,a_2,\dots,a_t\rangle}.
\end{equation}
\tag{4}
$$
The rule (3) has the following generalization:
$$
\begin{equation}
\begin{aligned} \, \notag &\langle a_1,\dots,a_t,a_{t+1},\dots,a_s\rangle =\langle a_1,\dots,a_{t}\rangle\langle a_{t+1},\dots,a_{s}\rangle+ \langle a_1,\dots,a_{t-1}\rangle\langle a_{t+2},\dots,a_{s}\rangle \\ &\qquad =\langle a_1,\dots,a_{t}\rangle\langle a_{t+1},\dots,a_{s}\rangle (1+[0;a_{t-1},\dots,a_1][0;a_{t+2},\dots,a_s]). \end{aligned}
\end{equation}
\tag{5}
$$
More information about the properties of continuants can be found in [5]. Given an irrational number $x=[0;a_1,a_2,\dots,a_n,\dots]$, we denote by $A_t$ the sequence of its first $t$ partial quotients and by $S_x(t)$ their sum: We also need the following constants:
$$
\begin{equation}
\Phi = \frac{1+\sqrt{5}}{2}\approx 1.618034, \qquad \kappa_1 = \frac{2\ln{\Phi}}{\ln{2}}\approx 1.3884838,
\end{equation}
\tag{6}
$$
$$
\begin{equation}
\widetilde{\lambda}= \frac{\sqrt{2}\,\lambda_4}{\lambda_5}\approx 1.1537043,
\end{equation}
\tag{8}
$$
$$
\begin{equation}
\gamma=\frac{2+\sqrt{5}}{\sqrt{20}}\cdot\frac{5+\sqrt{29}}{2\sqrt{29}} \bigl(1+[0;\overline{4}][0;\overline{5}]\bigr)^2 \approx 0.9982728,
\end{equation}
\tag{9}
$$
$$
\begin{equation}
\kappa_2 = \frac{4\ln\lambda_5-5\ln\lambda_4}{\ln \lambda_5-\ln\lambda_4-\ln\sqrt{2}} \approx 4.4010487,
\end{equation}
\tag{10}
$$
$$
\begin{equation}
\kappa_4 =\sqrt{\frac{\kappa_1-1}{\ln 2}}\approx 0.7486412,
\end{equation}
\tag{11}
$$
$$
\begin{equation}
\kappa_5=\sqrt{\frac{8(\kappa_2-4)(5-\kappa_2)\ln{(\gamma^{-1})}} {\ln{\widetilde{\lambda}}}}\approx 0.152427.
\end{equation}
\tag{12}
$$
For an arbitrary sequence $A$ of length $t$ we set
$$
\begin{equation*}
\varphi^{(1)}(A):=S(A)-\kappa_1t\quad\text{and} \quad\varphi^{(2)}(A):=\kappa_2t-S(A).
\end{equation*}
\notag
$$
For an irrational number $x=[0;a_1,\dots,a_t,\dots]$ we also use the notation
$$
\begin{equation*}
\varphi^{(1)}_{x}(t):=\varphi^{(1)}(A_t)=S_x(t)-\kappa_1t\quad\text{and} \quad\varphi^{(2)}_{x}(t):=\varphi^{(2)}(A_t)=\kappa_2t-S_x(t).
\end{equation*}
\notag
$$
1.3. The history of the question. Critical valuesIn [6] several theorems were proved concerning the relationships between the quantity $?'(x)$ and the limiting behaviour of the quotient $S_{x}(t) /t$. We present two of them. Theorem A (see [6], Theorem 1). (i) Let the inequality $S_x(t)<\kappa_1t$ hold for an irrational number $x$ in $(0,1)$ and for sufficiently large $t$. Then the derivative $?'(x)$ exists and is equal to $+\infty$. (ii) For every $\varepsilon>0$ there exists an irrational number $x\in (0,1)$ such that ${?'(x)=0}$ and the inequality $S_{x}(t)<(\kappa_1+\varepsilon)t$ holds for all sufficiently large $t$. Theorem B (see [6], Theorem 3). (i) Let $S_x(t)>(\kappa_2+\varepsilon)t$ for some irrational number $x$ in $(0,1)$, some $\varepsilon>0$ and all sufficiently large $t$. Then the derivative $?'(x)$ exists and is equal to $0$. (ii) For every $\varepsilon>0$ there exists an irrational number $x\in (0,1)$ such that ${?'(x)=+\infty}$ and the inequality $S_{x}(t)<(\kappa_2-\varepsilon)t$ holds for all sufficiently large $t$. Thus, the constants $\kappa_1$ and $\kappa_2$ in Theorems A and B are sharp. The question was also raised in [6] as to whether one could compare the limiting behaviour of the quantities $S_{x}(t) $ and $\kappa_2t$ in the case when, on the contrary, we have $?'({x})=+\infty$ instead of $?'({x})=0$. The assertion (ii) of Theorem B shows that in this case the quantity
$$
\begin{equation}
\frac{\kappa_2t-S_x(t)}{t}=\frac{\varphi^{(2)}_{x}(t)}{t}
\end{equation}
\tag{13}
$$
can be arbitrarily small. The first nontrivial result in this direction was obtained in [6]. Theorem C (see [6], Theorem 4). (i) Let the derivative $?'(x)$ exist for some irrational number $x$ in $(0,1)$ and let $?'(x)=+\infty$. Then for all sufficiently large $t$ the bound holds. (ii) There exists an irrational number $x\in (0,1)$ such that $?'(x)=+\infty$ and the following inequality holds for all sufficiently large $t$: Subsequently, inequalities (14) and (15) were repeatedly improved. The best bounds known today were obtained1 in [7]. Theorem D (see [7], Theorem 1.2). (i) Let the derivative $?'(x) =+\infty$ exist for an irrational number $x\in (0,1)$. Then the following estimate holds for all sufficiently large $t$: (ii) There exists an irrational number $x\in (0,1)$ such that $?'(x)=+\infty$ and the following inequality holds for all sufficiently large $t$: 1.4. Uniform and local boundsThe left-hand sides of inequalities (14) and (15) are obviously different. While the first inequality gives a uniform bound for $\varphi^{(2)}_{x}(t)$, the other, on the contrary, gives a local bound. The same is true for the pair of inequalities (16), (17). Therefore, if we want to obtain sharp constants in each case, we must estimate both the ‘uniform’ and ‘local’ behaviour of the quantity in question from above and below. For $\varphi^{(1)}_{x}(t) $ this problem was solved in [8]. Theorem E (see [8], Theorem 1). (i) Let $x=[0;a_1,\dots,a_n,\dots]$ be an irrational number such that $?'(x)=0$. Then for any $\varepsilon>0$, for infinitely many values of $t$. (ii) For every $\varepsilon>0$ there exists an irrational number $x\,{=}\,[0;a_1,\dots,a_n,\dots]$ such that $?'(x)=0$ and for all sufficiently large $t$.Theorem F (see [8], Theorem 2). (i) Let $x=[0;a_1,\dots,a_n,\dots]$ be an irrational number such that $?'(x)=0$. Then for any $\varepsilon>0$ and all sufficiently large $t$ (ii) For every $\varepsilon\,{>}\,0$ there exists an irrational number $x = [0;a_1,\dots,a_n,\dots]$ such that $?'(x)=0$ and for infinitely many values of $t$.We note that assertion (ii) of Theorem E was originally proved in [6]. 1.5. The main results of the paperThe following two theorems are analogues of Theorems E and F in [8]. Theorem 1. (i) Consider an arbitrary irrational number $x\in(0,1)$ such that $?'(x)=+\infty$. Then for any $\varepsilon>0$ for infinitely many values of $t$. (ii) For any $\varepsilon\!>\!0$ there exists an irrational number $x\!\in\!(0,1)$ such that ${?'(x)\!=\!+\infty}$ and the following inequality holds for all sufficiently large $t$: Theorem 2. (i) Consider an arbitrary irrational number $x\in(0,1)$ such that $?'(x)=+\infty$. Then for any $\varepsilon>0$ and all sufficiently large $t$ (ii) For every $\varepsilon>0$ there is an irrational number $x\in (0,1)$ such that $?'(x)=+\infty$ and the following inequality holds for infinitely many values of $t$: Note that, similarly to Theorems E and F, the optimal constants in Theorems 1 and 2 differ exactly twice. This is related to the fact that the corresponding constants in Lemmas 18 and 21 differ twice; these lemmas are proved below. § 2. Reduction to continuants of a special formLet $x=[0;a_1,a_2,\dots,a_t,\dots]$ be a number at which $?'(x)=+\infty$. We fix this number throughout §§ 2 and 3. We always denote the corresponding sequence $(a_1,a_2,\dots,a_t)$ of the first $t$ partial quotients of $x$ by $A_t$. Through the end of § 2 we also use the following notation: if $B_t$ is a sequence of $t$ elements and $\nu\leqslant t$, then $B_{\nu}$ is its subsequence of the first $\nu$ elements. The following auxiliary assertion (see [7], Lemma 2.2) is the main tool for investigating the derivative $?'(x)$. Lemma 1. (i) If for some irrational number $x\in(0,1)$, then $?'(x)=+\infty$. (ii) If $?'(x)=+\infty$ for some irrational number $x\in(0,1)$, then The key idea of the proof of bounds (namely, the first assertions of Theorems 1 and 2) is as follows. For example, suppose the first assertion of Theorem 2 fails to hold. This means that there exists $\varepsilon>0$ such that the following inequality holds for infinitely many values of $t$:
$$
\begin{equation}
\max_{u\leqslant t}\varphi^{(2)}_{x}(u)=\max_{u\leqslant t}\varphi^{(2)}_{x}(A_u) \leqslant\biggl(\frac{\kappa_5}{2}-\varepsilon\biggr)\sqrt{t}.
\end{equation}
\tag{26}
$$
As the next lemma states, we can assume without loss of generality that the continuant $\langle A_t\rangle$ consists of sufficiently long sequences of fours and fives. Lemma 2 (see [7], Lemma 6.2). Let $U$ be a fixed parameter.2 Let $x=[0;a_1,a_2,\dots, a_t,\dots]$ be an irrational number such that $?'(x)=+\infty$. Let condition (26) hold for some sufficiently large $t$. Then for $t'=[t-t^{2/3}]$ there exists a continuant $\langle D_{t'}\rangle=\langle d_1,d_2,\dots, d_{t'}\rangle$ that has the following properties. 1. $D_{t'}$ has the form
$$
\begin{equation}
( 5_{m_1},4_{n_1},\dots,5_{m_i},4_{n_i},\dots,5_{m_{\sigma}},4_{n_{\sigma}}),
\end{equation}
\tag{27}
$$
and each of the two tuples $(m_1, m_2,\dots,m_{\sigma})$ and $(n_1, n_2,\dots,n_{\sigma})$ contains at most one integer not exceeding $U$. 2. For any $\nu\leqslant t'$,
$$
\begin{equation}
\frac{\langle D_{\nu}\rangle}{\sqrt{2}^{\,S(D_{\nu})}} \geqslant C(U) \min\biggl(\frac{\langle A_{\nu}\rangle}{\sqrt{2}^{\,S_x(\nu)}},\, (1.05)^\nu\biggr)>1,
\end{equation}
\tag{28}
$$
where $C(U)$ is a constant independent of $x$. 3. The following inequality holds: The scheme of the proof below is as follows. By Lemma 1 and (28), for infinitely many values of $t'$ we have
$$
\begin{equation}
\frac{\langle D_{t'}\rangle}{\sqrt{2}^{\,S(D_{t'})}}>1.
\end{equation}
\tag{30}
$$
Recall that $S(D_{t'})=\kappa_2t'-\varphi^{(2)}(D_{t'})$, and therefore we obtain a lower bound for $S(D_{t'})$ from (29). Combining this with the fact that one can write a very precise upper bound for a continuant of the form (27), we arrive at a contradiction with (30). The aim of this section is to prove a statement similar to Lemma 2, with $\max\varphi^{(2)}$ replaced by $\varphi^{(2)}$ in part 3, under the assumption that bound (22) in Theorem 1 fails. Thus, let $\varepsilon>0$ be a number such that for any sufficiently large $t$ we have 2.1. Local maximaWe call $\nu\in\mathbb{N}$ a local maximum for $x$ if
$$
\begin{equation}
\varphi^{(2)}_{x}(\nu)>\max\bigl(\varphi^{(2)}_{x}(\nu+1), \varphi^{(2)}_{x}(\nu-1)\bigr).
\end{equation}
\tag{32}
$$
It can readily be seen that $\nu$ is a local maximum if and only if $a_{\nu}\leqslant 4$ and $a_{\nu+1}\geqslant 5$. For an arbitrary finite sequence $B_t$ a local maximum is defined similarly, except that for $\nu=t$ condition (32) is replaced by $\varphi^{(2)}(B_{\nu-1})<\varphi^{(2)}(B_{\nu})$. Without loss of generality we can assume that there exist infinitely many local maxima for $x$. Indeed, if all the $a_i$, starting from some index, are greater than or equal to $5$, then by the first assertion of Theorem B we have $?'(x)=0$, and we immediately arrive at a contradiction. If all the $a_i$, starting from some index, are smaller than or equal to $4$, then the assertions of Theorems 1 and 2 obviously hold. Lemma 3. Let (31) hold for all sufficiently large $t$ that are local maxima. Then (31) holds for all sufficiently large $t$. Proof. Assume that, on the contrary, for infinitely many values of $u$ we have
$$
\begin{equation}
\varphi^{(2)}_{x}(u)=\kappa_2u-S_x(u)\geqslant(\kappa_5-\varepsilon)\sqrt{u}.
\end{equation}
\tag{33}
$$
Suppose $u$ is not a local maximum. This is possible in two cases. First, $a_u\geqslant 5$ can hold; second, $a_u\leqslant 4$, but $a_{u+1}\leqslant 4$ can hold. In the first case we obtain
$$
\begin{equation*}
\varphi^{(2)}_{x}(u-1)>\varphi^{(2)}_{x}(u)\geqslant (\kappa_5-\varepsilon)\sqrt{u}>(\kappa_5-\varepsilon)\sqrt{u-1}.
\end{equation*}
\notag
$$
If $u-1$ is a local maximum, then we arrive at a contradiction with (31). Otherwise $a_{u-1}\geqslant 5$, and then we argue by induction until we meet a local maximum.
Consider the other case. Let $v$ be the local maximum closest to $u$ from the right. Then $a_{\nu}\leqslant 4$ for every $\nu$ such that $u\leqslant\nu\leqslant v$. We note that if (33) holds for some $u=\nu$ and $a_{\nu+1}\,{\leqslant}\, 4$, then the same inequality holds for $u=\nu+1$. Indeed, let (33) fail to hold for $u=\nu+1$, so that
$$
\begin{equation}
\varphi^{(2)}_{x}(\nu+1)=\kappa_2(\nu+1)-S_x(\nu+1)<(\kappa_5-\varepsilon)\sqrt{\nu+1}.
\end{equation}
\tag{34}
$$
We subtract (33) from this inequality for $u=\nu$. Then we obtain
$$
\begin{equation}
0.4<\kappa_2-a_{\nu+1}<(\kappa_5-\varepsilon)\bigl(\sqrt{\nu+1}-\sqrt{\nu}\,\bigr).
\end{equation}
\tag{35}
$$
Since the right-hand side of (35) tends to 0 as $\nu$ increases, we arrive at a contradiction. Hence inequality (33) holds for the local maximum $v$, and we have obtained a contradiction to the hypotheses of the lemma. This completes the proof. 2.2. Unit variations and the preservation of local maximaLemma 4. Let (31) hold for all sufficiently large $t$. Then there exists $T=T(x)$ such that for any $s\geqslant T$ the mean value of the partial quotients $a_T,\dots,a_s$ is at least $4$. Proof. We argue by contradiction. We construct increasing sequences $t_i$ and $s_i$ as follows. Let $t_1=1$. Next, suppose that the number $t_i$ has already been defined for some $i\geqslant1$. Then we choose $s_i$ so that the mean value of the partial quotients $a_{t_i},a_{t_i+1},\dots,a_{s_i}$ is less than $4$. Finally, we set $t_{i+1}=s_i+1$. It is clear from this construction that for every $i$ the arithmetic mean of $a_1,a_2,\dots,a_{s_i}$ is less than $4$. This means that $\varphi^{(2)}_{x}(s_i)>(\kappa_2-4)s_i$, which obviously contradicts (31). This completes the proof of the lemma. Since $T$ in the previous lemma is an absolute constant depending only on $x$, we assume below without loss of generality that $T= 1$. The following obvious corollary can be derived from the lemma just proved. Corollary 1. Under the assumptions of Lemma 4, for any $j\in\mathbb{N}$ such that $a_j\leqslant 3$ there exists an $i<j$ such that $a_i\geqslant 5$. Indeed, it suffices to apply Lemma 4 to the interval $(a_1,a_2,\dots,a_j)$. We call the following transformation of the continuant a unit variation of the first type:
$$
\begin{equation}
\begin{aligned} \, \notag \langle A_t\rangle &=\langle a_1,a_2,\dots,a_i,\underbrace{4,4,\dots,4}_{m \text{ numbers}},a_j,a_{j+1},\dots,a_t\rangle \\ &\to\langle a_1,a_2,\dots,a_i-1,\underbrace{4,4,\dots,4}_{m \text{ numbers}},a_j+1,a_{j+1},\dots,a_t\rangle=\langle A'_t\rangle, \end{aligned}
\end{equation}
\tag{36}
$$
where $a_i\geqslant 5$, $a_j\leqslant 3$ and $m\geqslant 0$. We denote the elements of $\langle A'_t\rangle$ by $a'_1,a'_2$, and so on. Here $a'_i=a_i-1$, $a'_j=a_j+1$, and $a'_k=a_k$ for $k\ne \{i,j\}$. Lemma 5. Under the assumptions of Lemma 4 let the sequence $(a_1,a_2,\dots,a_t)$ contain elements not exceeding $3$. Then the following hold. 1. A unit variation of the first type is defined for the continuant $\langle A_t\rangle$. 2. After a unit variation of the first type, the mean value of the partial quotients on any initial interval of the continuant $\langle A'_t\rangle$ remains at least $4$. 3. If $\nu_0$ is a local maximum of $A'_t$, then $\nu_0$ is also a local maximum of $A_t$. Moreover, $\varphi^{(2)}(A'_{\nu_0})=\varphi^{(2)}(A_{\nu_0})$. 4. The following inequality holds for every $\nu\leqslant t$: Proof. To prove part 1 consider the smallest $j$ such that $a_j\leqslant 3$ and the greatest $i<j$ such that $a_i\geqslant 5$. By Corollary 1 such $i$ exists. Now we note that if $j\ne i+1$, then $a_k\geqslant 4$ for all $k$ between $i$ and $j$ because of the minimality of $j$. On the other hand, $a_k\leqslant 4$ because of the maximality of $i$. The first assertion of the lemma is proved.
To prove part 2 we note that if $k\geqslant j$ or $k<i$, then after the transformation (36) the arithmetic mean of $a_1,a_2,\dots,a_k$ does not change. If $i\geqslant k<j$, then it can readily be seen that,after the transformation (36), the arithmetic mean of $a_1,a_2,\dots,a_{i-1}$ does not change and remains at least $4$. At the same time, all elements among $a_i-1,a_{i+1},\dots,a_k$ are greater than or equal to $4$. Therefore, the arithmetic mean of $a_1,a_2,\dots,a_k$ remains at least $4$ after the transformation (36). To prove part 3 recall that $\nu_0$ is a local maximum of $A'_t$ if and only if $a'_{\nu_0}\leqslant 4$ and $a'_{\nu_0+1}\geqslant 5$. Suppose $\nu_0$ is not a local maximum of $A_t$, that is, either $a_{\nu_0}\geqslant5$ or ${a_{\nu_0+1}\leqslant 4}$. Consider the first case, when $a_{\nu_0}=5$ and $a'_{\nu_0}=4$, that is, $\nu_0=i$ in (36). It can readily be seen that then $a'_{\nu_0+1}\leqslant 4$, which immediately leads to a contradiction. In the second case we have $a_{\nu_0+1}=4$, but $a'_{\nu_0+1}=5$, that is, $\nu_0+1=j$ in (36), which is impossible because $a_j\leqslant 3$. It remains to prove that $\varphi^{(2)}(A'_{\nu_0})=\varphi^{(2)}(A_{\nu_0})$. However, this is obvious for $\nu_0<i$ or $\nu_0 \geqslant j$, and there are no local maxima of $A'_t$ on the interval $(a_i,\dots,a_{j-1})$. The third assertion of the lemma is proved. Finally, we prove part 4. Clearly, (37) turns to equality for $\nu<i$. For $\nu\geqslant j$ it is easy to see that $S_a(\nu)=S_{a'}(\nu)$, but $\langle A'_{\nu}\rangle>\langle A_{\nu}\rangle$ by Lemma 6.1 in [7]. In the case when $i\leqslant\nu<j$ we use a bound which holds for arbitrary finite sequences of positive integers $A$ and $B$ (see, for example, [7], formula (2.4)):
$$
\begin{equation}
\frac{a}{a+1}\leqslant\frac{\langle A,a,B \rangle}{\langle A,a+1,B \rangle}.
\end{equation}
\tag{38}
$$
This implies that ${\langle A'_{\nu}\rangle}/{\langle A_{\nu}\rangle}\!\geqslant\!{4}/{5}$. Here ${\sqrt{2}^{\,S_a(\nu)}}\!/{\sqrt{2}^{\,S_{a'}(\nu)}}\!=\!\sqrt{2}$. Since ${\sqrt{2}\cdot{4}/{5}\!>\!1}$, the fourth assertion of the lemma is proved.
This completes the proof of the lemma. Applying Lemma 5 repeatedly until no partial quotients less than $4$ remain, we arrive at the following assertion. Lemma 6. Under the assumptions of Lemma 4, for every sufficiently large $t$ there exists a continuant $\langle B_t\rangle=\langle b_1,b_2,\dots,b_t\rangle$ satisfying the following conditions. 1. All elements of $\langle B_t\rangle$ are greater than or equal to $4$. 2. The following inequality holds for every $\nu\leqslant t$:
$$
\begin{equation}
\frac{\langle B_{\nu}\rangle}{\sqrt{2}^{\,S_b(\nu)}} \geqslant\frac{\langle A_{\nu}\rangle}{\sqrt{2}^{\,S_{a}(\nu)}}.
\end{equation}
\tag{39}
$$
3. If $\nu_0$ is a local maximum for $B_t$, then the same point is a local maximum of $A_t$. Moreover, $\varphi^{(2)}(B_{\nu_0})=\varphi^{(2)}(A_{\nu_0})$. A continuant $\langle X_t\rangle=\langle x_1,x_2,\dots,x_t\rangle$ is said to be right-balanced if for every $s\leqslant t$ the mean value of the partial quotients $x_s,x_{s+1},\dots,x_t$ does not exceed $5$. Lemma 7. Under the assumptions of Lemma 4 the continuant $\langle A_t\rangle$ is right balanced for infinitely many values of $t$. Moreover, if $\langle A_t\rangle$ is right-balanced, then so is also the continuant $\langle B_t\rangle$ in Lemma 6. Proof. We prove the first assertion of the lemma by contradiction. Let $T$ be such that the continuant $\langle A_t\rangle$ is not balanced for $t>T$. We choose $t=10T$ and construct decreasing sequences $s_i$ and $t_i$ as follows: $t_1=t$, and $s_i$ is the least integer such that the arithmetic mean of $a_{s_i},a_{s_i+1},\dots,a_{t_i}$ is greater than $5$. Finally, $t_{i+1}=s_i-1$. It can readily be seen that, if $t_i>T$, then the corresponding $s_i$ exists. Of course, the above recursive procedure terminates at some point and we obtain some $s_n<T$. It is clear from the construction that the arithmetic mean of the partial quotients $a_{s_n},a_{s_n+1},\dots,a_t$ is greater than $5$. On the other hand, the arithmetic mean of $a_1,a_2,\dots,a_{s_n-1}$ is easily estimated from below by 1. Therefore, $S(A_t)=S_x(t)>4.5t$ for any sufficiently large $t$. By the first assertion of Theorem B, this means that $?'(x)=0$, and we have arrived at a contradiction.
Now we prove part 2. Suppose that for each $s<t$ the arithmetic mean of the partial quotients $a_s,a_{s+1},\dots, a_t$ does not exceed $5$. We claim that the arithmetic mean of $b_s,b_{s+1},\dots, b_t $ does not exceed $5$ either. It suffices to show that this property remains valid after every transformation of the form (36). We argue by induction on $s$. Let $s=t$, that is, $a_t\leqslant 5$. We must show that $a'_t\leqslant 5$. This follows directly from the fact that the transformation (36) increases a partial quotient only when it does not exceed $3$. Now assume that the arithmetic mean of $a'_s,a'_{s+1},\dots, a'_t$ does not exceed $5$, but the means value of $a'_{s-1},a'_s,a'_{s+1},\dots, a'_t$ exceeds $5$. This means that $a'_{s-1}\geqslant 6$, and so $a_{s-1}\geqslant 6$. Since the arithmetic mean of $a_{s-1},a_s,a_{s+1},\dots, a_t$ does not exceed $5$, this means that the relation $i<s-1<j$ holds for the transformation (36), because the sum of partial quotients on a fixed interval decreases after this transformation. However, in this case, $a_{s-1}=4$ by the definition (36). This is a contradiction, which completes the proof of the lemma. The following transformation of a continuant is called a unit variation of the second type:
$$
\begin{equation}
\begin{aligned} \, \notag \langle B_t\rangle &=\langle b_1,b_2,\dots,b_i,\underbrace{5,5,\dots,5}_{m \text{ numbers}},b_j,b_{j+1}, \dots,b_t\rangle \\ &\to \langle b_1,b_2,\dots,b_i-1,\underbrace{5,5,\dots,5}_{m \text{ numbers}},b_j+1,b_{j+1},\dots,b_t\rangle=\langle B'_t\rangle, \end{aligned}
\end{equation}
\tag{40}
$$
where $b_i\geqslant 6$, $b_j=4$ and $m\geqslant 0$. We denote the elements involved in $\langle B'_t\rangle$ by $b'_1$, $b'_2$, and so on. Here $b'_i=b_i-1$, $b'_j=b_j+1$, and $b'_k=b_k$ for $k\ne \{ i,j\}$. Lemma 8. Let $\langle B_t\rangle$ be the right balanced continuant obtained in Lemma 6. Let $\langle B_t\rangle$ contain elements greater than or equal to $6$. Then the following assertions hold. 1. A unit variation of the second type is defined for $\langle B_t\rangle$. 2. The continuant $\langle B'_t\rangle$ obtained by a unit variation of the second type remains right balanced. 3. If $\nu_0$ is a local maximum for $B'_t$, then the same point is a local maximum for $B_t$. Moreover, $\varphi^{(2)}(B'_{\nu_0})=\varphi^{(2)}(B_{\nu_0})$. 4. The following inequality holds for every $\nu\leqslant t$: Proof. The proofs of the first, second and fourth assertions of the lemma are completely analogous to the proofs of the corresponding assertions of Lemma 5. Let us prove assertion 3.
Let $\nu_0$ be a local maximum for $B'_t$ but not a local maximum for $B_t$. That is, we have $b'_{\nu_0}\leqslant 4$ and $b'_{\nu_0+1}\geqslant 5$ and, moreover, either $b_{\nu_0}\geqslant5$ or $b_{\nu_0+1}\leqslant 4$. Consider the first case when $b_{\nu_0}=5$ and $b'_{\nu_0}=4$, that is, $\nu_0=i$. This is impossible since $b_i\geqslant 6$. In the second case we have $b_{\nu_0+1}=4$ and $b'_{\nu_0+1}=5$, and we see that $\nu_0+1=j$. However, it can readily be seen that $b'_{\nu_0}\geqslant 5$ in this case, which contradicts the fact that $\nu_0$ is a local maximum for $B'_t$. We claim now that $\varphi^{(2)}(B'_{\nu_0})=\varphi^{(2)}(B_{\nu_0})$. The considerations are perfectly similar to the proof of the corresponding assertion in Lemma 6. The equality to be proved is obvious for $\nu_0<i$ or $\nu_0 \geqslant j$, and there are no local maxima for $B_t$ on the interval $(b_i,\dots,b_{j-1})$. This completes the proof of the lemma. Applying Lemma 8 repeatedly until the continuant contains no partial quotients greater than $5$, we arrive at the following assertion, which sums up § 2.2. Lemma 9. Under the assumptions of Lemma 4, for infinitely many values of $t$ there exists a continuant $\langle C_t\rangle=\langle c_1,c_2,\dots,c_t\rangle$ satisfying the following conditions. 1. All elements involved in $\langle C_t\rangle$ are equal to $4$ or $5$. 2. The following inequality holds for every $\nu\leqslant t$:
$$
\begin{equation}
\frac{\langle C_{\nu}\rangle}{\sqrt{2}^{\,S_c(\nu)}} \geqslant\frac{\langle A_{\nu}\rangle}{\sqrt{2}^{\,S_{a}(\nu)}}.
\end{equation}
\tag{42}
$$
3. If $\nu_0$ is a local maximum for $C_t$, then the same point is a local maximum for $A_t$. Moreover, $\varphi^{(2)}(C_{\nu_0})=\varphi^{(2)}(A_{\nu_0})$. 2.3. Reflections and the two-sided lemmaIn the previous section we transformed an arbitrary continuant $\langle A_t\rangle$ satisfying the conditions of Lemma 4 into a continuant $\langle C_t\rangle$ consisting of fours and fives. In this section we transform $\langle C_t\rangle$ into a new continuant $\langle D_t\rangle$ consisting of sufficiently long sequences of the form $4,4,\dots,4$ and $5,5,\dots,5$. To transform $\langle C_t\rangle$ into $\langle D_t\rangle$ we use a transformation which we call a reflection. It is defined by
$$
\begin{equation}
\langle C_t\rangle=\langle \overbrace{c_1,\dots,5}^{P}, \overbrace{4_n,5_m}^{{\overrightarrow{Q}}},\overbrace{4,\dots,c_{t}}^{R}\,\rangle \to\langle \overbrace{c_1,\dots,5}^{P},\overbrace{5_m,4_n}^{\overleftarrow{ Q}}, \overbrace{4,\dots,c_{t}}^{ R}\,\rangle=\langle C'_t \rangle,
\end{equation}
\tag{43}
$$
where $m\geqslant 1$ and $n\geqslant 1$. We note that, according to [9] (the main lemma) the transformation (43) increases the continuant $\langle C_t\rangle$. It is also easy to see that $S(C'_{\nu})\geqslant S(C_{\nu})$ for all $\nu\leqslant t$. Lemma 10. Under the assumptions of Lemma 4 let $U$ be a fixed positive integer. Then for infinitely many values of $t$ there exists a continuant $\langle D_t\rangle=\langle d_1,d_2,\dots,d_t\rangle$ satisfying the following conditions. 1. $D_{t}$ has the form (27), and $m_i \geqslant U$ and $n_i\geqslant U$ for all $i\geqslant 2$. 2. There exists a positive constant $F(U)$ depending only on $U$ such that for any ${\nu\leqslant t}$
$$
\begin{equation}
\frac{\langle D_{\nu}\rangle}{\sqrt{2}^{\,S(D_{\nu})}} \geqslant F(U)\frac{\langle A_{\nu}\rangle}{\sqrt{2}^{\,S(A_{\nu})}}.
\end{equation}
\tag{44}
$$
3. If $\nu_0$ is a local maximum for $D_t$, then the same point is a local maximum for $A_t$. Moreover, $\varphi^{(2)}(D_{\nu_0})=\varphi^{(2)}(A_{\nu_0})$. Proof. We carry out the proof using induction on $U$. The case $U=1$ is obviously a tautology. Let all the $m_i$ and $n_i$ for $i\geqslant 2$ be no smaller than $U$. Our aim is to make transformations of the form (43) that take the continuant $\langle C_t\rangle$ to another continuant in which all the $m_i$ and $n_i$ for $i\geqslant 2$ are no smaller than $U+1$. To do this we need two chains of transformations. First we achieve that $m_i\geqslant U+1$ for all $i\geqslant 2$. We find the largest $i\geqslant2$ such that $m_i=U$ and consider the reflection
$$
\begin{equation}
\begin{aligned} \, \notag \langle C_t\rangle &=\langle \overbrace{c_1,\dots,5_{m_{i-1}}}^{P}, \overbrace{4_{n_{i-1}},5_{m_i}}^{{Q}}, \overbrace{4_{n_i},\dots,c_t}^{R}\, \rangle \\ &\to\langle \overbrace{c_1,\dots,5_{m_{i-1}}}^{P}, \overbrace{5_{m_i},4_{n_{i-1}}}^{\overleftarrow{Q}},\overbrace{4_{n_i}, \dots,c_t}^{R}\, \rangle =\langle C'_t\rangle. \end{aligned}
\end{equation}
\tag{45}
$$
It is clear that $m_i+m_{i-1}\geqslant 2U\geqslant U+1$. The property that $n_j\geqslant U$ for all $j$ is obviously preserved by the transformation (45). Repeating this transformation we achieve the situation where all the $m_i$ for $i\geqslant 2$ are not smaller than $U+1$. In exactly the same way we ensure that all the $n_i$ for $i\geqslant 2$ are not smaller than $U+1$. Thus, applying this chains of transformations $U$ times we can reduce the continuant $\langle C_t\rangle$ to the form (27) for which the first assertion of the lemma holds. It remains to show that the continuant thus obtained satisfies the second and third assertions of the lemma.
We prove the second assertion. It can readily be seen that, if in the transformation (45) an index $\nu$ occurs in the interval $P$ or $R$, then $\langle C'_{\nu}\rangle\geqslant\langle C_{\nu}\rangle$ and $S(C'_{\nu})=S(C_{\nu})$. If $\nu$ occurs in the part $Q$, then it can readily be seen that On the other hand, since $\langle C'_{\nu}\rangle\geqslant\langle C_{\nu}\rangle$, we obtain the bound
$$
\begin{equation}
\frac{\langle C'_{\nu}\rangle}{\sqrt{2}^{\,S(C'_{\nu})}} \geqslant \frac{1}{\sqrt{2}^{\,U}}\biggl(\frac{\langle C_{\nu}\rangle}{\sqrt{2}^{\,S(C_{\nu})}}\biggr).
\end{equation}
\tag{46}
$$
Further, it is obvious that every index $\nu$ can occur in $Q$ only once in both the first and second chains of transformations. As shown already, to reduce $\langle C_t\rangle$ to the form (27) we need at most $U$ pairs of chains of transformations. Hence
$$
\begin{equation}
\frac{\langle D_{\nu}\rangle}{\sqrt{2}^{\,S(D_{\nu})}} \geqslant \biggl(\frac{1}{\sqrt{2}^{\,U}}\biggr)^{2U} \biggl(\frac{\langle C_{\nu}\rangle}{\sqrt{2}^{\,S(C_{\nu})}}\biggr).
\end{equation}
\tag{47}
$$
This completes the proof of the second assertion of the lemma .
The third assertion follows directly from the form of the transformation (43). Indeed, if $\nu_0$ is a local maximum for $C'_t$, then, since $c'_{\nu_0}=4$ and $c'_{\nu_0+1}=5$, we can immediately conclude that $\nu_0$ does not lie in $\overleftarrow{Q}$ and is not the last element of $P$. Hence $c_{\nu_0}=4$ and $c_{\nu_0+1}=5$, that is, $\nu_0$ is a local maximum for $C_t$. Since $\nu_0$ does not lie in $\overleftarrow{Q}$, it also follows that $S(C'_{\nu})=S(C_{\nu})$. This completes the proof of the lemma. In what follows we assume without loss of generality that $m_i \geqslant U$ and $n_i\geqslant U$ in (27) for all $i$. Combining Lemmas 3 and 10 we obtain the following assertion, which is an analogue of Lemma 2. It summarizes the use of the above transformations. Lemma 11. Let $U$ be a fixed parameter. Let $x$ be an irrational number for which $?'(x)=+\infty$. Let $\varepsilon$ be a positive number such that 1. For all $\nu$ such that $\sqrt{t}<\nu\leqslant t$
$$
\begin{equation}
\frac{\langle D_{\nu}\rangle}{\sqrt{2}^{\,S(D_\nu)}} >1.
\end{equation}
\tag{49}
$$
2. For all $\nu$ such that $\sqrt{t}<\nu\leqslant t$ Proof. The first assertion follows directly from the fact that ${\langle A_{t}\rangle}/{\sqrt{2}^{\,S(A_{t})}}$ tends to infinity and from Lemmas 1 and 10. The proof of the second part is also not difficult. Lemma 3 implies that it suffices to verify (50) only for the local maxima of $D_t$. From the third assertion of Lemma 10 we know that, if $\nu_0$ is a local maximum, then This completes the proof of the lemma.
§ 3. Upper bounds3.1. A two-sided lemmaLemmas 2 and 11 reduce actually the original problem to the consideration of continuants of the form (27). We denote by $\sigma(D_t)$ the number of intervals $(4,4,\dots,4)$ for a continuant $\langle D_t\rangle$ of this form. We also recall the previously introduced constants $\widetilde{\lambda}= {\sqrt{2}\,\lambda_4}/{\lambda_5}\approx 1.1537043$ and
$$
\begin{equation*}
\gamma=\frac{2+\sqrt{5}}{\sqrt{20}}\,\frac{5+\sqrt{29}}{2\sqrt{29}} \bigl(1+[0;\overline{4}][0;\overline{5}]\bigr)^2\approx 0.9982728.
\end{equation*}
\notag
$$
We need the following lemma from [7] (Lemma 8.3). Lemma 12. Let $\langle D_t\rangle$ be a continuant of the form (27). Then there exists a constant $\varepsilon_0=\varepsilon_0(U)>0$ tending to $0$ as $U$ tends to infinity and such that for all sufficiently large (depending on $\varepsilon_0$) positive integers $t$ and all integers $\nu$ in the interval $t^{1/3}\leqslant \nu\leqslant t$, This lemma shows already the way to the proof of the first assertions of Theorems 1 and 2. Suppose inequality (22) in the first assertion of Theorem 2 fails to hold. Then it follows from Lemmas 11 and 12 that if $?'(x)=+\infty$, then for infinitely many values of $\nu$ there exists a continuant $\langle D_{\nu}\rangle$ for which
$$
\begin{equation}
(\gamma+\varepsilon_0)^{\sigma(D_{\nu})}\widetilde{\lambda}^{\varphi^{(2)}(D_{\nu})} >\frac{\langle D_{\nu}\rangle}{\sqrt{2}^{\,S(D_{\nu})}}>1.
\end{equation}
\tag{52}
$$
From (52) we will obtain a lower bound for $\varphi^{(2)}(D_{\nu})$, which, since $\varphi^{(2)}(D_{\nu})<\varphi^{(2)}_x(\nu)<(\kappa_5-\varepsilon)\sqrt{\nu}$, will contradict inequality (31) and thus lead to a contradiction. The proof of Theorem 1 is similarly structured. Finding a lower bound for the quantities $\varphi^{(2)}(D_{t})$ and $\max_{\nu\leqslant t}\varphi^{(2)}(D_{\nu})$ is the subject of the following subsection. 3.2. A lower bound for $\varphi^{(2)}(D_{t})$To begin with, we introduce some new parameters. Let $\varepsilon>0$ be the fixed number from the hypotheses of Theorems 1 and 2. Let $\lambda=\lambda(\varepsilon)$ be an arbitrary rational number that satisfies $1>\lambda>1-\varepsilon^6 10^{-6}$. We need the following constants:
$$
\begin{equation}
M=\frac{10\ln{\varepsilon}}{\ln{\lambda}}, \qquad P=\biggl[\frac{\ln{6}}{\ln{(1+\varepsilon^2)}}\biggr]+1\quad\text{and} \quad N=2M(P+2).
\end{equation}
\tag{53}
$$
Now let $t$ be sufficiently large so as to satisfy $\lambda^Nt>{t}/{\ln{t}}$ and let $\lambda^Nt\in\mathbb{Z}$. We set $t_i=\lambda^ i t$. We call the interval of partial quotients $(d_{t_i+1},d_{t_i+2},\dots,d_{t_{i-1}})$, $1\leqslant i\leqslant N$, the $i$th block $B^{(i)}$. We set $t_{N+1}=0$; then $B^{(N+1)}=(d_1,\dots,d_{t_N})$ is the initial interval of the continuant $\langle D_t\rangle$. Thus,
$$
\begin{equation*}
\langle d_1,\dots,d_t\rangle=\langle B^{(1)},B^{(2)},\dots,B^{(N)}\rangle.
\end{equation*}
\notag
$$
In every block $B^{(i)}$ we choose the longest sequence of the form $(4,4,\dots,4)$ (if there are many such sequences, we choose the leftmost one). We denote by $B'^{(i)}$ the initial interval of the block $B^{(i)}$ up to the beginning of this sequence, and we denote the index of the last element of $B'^{(i)}$ by $m'_{i }$. We define the numbers $f_i$ and $f'_i$ by the equalities
$$
\begin{equation}
\langle B^{(i)}\rangle=\sqrt{2}^{\,S(B^{(i)})+f_i\sqrt{t_{i-1}}}\quad\text{and} \quad \langle B'^{(i)}\rangle=\sqrt{2}^{\,S(B'^{(i)})+f'_i\sqrt{t_{i-1}}},
\end{equation}
\tag{54}
$$
or, equivalently, by
$$
\begin{equation}
\begin{gathered} \, \langle B^{(i)}\rangle=\sqrt{2}^{\,\kappa_2(t_{i-1}-t_i)-\varphi^{(2)}(B^{(i)}) +f_i\sqrt{t_{i-1}}}, \\ \langle B'^{(i)}\rangle=\sqrt{2}^{\,\kappa_2(t_{i-1}-t'_i)-\varphi^{(2)}(B'^{(i)}) +f'_i\sqrt{t_{i-1}}}. \end{gathered}
\end{equation}
\tag{55}
$$
Lemma 13. Let $\langle D_t\rangle$ be a continuant of the form (27), where for all $\sqrt{t}<\nu\leqslant t$. Then $\varphi^{(2)}(D_{\nu})>0$ for $\sqrt{t}<\nu\leqslant t$. Proof. Note that if a continuant $\langle D_t\rangle$ has the form (27), then any initial interval $\langle D_{\nu}\rangle$ of $\langle D_t\rangle$ has the same form. Applying Lemma 12 to $D_{\nu}$ we obtain Since $\sigma(D_{\nu})$ is always a positive number, we see that $\varphi^{(2)}(D_{\nu})>0$. This completes the proof of the lemma. Lemma 14. Under the assumptions of Lemma 13, assume that at least one of the following two conditions hold: 1)
$$
\begin{equation}
\max_{u\leqslant t}\varphi^{(2)}(D_u)\leqslant\biggl(\frac{\kappa_5}{2}-\varepsilon\biggr)\sqrt{t};
\end{equation}
\tag{57}
$$
2) for all $\nu$ such that $\sqrt{t}<\nu\leqslant t$
$$
\begin{equation}
\varphi^{(2)}(D_{\nu})<(\kappa_5-\varepsilon)\sqrt{\nu}.
\end{equation}
\tag{58}
$$
Then for any $i \leqslant N+1$ Proof. Let (57) hold. Then by Lemma 13 we have
$$
\begin{equation*}
0<\varphi^{(2)}(B^{(N+1)})=\varphi^{(2)}(D_{t_N}) <\biggl(\frac{\kappa_5}{2}-\varepsilon\biggr)\sqrt{t}< t^{0.9}.
\end{equation*}
\notag
$$
Now, by (92), Taking into account that for the same reasons, we obtain the assertion of the lemma.
The case when condition (58) holds is considered perfectly similarly. This completes the proof of the lemma. We determine the value of $c_k$ by the following condition: the length of the longest part of the form $(4,4,\dots,4)$ of the block $B^{(k)}$ for $k\leqslant N$ is $c_k\sqrt{t_k}$. In the following lemma we obtain a lower bound for $\varphi^{(2)}(B^{(k)})$ in terms of the quantity $c_k$, using Lemma 12. Lemma 15. Under the assumptions of Lemma 14 the following inequality holds: Proof. Since
$$
\begin{equation*}
S(B^{(k)})=\kappa_2(1-\lambda)t_k(1+o(\varepsilon^4))
\end{equation*}
\notag
$$
by Lemma 14 and the length of $B^{(k)}$ is $(1-\lambda)t_k$, it is easy to see that the total number of fours in $B^{(k)}$ is $(1-\lambda)(5-\kappa_2)t_k(1+o(\varepsilon^4))$. Hence $\sigma(B^{(k)})$ can be estimated from below as follows:
$$
\begin{equation}
\sigma(B^{(k)})\geqslant\frac{(1-\lambda)(5-\kappa_2)\sqrt{t_k}}{c_k}(1+o(\varepsilon^4)).
\end{equation}
\tag{61}
$$
Recall that $\sigma(B^{(k)})$ is the number of intervals $(4,4,\dots,4)$ in the block $B^{(k)}$. Clearly, the continuant $\langle B^{(k)}\rangle$ also has the form (27). Using the bound (51) we see that
$$
\begin{equation}
\begin{aligned} \, \notag & \sqrt{2}^{\,S(B^{(k)})+f_k\sqrt{t_{k-1}}} \\ &\qquad =\langle B^{(k)}\rangle\leqslant(\gamma+\varepsilon_0)^{({(1-\lambda)(5-\kappa_2)\sqrt{t_k}}/{c_k} (1+o(\varepsilon^4)))} \widetilde{\lambda}^{\varphi^{(2)}(B^{(k)})}\sqrt{2}^{\,S(B^{(k)})}. \end{aligned}
\end{equation}
\tag{62}
$$
Taking the logarithm of (62) and taking the relation ${t_k}/{t_{k-1}}=\lambda=1+o(\varepsilon^4)$ into account we obtain the assertion of the lemma. For the initial block $B^{(N+1)}$ the trivial bound from Lemma 13 is sufficient: Now we estimate the quantity $\varphi^{(2)}(B'^{(k)})$. This is perfectly similar to the argument in Lemma 15 and yields the bound
$$
\begin{equation}
\varphi^{(2)}(B'^{(k)})\geqslant\biggl(\frac{(5-\kappa_2)S(B'^{(k)}) \ln{((\gamma+\varepsilon_0)^{-1})}}{c_k\ln{\widetilde{\lambda}} \sqrt{t_{k-1}}}+f'_k\frac{\ln{\sqrt{2}}}{\ln{\widetilde{\lambda}}} \sqrt{t_{k-1}}\biggr)(1+o(\varepsilon^4)).
\end{equation}
\tag{64}
$$
A coarse version of this bound follows trivially and is sufficient for us:
$$
\begin{equation}
\varphi^{(2)}(B'^{(k)})>f'_k\frac{\ln{\sqrt{2}}}{\ln{\widetilde{\lambda}}} \sqrt{t_{k-1}}(1+o(\varepsilon^4)).
\end{equation}
\tag{65}
$$
We recall that we let $m'_k$ denote the length of the continuant $\langle B^{(N+1)},\dots,B'^{(k)}\rangle$, and the length of the largest sequence $(4,4,\dots,4)$ in the block $B^{(k)}$ is equal to $c_k\sqrt{t_k}$. We set $m_k=m'_k+c_k\sqrt{t_k}$. It can readily be seen that
$$
\begin{equation}
\varphi^{(2)}(D_{m_k})=\sum_{i=k+1}^{N+1} \varphi^{(2)}(B^{(i)}) + \varphi^{(2)}(B'^{(k)})+(\kappa_2-4)c_k\sqrt{t_k}.
\end{equation}
\tag{66}
$$
Lemma 16. Under the assumptions of Lemma 14, for any $k\leqslant N$ Proof. Using (56) for $\nu=m'_k$, we obtain
$$
\begin{equation}
\frac{\langle B^{(N+1)},B^{(N)},\dots,B^{(k+1)},B'^{(k)}\rangle} {\sqrt{2}^{\,S(B^{(n+1)})+\dots+S(B^{(k+1)})+S(B'^{(k)})}}>1.
\end{equation}
\tag{68}
$$
Since $N$ does not exceed $O(\ln{t})$, we expand the continuant in the numerator in accordance with (5):
$$
\begin{equation}
\frac{\langle B_{N+1}\rangle\langle B_N\rangle\dotsb\langle B_{k+1}\rangle \langle B'_k\rangle\sqrt{2}^{\,O(\ln{t})}} {\sqrt{2}^{\,S(B_{N+1})+S(B_N)+\dots+S(B_{k+1})+S(B'_k)}}>1.
\end{equation}
\tag{69}
$$
Substituting identities (55) into (69) we obtain the assertion of the lemma. Lemma 17. Let the continuant $\langle D_t\rangle$ have the form (27), where (56) holds for $\sqrt{t}<\nu\leqslant t$. Assume that at least one of the two assertions in the statement of Lemma 14 holds. Then the following inequality holds for every $k<N$: Proof. Substituting (60), (63) and (65) into (66) and also using Lemma 16 we immediately obtain the required assertion. 3.3. Two combinatorial lemmasWe set
$$
\begin{equation}
\alpha=\frac{(5-\kappa_2)\ln{((\gamma+\varepsilon_0)^{-1})}}{\ln{\widetilde{\lambda}}}\quad\text{and} \quad \eta=\frac{\kappa_2-4}{\alpha}.
\end{equation}
\tag{71}
$$
Inequality (70) can be written as
$$
\begin{equation}
\varphi^{(2)}(D_{m_k})\geqslant\biggl((1-\lambda)\sum_{i=k+1}^{N} \frac{(\sqrt{\lambda})^{i-k}}{c_i}+\eta c_k\biggr)\alpha\sqrt{t_k}(1+o(\varepsilon^4)).
\end{equation}
\tag{72}
$$
Here we have used the fact that $t_{k+i}=\lambda^it_k$. To complete the proofs of the first assertions of Theorems 1 and 2 we need the following two combinatorial statements from [8]. For a self-contained presentation we include here their full proofs. Lemma 18. Let $\eta, c_1,c_2,\dots, c_N$ be arbitrary positive real numbers. Let $\varphi_i$ be the numbers defined according to the following rule: Then The proof of Lemma 18 consists of several steps. We recall that the constants $M$, $N$ and $P$ used in the argument are defined in (53). First of all, we note the following fact. Lemma 19. Let (74) fail to hold. Then there exists $i_1\,{\leqslant}\, M$ such that ${c_i\,{\geqslant}\,{1}/(2\sqrt{\eta}\,)}$. Moreover, for all $i\leqslant N$ Proof. We argue by contradiction. Let us estimate $\varphi_1$ as follows:
$$
\begin{equation}
\begin{aligned} \, \notag \varphi_1 &\geqslant(1-\lambda)\sum_{i=1}^{N}\frac{\sqrt{\lambda}^{\,i}}{c_{i+1}} +\eta c_1\geqslant(1-\lambda)\sum_{i=1}^{M}\frac{\sqrt{\lambda}^{\,i}}{c_{i+1}} \geqslant 2\sqrt{\eta}(1-\lambda)\sum_{i=1}^{M}\sqrt{\lambda}^{\,i} \\ &= 2\sqrt{\eta}(1-\lambda)\sum_{i=1}^{\infty}\sqrt{\lambda}^{\,i}(1-\lambda^{M}) =2(1+\sqrt{\lambda})\sqrt{\eta}(1+o(\varepsilon^4))>\sqrt{8\eta}(1+o(\varepsilon^4)). \end{aligned}
\end{equation}
\tag{76}
$$
We arrive at a contradiction with (74). The bound (75) follows from the obvious inequality $\varphi_k>\eta c_k$. This completes the proof of the lemma. Lemma 20. Let (74) fail to hold. Then for every $i_k<N-M$, $k\geqslant 1$, there is a number $i_{k+1}$, $i_k<i_{k+1}<i_k+M$, such that $c_{i_{k+1}}>(1+\varepsilon^2)c_{i_k}.$ Proof. We argue by contradiction. Let $c_{i_k+j}<c_{i_k}(1+\varepsilon^2)$ for all $1\leqslant j\leqslant M$. Then, arguing similarly to (76) we obtain
$$
\begin{equation}
\begin{aligned} \, \notag & \varphi_{i_k}\geqslant(1-\lambda)\sum_{j=1}^{M}\frac{\sqrt{\lambda}^{\,j}}{c_{i_k+j}}+\eta c_{i_k}\geqslant\frac{1-\lambda}{1+\varepsilon^2} \sum_{j=1}^{M}\frac{\sqrt{\lambda}^{\,j}}{c_{i_k}}+\eta c_{i_k} \\ &\qquad\geqslant\biggl(\frac{2}{(1+\varepsilon^2)c_{i_k}} +\eta c_{i_k}\biggr)(1+o(\varepsilon^4)) \geqslant \sqrt{\frac{8\eta}{1+\varepsilon^2}}(1+o(\varepsilon^4)). \end{aligned}
\end{equation}
\tag{77}
$$
In the last inequality in (77) we used Cauchy’s inequality. Thus, (74) holds and we have arrived at a contradiction. This completes the proof of the lemma. Proof of Lemma 18. We argue by contradiction. If the statement fails, then by Lemma 19 there exists $i_1<M$ such that $c_i\geqslant{1}/(2\sqrt{\eta}\,)$. Using Lemma 20 ${P=[{\ln{6}}/{\ln{(1+\varepsilon^2)}}]+1}$ times we see that
$$
\begin{equation*}
c_{i_{P+1}}>\frac{1}{2\sqrt{\eta}}(1+\varepsilon^2)^P \eta>\frac{3}{\sqrt{\eta}}.
\end{equation*}
\notag
$$
We arrive at a contradiction with (75). This completes the proof of the lemma. Lemma 21. Let $C=(c_1,c_2,\dots, c_N)$ be an arbitrary sequence of nonnegative real numbers. Let $\eta$ be an arbitrary positive number. Let the numbers $\varphi_i$ be defined by (73). Define the numbers $\varphi'_i$ as follows: The proof of Lemma 21 consists of several steps. First of all, we make the change $d_i=\sqrt{\lambda}^{\,i}c_i$. We denote the family $(d_1,d_2,\dots,d_N)$ by $D$. Then (78) becomes
$$
\begin{equation}
\varphi'_k(D)=(1-\lambda)\sum_{i=k+1}^{N}\frac{\lambda^{i}}{d_i}+\eta d_k.
\end{equation}
\tag{81}
$$
We set $\varphi'_{\max}(D)=\max_{1\leqslant k\leqslant N}\varphi'_k(D)$ and claim that
$$
\begin{equation}
\min_{D\in\mathbb{R}^N_{+}}\varphi'_{\max}(D)=\sqrt{2\eta}(1+o(\varepsilon)).
\end{equation}
\tag{82}
$$
Let us denote the conjectural minimum in (82) by $y_{\min}$. Lemma 22. If $\varphi'_{\max}(D)=y_{\min}$, then $\varphi'_k(D)\!=\!y_ {\min}$ for every $k$ such that ${1\!\leqslant\! k\!\leqslant\! N}$. Proof. We argue by contradiction. Suppose that $\varphi'_{\max}(D)=y_{\min}$ and $\varphi'_i(D)<y_{\min}$ for some $i$. We say that an index $n$ is minimizing if $\varphi'_n(D)=y_{\min}$. Let $k$ be the greatest nonminimizing index. Without loss of generality we can assume that all $i<k$ are not minimizing either. Indeed, for any $\delta>0$, for the family we have $\varphi'_k(D')<\varphi'_k(D)$ for $k<i$, that is, the index $k$ is not minimizing. On the other hand $\varphi'_k(D')=\varphi'_k(D)$ for $k>i$. Therefore, for sufficiently small $\delta$ we can achieve the relation $y_{\min}\leqslant\varphi'_{\max}(D')\leqslant\varphi'_{\max}(D)=y_{\min}$, where the first inequality follows from the definition of $y_{\min}$. That is, $\varphi'_{\max}(D')=y_{\min}$.
Thus we can assume that $k+1$ is the least minimizing index. Therefore, by continuity $d_{k+1}$ can be reduced so that all the $\varphi'_i(D)$ for $i<k+1$ remain less than $y_{\min}$. Now we have obtained a family in which $k+2$ is the smallest minimizing index. Repeating this procedure we arrive at the situation where $N$ is a unique minimizing number. Thus, by reducing $d_N$ we obtain a set $D'$ such that $\varphi'_{\max}(D')<\varphi'_{\max}(D)=y_{\min}$, and we have arrived at a contradiction. This completes the proof of the lemma. Lemma 23. There exists a unique family $D=(d_1,d_2,\dots, d_N)$ such that ${\varphi'_{\max}(D)=y_{\min}}$. It satisfies the recurrence relation and the initial condition $d_1=0$. Proof. We use the previous lemma. Note that the equation $\varphi'_N(D)=y_{\min}$ takes us to a linear equation $\eta d_N=y_{\min}$, from which $d_N$ is uniquely found. Now, substituting this value into the equation $\varphi'_N(D)=y_{\min}$ we find $D_{N-1}$ and so on. The uniqueness of the family $D$ is proved.
Let $d_1>0$. Then by reducing $d_1$ we also reduce $\varphi'_1(D)$, but $\varphi'_{\max}(D)$ remains equal to $y_{\min}$, which contradicts the uniqueness of the minimum. Finally, from the equality $\varphi'_{k+1}(D)=\varphi'_{k}(D)$ we derive
$$
\begin{equation}
\eta(d_{k+1}-d_k)=(1-\lambda)\frac{\lambda^{k+1}}{d_{k+1}}.
\end{equation}
\tag{84}
$$
Solving (84) as a quadratic equation with respect to $d_{k+1}$ and choosing the positive root we obtain identity (83). This completes the proof of the lemma. Thus, $y_{\min}={d_N}/{\eta}$, where $d_N$ is found from recurrence equation (83). Multiplying (83) by $\sqrt{\eta}$ we obtain
$$
\begin{equation}
\sqrt{\eta}\,d_{k+1}=\frac{\sqrt{\eta}\,d_k+\sqrt{(\sqrt{\eta}\,d_k)^2 +4(1-\lambda)\lambda^{k+1}}}{2}.
\end{equation}
\tag{85}
$$
It follows from the condition $d_1=0$ that for $e_k=\sqrt{\eta}\, d_k$ we have
$$
\begin{equation}
e_{k+1}=\frac{e_k+\sqrt{e_k^2+4(1-\lambda)\lambda^{k+1}}}{2}.
\end{equation}
\tag{86}
$$
We set3 $\delta_k=(1-\lambda)\lambda^{k}$ and $X_n=\sum_{k=1}^n\delta_n$. Lemma 24. For all $n\geqslant 1$ the inequalities $n<\sqrt{2X_n}$ and $e_{n+1}-e_n>\sqrt{2X_{n+2}}-\sqrt{2X_{n+1}}$ hold. Proof. The first assertion is proved using induction. For $n=1$ the inequality is verified directly. Since the function $\sqrt{x}$ is convex, it can readily be seen that as required.
Let us prove the second assertion. It can readily be seen that On the other hand $\sqrt{2X_{n+2}}>\sqrt{2X_{n+1}}>e_{n+1}$ as proved above, and therefore as required. This completes the proof of the lemma.Now we are ready to prove Lemma 21. Proof of Lemma 21. By Lemma 24 $\sqrt{2X_{n+1}}-e_n$ is a decreasing positive sequence. That is,
$$
\begin{equation*}
0\,{<}\,\sqrt{2X_N}-e_{N-1}<\sqrt{2X_2}-e_1<2\sqrt{X_2} =\sqrt{2(1-\lambda)(\lambda+\lambda^2)} <2\sqrt{1-\lambda}=o(\varepsilon^{2}).
\end{equation*}
\notag
$$
On the other hand $2X_N=2\sum_{k=1}^N (1-\lambda)\lambda^{k}=2(1-\lambda^{N+1})=2+o(\varepsilon^5)$. Hence $e_{N-1}=\sqrt{2}+o(\varepsilon^{2})$. It follows from (86) that $e_{N}$ is also equal to $\sqrt{2}+o(\varepsilon^{2})$. Therefore, This completes the proof of the lemma. 3.4. The proof of the first assertion of Theorem 1We argue by contradiction. Let the inequality
$$
\begin{equation*}
\varphi^{(2)}_x(t)=\kappa_2t-S_x(t)<(\kappa_5-\varepsilon)\sqrt{t}
\end{equation*}
\notag
$$
hold for all sufficiently large $t$. Then by Lemma 11, for infinitely many values of $t$ there exists a continuant $\langle D_t\rangle$ of the form (27) such that for all $\nu$ satisfying $\sqrt{t}<\nu\leqslant t$ we have
$$
\begin{equation*}
\frac{\langle D_{\nu}\rangle}{\sqrt{2}^{\,S(D_{\nu})}} >1
\end{equation*}
\notag
$$
and
$$
\begin{equation}
\varphi^{(2)}(D_\nu)<(\kappa_5-\varepsilon)\sqrt{\nu}.
\end{equation}
\tag{90}
$$
Applying Lemma 18 to (72) we see that there exists an integer $m_k$, ${t}/{\ln{t}}<m_k\leqslant t$, such that
$$
\begin{equation}
\begin{aligned} \, \notag \varphi^{(2)}(D_{m_k}) &\geqslant\sqrt{8\eta}\, \alpha\sqrt{t_k}(1+o(\varepsilon)) =\sqrt{8\eta}\,\alpha\sqrt{m_k}(1+o(\varepsilon)) \\ &=\sqrt{\frac{8(5-\kappa_2)(\kappa_2-4)\ln{(\gamma^{-1})}}{\ln{\widetilde{\lambda}}}} \sqrt{m_k}(1+o(\varepsilon))=\kappa_5\sqrt{m_k}(1+o(\varepsilon)). \end{aligned}
\end{equation}
\tag{91}
$$
We have arrived at a contradiction with (90). The first assertion of Theorem 1 is proved. 3.5. The proof of the first assertion of Theorem 2We argue by contradiction again. Let the inequality
$$
\begin{equation*}
\max_{u\leqslant t}\varphi^{(2)}_x(u)<\biggl(\frac{\kappa_5}{2}-\varepsilon\biggr)\sqrt{t}
\end{equation*}
\notag
$$
hold for infinitely many values of $t$. Then by Lemma 2, for infinitely many values of $t_0$ there exists a continuant $\langle D_{t_0}\rangle$ such that
$$
\begin{equation*}
\frac{\langle D_{t_0}\rangle}{\sqrt{2}^{\,S(D_{t_0})}} >1
\end{equation*}
\notag
$$
and
$$
\begin{equation}
\max_{\nu \leqslant t_0}\varphi^{(2)}(D_{\nu})\leqslant\biggl(\frac{\kappa_5}{2} -\varepsilon\biggr)\sqrt{t_0}.
\end{equation}
\tag{92}
$$
It follows from (72) and Lemma 21 that for all sufficiently large $t_0$ there exist integers $k$ and $m_k$ satisfying $1\leqslant k\leqslant N$ and ${t_0}/{\ln{t_0}}<m_k\leqslant t_0$ for which
$$
\begin{equation*}
\sqrt{\lambda}^{\,k}\varphi^{(2)}(D_{m_k})\geqslant\sqrt{2\eta}\, \alpha\sqrt{t_k}(1+o(\varepsilon)) =\frac{\kappa_5}{2}\sqrt{t_k}(1+o(\varepsilon)).
\end{equation*}
\notag
$$
In other words,
$$
\begin{equation*}
\max_{\nu\leqslant t_0}\varphi^{(2)}(D_{\nu})\geqslant\varphi^{(2)}(D_{m_k}) \geqslant\frac{\kappa_5}{2}\sqrt{t_0}(1+o(\varepsilon)),
\end{equation*}
\notag
$$
which contradicts (92). The first assertion of Theorem 2 is proved.
§ 4. Construction of examples4.1. The proof of the second assertion of Theorem 14.1.1. The block structureIn this subsection, we construct explicitly a continued fraction for an irrational number $x$ such that $?'(x)=+\infty$ and (23) holds for all sufficiently large $t$. First of all, we define the positive constant
$$
\begin{equation}
c_0:=\sqrt{\frac{2(5-\kappa_2)\ln({\gamma^{-1})}}{(\kappa_2-4) \ln{\widetilde{\lambda}}}}=\frac{\kappa_5}{2(\kappa_2-4)},
\end{equation}
\tag{93}
$$
which plays a key role in our constructions. Lemma 25. For any sufficiently large positive integer $s_{k}$ there exist positive integers $s_{k+1}$, $L^{(4)}_k$ and $L^{(5)}_k$ satisfying the following conditions: Proof. The first and second inequalities in (94) mean that the numbers $L^{(4)}_k$ and $L^{(5)}_k$ range over an interval of length $\sim \sqrt{s_k}$. We recall that $\varepsilon$ is an absolute constant fixed in the statement of the theorem under consideration. Therefore, for sufficiently large $s_k$ the linear form has values in any interval of fixed length. The existence of a number $s_{k+1}$ satisfying (94) is obvious: it is sufficient to consider the integer multiple of $L^{(4)}_k+L^{(5)}_k$ closest to ${s_0}/{\lambda^{k+1}}$. This completes the proof of the lemma. By the block $C^{(k)}$, where $k\geqslant 0$, we mean the sequence
$$
\begin{equation}
C^{(k)}=\bigl(4_{L^{(4)}_k},5_{L^{(5)}_k},4_{L^{(4)}_k},5_{L^{(5)}_k},\dots,4_{L^{(4)}_k}, 5_{L^{(5)}_k}\bigr)
\end{equation}
\tag{95}
$$
of length $s_{k+1}-s_k$. It consists of $(s_{k+1}-s_k)/(L^{(4)}_k+L^{(5)}_k)\in\mathbb{N}$ sequences of the form $(4_{L^{(4)}_k},5_{L^{(5)}_k})$. We denote the sequence $5_{s_0-1}$ by $C^{(-1)}$. Finally, we set
$$
\begin{equation}
x:=[0;C^{(-1)},C^{(0)},C^{(1)},\dots,C^{(k)},\dots]:=[0;a_1,a_2,\dots,a_t,\dots],
\end{equation}
\tag{96}
$$
where the infinite sequence of blocks $C^{(k)}$ is defined by (95) and (94). It remains to show that the irrational number $x$ constructed in this way satisfies the condition in the second assertion of Theorem 1. 4.1.2. $?'(x)=+\infty$Recall that $A_{\nu}=(a_1,a_2,\dots,a_{\nu})$ is the sequence of $\nu$ first partial quotients of the irrational number $x$. We introduce the function
$$
\begin{equation*}
f_x(t)=\frac{\langle A_t \rangle}{\sqrt{2}^{\,S_x(t)}}.
\end{equation*}
\notag
$$
It follows from Lemma 1 and the boundedness of the partial quotients of $x$ that if then $?'(x)=+\infty$. It can also readily be seen that $f_x(t)>f_x(t-1)$ for $a_{t}=4$ and $f_x(t)<f_x(t-1)$ for $a_{t}=5$. Therefore, it suffices to consider only the values of $t$ for which $a_{t}=5$ and $a_{t+1}=4$. It follows from the inequality $\langle A,B\rangle>\langle A\rangle\langle B\rangle$ that it suffices to show that
$$
\begin{equation}
\frac{\bigl\langle 4_{L^{(4)}_k},5_{L^{(5)}_k}\bigr\rangle}{\sqrt{2}^{\,L^{(4)}_k +L^{(5)}_k}}>1+\delta
\end{equation}
\tag{97}
$$
for some $\delta>0$ and all $k\geqslant 1$. We denote the continuant in the numerator of (97) by $\langle E\rangle$. We can apply the bound from Lemma 12 to this continuant, taking into account that $\sigma(E)=1$ and
$$
\begin{equation*}
\varphi^{(2)}(E)=\frac{\ln{(\gamma^{-1})}}{\ln{\widetilde{\lambda}}} \biggl(1+\frac{\varepsilon}{8}+o(\varepsilon^2)\biggr)
\end{equation*}
\notag
$$
by (94). Hence
$$
\begin{equation}
\frac{\langle E\rangle}{\sqrt{2}^{\,S(E)}} >(\gamma-\varepsilon_0)\widetilde{\lambda}^{\ln(\gamma^{-1})(1+{\varepsilon}/{8} +o(\varepsilon^2))/\ln{\widetilde{\lambda}}}= \frac{\gamma-\varepsilon_0}{\gamma^{1+\varepsilon/8+o(\varepsilon^2)}}.
\end{equation}
\tag{98}
$$
Since the quantity $\varepsilon_0$ can be made arbitrarily small for fixed $\varepsilon$, the right-hand side of (98) is strictly greater than 1. Hence the quantity $f_x(t)$ tends to infinity, and thus we also have $?'(x)=+\infty$. 4.1.3. The verification of (23)To complete the proof of the second assertion of Theorem 1 it remains to show that (23) holds for all sufficiently large $t$. By Lemma 3 it suffices to establish this inequality for all $t$ such that $a_t=4$ and $a_{t+1}=5$. To formulate the following lemma we introduce the notation $s'_k=s_k+L^{(4)}_k$. Proof. Note that
$$
\begin{equation}
\varphi^{(2)}(A_{s'_k})= (\kappa_2-4)L^{(4)}_k+\sum_{i=-1}^{k-1}\varphi^{(2)}(C^{(i)}).
\end{equation}
\tag{100}
$$
Let us estimate the quantity $\varphi^{(2)}(C^{(i)})$ from above for $i\geqslant 0$. By (94) we have
$$
\begin{equation}
\varphi^{(2)}(C^{(i)})=\frac{s_{i+1}-s_i}{L^{(4)}_i+L^{(5)}_i}\, \frac{\ln{(\gamma^{-1})}}{\ln{\widetilde{\lambda}}} \biggl(1+\frac{\varepsilon}{8}+o(\varepsilon^2)\biggr).
\end{equation}
\tag{101}
$$
Further, since
$$
\begin{equation}
\frac{s_{i+1}-s_i}{L^{(4)}_i+L^{(5)}_i}=\frac{s_{i+1}(1-\lambda)(5-\kappa_2)} {c_0\sqrt{s_i}}(1+o(\varepsilon^2))=\frac{\sqrt{s_{i}}\,(1-\lambda)(5-\kappa_2)} {c_0}(1+o(\varepsilon^2)),
\end{equation}
\tag{102}
$$
we obtain the bound
$$
\begin{equation}
\varphi^{(2)}(C^{(i)})=\frac{\sqrt{s_{i}}\,(1-\lambda)(5-\kappa_2)}{c_0}\, \frac{\ln{(\gamma^{-1})}}{\ln{\widetilde{\lambda}}}\biggl(1+\frac{\varepsilon}{8} +o(\varepsilon^2)\biggr).
\end{equation}
\tag{103}
$$
We also note that, since $(1-\lambda)=o(\varepsilon^5)$, we have Summing the bounds (103) as in (100) and combining the higher-order terms into $o(\varepsilon^2)$ we obtain
$$
\begin{equation}
\varphi^{(2)}(A_{s'_k})= \biggl((\kappa_2-4)c_0+\biggl(1+\frac{\varepsilon}{8}\biggr) \frac{(1-\lambda)(5-\kappa_2)}{c_0}\,\frac{\ln{(\gamma^{-1})}} {\ln{\widetilde{\lambda}}}\sum_{i=0}^{k-1}\sqrt{\lambda}^{\,i}\biggr) \sqrt{s_k}(1+o(\varepsilon^2)).
\end{equation}
\tag{105}
$$
Replacing $\sum_{i=0}^{k-1}\sqrt{\lambda}^{\,i}$ by an infinite sum yields
$$
\begin{equation}
\varphi^{(2)}(A_{s'_k})<\biggl((\kappa_2-4)c_0+ \biggl(1+\frac{\varepsilon}{4}\biggr)\frac{2(5-\kappa_2)}{c_0}\, \frac{\ln{(\gamma^{-1})}}{\ln{\widetilde{\lambda}}}\biggr)\sqrt{s_k}.
\end{equation}
\tag{106}
$$
Finally, substituting in $c_0$ from (93) we see that as required. Thus, we have shown that if $t$ is such that $a_t$ is the last element in the first part of the form $4_{L^{(4)}_k}$ in the block $C^{(k)}$, then (23) holds for this $t$. By Lemma 3 it remains to consider the case when $a_t$ is the last element in an arbitrary part of the form $4_{L^{(4)}_k}$ in $C^{(k)}$. We denote the sequence $(5_{L^{(5)}_k}, 4_{L^{(4)}_k})$ by $F$. Proof. In the case when $n=0$ the assertion follows immediately from Lemma 26. In the other cases we obtain
$$
\begin{equation}
\varphi^{(2)}\bigl(C^{(-1)},C^{(0)},C^{(1)},\dots,C^{(k-1)},4_{L^{(4)}_k}, \underbrace{F,F,\dots,F}_{n \text{ times}}\, \bigr)=\varphi^{(2)}(A_{s'_k})+n\varphi^{(2)}(F).
\end{equation}
\tag{108}
$$
Since $\varphi^{(2)}(F)\!>\!0$ by (94), it suffices to establish (108) for $n\!=\!(s_{k+1}-s_k)/(L^{(4)}_k+L^{(5)}_k)$. We note that $(\underbrace{F,F,\dots,F}_{n \text{ times}})=C^{(k)}$ in this case. Substituting (99) and (104) into (108) we obtain
$$
\begin{equation}
\varphi^{(2)}\bigl(C^{(-1)},C^{(0)},C^{(1)},\dots,C^{(k-1)},4_{L^{(4)}_k}, \underbrace{F,F,\dots,F}_{n \text{ times}}\, \bigr) <\biggl(\kappa_5+\frac{\varepsilon}{4}+\varepsilon^4\biggr)\sqrt{s_{k}},
\end{equation}
\tag{109}
$$
as required. This completes the proof of the lemma, and thus the proof of the second assertion of Theorem 1 is also complete. 4.2. The proof of the second assertion of Theorem 2In this subsection we construct explicitly a continued fraction for an irrational number $x=[0;a_1,a_2,\dots,a_t,\dots]$ such that $?'(x)=+\infty$ and, at the same time, inequality (25) holds for infinitely many values of $t$. The construction is similar to the one in the proof of Theorem 1 but it is slightly more complicated. 4.2.1. SuperblocksWe recall that the number $N$ was defined in (53). Fix an arbitrary number $S_0$ such that $S_0(1-\lambda)^N\lambda^N>S_0/\ln{S_0}$. We define the numbers $S_i$ in accordance with the formula $S_i=[{S_{i-1}}/{\lambda^N}]$. For all $i\geqslant 0$ we set $s^{(i)}_0=S_i$. We recall that the numbers $d_1,d_2,\dots, d_N$ are found using (80), where $\eta$ is given in (71). However, it is convenient to set $d_1=d_2=\sqrt{{(1-\lambda)\lambda}/{\eta}} =o(\varepsilon^5)$ and leave the other elements of the sequence unchanged. Lemma 28. For every number $s^{(i)}_{k}$, where $0\leqslant k<N$, $i\geqslant 1$, there exist positive integers $s^{(i)}_{k+1}$, $L^{(4,i)}_k$ and $L^{(5,i)}_k$ satisfying the following conditions: The proof is perfectly similar to the proof of Lemma 25. Now we proceed to the construction of the continued fraction of an irrational number $x$. By the block $C^{(i)}_k$ of the $i$th level, where $i\geqslant 0$ and $1\leqslant k\leqslant N$, we mean the sequence
$$
\begin{equation}
C^{(i)}_k=\bigl(4_{L^{(4,i)}_k},5_{L^{(5,i)}_k},4_{L^{(4,i)}_k},5_{L^{(5,i)}_k}, \dots,4_{L^{(4,i)}_k},5_{L^{(5,i)}_k}\bigr)
\end{equation}
\tag{111}
$$
of length $s^{(i)}_{k}-s^{(i)}_{k+1}$. For every $i\geqslant 0$, by the superblock $\mathcal{C}^{(i)}$ we mean the sequence of all blocks of the $i$th level:
$$
\begin{equation*}
\mathcal{C}^{(i)}=(C^{(i)}_N,C^{(i)}_{N-1},\dots,C^{(i)}_1).
\end{equation*}
\notag
$$
Finally, we define the number $x$ as follows:
$$
\begin{equation*}
x=[0;\mathcal{C}^{(-1)},\mathcal{C}^{(0)},\mathcal{C}^{(0)},\dots,\mathcal{C}^{(n)},\dots]).
\end{equation*}
\notag
$$
Here the initial superblock $\mathcal{C}^{(-1)}$ has the form $5_{S^{(0)}_N}$. Thus, the construction of the irrational number $x$ is complete. The proof of the fact that $?'(x)=+\infty$ is completely analogous to the corresponding proof in § 4.1.2: it immediately follows from (110) that
$$
\begin{equation*}
\frac{\bigl\langle 4_{L^{(4,i)}_k},5_{L^{(5,i)}_k}\bigr\rangle}{\sqrt{2}^{\,L^{(4,i)}_k +L^{(5,i)}_k}}>1+\delta,
\end{equation*}
\notag
$$
where $\delta$ is a positive constant independent of $i$ and $k$. It remains to establish inequality (25). 4.2.2. The verification of inequality (25)In this subsection we show that (25) holds for all $S_i$, $i\geqslant 0$. First we prove this for $i=0$. We must show that for any $\nu\leqslant S_0$ we have
$$
\begin{equation*}
\varphi^{(2)}(A_{\nu})\leqslant\biggl(\frac{\kappa_5}{2}+\varepsilon\biggr)\sqrt{S_0}.
\end{equation*}
\notag
$$
Recall that $A_{\nu}=(a_1,a_2,\dots,a_{\nu})$ is the sequence of first $\nu$ partial quotients of $x$. As mentioned already, it suffices to consider the case when $a_{\nu}=4$ and $a_{\nu+1}\,{=}\,5$. First of all, we prove an analogue of Lemma 26. To formulate it we introduce the notation $s'^{(0)}_k=s^{(0)}_k+L^{(4,0)}_k$. Proof. It can readily be seen from (110) that
$$
\begin{equation}
\begin{aligned} \, \notag &\varphi^{(2)}(A_{s'^{(0)}_k})=\varphi^{(2)}(\mathcal{C}^{(-1)})+(\kappa_2-4)L^{(4,0)}_k +\sum_{i=N}^{k+1}\varphi^{(2)}(C^{(0)}_i) \\ &\qquad<(\kappa_2-4) d_k\sqrt{S_0}(1+\varepsilon^4) +\frac{\ln{(\gamma^{-1})}}{\ln{\widetilde{\lambda}}} \biggl(1+\frac{\varepsilon}{8}+o(\varepsilon^2)\biggr) \sum_{i=N}^{k+1}\frac{s^{(0)}_{i}-s^{(0)}_{i+1}}{L^{(4,0)}_i+L^{(5,0)}_i}. \end{aligned}
\end{equation}
\tag{113}
$$
Now, since
$$
\begin{equation}
\frac{s^{(0)}_{i}-s^{(0)}_{i+1}}{L^{(4,0)}_i+L^{(5,0)}_i}= \frac{\lambda^i(1-\lambda)(5-\kappa_2)}{d_i}\sqrt{S_0}(1+o(\varepsilon^2)),
\end{equation}
\tag{114}
$$
we obtain the bound
$$
\begin{equation}
\begin{aligned} \, \notag \varphi^{(2)}(A_{s'^{(0)}_k}) &< \biggl(d_k+\biggl(1+\frac{\varepsilon}{8}\biggr)\frac{(1-\lambda)(5-\kappa_2) \ln{(\gamma^{-1})}}{(\kappa_2-4)\ln{\widetilde{\lambda}}} \sum_{i=N}^{k+1}\frac{\lambda^{k-i}}{d_i}\biggr) \\ &\qquad\times(\kappa_2-4)\sqrt{S_0}(1+o(\varepsilon^2)). \end{aligned}
\end{equation}
\tag{115}
$$
We set
$$
\begin{equation*}
\alpha=\biggl(1+\frac{\varepsilon}{8}\biggr) \frac{(5-\kappa_2)\ln{(\gamma^{-1})}}{(\kappa_2-4)^2\ln{\widetilde{\lambda}}} =\biggl(1+\frac{\varepsilon}{8}\biggr)\frac{\kappa_5^2}{8(\kappa_2-4)}, \qquad \eta=\frac{1}{\alpha}.
\end{equation*}
\notag
$$
Then (115) can be written in the form
$$
\begin{equation}
\varphi^{(2)}(A_{s'^{(0)}_k}) <\biggl(\eta d_k+(1-\lambda)\sum_{i=N}^{k+1}\frac{\lambda^{k-i}}{d_i}\biggr) (\kappa_2-4)\alpha\sqrt{S_0}(1+o(\varepsilon^2)).
\end{equation}
\tag{116}
$$
Hence from Lemma 21 we obtain
$$
\begin{equation}
\begin{aligned} \, \notag \varphi^{(2)}(A_{s'^{(0)}_k}) &<\sqrt{2\eta}(\kappa_2-4)\alpha\sqrt{S_0}(1+o(\varepsilon)) \\ &=(\kappa_2-4)\sqrt{2\alpha}\sqrt{S_0}(1+o(\varepsilon))< \biggl(1+\frac{\varepsilon}{4}\biggr)\frac{\kappa_5}{2}\sqrt{S_0}, \end{aligned}
\end{equation}
\tag{117}
$$
as required. To formulate the following lemma we set $F=(5_{L^{(5,0)}_k}, 4_{L^{(4,0)}_k})$. The proof is completely analogous to Lemma 27. Thus, inequality (25) holds for $t=S_0$. Arguing in just the same way we can show that for any $n\geqslant 0$ we have
$$
\begin{equation}
\varphi^{(2)}(\mathcal{C}^{(n)})\leqslant \biggl(\frac{\kappa_5}{2} +\frac{\varepsilon}{4}\biggr)\sqrt{S_n}.
\end{equation}
\tag{119}
$$
Now let us show that inequality (25) holds for $t=S_n$ for all $n\geqslant 1$. Arguing similarly to $n=0$ we see that it suffices to show that
$$
\begin{equation}
\varphi^{(2)}(A_{\nu})\leqslant \biggl(\frac{\kappa_5}{2}+\frac{\varepsilon}{2}\biggr)\sqrt{S_n},
\end{equation}
\tag{120}
$$
where $S_{n-1}<\nu\leqslant S_n$ and
$$
\begin{equation*}
A_{\nu}=\bigl(\mathcal{C}^{(-1)},\mathcal{C}^{(0)},\dots, \mathcal{C}^{(n-1)},C^{(n)}_N,C^{(n)}_{N-1},\dots,C^{(n)}_{k+1},4_{L^{(4,n)}_k}\bigr)
\end{equation*}
\notag
$$
for some $0\leqslant k<N$. We express $\varphi^{(2)}(A_{\nu})$ as the sum
$$
\begin{equation}
\varphi^{(2)}(A_{\nu})=\sum_{i=-1}^{n-1}\varphi^{(2)}(\mathcal{C}^{(i)}) +\sum_{i=N}^{k+1}\varphi^{(2)}(C^{(n)}_i)+(\kappa_2-4)L^{(4,n)}_k.
\end{equation}
\tag{121}
$$
We estimate the first sum in (121) using (119):
$$
\begin{equation}
\sum_{i=-1}^{n-1}\varphi^{(2)}(\mathcal{C}^{(i)})<\sum_{i=0}^{n-1}\varphi^{(2)} (\mathcal{C}^{(i)})\leqslant \sqrt{S_n}\sum_{i=1}^{n}\sqrt{\lambda}^{\,Ni} < \sqrt{S_n}\sum_{i=1}^{\infty}\sqrt{\lambda}^{\,Ni}= \sqrt{S_n} o(\varepsilon^4).
\end{equation}
\tag{122}
$$
In exactly the same way as in Lemma 29, the other half of the sum in (121) is estimated as follows:
$$
\begin{equation}
\sum_{i=N}^{k+1}\varphi^{(2)}(C^{(n)}_i)+(\kappa_2-4)L^{(4,n)} \leqslant\biggl(\frac{\kappa_5}{2}+\frac{\varepsilon}{4}\biggr)\sqrt{S_n}.
\end{equation}
\tag{123}
$$
Substituting (122) and (123) into (121) we see that inequality (120) holds for all ${n\geqslant 1}$. The second assertion of Theorem 2 is completely proved.
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