Some trigonometric polynomials with extremely small uniform norm and their applications

We begin with some necessary definitions. The sets of positive integers, integers and real numbers are denoted by $\mathbb{N}$, $\mathbb{Z}$ and $\mathbb{R}$, respectively. Given $f\in L^{p}(0,2\pi)$, put

$$\begin{equation*} \begin{alignedat}{2} \|f\|_{p}&=\biggl(\int_{0}^{2\pi}|f(x)|^{p}\, dx\biggr)^{1/p} &\text{when} \quad 1&\leq p <\infty, \\ \|f\|_{\infty}&= \mathrm{ess}\sup_{[0, 2\pi]} |f(x)| &\text{when}\quad p &= \infty. \end{alignedat} \end{equation*}$$

Given $f\in L^{1}(0,2\pi)$, we consider the Fourier coefficients

$$\begin{equation*} \begin{aligned} \, a_{k}(f)&=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\cos kx\, dx, \\ b_{k}(f)&=\frac{1}{\pi}\int_{0}^{2\pi}f(x)\sin kx\, dx,\qquad k=0, 1, 2, \dots, \\ c_{n}(f)&=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)e^{-i n x}\, dx,\qquad n\in \mathbb{Z}. \end{aligned} \end{equation*}$$

Given a real number $x$, we denote by $[x]$ its integer part and by $\lceil x\rceil$ the least integer $n$ such that $n\geq x$. We write $|A|$ for the cardinality of a finite set $A$. We denote the exact order of a non-zero trigonometric polynomial $T(x)$ by $\deg(T)$. The Lebesgue measure of $E\subset \mathbb{R}^{1}$ is denoted by $\operatorname{mes}(E)$.

If $x, y\in (0, 2\pi]$, then $\operatorname{dist}(x,y)$ stands for the distance around the circle, that is,

$$\begin{equation*} \operatorname{dist}(x,y)= \begin{cases} |x-y|, &\text{when }|x-y|\leq\pi, \\ 2\pi - |x-y|, &\text{when }|x-y|>\pi. \end{cases} \end{equation*}$$

Let $\varepsilon >0$ be a real number and $X$ a subset of $(0,2\pi]$. We define $O_{\varepsilon}(X)$ as follows. Put

$$\begin{equation*} O_{\varepsilon}(X)= \{ y\in (0,2\pi]\colon \exists\, x\in X \text{ such that }\operatorname{dist}(x,y)<\varepsilon\} \end{equation*}$$

when $\varnothing \neq X\subset (0,2\pi]$ and $O_{\varepsilon}(X)=\varnothing$ when $X=\varnothing$. We also define $\operatorname{Conn}(X)$ for $X\subset (0,2\pi]$: put $\operatorname{Conn}(X)=k$ when $X$ is a non-empty subset of $(0,2\pi]$ having $k\in \mathbb{N}$ components with respect to the topology of the circle and $\operatorname{Conn}(X)=0$ when $X=\varnothing$.

Let $K_{0}(x)\equiv 1$. For a positive integer $n\geq 2$ put

$$\begin{equation*} K_{n-1}(x)=\sum_{j=-(n-1)}^{n-1} \biggl(1-\frac{|j|}{n}\biggr)e^{i j x}=1+2\sum_{j=1}^{n-1}\biggl(1-\frac{j}{n}\biggr)\cos jx =\frac{(\sin ({nx}/{2}))^{2}}{n(\sin({x}/{2}))^{2}}. \end{equation*}$$

The function $K_{n-1}(x)$ is called the Fejér kernel of order $n-1$. We mention some properties of the Fejér kernel. If $n$ is a positive integer, then $K_{n-1}(x)$ is a non-zero real positive even trigonometric polynomial, $\deg(K_{n-1})= n-1$, $K_{n-1}(0)=n$, $\|K_{n-1}\|_{\infty}=n$, $\|K_{n-1}\|_{1}=2\pi$, and the following inequality holds:

$$\begin{equation} \frac{1}{n}K_{n-1}(x-y)\leq \min\biggl(1,\frac{{\pi}^{2}}{n^{2}\operatorname{dist}^{2}(x,y)}\biggr)\quad \forall\,x, y\in(0,2\pi]. \end{equation} \tag{ 1.1 }$$

Let $N$ be a positive integer. Given $x=(x_1,\dots, x_N)\in \mathbb{R}^{N}$, put

$$\begin{equation*} \|x\|_{l_{p}^{N}}=\begin{cases} \biggl({\displaystyle\sum_{j=1}^{N}|x_j|^{p}}\biggr)^{1/p}, &\text{when }1\leq p < \infty, \\ {\displaystyle\max_{1\leq j \leq N} |x_j|}, &\text{when }p=\infty. \end{cases} \end{equation*}$$

We denote by $(v)_i$, $1\leq i \leq N$, the $i$th coordinate of $v\in \mathbb{R}^{N}$. We write $\dim V$ for the dimension of a finite-dimensional vector space $V$ over $\mathbb{R}$.

Put

$$\begin{equation*} \mathrm{T}(0)=E_0=\{f(x)=c,\, x\in \mathbb{R},\, c\in \mathbb{R}\}. \end{equation*}$$

Let $n$ be a positive integer. We denote by $\mathrm{T}(n)$ the vector space of real trigonometric polynomials of the form

$$\begin{equation*} t(x)=A+\sum_{k=1}^{n} (a_k \cos kx + b_k\sin kx) \end{equation*}$$

and by $E_n$ the vector space of real trigonometric polynomials of the form

$$\begin{equation*} t(x)=\sum_{k=2^{n-1}}^{2^n - 1}(a_k \cos kx + b_k \sin kx). \end{equation*}$$

It easy to see that $\mathrm{T}(n)$ and $E_n$ are finite dimensional and that $\dim \mathrm{T}(n)=2n+1$ and $\dim E_n=2^n$. It is also evident that

$$\begin{equation*} \mathrm{T}(2^n - 1)= E_0\oplus E_1 \oplus\dots\oplus E_n. \end{equation*}$$

We first note the following fact. Let $\alpha>0$ be a real number and $\{t_n(x)\}_{n=1}^{\infty}$ a sequence of orthogonal trigonometric polynomials such that $\|t_n\|_{1}\geq \alpha$, $n=1, 2, \dots$ . Then

$$\begin{equation*} \begin{aligned} \, \biggl\|\sum_{j=1}^{n}t_{j}\biggr\|_{\infty}&\geq\frac{1}{\sqrt{2\pi}}\biggr\| \sum_{j=1}^{n}t_{j}\biggr\|_{2}= \frac{1}{\sqrt{2\pi}}\biggl(\,\sum_{j=1}^{n}\|t_{j}\|_{2}^{2}\biggr)^{1/2} \\ &\geq\frac{1}{\sqrt{2\pi}}\biggl(\frac{1}{2\pi}\sum_{j=1}^{n}\|t_{j}\|_{1}^{2} \biggr)^{1/2}\geq\frac{\alpha}{2\pi}\sqrt{n},\qquad n=1,2,\dots\,. \end{aligned} \end{equation*}$$

Thus, under the assumptions above, the uniform norm of $\sum_{j=1}^{n}t_{j}(x)$ cannot have order of growth smaller than $\sqrt{n}$. We have the following result.

Theorem 1.  Let $\lambda>1$ be a real number, $f\colon [1, +\infty)\to \mathbb{R}$ a positive non-increasing function, and $\{m_n\}_{n=1}^{\infty}$ a sequence of positive integers such that

$$\begin{equation*} \frac{m_{n+1}}{m_n}\geq \lambda^{f(n)}=:\lambda_n,\qquad n=1, 2, \dots\,. \end{equation*}$$

Also, let a sequence $\{d_n\}_{n=1}^{\infty}$ of positive integers satisfy the condition $1\leq d_n\leq m_{n+1}-m_n$, $n=1,2,\dots$ . Then there is a sequence $\{\delta_{n}(x)\}_{n=1}^{\infty}$ of complex trigonometric polynomials such that

$$\begin{equation*} \begin{gathered} \, \delta_{n}(x)=\sum_{m_n\leq s < m_n+d_n} c_s e^{i s x},\qquad \|\delta_n\|_1\geq \frac{5\pi}{8},\quad \|\delta_n\|_{\infty}\leq 6,\qquad n=1, 2,\dots, \\ \biggl\|\sum_{j=1}^{n}\delta_{j}\biggr\|_{\infty}\leq 277 + \alpha\sqrt{n}+180\max_{1\leq j\leq n}\log_{\lambda_j}\max\biggl(\frac{m_j}{d_j}, \frac{1}{\ln \lambda_j}, 1\biggr),\qquad n=1, 2,\dots, \end{gathered} \end{equation*}$$

where $\alpha>0$ is an absolute constant.

Remark 1.  Theorem 1 is a generalization of a result in [1], where the corresponding assertion is obtained in the case when $f(n)\equiv1$. The proof of Theorem 1 is based on ideas in [1]–[3].

Proof of Theorem 1. A celebrated result of Carleson and Hunt (see [4]) implies that if $g\in L^{2}(0,2\pi)$, and

$$\begin{equation*} \begin{aligned} \, \mathcal{S}_{N} (g, x)&=\sum_{|k|\leq N} c_{k}(g)e^{i k x},\qquad N=0, 1, 2, \dots, \\ \mathcal{S}^{*}(g, x)&=\sup_{N\geq 0} |\mathcal{S}_{N}(g, x)|, \end{aligned} \end{equation*}$$

then

$$\begin{equation} \|\mathcal{S}^{*}(g)\|_{2}^{2}\leq c_{H}\|g\|_{2}^{2}, \end{equation} \tag{ 2.1 }$$

where $c_{H}>0$ is an absolute constant. It is easy to see that $c_{H}\geq 1$. Note that if $n>j$ are positive integers, then

$$\begin{equation} \frac{m_n}{m_j}=\frac{m_{j+1}}{m_j}\cdots\frac{m_n}{m_{n-1}}\geq \lambda^{f(j)+\dots+f(n-1)}\geq \lambda^{f(n)(n-j)}=\lambda_{n}^{n-j}. \end{equation} \tag{ 2.2 }$$

Put $\alpha=\sqrt{72c_{H}}$. We will construct trigonometric polynomials $\delta_n(x)$, $n=1, 2, \dots$, such that

$$\begin{equation} \delta_{n}(x)\,{=}\,\sum_{m_n\leq s \leq m_n+2[(d_n-1)/2]} c_s e^{i s x},\qquad \|\delta_n\|_1\,{\geq}\, \frac{5\pi}{8},\quad \|\delta_n\|_{\infty}\,{\leq}\, 6,\qquad n\,{=}\,1, 2,\dots, \end{equation} \tag{ 2.3 }$$

$$\begin{equation} \biggl\|\sum_{j=1}^{n}\delta_{j}\biggr\|_{\infty}\leq 277 + \alpha\sqrt{n}+180\max_{1\leq j\leq n}\log_{\lambda_j}\max\biggl(\frac{m_j}{d_j}, \frac{1}{\ln \lambda_j}, 1\biggr),\qquad n=1, 2,\dots\,. \end{equation} \tag{ 2.4 }$$

It is obvious that this will prove the theorem. There is no loss of generality in assuming that $d_j \leq m_j$, $j=1,2,\dots$ . Put

$$\begin{equation*} \begin{aligned} \, \rho_{j}&= \max\biggl(\frac{m_j}{d_j}, \frac{1}{\ln \lambda_j}, 1\biggr)=\max\biggl(\frac{m_j}{d_j}, \frac{1}{\ln \lambda_j}\biggr),\qquad j=1, 2, \dots, \\ \tau_n&=\Bigl\lceil45+30\max_{1\leq j\leq n}\log_{\lambda_{j}}\rho_{j}\Bigr\rceil,\qquad n=1, 2, \dots, \\ \widetilde{d}_{j}&=\biggl[\frac{d_j - 1}{2}\biggr]+1,\qquad j=1,2,\dots\,. \end{aligned} \end{equation*}$$

Note that $\widetilde{d}_{j}$ is a positive integer and

$$\begin{equation} 1\leq \frac{d_j}{\widetilde{d}_{j}} \leq 2,\qquad j=1,2,\dots\,. \end{equation} \tag{ 2.5 }$$

We proceed to construct by induction trigonometric polynomials $\delta_{n}(x)$, $n=1, 2, \dots$, satisfying (2.3) and then show that (2.4) also holds. Put $\delta_{1}(x)=e^{i m_1 x}$. It is clear that it satisfies (2.3). Assume that $n\geq 2$ is a positive integer and $\delta_{1}(x),\dots, \delta_{n-1}(x)$ have been constructed. We introduce the notation

$$\begin{equation*} S_{j}(x):= \sum_{k=1}^{j}\delta_k(x). \end{equation*}$$

Put

$$\begin{equation} A_{n}^{j}=\{x\in (0,2\pi]\colon |S_{j}(x)|>\alpha\sqrt{n}\,\},\qquad j=1, \dots, n-1, \end{equation} \tag{ 2.6 }$$

$$\begin{equation} B_{n}=\bigcup_{j=1}^{n-1} A_{n}^{j}, \end{equation} \tag{ 2.7 }$$

$$\begin{equation} \widetilde{B}_{n}=\bigcup_{1\leq j\leq n-\tau_n}O_{{2\pi(n-j)^{2}}/{d_n}}(A_{n}^{j}), \end{equation} \tag{ 2.8 }$$

$$\begin{equation} \Lambda_{n}=\biggl\{s\in\{1,\dots,d_n\}\colon \frac{2\pi s}{d_n}\notin \widetilde{B}_{n}\biggr\}, \end{equation} \tag{ 2.9 }$$

$$\begin{equation} \delta_{n}(x)=\frac{1}{d_n}e^{i(m_n + {(d_n-1)}/{2})x}\sum_{s\in \Lambda_{n}} K_{\widetilde{d}_{n}-1}\biggl(x-\frac{2\pi s}{d_n}\biggr). \end{equation} \tag{ 2.10 }$$

When $n-\tau_n<1$, the union in (2.8) is defined as the empty set.

We claim that

$$\begin{equation} |\Lambda_n|\geq\frac{5}{16}d_n. \end{equation} \tag{ 2.11 }$$

Indeed, if $\widetilde{B}_{n}=\varnothing$, then $|\Lambda_n|=d_n$ and so (2.11) holds. Suppose that $\widetilde{B}_{n}\neq\varnothing$. Then

$$\begin{equation*} W= \{ 1\leq j \leq n-\tau_n\colon A_{n}^{j}\neq\varnothing\} \end{equation*}$$

is not empty and

$$\begin{equation*} \widetilde{B}_{n}=\bigcup_{j\in W}O_{{2\pi(n-j)^{2}}/{d_n}}(A_{n}^{j}). \end{equation*}$$

We will first find an upper bound for $\operatorname{mes}(B_n)$. Define

$$\begin{equation*} T(x)=\sum_{k=1}^{n-1}\delta_{k}(x),\qquad F_{n-1}^{*}(x)=\max_{1\leq j\leq n-1}\biggl|\sum_{k=1}^{j}\delta_{k}(x)\biggr|. \end{equation*}$$

It is evident that

$$\begin{equation*} F_{n-1}^{*}(x)\leq \mathcal{S}^{*}(T,x)\quad \forall\, x\in \mathbb{R}. \end{equation*}$$

Then we have (see (2.6), (2.7))

$$\begin{equation*} B_n= \bigl\{x\in(0,2\pi]\colon F_{n-1}^{*}(x)>\alpha\sqrt{n}\bigr\}=\bigl\{x\in(0,2\pi]\colon (F_{n-1}^{*}(x))^{2}>\alpha^{2}n\bigr\}. \end{equation*}$$

Applying Chebyshev's inequality and (2.1), and then using the orthogonality of the polynomials $\delta_j$ and the bound $\|\delta_j\|_{\infty}\leq 6$, $j=1,\dots,n-1$, we obtain

$$\begin{equation} \begin{aligned} \, \operatorname{mes}(B_n)&\leq \frac1{{\alpha}^{2}n}\int_{0}^{2\pi}(F_{n-1}^{*}(x))^{2}\,dx\leq \frac1{{\alpha}^{2}n}\int_{0}^{2\pi}(\mathcal{S}^{*}(T,x))^{2}\,dx =\frac{\|\mathcal{S}^{*}(T)\|_{2}^{2}}{{\alpha}^{2}n} \notag \\ &\leq\frac{c_H \|T\|_{2}^{2}}{{\alpha}^{2}n} =\frac{c_H}{{\alpha}^{2}n}\int_{0}^{2\pi}T(x)\,\overline{T}(x)\,dx =\frac{c_H}{{\alpha}^{2}n}\sum_{j=1}^{n-1}\int_{0}^{2\pi}|\delta_{j}(x)|^{2}\, dx \notag \\ &\leq \frac{c_H\cdot 36\cdot 2\pi(n-1)}{{\alpha}^{2}n}<\frac{72\pi c_H}{{\alpha}^{2}}=\pi. \end{aligned} \end{equation} \tag{ 2.12 }$$

Let $j\in W$. Then

$$\begin{equation*} \begin{aligned} \, S_{j}(x)&=\sum_{k=1}^{j}\delta_{k}(x)=\sum_{m_1\leq s \leq m_j + 2[(d_j-1)/2]} c_s e^{isx}, \\ |S_{j}(x)|^{2}&=S_{j}(x)\overline{S_{j}(x)} \\ &=\Biggl(\,\sum_{m_1\leq s \leq m_j + 2[(d_j-1)/2]} c_s e^{isx}\Biggr)\Biggl(\,\sum_{m_1\leq s \leq m_j + 2[(d_j-1)/2]} \overline{c_s} e^{-isx}\Biggr). \end{aligned} \end{equation*}$$

The trigonometric polynomial $S_j (x)$ is non-zero since the spectra of $\delta_{k}(x)$, $k=1,\dots, j$, are pairwise disjoint and $\delta_{1}(x)=e^{i m_1 x}$. Thus there is an $x_0 \in (0,2\pi]$ such that $S_{j}(x_0)\neq 0$. Therefore $|S_{j}(x_0)|>0$ and so $|S_{j}(x_0)|^{2}>0$. Hence $|S_{j}(x)|^{2}$ is a non-zero real trigonometric polynomial. It is easy to see that $\deg (|S_j|^2)\leq m_j +2[(d_j-1)/2]$. Put $T_{j}(x)=|S_{j}(x)|^{2}-\alpha^{2}n$. It is obvious that

$$\begin{equation} A_{n}^{j}=\{x\in (0,2\pi]\colon T_{j}(x)>0\} \end{equation} \tag{ 2.13 }$$

and $A_{n}^{j}\neq\varnothing$ since $j\in W$. Therefore there is an $x_1\in (0,2\pi]$ such that $T_{j}(x_1)>0$. Hence $T_{j}(x)$ is a non-zero real trigonometric polynomial and

$$\begin{equation*} \deg(T_j)=\deg(|S_{j}|^{2})\leq m_j +2\biggl[\frac{d_j-1}2\biggr]. \end{equation*}$$

It follows from well-known facts about trigonometric polynomial (see, for example, [5], V. 2, Ch. X, Theorem 1.7) that the number of zeros of $T_{j}(x)$ in $(0,2\pi]$ (counted with multiplicities) is equal to $R_j\leq 2(m_j +2[(d_j-1)/2])$. Note that

$$\begin{equation*} \begin{aligned} \, &\int_{0}^{2\pi} T_{j}(x)\, dx= \int_{0}^{2\pi}S_{j}(x)\overline{S_{j}(x)}\,dx-2\pi\alpha^{2}n =\sum_{k=1}^{j}\int_{0}^{2\pi}|\delta_{k}(x)|^{2}\,dx - 2\pi\alpha^{2}n \\ &\leq 72\pi j - 2\pi\alpha^{2}n<72\pi n - 2\pi\alpha^{2}n=72\pi n - 144\pi c_{H}n\leq -72\pi n<0. \end{aligned} \end{equation*}$$

Therefore we can find an $x_2\in(0,2\pi]$ such that $T_j(x_2)<0$. Since there is an $x_1\in (0,2\pi]$ such that $T_{j}(x_1)>0$, there is an $x_3\in(0,2\pi]$ such that $T_{j}(x_3)=0$. Hence $R_j>0$. Applying (2.13), we obtain that $A_{n}^{j}$ is the union of finitely many pairwise-disjoint open intervals on the circle $(0,2\pi]$ and

$$\begin{equation*} 2\cdot\operatorname{Conn}(A_{n}^{j})\leq R_j \leq 2\biggl(m_j +2\biggl[\frac{d_j-1}{2}\biggr]\biggr). \end{equation*}$$

Since

$$\begin{equation*} 2\biggl[\frac{d_j-1}{2}\biggr]\leq 2\cdot\frac{d_j-1}{2}=d_j-1< d_j\leq m_j, \end{equation*}$$

we have

$$\begin{equation} \operatorname{Conn}(A_{n}^{j}) \leq 2 m_j. \end{equation} \tag{ 2.14 }$$

Combining (2.7), (2.8), (2.12), (2.14) and (2.2), we obtain

$$\begin{equation*} \begin{aligned} \, &\operatorname{mes}(\widetilde{B}_{n})\leq \operatorname{mes}(B_n) + \sum_{j\in W} 2\cdot \frac{2\pi(n-j)^{2}}{d_n}\cdot \operatorname{Conn}(A_{n}^{j}) \leq \pi + \sum_{j\in W}\frac{8\pi (n-j)^{2}}{d_n} m_j \\ &\ \leq \pi + \sum_{1 \leq j \leq n-\tau_n}\frac{8\pi (n-j)^{2}}{d_n} m_j \leq \pi + \frac{8\pi m_n}{d_n} \sum_{j=1}^{n-\tau_n}\frac{(n-j)^{2}}{\lambda_{n}^{n-j}}\leq \pi + \frac{8\pi m_n}{d_n} \sum_{s=\tau_n}^{\infty}\frac{s^{2}}{\lambda_{n}^{s}}. \end{aligned} \end{equation*}$$

To continue estimating $\operatorname{mes}(\widetilde{B}_{n})$ we need the inequality

$$\begin{equation} \sum_{s=k}^{\infty} s^{2}q^{-s}\leq \frac{2k^{2}q^{3-k}}{(q-1)^{3}}\quad \text{when }q>1\text{ and }k=1,2,\dots\, \end{equation} \tag{ 2.15 }$$

(with $q=\lambda_n$ and $k = \tau_n$). This bound follows easily from the equality

$$\begin{equation*} \sum_{s=k}^{\infty}s^2 q^{-s}=\frac{q^{3-k}}{(q-1)^{3}}\bigl(k^{2}+q^{-1}(1+2k-2k^2)+q^{-2}(k-1)^{2}\bigr), \end{equation*}$$

for all $q>1$ and $k=0, 1, 2,\dots$, as is easily checked. Also, using the fact that

$$\begin{equation} \ln x\leq x-1\quad \text{when}\quad x>1, \end{equation} \tag{ 2.16 }$$

we have

$$\begin{equation*} \operatorname{mes}(\widetilde{B}_{n})\leq \pi + \frac{16\pi m_n}{d_n}\cdot\frac{{\tau_n}^{2}{\lambda_{n}}^{3-\tau_n}}{(\lambda_n - 1)^{3}}\leq \pi + \frac{16\pi m_n}{d_n}\cdot\frac{{\tau_n}^{2}{\lambda_{n}}^{3-\tau_n}}{\ln^{3} \lambda_n}. \end{equation*}$$

We now claim that

$$\begin{equation} 45+ 30\log_{\lambda_n}\rho_{n}\geq \frac{2}{\ln \lambda_n}. \end{equation} \tag{ 2.17 }$$

Two cases may occur.

1) Let $\ln \lambda_n \leq e^{-1/15}$. Then $\rho_n\geq 1/ \ln \lambda_n\geq e^{1/15}$. Hence,

$$\begin{equation*} \log_{\lambda_n}\rho_n\geq \frac{1}{15}\log_{\lambda_n} e=\frac{1}{15\ln \lambda_n}. \end{equation*}$$

Therefore,

$$\begin{equation*} 45+ 30\log_{\lambda_n}\rho_{n}\geq 45+ \frac{2}{\ln \lambda_n}> \frac{2}{\ln \lambda_n}. \end{equation*}$$

2) Let $\ln \lambda_n > e^{-1/15}$. Then $2/\ln \lambda_n < 2 e^{1/15}=2.1378\dots$ . Therefore,

$$\begin{equation*} 45+ 30\log_{\lambda_n}\rho_{n}\geq 45> \frac{2}{\ln \lambda_n}. \end{equation*}$$

Thus (2.17) is proved.

It is not hard to see that $f(x)=x^2 q^{-x}$ is decreasing when $x\geq 2/ \ln q$, where $q>1$ is a real number. It follows from the inequality

$$\begin{equation*} \tau_n\geq 45 + 30 \log_{\lambda_n}\rho_n \geq \frac{2}{\ln \lambda_n}, \end{equation*}$$

that

$$\begin{equation*} \begin{aligned} \, \tau_{n}^{2}\lambda_{n}^{3-\tau_n}&=\tau_{n}^{2}\lambda_{n}^{-\tau_n}\lambda_{n}^{3} \leq \lambda_{n}^{3}(45+30\log_{\lambda_n}\rho_n)^{2} \lambda_{n}^{-(45+30\log_{\lambda_n}\rho_n)} \\ &=(45+30\log_{\lambda_n}\rho_n)^{2}\lambda_{n}^{-42}\rho_{n}^{-30}. \end{aligned} \end{equation*}$$

Because

$$\begin{equation*} \frac{m_n} {d_n}\leq\rho_n\quad \text{and}\quad \frac{1}{\ln \lambda_n}\leq \rho_n, \end{equation*}$$

and also $(a+b)^2\leq 2 (a^2 + b^2)$, $a, b\in \mathbb{R}$, it follows that

$$\begin{equation*} \begin{aligned} \, \operatorname{mes}(\widetilde{B}_{n})&\leq \pi + 16\pi\frac{m_n}{d_n}\frac{(45+30\log_{\lambda_n} \rho_n)^{2}\lambda_{n}^{-42}\rho_{n}^{-30}}{\ln^{3}\lambda_n} \\ &\leq \pi+ 16\pi(45+30\log_{\lambda_n} \rho_n)^{2}\lambda_{n}^{-42}\rho_{n}^{-26} \\ &\leq \pi+ 32\pi\bigl(45^{2}+30^{2}(\log_{\lambda_n} \rho_n)^{2}\bigr)\lambda_{n}^{-42}\rho_{n}^{-26}. \end{aligned} \end{equation*}$$

Since $\ln x\leq x$ when $x>0$, we have

$$\begin{equation*} 0\leq \log_{\lambda_n}\rho_n=\frac{\ln \rho_n}{\ln \lambda_n}\leq \rho_n \ln \rho_n\leq \rho_n^{2}. \end{equation*}$$

Thus,

$$\begin{equation*} \operatorname{mes}(\widetilde{B}_{n})\leq \pi +32\pi (45^2\rho_{n}^{-26}+30^2\rho_{n}^{-22})\lambda_{n}^{-42}. \end{equation*}$$

We claim further that

$$\begin{equation} (45^2\rho_{n}^{-26}+30^2\rho_{n}^{-22})\lambda_{n}^{-42}\leq 2^{-8}. \end{equation} \tag{ 2.18 }$$

Two cases may occur.

1) Let $\lambda_n\geq e^{1/2}$. Then $\lambda_{n}^{-42}\leq e^{-21}$. It follows from $\rho_n \geq 1$ that

$$\begin{equation*} (45^2\rho_{n}^{-26}+30^2\rho_{n}^{-22})\lambda_{n}^{-42}\leq \frac{45^2 + 30^2}{e^{21}}<\frac{64^2+64^2}{2^{21}}=\frac{2(2^6)^2}{2^{21}}=2^{-8}. \end{equation*}$$

2) Let $1<\lambda_n < e^{1/2}$. Then $\rho_n\geq 1/\ln \lambda_n \geq 2$. Therefore,

$$\begin{equation*} (45^2\rho_{n}^{-26}+30^2\rho_{n}^{-22})\lambda_{n}^{-42}\leq \frac{45^2}{2^{26}}+\frac{30^2}{2^{22}}<\frac{45^2+30^2}{2^{22}}< \frac{64^2+64^2}{2^{22}}=2^{-9}<2^{-8}. \end{equation*}$$

Thus (2.18) is proved.

Consequently,

$$\begin{equation} \operatorname{mes}(\widetilde{B}_{n})\leq \pi +\frac{32\pi}{2^{8}}=\frac{9\pi}{8}. \end{equation} \tag{ 2.19 }$$

This bound implies, in particular, that $\widetilde{B}_{n}\neq (0,2\pi]$. Thus $\widetilde{B}_{n}$ is the union of finitely many pairwise-disjoint open intervals on the circle $(0,2\pi]$.

We claim, in addition, that

$$\begin{equation} \operatorname{Conn}(\widetilde{B}_{n})\leq\frac{d_n}{16}. \end{equation} \tag{ 2.20 }$$

We have (see (2.14), (2.2), (2.16)) that

$$\begin{equation*} \begin{aligned} \, &\operatorname{Conn}(\widetilde{B}_{n})\leq \sum_{j\in W} \operatorname{Conn}(A_{n}^{j})\leq \sum_{j\in W} 2 m_j \leq \sum_{1\leq j\leq n - \tau_n}2m_j \\ &\ \leq 2 \sum_{1\leq j\leq n - \tau_n}\frac{m_n}{\lambda_{n}^{n-j}}\leq 2m_n\sum_{s=\tau_n}^{\infty}\frac{1}{\lambda_{n}^{s}}= 2m_n \lambda_{n}^{-\tau_n}\frac{\lambda_n}{\lambda_n -1}\leq 2m_n \lambda_{n}^{-\tau_n}\frac{\lambda_n}{\ln\lambda_n}. \end{aligned} \end{equation*}$$

It follows from

$$\begin{equation*} \frac{1}{\ln \lambda_n}\leq \rho_n,\quad \tau_n\geq 45+30\log_{\lambda_n}\rho_n\quad\text{and}\quad \rho_n\geq \frac{m_n}{d_n} \end{equation*}$$

that

$$\begin{equation*} \begin{aligned} \, \operatorname{Conn}(\widetilde{B}_{n})\leq 2m_n \lambda_{n}^{-(45+30\log_{\lambda_n}\rho_n)}\lambda_n\rho_n=2m_n \lambda_{n}^{-44}\rho_{n}^{-29}\leq 2 d_n\lambda_{n}^{-44}\rho_{n}^{-28}. \end{aligned} \end{equation*}$$

Two cases may occur.

1) Let $\lambda_n\geq \sqrt{e}$. Then $\lambda_{n}^{-44}\leq e^{-22}$. It follows from $\rho_n \geq 1$ that

$$\begin{equation*} 2 d_n\lambda_{n}^{-44}\rho_{n}^{-28} \leq 2 d_n e^{-22}<2d_n 2^{-22}<\frac{d_n}{16}. \end{equation*}$$

2) Let $1< \lambda_n < \sqrt{e}$. Then $\rho_n \geq 1/\ln \lambda_n > 2$ and so

$$\begin{equation*} 2 d_n\lambda_{n}^{-44}\rho_{n}^{-28} \leq 2 d_n 2^{-28}<\frac{d_n}{16}. \end{equation*}$$

Thus (2.20) is proved.

Put

$$\begin{equation*} I_s=O_{{\pi}/{d_n}}\biggl(\frac{2\pi s}{d_n}\biggr),\qquad s=1,\dots,d_n. \end{equation*}$$

It is obvious that the length of $I_s$ is $2\pi/d_n$ and the intervals are pairwise disjoint. Consider

$$\begin{equation*} \Lambda_{n}^{c}=\biggl\{s\in\{1,\dots,d_n\}\colon \frac{2\pi s}{d_n}\in \widetilde{B}_{n}\biggr\}, \end{equation*}$$

the complement of the subset $\Lambda_n$ (see (2.9)). We will find an upper bound for $|\Lambda_{n}^{c}|$. It is not hard to see that $|\Lambda_{n}^{c}|$ is at most the number of intervals $I_s$, $s=1,\dots, d_n$, intersecting $\widetilde{B}_{n}$ non-trivially. Such intervals are of two types: those contained in $\widetilde{B}_{n}$ and those containing a boundary point of $\widetilde{B}_{n}$. Denote by $N_1$ and $N_2$ the numbers of intervals of the first and second types, respectively. Then (see (2.19))

$$\begin{equation*} N_1 \frac{2\pi}{d_n} \leq \operatorname{mes}(\widetilde{B}_{n})\leq \frac{9\pi}{8}. \end{equation*}$$

Therefore,

$$\begin{equation*} N_1\leq \frac{9}{16}d_n. \end{equation*}$$

It is clear (see (2.20)) that

$$\begin{equation*} N_2\leq 2 \operatorname{Conn}(\widetilde{B}_{n})\leq\frac{d_n}{8}. \end{equation*}$$

Hence,

$$\begin{equation*} |\Lambda_{n}^{c}|\leq \frac{9}{16}d_n +\frac{d_n}{8}=\frac{11}{16}d_n. \end{equation*}$$

Thus,

$$\begin{equation*} |\Lambda_n|=d_n-|\Lambda_{n}^{c}|\geq\frac{5}{16}d_n \end{equation*}$$

and (2.11) is proved.

We have (see (2.10))

$$\begin{equation*} \begin{gathered} \, |\delta_{n}(x)|=\frac{1}{d_n}\sum_{s\in \Lambda_{n}} K_{\widetilde{d}_{n}-1}\biggl(x-\frac{2\pi s}{d_n}\biggr), \\ \|\delta_n\|_1=\int_{0}^{2\pi}|\delta_{n}(x)|\,dx=\frac{1}{d_n}\sum_{s\in \Lambda_{n}}\int_{0}^{2\pi} K_{\widetilde{d}_{n}-1}\biggl(x-\frac{2\pi s}{d_n}\biggr)\, dx=\frac{1}{d_n}\cdot2\pi |\Lambda_n|\geq\frac{5\pi}{8}. \end{gathered} \end{equation*}$$

Let $x\in (0,2\pi]$. Then (see (2.10), (1.1), (2.5))

$$\begin{equation*} \begin{aligned} \, &|\delta_{n}(x)|=\frac{1}{d_n}\sum_{s\in \Lambda_{n}} K_{\widetilde{d}_{n}-1}\biggl(x-\frac{2\pi s}{d_n}\biggr)\leq \frac{1}{d_n}\sum_{s\in \Lambda_{n}} \widetilde{d}_{n} \min\biggl(1,\frac{{\pi}^{2}}{{\widetilde{d}_{n}}^{2}\operatorname{dist}^{2}(x,2\pi s/d_n)}\biggr) \\ &\ \leq \frac{1}{d_n}\sum_{1\leq s\leq d_n} \widetilde{d}_{n} \min\biggl(1,\frac{{\pi}^{2}}{{\widetilde{d}_{n}}^{2}\operatorname{dist}^{2}(x,2\pi s/d_n)}\biggr)\leq \frac{\widetilde{d}_{n}}{d_n}\cdot 2\biggl(1+\sum_{s=1}^{\infty}\frac{\pi^2}{\widetilde{d}_{n}^{2}(2\pi s/d_n)^{2}}\biggr) \\ &\ \leq 2\biggl(1+\frac{1}{4}\cdot \frac{d_{n}^{2}}{\widetilde{d}_{n}^{2}}\sum_{s=1}^{\infty}\frac{1}{s^{2}}\biggr)\leq 2\biggl(1+\sum_{s=1}^{\infty}\frac{1}{s^2}\biggr)=2\biggl(1+\frac{\pi^2}{6}\biggr)<6. \end{aligned} \end{equation*}$$

Hence $\|\delta_n\|_{\infty}\leq 6$. It is clear that $\delta_n (x)$ has the form (see (2.10))

$$\begin{equation*} \delta_{n}(x)=\sum_{m_n\leq s\leq m_n +2[(d_n-1)/2]} c_s e^{i s x}. \end{equation*}$$

Thus we have constructed trigonometric polynomials $\delta_{n}(x)$, $n=1, 2, \dots$, satisfying the conditions in (2.3).

We now claim that these trigonometric polynomials satisfy (2.4). Let $n$ be a positive integer. If $n=1$, then (2.4) holds. Suppose that $n\geq 2$. Put

$$\begin{equation*} \omega (x):=\max\big\{1\leq j \leq n-1\colon |S_{j}(x)|\leq \alpha\sqrt{n}\big\}. \end{equation*}$$

Since

$$\begin{equation*} |S_{1}(x)|=|e^{i m_1 x}|=1< \alpha\leq \alpha\sqrt{n}, \end{equation*}$$

the function $\omega (x)$ is well defined. Let $x\in (0,2\pi]$. Two cases may occur.

1) Let $\omega (x)\geq n- \tau_n$. Then

$$\begin{equation*} \begin{aligned} \, |S_{n}(x)|&\leq |S_{\omega(x)}(x)|+\sum_{j=\omega(x)+1}^{n}|\delta_j(x)|\leq \alpha\sqrt{n} + 6(n-\omega (x))\leq\alpha\sqrt{n}+6\tau_{n} \\ &\leq \alpha\sqrt{n}+6\Bigl(46+30\max_{1\leq j\leq n}\log_{\lambda_{j}}\rho_{j}\Bigr) \\ &=276 + \alpha\sqrt{n}+180\max_{1\leq j\leq n}\log_{\lambda_j}\max\biggl(\frac{m_j}{d_j}, \frac{1}{\ln \lambda_j}, 1\biggr). \end{aligned} \end{equation*}$$

2) Let $\omega (x)< n- \tau_n$. Then

$$\begin{equation*} \begin{aligned} \, |S_{n}(x)|&\leq |S_{\omega(x)}(x)|+\sum_{j=\omega(x)+1}^{\omega (x)+\tau_n}|\delta_j(x)|+ \sum_{j=\omega(x)+\tau_n +1}^{n}|\delta_j(x)| \\ &\leq \alpha\sqrt{n} + 6\tau_n + \sum_{j=\omega(x)+\tau_n +1}^{n}|\delta_j(x)| \\ &\leq 276 + \alpha\sqrt{n}+180\max_{1\leq j\leq n}\log_{\lambda_j}\max\biggl(\frac{m_j}{d_j}, \frac{1}{\ln \lambda_j}, 1\biggr) + \sum_{j=\omega(x)+\tau_n +1}^{n}|\delta_j(x)|. \end{aligned} \end{equation*}$$

We will estimate the last summand. Suppose that $\omega(x)+\tau_n +1 \leq j \leq n$. It follows from $\tau_j \leq \tau_n$ that $\omega(x)+\tau_j +1 \leq j \leq n$. We claim that

$$\begin{equation} x\in A_{j}^{\omega(x)+1}\subset \widetilde{B}_{j}. \end{equation} \tag{ 2.21 }$$

Indeed, since

$$\begin{equation*} |S_{\omega(x)+1}(x)|>\alpha\sqrt{n}\geq\alpha \sqrt{j} \end{equation*}$$

and $1\leq \omega(x)+1\leq j-\tau_j$, the inclusion (2.21) holds (see (2.6), (2.8)). Therefore (see (2.8), (2.9))

$$\begin{equation*} \operatorname{dist}\biggl(x,\frac{2\pi s}{d_j}\biggr)\geq \frac{2\pi (j-\omega(x)-1)^{2}}{d_j} \end{equation*}$$

for any $s\in\Lambda_j$. We have (see (2.10) and also (1.1), (2.5))

$$\begin{equation*} \begin{aligned} \, |\delta_j (x)|&= \frac{1}{d_j}\sum_{s\in \Lambda_{j}} K_{\widetilde{d}_{j}-1}\biggl(x-\frac{2\pi s}{d_j}\biggr)\leq \frac{1}{d_j}\sum_{s\in \Lambda_{j}} \widetilde{d}_{j} \min\biggl(1,\frac{{\pi}^{2}}{{\widetilde{d}_{j}}^{2}\operatorname{dist}^{2}(x,2\pi s/d_j)}\biggr) \\ &\leq \frac{\pi^2}{d_j \widetilde{d}_{j}}\sum_{s\in \Lambda_{j}} \frac{1}{\operatorname{dist}^{2}(x,2\pi s/d_j)}\leq \frac{2\pi^2}{d_j \widetilde{d}_{j}}\sum_{s=(j-\omega(x)-1)^{2}}^{\infty}\frac{1}{(2\pi s/d_j)^{2}} \\ &=\frac{d_j}{2\widetilde{d}_{j}}\sum_{s=(j-\omega(x)-1)^{2}}^{\infty}\frac{1}{s^2}\leq \sum_{s=(j-\omega(x)-1)^{2}}^{\infty}\frac{1}{s^2}\leq \frac{2}{(j-\omega(x)-1)^{2}}. \end{aligned} \end{equation*}$$

Hence,

$$\begin{equation*} \sum_{j=\omega(x)+\tau_n+1}^{n}|\delta_{j}(x)|\leq \sum_{j=\omega(x)+\tau_n+1}^{n} \frac{2}{(j-\omega(x)-1)^{2}}\leq 2\sum_{s=\tau_n}^{\infty}\frac{1}{s^2}\leq\frac{4}{\tau_n}<1. \end{equation*}$$

Therefore,

$$\begin{equation*} |S_{n}(x)|\leq 277 + \alpha\sqrt{n}+180\max_{1\leq j\leq n}\log_{\lambda_j}\max\biggl(\frac{m_j}{d_j}, \frac{1}{\ln \lambda_j}, 1\biggr) \end{equation*}$$

and inequality (2.4) is proved. This completes the proof of Theorem 1. $\Box$

Corollary 1.  Let $\lambda>1$, $0\leq \varepsilon<1$, $0<\theta < 1-\varepsilon$ be real numbers and

$$\begin{equation*} m_n=\big[\lambda^{n^{1-\varepsilon}}\big],\quad d_n=\big[\lambda^{n^{1-\varepsilon}-n^{\theta}}\big],\qquad n=1,2,\dots\,. \end{equation*}$$

Then there are complex trigonometric polynomials $\delta_{n}(x)$, $n=1,2,\dots$, such that

$$\begin{equation*} \begin{gathered} \, \delta_{n}(x)=\sum_{m_n\leq s < m_n+d_n} c_s e^{i s x},\qquad \|\delta_n\|_1\geq \frac{5\pi}{8},\quad \|\delta_n\|_{\infty}\leq 6,\qquad n=1, 2,\dots, \\ \biggl\|\sum_{j=1}^{n}\delta_{j}\biggr\|_{\infty}\leq C(\lambda, \varepsilon, \theta)\,n^{\max(1/2,\,\theta+\varepsilon)},\qquad n=1, 2,\dots, \end{gathered} \end{equation*}$$

where $C(\lambda,\varepsilon,\theta)>0$ depends only on $\lambda$, $\varepsilon$ and $\theta$.

Let $f\in L^{1}(0,2\pi)$ have the Fourier series $f\sim \sum_{j=0}^{\infty}\sigma_{j}(f)$, where $\sigma_{0}(f)=a_0 (f)/2$, and

$$\begin{equation*} \sigma_{j}(f, x)=\sum_{n=2^{j-1}}^{2^j - 1} a_n (f)\cos nx + b_n(f) \sin nx,\qquad j=1,2,\dots\,. \end{equation*}$$

We put

$$\begin{equation} \|f\|_{\mathrm{QC}}\equiv \int_{0}^{1} \biggl\|\sum_{j=0}^{\infty}r_j (\omega)\sigma_{j}(f,\cdot\,)\biggr\|_{\infty}\, d\omega, \end{equation} \tag{ 3.1 }$$

where $\{r_j (\omega)\}_{j=0}^{\infty}$ is the Rademacher system (see, for example, [6]). The norm in (3.1) is called the $\mathrm{QC}$-norm. The closure of the set of trigonometric polynomials with respect to the $\mathrm{QC}$-norm is called the space of quasi-continuous functions. This norm and space were introduced by Kashin and Temlyakov in [7], [8], where they obtained many interesting results on the properties of this norm. In particular, the following theorem was proved in [8] (see [8], Theorem 2.1). If $n$ is a positive integer and $f(x)=\sum_{j=0}^{n} t_j (x)$, where $t_j\in E_j$, $j=0,\dots,n$, then

$$\begin{equation} \|f\|_{\mathrm{QC}}\geq \frac{1}{48\pi}\sum_{j=0}^{n}\|t_j\|_1. \end{equation} \tag{ 3.2 }$$

In connection with the problems of approximation theory (see the details in [8]), it is interesting to relate the $\mathrm{C}$- and $\mathrm{QC}$-norms. It follows from the inequality (3.2) and a result of Grigor'ev [2] that

$$\begin{equation*} \sup_{t\in \mathrm{T}(2^n)} \frac{\|t\|_{\mathrm{QC}}}{\|t\|_{\infty}}\geq c_1\sqrt{n}, \end{equation*}$$

where $c_1> 0$ is an absolute constant. Oskolkov proved (see [8]) that

$$\begin{equation*} \sup_{t\in \mathrm{T}(2^n)} \frac{\|t\|_{\infty}}{\|t\|_{\mathrm{QC}}}\geq c_2\sqrt{n}, \end{equation*}$$

where $c_2>0$ is an absolute constant. Oskolkov's example was generalized in [9]. The following question was open for a long time: do there exist subspaces $L_n\subset E_n$, $n=1,2,\dots$, such that $\dim L_n \geq \alpha \dim E_n$, $n=1,2,\dots$, and

$$\begin{equation*} A\|t\|_{\infty}\leq \|t\|_{\mathrm{QC}}\leq B\|t\|_{\infty} \end{equation*}$$

for any $n$ and $t\in L_1\oplus \dots \oplus L_n$, where $\alpha\in (0,1)$, $A>0$, $B>0$ are absolute constants? It was shown in [10] that the answer is negative. In the present paper, we strengthen Theorem 2 in [10] (see Corollary 2 below). Our result is obtained by specifying functions of the variable $\varepsilon$ considered in Theorems 1 and 2 in [10]. This is accomplished by a significant improvement of the argument in several places; in particular, Lemmas 1, 2 and 4 of [10] are refined. Note also that our Corollary 2 is stronger than results in [11]. The following theorem is central in this section.

Theorem 2.  Let $\varepsilon\,{\in}\,(0,1)$ be a real number, $1\,{\leq}\,k_1\,{<}\,\cdots\,{<}\,k_n\,{<}\,\cdots$ a sequence of positive integers, and $L_n$ a subspace of $E_{k_n}$ such that $\dim L_n \geq \varepsilon \dim E_{k_n}$, $n=1,2,\dots$ . Then there are trigonometric polynomials $t_n \in L_n$, $n=1, 2, \dots$, such that

$$\begin{equation*} \begin{gathered} \, \|t_n\|_{\infty}\leq 1,\quad \|t_n\|_1\geq c\cdot\varepsilon,\qquad n=1,2,\dots, \\ \biggl\|\sum_{j=1}^{n} t_j\biggr\|_{\infty}\leq \frac{a}{\sqrt{\varepsilon}}\sqrt{n}+\frac{b}{\varepsilon^{3/2}},\qquad n=1,2,\dots, \end{gathered} \end{equation*}$$

where $a$, $b$, $c$ are positive absolute constants.

Proof.  We need the following lemmas, of which Lemma 2 is well known. Lemma 1 can also be considered as well known, but we include a proof for completeness.

Lemma 1.  Let $1\leq k \leq N$ be positive integers and $L$ a subspace of $\mathbb{R}^{N}$, $\dim L = k$. Then there is a non-zero $x\in L$ such that

$$\begin{equation*} \|x\|_{l_{1}^{N}}\geq k \|x\|_{l_{\infty}^{N}}. \end{equation*}$$

Proof.  The case when $N=1$ is trivial. Suppose that $N\geq 2$ and put

$$\begin{equation} \begin{gathered} \, B_{\infty}^{N} = \bigl\{x\in \mathbb{R}^{N}\colon \|x\|_{l_{\infty}^{N}} \leq 1 \bigr\},\qquad B=L\cap B_{\infty}^{N}, \nonumber \\ \rho = \sup_{x\in B} \|x\|_{l_{2}^{N}}. \end{gathered} \end{equation} \tag{ 3.3 }$$

Since $B$ is compact, we can replace $\sup$ by $\max$ in (3.3). Let $x_0$ be an element of $B$ such that $\|x_0\|_{l_{2}^{N}}=\rho$. Denote by $s=s(x_0)$ the number of coordinates of $x_0$ having absolute value $1$.

We claim that $s\,{\geq}\, k$. Assume the opposite, that is, $s\,{\leq}\, k-1$. Let $J\,{=}\,\{j_1,\dots, j_s\}$ and $I=\{i_1,\dots, i_{N-s}\}$ be tuples of positive integers such that $j_1<\dots < j_s$, $i_1<\dots < i_{N-s}$, $\{1,\dots, N\}=\{j_1,\dots, j_s\}\sqcup \{i_1,\dots, i_{N-s}\}$,

$$\begin{equation*} \begin{alignedat}{3} |(x_0)_{j_n}|&=1,&\qquad 1 &\leq n \leq s, \\ |(x_0)_{i_n}|&<1,&\qquad 1 &\leq n \leq N-s. \end{alignedat} \end{equation*}$$

If $s=0$, then $J$ is empty. Put

$$\begin{equation*} L'=\{x\in\mathbb{R}^{N}\colon (x)_{j_n}=0,\, 1\leq n\leq s\}. \end{equation*}$$

It is clear that $L'$ is a subspace of $\mathbb{R}^N$ with $\dim L'=N-s$. We have

$$\begin{equation*} \begin{aligned} \, N&\geq \dim (L+ L')=\dim L + \dim L' - \dim (L\cap L')=k+N-s- \dim (L\cap L') \\ &\geq k+ N- k +1 - \dim (L\cap L')= N+1 - \dim (L\cap L'). \end{aligned} \end{equation*}$$

Therefore,

$$\begin{equation*} \dim (L\cap L') \geq 1. \end{equation*}$$

Put $U=L\cap L'$. If $\delta >0$ is a real number and $y\in \mathbb{R}^{N}$, define

$$\begin{equation*} S_{\delta}(y)=\bigl\{x\in \mathbb{R}^{N}\colon \|x-y\|_{l_{2}^{N}}\leq \delta\bigr\}. \end{equation*}$$

Put

$$\begin{equation*} \begin{gathered} \, \delta = 1- \max_{1\leq n\leq N-s} |(x_0)_{i_n}| >0, \\ M=S_{\delta/ 2}(0)\cap U,\qquad M'= x_0+M. \end{gathered} \end{equation*}$$

We claim further that

$$\begin{equation} M'\subset B_{\infty}^{N}. \end{equation} \tag{ 3.4 }$$

Indeed, let $y\in M'$. Then $y=x_0+x$, where $x\in M$. Hence $x\in L'$ and so

$$\begin{equation*} (x)_{j_n}=0,\qquad 1\leq n\leq s. \end{equation*}$$

We have

$$\begin{equation*} |(y)_{j_n}|=|(x_0+x)_{j_n}|=|(x_0)_{j_n}|=1,\qquad 1\leq n\leq s. \end{equation*}$$

Note that

$$\begin{equation*} |(x)_{n}|\leq \|x\|_{l_{2}^{N}}\leq \frac{\delta}{2},\qquad 1\leq n\leq N. \end{equation*}$$

Therefore,

$$\begin{equation*} \begin{aligned} \, |(y)_{i_{n}}|&=|(x_0+x)_{i_n}|=|(x_0)_{i_n}+(x)_{i_n}|\leq |(x_0)_{i_n}| + |(x)_{i_n}| \\ &\leq|(x_0)_{i_n}| + \frac{\delta}{2}\leq 1,\qquad 1\leq n\leq N-s. \end{aligned} \end{equation*}$$

Thus,

$$\begin{equation*} \|y\|_{l_{\infty}^{N}}\leq 1 \end{equation*}$$

and (3.4) is proved.

Since $x_0\in L$ and $M\subset L$, we have $M' \subset L$. Hence $M' \subset B$. Note that

$$\begin{equation*} M' = S_{\delta/2}(x_0)\cap (x_0 + U). \end{equation*}$$

Consider the ball $S_{\rho}(0)$ with centre $0$ and radius $\rho$. Then $x_0$ lies on the boundary of the ball. Also, $M'$ is either tangent to the boundary of $S_{\rho}(0)$ at $x_0$ or intersects it transversally at $x_0$. In both cases, there is a $z\in M'$ such that $z\notin S_{\rho}(0)$. Hence $\|z\|_{l_{2}^{N}}>\rho$ and $z\in B$. But this is not the case because of (3.3). This contradiction implies that $s(x_0)\geq k$, as is claimed above.

Thus $\|x_0\|_{l_{1}^{N}}\geq k$, whence $x_0 \neq 0$. Since $x_0\in B$, we have $x_0\in L$ and $\|x_0\|_{l_{\infty}^{N}}\leq 1$. Therefore,

$$\begin{equation*} \|x_0\|_{l_{1}^{N}}\geq k\geq k\|x_0\|_{l_{\infty}^{N}}. \end{equation*}$$

So we can take $x_0$ as the desired vector $x$. The proof of Lemma 1 is complete. $\Box$

Lemma 2  (see [12], p. 40, Theorem 2.5). Let $n$ be a positive integer, $N=4n$, $t\in \mathrm{T}(n)$, and

$$\begin{equation*} v(t):= \biggl(t\biggl(\frac{2\pi}{N}\biggr),\dots,t\biggl(\frac{2\pi j}{N}\biggr),\dots, t(2\pi) \biggr)\in \mathbb{R}^{N}. \end{equation*}$$

Then

$$\begin{equation*} \frac{c_1}{N}\|v(t)\|_{l_{1}^{N}}\leq \|t\|_{1}\leq \frac{c_2}{N}\|v(t)\|_{l_{1}^{N}},\qquad \|v(t)\|_{l_{\infty}^{N}}\leq \|t\|_{\infty} \leq c_3 \|v(t)\|_{l_{\infty}^{N}}, \end{equation*}$$

where $c_1$, $c_2$, $c_3$ are positive absolute constants.

The proof of Lemma 3 given below is based on Lemmas 1 and 2. It was communicated to the author by Y. V. Malykhin. We notice that, when one does not specify $c(\varepsilon)$, the conclusion of the lemma follows from results in [13].

Lemma 3.  Let $\varepsilon\in (0,1)$ be a real number, $n$ a positive integer, and $L$ a subspace of $\mathrm{T}(n)$ such that $\dim L \geq \varepsilon (2n+1)$. Then there is a trigonometric polynomial $t\in L$ such that

$$\begin{equation*} \|t\|_{\infty} \leq 1,\qquad \|t\|_{1}\geq c_4\cdot\varepsilon, \end{equation*}$$

where $c_4 >0$ is an absolute constant.

Proof.  Put $N=4n$. Since $\varepsilon(2n+1)>0$, the subspace $L$ is not trivial, $\dim L= k$, $k\in \mathbb{N}$ and $k\geq \varepsilon(2n+1)$. Consider the map

$$\begin{equation*} \varphi\colon \mathrm{T}(n)\to \mathbb{R}^{N},\qquad t \to \biggl(t\biggl(\frac{2\pi}{N}\biggr),\dots, t\biggl(\frac{2\pi j}{N}\biggr),\dots, t(2\pi)\biggr). \end{equation*}$$

It is not hard to show that $\varphi$ is linear and injective. Therefore $U=\varphi (L)$ is a subspace of $\mathbb{R}^{N}$ and $\dim U = k$. By Lemma 1, there is a non-zero $x\in U$ such that $\|x\|_{l_{1}^{N}}\geq k \|x\|_{l_{\infty}^{N}}$. Then $x=\varphi (t)$, $t\in L$ and $t\neq 0$ (because otherwise $x=0$). Applying Lemma 2, we have

$$\begin{equation*} \frac{c_1}{N}\|x\|_{l_{1}^{N}}\leq \|t\|_{1}\leq \frac{c_2}{N}\|x\|_{l_{1}^{N}},\qquad \|x\|_{l_{\infty}^{N}}\leq \|t\|_{\infty} \leq c_3 \|x\|_{l_{\infty}^{N}}. \end{equation*}$$

Since

$$\begin{equation*} k\geq \varepsilon (2n+1)\geq \frac{\varepsilon}{2} N, \end{equation*}$$

we obtain

$$\begin{equation*} \frac{1}{c_1}\|t\|_1 \geq \frac{1}{N}\|x\|_{l_{1}^{N}} \geq\frac{k}{N}\|x\|_{l_{\infty}^{N}}\geq\frac{\varepsilon}{2} \|x\|_{l_{\infty}^{N}}\geq \frac{\varepsilon}{2 c_3}\|t\|_{\infty}. \end{equation*}$$

Therefore,

$$\begin{equation*} \|t\|_1 \geq c_4\cdot \varepsilon\|t\|_{\infty}, \end{equation*}$$

where $c_4=c_1/ (2 c_3) >0$ is an absolute constant. Since $t\neq 0$, we have $\|t\|_{\infty} > 0$. Put $t_0=t/\|t\|_{\infty}$. Then $t_0\in L$, $\|t_0\|_{\infty}=1$ and $\|t_0\|_1 \geq c_4\cdot \varepsilon$. We can take $t_0$ as the desired trigonometric polynomial $t$. Lemma 3 is proved. $\Box$

Lemma 4.  Let $\varepsilon\in (0,1)$ a real number, $n$ a positive integer, and $L$ a subspace of $E_n$ such that $\dim L \geq \varepsilon 2^n=\varepsilon \dim E_n$. Then there is a trigonometric polynomial $t\in L$ such that

$$\begin{equation*} \|t\|_{\infty} \leq 1,\qquad \|t\|_{1}\geq c_5\cdot\varepsilon, \end{equation*}$$

where $c_5 >0$ is an absolute constant.

Proof.  Put $m=2^n - 1$. Then $m$ is a positive integer, $L$ is a subspace of $\mathrm{T}(m)$ and

$$\begin{equation*} \dim L\geq \varepsilon 2^n =\frac{\varepsilon}{2}\, 2^{n+1}>\frac{\varepsilon}{2} (2^{n+1}-1)=\frac{\varepsilon}{2} (2m+1). \end{equation*}$$

Since $\varepsilon/2 \in (0,1)$, it follows from Lemma 3 that there is a trigonometric polynomial $t\in L$ such that $\|t\|_{\infty}\leq 1$ and

$$\begin{equation*} \|t\|_{1}\geq c_4\cdot\frac{\varepsilon}{2}= c_5\cdot \varepsilon, \end{equation*}$$

where $c_5 = c_4 / 2 >0$ is an absolute constant. Lemma 4 is proved. $\Box$

Lemma 5.  Let $0\,{<}\,\varepsilon \,{<}\,0.03$ be a real number, $r\,{=}\,1-0.5 \varepsilon$, $m\,{\geq}\, \lceil50/\varepsilon^{3/2}\rceil\,{=:}\,\eta_{0}(\varepsilon)$ a positive integer, and $l\in\{1,\dots, 2\lceil rm\rceil\}$. Then

$$\begin{equation*} \begin{gathered} \, \sum_{1\leq s\leq 2\lceil rm\rceil,\, s\neq l}\frac{1}{m}K_{m-1} \biggl(\frac{\pi l}{\lceil rm\rceil} - \frac{\pi s}{\lceil rm\rceil}\biggr) \leq d(\varepsilon), \\ d(\varepsilon)=1-\frac{1}{5}\varepsilon +\frac{1}{25}\varepsilon^{3/2} -\frac{9}{25}\varepsilon^2+\frac{2}{125}\varepsilon^{5/2}-\frac{1}{25}\varepsilon^3+ \frac{1}{625}\varepsilon^{7/2}=1-\frac{1}{5}\varepsilon +d_1 (\varepsilon), \\ 0< d(\varepsilon)<1,\qquad |d_{1}(\varepsilon)|<\frac{1}{50}\varepsilon. \end{gathered} \end{equation*}$$

Proof.  Put $\Delta_1 = 0.5 \varepsilon$, $\Delta_2 = 0.2 \varepsilon$, $\Delta_3 = 0.2 \varepsilon$,

$$\begin{equation*} \gamma = \frac{1}{1+\Delta_2},\qquad \nu=\biggl\lceil\frac{1}{\Delta_3}\biggr\rceil. \end{equation*}$$

Then $r=1-\Delta_1$. Note that

$$\begin{equation} 1 \leq \lceil rm \rceil \leq m\quad \text{(in particular, }\lceil rm \rceil \in \mathbb{N}) \end{equation} \tag{ 3.5 }$$

because $0< rm < m$. We introduce the sets $G=\big\{1\leq s \leq 2 \lceil rm\rceil,\, s\neq l \big\}$,

$$\begin{equation*} \begin{aligned} \, G_1&=\biggl\{1\leq s \leq 2 \lceil rm\rceil,\, s\neq l \colon \operatorname{dist}\biggl(\frac{\pi s}{\lceil rm\rceil}, \frac{\pi l}{\lceil rm\rceil}\biggr)\leq \frac{\pi \nu}{\lceil rm\rceil}\biggr\}, \\ G_2&=\biggl\{1\leq s \leq 2 \lceil rm\rceil,\, s\neq l \colon \operatorname{dist}\biggl(\frac{\pi s}{\lceil rm\rceil}, \frac{\pi l}{\lceil rm\rceil}\biggr)> \frac{\pi \nu}{\lceil rm\rceil}\biggr\}. \end{aligned} \end{equation*}$$

It is clear that $G=G_1 \sqcup G_2$.

Note that

$$\begin{equation} \nu=\biggl\lceil\frac{1}{0.2 \varepsilon}\biggr\rceil \leq \frac{1}{0.2 \varepsilon} + 1 \leq \frac{1.2}{0.2\varepsilon} = \frac{6}{\varepsilon}. \end{equation} \tag{ 3.6 }$$

We claim that

$$\begin{equation} \frac{\pi \nu}{\lceil rm\rceil} \leq \frac{\pi}{20}. \end{equation} \tag{ 3.7 }$$

Since $20 \nu \leq 120/\varepsilon$, we have

$$\begin{equation*} \lceil rm\rceil \geq rm> \frac{m}{2}\geq \frac{1}{2}\eta_{0}(\varepsilon) \geq \frac{25}{\varepsilon^{3/2}}, \end{equation*}$$

and it suffices to show that

$$\begin{equation*} \frac{25}{\varepsilon^{3/2}} \geq \frac{120}{\varepsilon}. \end{equation*}$$

But this inequality holds because $0< \varepsilon < 0.03$. Thus (3.7) is proved.

Consider the sum

$$\begin{equation*} \begin{aligned} \, S&=\sum_{s\in G} \frac{1}{m}K_{m-1}\biggl(\frac{\pi s}{\lceil rm\rceil} - \frac{\pi l}{\lceil rm\rceil}\biggr)= \sum_{s\in G_1} \frac{1}{m}K_{m-1}\biggl(\frac{\pi s}{\lceil rm\rceil} - \frac{\pi l}{\lceil rm\rceil}\biggr) \\ &\qquad +\sum_{s\in G_2} \frac{1}{m}K_{m-1}\biggl(\frac{\pi s}{\lceil rm\rceil} - \frac{\pi l}{\lceil rm\rceil}\biggr)= S_1 + S_2. \end{aligned} \end{equation*}$$

We have

$$\begin{equation*} \begin{aligned} \, S_1&= \sum_{s\in G_1} \frac{1}{m}K_{m-1}\biggl(\frac{\pi (s-l)}{\lceil rm\rceil} \biggr)= 2\sum_{s=1}^{\nu}\frac{1}{m} K_{m-1}\biggl(\frac{\pi s}{\lceil rm\rceil}\biggr) \\ &= 2\sum_{s=1}^{\nu}\biggl(\frac{\sin(m\pi s/ (2 \lceil rm\rceil))}{m\sin (\pi s/ (2 \lceil rm\rceil))}\biggr)^{2}. \end{aligned} \end{equation*}$$

We claim that if $0< h< \pi$ is a real number, then

$$\begin{equation} \sum_{s=1}^{\infty} \biggl(\frac{\sin s h}{s h}\biggr)^{2}=\frac{\pi}{2 h} - \frac{1}{2}. \end{equation} \tag{ 3.8 }$$

Indeed, it suffices to apply Parseval's identity to the function

$$\begin{equation*} \theta (x)= \begin{cases} 1, &\text{when }|x|\leq h, \\ 0, &\text{when }h< |x|\leq \pi. \end{cases} \end{equation*}$$

We claim further that if $0 \leq x \leq \sqrt{\Delta_2}$, then

$$\begin{equation} \sin x \geq \gamma x. \end{equation} \tag{ 3.9 }$$

Indeed, if $x=0$, then (3.9) is trivial. Suppose that $0< x \leq \sqrt{\Delta_2}$. In view of the well-known inequality $\sin t \geq t- t^3/6$ when $t\geq 0$, it suffices to prove that $x- x^3/6 \geq \gamma x$. The last is equivalent to

$$\begin{equation*} x^2 \leq 6(1-\gamma)=6\biggl(1-\frac{1}{1+\Delta_2}\biggr)=\frac{6 \Delta_2}{1+\Delta_2}. \end{equation*}$$

Since

$$\begin{equation*} \frac{6 \Delta_2}{1+\Delta_2}=\frac{6 \Delta_2}{1+0.2 \varepsilon}\geq \frac{6 \Delta_2}{1.2}=5 \Delta_2, \end{equation*}$$

it is enough to show that $x^2 \leq 5 \Delta_2$. But this inequality holds because $0 < x \leq \sqrt{\Delta_2}$. Thus (3.9) is proved.

We now claim that

$$\begin{equation} \frac{\pi \nu}{2 \lceil rm \rceil} \leq \sqrt{\Delta_2}. \end{equation} \tag{ 3.10 }$$

It suffices to prove that $r m \geq \pi \nu / (2 \sqrt{\Delta_2})$. Since $rm >m/2\geq \eta_{0}(\varepsilon)/2$, it is enough to show that $\eta_{0}(\varepsilon)\geq \pi \nu/\sqrt{\Delta_2}$. It is enough to show, in turn, that

$$\begin{equation*} \frac{50}{\varepsilon^{3/2}}\geq \frac{\pi \nu}{\sqrt{\Delta_2}}. \end{equation*}$$

Applying (3.6), we obtain

$$\begin{equation*} \frac{\pi \nu}{\sqrt{\Delta_2}}\leq \frac{6\pi}{\varepsilon \sqrt{\Delta_2}}=\frac{6\pi}{\sqrt{0.2}\varepsilon^{3/2}} <\frac{50}{\varepsilon^{3/2}}. \end{equation*}$$

Thus the inequality (3.10) is proved.

Therefore,

$$\begin{equation*} 0< \frac{\pi s}{2 \lceil rm \rceil} \leq \sqrt{\Delta_2} \end{equation*}$$

when $s\in \{1,\dots, \nu\}$ and hence (see (3.9))

$$\begin{equation*} \sin\biggl(\frac{\pi s}{2 \lceil rm \rceil}\biggr)\geq \gamma\cdot \frac{\pi s}{2 \lceil rm \rceil}>0. \end{equation*}$$

Then

$$\begin{equation*} S_1 \leq 2\sum_{s=1}^{\nu}\biggl(\frac{\sin (m\pi s / (2 \lceil rm\rceil))}{m \gamma \pi s / (2 \lceil rm \rceil)}\biggr)^2 \leq \frac{2}{\gamma^2}\sum_{s=1}^{\infty}\biggl(\frac{\sin (m\pi s / (2 \lceil rm\rceil))}{m \pi s / (2 \lceil rm \rceil)}\biggr)^2. \end{equation*}$$

Applying (3.8) with $h=\pi m / (2 \lceil rm\rceil)$ (it is clear that $0< h < \pi$) and using the fact that $m\geq \eta_0 (\varepsilon)\geq 50/\varepsilon^{3/2}$, we have

$$\begin{equation*} \begin{aligned} \, S_1 &\leq \frac{1}{\gamma^2}\biggl( \frac{2 \lceil rm\rceil}{m} - 1 \biggr)\leq \frac{1}{\gamma^2}\biggl( \frac{2 (rm +1)}{m} - 1 \biggr)= \frac{1}{\gamma^2}\biggl( 2r-1+ \frac{2}{m}\biggr) \\ &\leq \frac{1}{\gamma^2}\biggl( 2r-1+ \frac{2}{50/\varepsilon^{3/2}}\biggr)= \biggl(1+\frac{1}{5}\varepsilon\biggr)^{2} \biggl( 1-\varepsilon +\frac{1}{25}\varepsilon ^{3/2} \biggr) \\ &=1 - \frac{3}{5}\varepsilon+\frac{1}{25}\varepsilon^{3/2} - \frac{9}{25} \varepsilon^2 +\frac{2}{125}\varepsilon^{5/2}-\frac{1}{25}\varepsilon^3 +\frac{1}{625}\varepsilon^{7/2}. \end{aligned} \end{equation*}$$

Next, we will find a bound for $S_2$. Applying (1.1), (3.5) and $\nu\geq 5/\varepsilon$, we deduce (see also the definition of $G_2$) that

$$\begin{equation*} \begin{aligned} \, S_2&\leq \sum_{s\in G_2} \min\biggl(1,\frac{{\pi}^{2}}{m^{2}\operatorname{dist}^{2}(\pi s / \lceil rm \rceil,\pi l/\lceil rm \rceil)} \biggr) \\ &\leq \sum_{s\in G_2} \frac{{\pi}^{2}}{m^{2}\operatorname{dist}^{2}(\pi s / \lceil rm \rceil,\pi l/\lceil rm \rceil)}\leq 2\sum_{s=\nu+1}^{\infty}\frac{\pi^2}{m^2 (\pi s/ \lceil rm\rceil)^{2}} \\ &=2\frac{\lceil rm\rceil^{2}}{m^2}\sum_{s=\nu+1}^{\infty}\frac{1}{s^2} \leq 2 \sum_{s=\nu+1}^{\infty}\frac{1}{s^2}\leq \frac{2}{\nu}\leq \frac{2}{5}\varepsilon . \end{aligned} \end{equation*}$$

We finally have

$$\begin{equation*} S=S_1+ S_2 \leq 1 - \frac{1}{5}\varepsilon+\frac{1}{25}\varepsilon^{3/2} - \frac{9}{25} \varepsilon^2 +\frac{2}{125}\varepsilon^{5/2}-\frac{1}{25}\varepsilon^3 +\frac{1}{625}\varepsilon^{7/2}= d(\varepsilon). \end{equation*}$$

Since $0< \varepsilon < 0.03$, we obtain

$$\begin{equation*} \begin{aligned} \, |d_1 (\varepsilon)|&\leq \frac{1}{25}\varepsilon^{3/2} + \frac{9}{25} \varepsilon^2 +\frac{2}{125}\varepsilon^{5/2}+\frac{1}{25}\varepsilon^3 +\frac{1}{625}\varepsilon^{7/2} \\ &=\varepsilon \biggl(\frac{1}{25}\sqrt{\varepsilon} + \frac{9}{25} \varepsilon +\frac{2}{125}\varepsilon^{3/2}+\frac{1}{25}\varepsilon^2 +\frac{1}{625}\varepsilon^{5/2}\biggr) \\ &\leq \varepsilon \biggl(\frac{1}{25}\sqrt{0.03} + \frac{9}{25} 0.03 +\frac{2}{125}0.03^{3/2}+\frac{1}{25}0.03^2 +\frac{1}{625}0.03^{5/2}\biggr)<\frac{\varepsilon}{50}. \end{aligned} \end{equation*}$$

Therefore,

$$\begin{equation*} \begin{aligned} \, d(\varepsilon)&= 1-\frac{\varepsilon}{5}+d_1(\varepsilon)> 1-\frac{\varepsilon}{5}-\frac{\varepsilon}{50}= 1-\frac{11}{50} \varepsilon > 1 - \frac{11}{50}>0, \\ d(\varepsilon)&= 1-\frac{\varepsilon}{5}+d_1(\varepsilon)< 1-\frac{\varepsilon}{5}+\frac{\varepsilon}{50}= 1-\frac{9}{50} \varepsilon <1 . \end{aligned} \end{equation*}$$

Lemma 5 is proved. $\Box$

Lemma 6.  Let $\varepsilon$, $0<\varepsilon <0.03$, be a real number, $r=1-0.5 \varepsilon$, $m\geq \lceil50/\varepsilon^{3/2}\rceil$ a positive integer, $\alpha_s, \beta_s \in \mathbb{R}$, $s=1,\dots, 2 \lceil rm \rceil$, and

$$\begin{equation*} \begin{aligned} \, \Biggl\| \Biggl(\,\sum_{s=1}^{2 \lceil rm \rceil} \alpha_s &\frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \Biggr)\cos (3mx) \\ &\ + \Biggl(\,\sum_{s=1}^{2 \lceil rm \rceil} \beta_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \Biggr)\sin (3mx) \Biggr\|_{\infty} \leq 1 . \end{aligned} \end{equation*}$$

Then

$$\begin{equation*} |\alpha_s|\leq \frac{A}{\varepsilon},\quad |\beta_s|\leq \frac{A}{\varepsilon},\qquad s=1,\dots, 2 \lceil rm \rceil, \end{equation*}$$

where $A>0$ is an absolute constant.

Proof.  We first claim that if the real trigonometric polynomial

$$\begin{equation*} T(x) = \sum_{s=1}^{2 \lceil rm \rceil} \gamma_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \end{equation*}$$

satisfies $\|T\|_{\infty} \leq 1$, then

$$\begin{equation} \max_{1\leq s \leq 2 \lceil rm\rceil} |\gamma_s| \leq \frac{1}{1- d(\varepsilon)}, \end{equation} \tag{ 3.11 }$$

where $d(\varepsilon)$ is as in Lemma 5. Let the absolute value of $\gamma_l$ be the largest for the $\gamma_s$, $s=1,\dots, 2 \lceil rm\rceil$. If $|\gamma_l| = 0$, then (3.11) is trivial. Suppose that $|\gamma_l| > 0$. Put

$$\begin{equation*} \begin{aligned} \, T_{1}(x)&= \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi l}{\lceil rm \rceil} \biggr) + \sum_{1 \leq s \leq 2 \lceil rm\rceil,\, s\neq l}\frac{\gamma_s}{\gamma_l} \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \\ &=\frac{1}{m} K_{m-1} \biggl( x- \frac{\pi l}{\lceil rm \rceil} \biggr) + T_2 (x). \end{aligned} \end{equation*}$$

Then $T(x)= \gamma_l \cdot T_{1}(x)$. Hence,

$$\begin{equation} |\gamma_l|\cdot \biggl|T_1 \biggl(\frac{\pi l}{\lceil rm \rceil}\biggr)\biggr| = \biggl|T \biggl(\frac{\pi l}{\lceil rm \rceil}\biggr)\biggr| \leq 1. \end{equation} \tag{ 3.12 }$$

We have

$$\begin{equation*} T_1 \biggl(\frac{\pi l}{\lceil rm \rceil}\biggr)= 1+ T_2 \biggl(\frac{\pi l}{\lceil rm \rceil}\biggr). \end{equation*}$$

Applying Lemma 5, we obtain

$$\begin{equation*} \biggl|T_2 \biggl(\frac{\pi l}{\lceil rm \rceil}\biggr)\biggr| \leq \sum_{1 \leq s \leq 2 \lceil rm\rceil,\, s\neq l} \frac{1}{m} K_{m-1} \biggl( \frac{\pi l}{\lceil rm \rceil} - \frac{\pi s}{\lceil rm \rceil} \biggr)\leq d(\varepsilon). \end{equation*}$$

Therefore,

$$\begin{equation*} \biggl|T_1 \biggl(\frac{\pi l}{\lceil rm \rceil}\biggr)\biggr| \geq 1- d(\varepsilon)> 0 . \end{equation*}$$

Thus (see (3.12))

$$\begin{equation*} |\gamma_l| \leq \frac{1}{1-d(\varepsilon)},\qquad |\gamma_l|=\max_{1 \leq s \leq 2 \lceil rm\rceil} |\gamma_s| \end{equation*}$$

and (3.11) is proved.

We continue with the proof of the lemma. Put

$$\begin{equation*} \begin{gathered} \, P_1(x)=\sum_{s=1}^{2 \lceil rm \rceil} \alpha_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr),\qquad P_2(x)=\sum_{s=1}^{2 \lceil rm \rceil} \beta_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr), \\ P(x)=P_1(x)\cos (3mx)+P_2 (x)\sin (3mx) . \end{gathered} \end{equation*}$$

By hypothesis, $\|P\|_{\infty}\leq 1$. Consider the points

$$\begin{equation*} x_\nu = \frac{2\pi \nu}{6m},\qquad \nu = 1,\dots, 6m. \end{equation*}$$

Then $\cos (3m x_\nu)=\pm 1$, $\sin (3m x_\nu) = 0$, $\nu = 1,\dots, 6 m$, and so

$$\begin{equation*} |P_1 (x_\nu)| = |P(x_\nu)| \leq 1,\qquad \nu = 1,\dots, 6 m. \end{equation*}$$

Since $\deg(K_{m-1}) = m-1$, it follows from the celebrated theorem of Marcinkiewicz (see [5], V. II, Russian p. 54, Theorem 7.28) that

$$\begin{equation*} \|P_1\|_{\infty} \leq A_2\cdot \max_{1\leq \nu \leq 6 m} |P_1 (x_\nu)|\leq A_2, \end{equation*}$$

where $A_2 > 0$ is an absolute constant. Therefore,

$$\begin{equation*} \biggl\|\frac{1}{A_2} P_1\biggr\|_{\infty}\leq 1. \end{equation*}$$

It follows from the argument above that $|\alpha_s / A_2|\leq 1/ (1-d(\varepsilon))$, $s=1,\dots$, $2 \lceil rm \rceil$, that is,

$$\begin{equation*} |\alpha_s|\leq \frac{A_2}{1-d(\varepsilon)},\qquad s=1,\dots, 2 \lceil rm \rceil . \end{equation*}$$

Put

$$\begin{equation*} \widetilde{x}_\nu = - \frac{\pi}{6 m}+\frac{2\pi \nu}{6 m},\qquad \nu = 1,\dots, 6 m. \end{equation*}$$

Then $\cos (3m \widetilde{x}_\nu)=0$, $\sin (3m \widetilde{x}_\nu)=\pm 1$, $\nu = 1, \dots, 6m$, and so

$$\begin{equation*} |P_2 (\widetilde{x}_\nu)| = |P (\widetilde{x}_\nu)| \leq 1,\qquad \nu =1,\dots, 6 m. \end{equation*}$$

Again applying Marcinkiewicz' theorem, we obtain

$$\begin{equation*} \|P_2\|_{\infty} \leq A_2\cdot \max_{1 \leq \nu \leq 6m} |P_2 (\widetilde{x}_{\nu})|\leq A_2. \end{equation*}$$

Therefore, $\|P_2 / A_2\|_{\infty} \leq 1$ and so $|\beta_s / A_2|\leq 1/(1-d(\varepsilon))$, $s=1,\dots, 2 \lceil rm\rceil$, that is,

$$\begin{equation*} |\beta_s| \leq \frac{A_2}{1- d(\varepsilon)},\qquad s=1,\dots, 2 \lceil rm\rceil . \end{equation*}$$

By Lemma 5, we have

$$\begin{equation*} \frac{A_2}{1- d(\varepsilon)}=\frac{A_2}{(1/5)\varepsilon - d_1 (\varepsilon)}\leq \frac{A_2}{(1/5)\varepsilon - (1/50)\varepsilon}=\frac{50A_2}{9 \varepsilon}=\frac{A}{\varepsilon}, \end{equation*}$$

where $A=50A_2 / 9 > 0$ is an absolute constant. Lemma 6 is proved. $\Box$

Lemma 7.  Let $0<\varepsilon <0.03$ be a real number, $r=1-0.5 \varepsilon$, $m\geq \lceil50/\varepsilon^{3/2}\rceil$ a positive integer, $\alpha_s, \beta_s \in \mathbb{R}$, $s=1,\dots, 2 \lceil rm \rceil$, and

$$\begin{equation} \begin{aligned} \, &\biggl(\,\sum_{s=1}^{2 \lceil rm \rceil} \alpha_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \biggr)\cos (3mx) \notag \\ &\qquad + \biggl(\,\sum_{s=1}^{2 \lceil rm \rceil} \beta_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \biggr)\sin (3mx) \equiv 0. \end{aligned} \end{equation} \tag{ 3.13 }$$

Then $\alpha_s=0$, $\beta_s = 0$, $s=1,\dots, 2 \lceil rm \rceil$.

Proof.  If $\alpha_s=0$, $\beta_s = 0$, $s=1,\dots, 2 \lceil rm \rceil$, then (3.13) holds. Suppose that $(\alpha_1,\beta_1,\dots,\alpha_{2 \lceil rm \rceil}, \beta_{2 \lceil rm \rceil})$ is a non-zero tuple of real numbers such that (3.13) holds. Suppose for definiteness that $\alpha_l \neq 0$. Put

$$\begin{equation*} \begin{aligned} \, T(x)&=\biggl(\,\sum_{s=1}^{2 \lceil rm \rceil} \alpha_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \biggr)\cos (3mx) \\ &\qquad + \biggl(\,\sum_{s=1}^{2 \lceil rm \rceil} \beta_s \frac{1}{m} K_{m-1} \biggl( x- \frac{\pi s}{\lceil rm \rceil} \biggr) \biggr)\sin (3mx). \end{aligned} \end{equation*}$$

Let $\lambda > 0$ be a real number. Then $\lambda T(x) \equiv 0$. By Lemma 6, we have $|\lambda \alpha_l|\leq A/\varepsilon$. But this is not the case when $\lambda$ is sufficiently large (for example, when $\lambda = (2A)/ (\varepsilon |\alpha_l|)$). We have a contradiction. Thus (3.13) holds only when $\alpha_s = 0$, $\beta_s=0$, $s=1,\dots, 2 \lceil rm\rceil$. Lemma 7 is proved. $\Box$

We continue with the proof of the theorem. Suppose that $0 < \varepsilon < 0.03$. Put

$$\begin{equation} \begin{gathered} \, r=1-\frac{\varepsilon}2,\qquad H(\varepsilon)=\frac{A}\varepsilon, \nonumber \\ \eta_{0}= \eta_{0}(\varepsilon)=\biggl\lceil\frac{50}{\varepsilon^{3/2}}\biggr\rceil,\qquad \tau = \tau (\varepsilon)=\biggl\lceil3\log_{2} \frac{5440}{\varepsilon}\biggr\rceil,\qquad \chi=\chi (\varepsilon) = \frac{5\cdot \sqrt{c_H}}{\sqrt{\varepsilon}}, \end{gathered} \end{equation} \tag{ 3.14 }$$

where $A$ is the absolute constant in Lemma 6 and $c_H \geq 1$ is the absolute constant in (2.1). Then

$$\begin{equation} \frac{c_H}{\chi^{2}}=\frac{c_H}{25 c_H} \varepsilon = \frac{1}{25} \varepsilon < 0.05 \varepsilon. \end{equation} \tag{ 3.15 }$$

We have

$$\begin{equation*} \tau \geq 3 \log_{2}\frac{5440}{\varepsilon}> 3\log_{2} 5440>37. \end{equation*}$$

Hence $\tau \geq 30$ is a positive integer.

We claim that

$$\begin{equation} \frac{128}{r}\cdot\frac{\tau^{2}}{2^{\tau}} + \frac{16}{2^{\tau}} \leq 0.05 \varepsilon. \end{equation} \tag{ 3.16 }$$

It is not hard to see that $2^{x/2}\geq x^2$ when $x\geq 16$. Hence,

$$\begin{equation*} \frac{\tau^{2}}{2^{\tau}} \leq \frac{1}{2^{\tau / 2}}. \end{equation*}$$

Therefore,

$$\begin{equation*} \frac{128}{r}\cdot\frac{\tau^{2}}{2^{\tau}} + \frac{16}{2^{\tau}}\leq 256\, \frac{\tau^{2}}{2^{\tau}} + \frac{16}{2^{\tau}} \leq \frac{256}{2^{\tau / 2}} +\frac{16}{2^{\tau/2}}=\frac{272}{2^{\tau/2}} < 0.05 \varepsilon. \end{equation*}$$

The last inequality is equivalent to $\tau> 2 \log_{2} (5440 / \varepsilon)$, which obviously holds. Thus (3.16) is proved.

Note that if $j< n$ are positive integers, then

$$\begin{equation} k_n - k_j \geq n- j,\qquad k_j \geq j. \end{equation} \tag{ 3.17 }$$

We define

$$\begin{equation*} m_j = 2^{k_j - 2},\qquad j\geq \tau+\eta_0 +4. \end{equation*}$$

It is clear that $m_j\in \mathbb{N}$, $j\geq\tau+ \eta_0+4$.

We write

$$\begin{equation} \widetilde{a} = 5\cdot \sqrt{c_{H}},\qquad \widetilde{b} = 22055 +\frac{8}{3}A,\qquad \widetilde{c} = 0.4\cdot c_5, \end{equation} \tag{ 3.18 }$$

where $c_5$ is the absolute constant in Lemma 4. Thus $\widetilde{a}$, $\widetilde{b}$ are $\widetilde{c}$ are positive absolute constants. We put

$$\begin{equation} a=\frac{\widetilde{a}}{\sqrt{0.015}},\qquad b=\frac{\widetilde{b}}{0.015^{3/2}},\qquad c=\widetilde{c}\cdot 0.015. \end{equation} \tag{ 3.19 }$$

Hence $a$, $b$ are $c$ are positive absolute constants.

We will construct trigonometric polynomials $t_{n}(x)$, $n=1, 2, \dots$, such that

$$\begin{equation} t_n\in L_n,\qquad \|t_n\|_{\infty} \leq 1,\quad \|t_n\|_{1}\geq 0.4 c_5\cdot \varepsilon,\qquad n=1,2,\dots, \end{equation} \tag{ 3.20 }$$

$$\begin{equation} \biggl\|\sum_{j=1}^{n}t_j \biggr\|_{\infty}\leq \chi \sqrt{n} + \tau +\eta_0 + \frac{8}{3} H(\varepsilon) + 4,\qquad n=1,2,\dots\,. \end{equation} \tag{ 3.21 }$$

We use induction to construct trigonometric polynomials $t_{n}(x)$, $n=1, 2, \dots$, satisfying the conditions in (3.20) and then show that (3.21) also holds. Suppose that $j\in\{1,\dots, \tau + \eta_0 +4 \}$. By Lemma 4, there is a trigonometric polynomial $t\in L_j$ such that $\|t\|_{\infty} \leq 1$, $\|t\|_1 \geq c_5 \varepsilon\geq 0.4c_5\varepsilon$. We define $t_j = t$. Thus $t_{j}(x)$, $j=1,\dots, \tau+\eta_0 +4$ have been constructed. We now proceed by induction. Suppose that $n\geq \tau + \eta_0 +5$ is a positive integer and $t_1 (x),\dots,t_{n-1} (x)$ are given. We shall now construct $t_n (x)$. Write

$$\begin{equation*} S_j(x)= \sum_{s=1}^{j}t_{s}(x). \end{equation*}$$

Put

$$\begin{equation} A_{n}^{j}=\bigl\{x\in (0,2\pi]\colon |S_{j}(x)|>\chi\sqrt{n}\bigr\},\qquad j=1, \dots, n-1, \end{equation} \tag{ 3.22 }$$

$$\begin{equation} B_{n}=\bigcup_{j=1}^{n-1} A_{n}^{j}, \end{equation} \tag{ 3.23 }$$

$$\begin{equation} \widetilde{B}_{n}=\bigcup_{j=1}^{n-\tau}O_{{\pi(n-j)^{2}}/{\lceil r m_n \rceil}}(A_{n}^{j}), \end{equation} \tag{ 3.24 }$$

$$\begin{equation} \Lambda_{n}=\biggl\{s\in\{1,\dots,2 \lceil r m_n \rceil\}\colon \frac{\pi s}{\lceil r m_n \rceil}\notin \widetilde{B}_{n}\biggr\}. \end{equation} \tag{ 3.25 }$$

We claim that

$$\begin{equation} |\Lambda_{n}|\geq (1-0.6 \varepsilon) 2^{k_n - 1}. \end{equation} \tag{ 3.26 }$$

Indeed, if $\widetilde{B}_n = \varnothing$, then

$$\begin{equation*} |\Lambda_n| = 2\lceil r m_n \rceil \geq 2 r m_n = (1 - 0.5 \varepsilon) 2^{k_n - 1}> (1-0.6 \varepsilon) 2^{k_n - 1} \end{equation*}$$

and (3.26) is proved.

Let $\widetilde{B}_n \neq \varnothing$. Then the set

$$\begin{equation*} W=\bigl\{j\in \{1,\dots, n- \tau\}\colon A_{n}^{j} \neq \varnothing\bigr\} \end{equation*}$$

is not empty and

$$\begin{equation*} \widetilde{B}_{n}=\bigcup_{j\in W}O_{{\pi(n-j)^{2}}/{\lceil r m_n \rceil}}(A_{n}^{j}) . \end{equation*}$$

We will first find an upper bound for $\operatorname{mes}(B_n)$. Put

$$\begin{equation*} T(x)=\sum_{j=1}^{n-1}t_{j}(x),\qquad F_{n-1}^{*}(x)=\max_{1\leq j\leq n-1}\biggl|\sum_{s=1}^{j}t_s (x)\biggr|. \end{equation*}$$

It is obvious that

$$\begin{equation*} F_{n-1}^{*}(x)\leq \mathcal{S}^{*}(T,x)\quad \forall\, x\in \mathbb{R}. \end{equation*}$$

We have (see (3.22), (3.23))

$$\begin{equation*} B_n= \{x\in(0,2\pi]\colon F_{n-1}^{*}(x)>\chi\sqrt{n}\}=\{x\in(0,2\pi]\colon (F_{n-1}^{*}(x))^{2}>\chi^{2}n\}. \end{equation*}$$

Applying Chebyshev's inequality and (2.1), and then using the orthogonality of the polynomials $t_j$ and the bound $\|t_j\|_{\infty}\leq 1$, $j=1,\dots,n-1$, we obtain

$$\begin{equation} \begin{aligned} \, \operatorname{mes}(B_n)&\leq \frac1{{\chi}^{2}n}\int_{0}^{2\pi}(F_{n-1}^{*}(x))^{2}\,dx\leq \frac1{{\chi}^{2}n}\int_{0}^{2\pi}(\mathcal{S}^{*}(T,x))^{2}\,dx =\frac{\|\mathcal{S}^{*}(T)\|_{2}^{2}}{{\chi}^{2}n} \notag \\ &\leq\frac{c_H \|T\|_{2}^{2}}{{\chi}^{2}n} =\frac{c_H}{{\chi}^{2}n}\sum_{j=1}^{n-1}\int_{0}^{2\pi}t_{j}^{2}(x)\, dx\leq \frac{2\pi(n-1)c_H}{{\chi}^{2}n} < \frac{2\pi c_H}{{\chi}^{2}}. \end{aligned} \end{equation} \tag{ 3.27 }$$

Let $j\in W$. As in the proof of Theorem 1, one can show that $A_{n}^{j}$ is the union of finitely many pairwise-disjoint open intervals on the circle $(0,2\pi]$ and

$$\begin{equation} \operatorname{Conn}(A_{n}^{j}) \leq 2\cdot 2^{k_j}. \end{equation} \tag{ 3.28 }$$

Taking (3.23), (3.24), (3.27) and (3.28) into account, we obtain

$$\begin{equation*} \begin{aligned} \, \operatorname{mes}(\widetilde{B}_{n})&\leq \operatorname{mes}(B_n) + \sum_{j\in W} \frac{2\pi(n-j)^{2}}{\lceil r m_n \rceil}\cdot \operatorname{Conn}(A_{n}^{j}) \leq \frac{2\pi c_H}{\chi^2} \\ &\qquad + \sum_{j\in W} \frac{2\pi(n-j)^{2}}{\lceil r m_n \rceil}\cdot 2\cdot 2^{k_j}\leq \frac{2\pi c_H}{\chi^2} + \sum_{j=1}^{n- \tau} \frac{4\pi(n-j)^{2}}{\lceil r m_n \rceil}\cdot 2^{k_j} \\ &\leq \frac{2\pi c_H}{\chi^2} + \sum_{j=1}^{n- \tau} \frac{4\pi(n-j)^{2}}{ r m_n }\cdot 2^{k_j}= \frac{2\pi c_H}{\chi^2} + \frac{16 \pi}{r}\sum_{j=1}^{n- \tau} \frac{(n-j)^{2}}{ 2^{k_n - k_j}}. \end{aligned} \end{equation*}$$

Applying (3.17) and (2.15) with $q= 2$ and $k=\tau$, we have

$$\begin{equation} \begin{aligned} \, \operatorname{mes}(\widetilde{B}_{n})&\leq \frac{2\pi c_H}{\chi^2} + \frac{16 \pi}{r}\sum_{j=1}^{n- \tau} \frac{(n-j)^{2}}{ 2^{n - j}}\leq \frac{2\pi c_H}{\chi^2} + \frac{16 \pi}{r}\sum_{s=\tau}^{\infty} \frac{s^{2}}{ 2^{s}} \notag \\ &\leq \frac{2\pi c_H}{\chi^2} + \frac{256 \pi}{r}\frac{\tau^{2}}{2^{\tau}}. \end{aligned} \end{equation} \tag{ 3.29 }$$

This bound implies, in particular, that $\operatorname{mes}(\widetilde{B}_{n}) < 2\pi$ and so $\widetilde{B}_{n} \neq (0, 2\pi]$. Indeed (see (3.15), (3.16)),

$$\begin{equation*} \frac{2\pi c_H}{\chi^2} + \frac{256 \pi}{r}\frac{\tau^{2}}{2^{\tau}}=2\pi\biggl(\frac{c_H}{\chi^2} + \frac{128}{r}\frac{\tau^{2}}{2^{\tau}}\biggr)\leq 2\pi\cdot 0.1 \varepsilon < 2\pi\cdot 0.1 < 2\pi. \end{equation*}$$

Thus $\widetilde{B}_{n}$ is the union of finitely many pairwise-disjoint open intervals on the circle $(0,2\pi]$. We have (see (3.28), (3.17))

$$\begin{equation} \begin{aligned} \, \operatorname{Conn}(\widetilde{B}_{n})&\leq \sum_{j\in W} \operatorname{Conn}(A_{n}^{j})\leq \sum_{j\in W} 2\cdot 2^{k_j}\leq 2 \sum_{j=1}^{n- \tau} 2^{k_j}< 2\cdot 2^{k_{n-\tau}+1} \notag \\ &\ =4\cdot 2^{k_{n-\tau}}\leq 4 \cdot 2^{k_n - \tau}=\frac{4}{2^{\tau}}\cdot 2^{k_n}. \end{aligned} \end{equation} \tag{ 3.30 }$$

Put

$$\begin{equation*} I_s = O_{\pi/ (2 \lceil rm_n \rceil)}\biggl(\frac{\pi s}{\lceil r m_n \rceil}\biggr),\qquad s=1,\dots, 2 \lceil rm_n \rceil . \end{equation*}$$

It is obvious that the length of $I_s$ is $\pi / \lceil r m_n\rceil$ and the intervals are pairwise disjoint. Consider

$$\begin{equation*} \Lambda_{n}^{c}=\biggl\{ s\in \{1,\dots, 2 \lceil r m_n \rceil\} \colon \frac{\pi s}{\lceil r m_n \rceil} \in \widetilde{B}_n \biggr\}, \end{equation*}$$

the complement of the subset $\Lambda_n$ (see (3.25)). We will find an upper bound for $|\Lambda_{n}^{c}|$. It is not hard to see that $|\Lambda_{n}^{c}|$ is at most the number of $I_s$, $s=1,\dots,2 \lceil r m_n\rceil$, intersecting $\widetilde{B}_{n}$ non-trivially. Such intervals are of two types: those contained in $\widetilde{B}_{n}$ and those containing a boundary point of $\widetilde{B}_{n}$. Denote by $N_1$ and $N_2$ the numbers of intervals of the first and second types, respectively. Then (see (3.29))

$$\begin{equation*} N_1 \frac{\pi}{\lceil r m_n \rceil} \leq \operatorname{mes}(\widetilde{B}_{n})\leq \frac{2\pi c_H}{\chi^2} + \frac{256 \pi}{r}\frac{\tau^{2}}{2^{\tau}}. \end{equation*}$$

Since $\lceil rm_n\rceil \leq m_n$, we obtain

$$\begin{equation*} N_1\leq \biggl(\frac{2 c_H}{\chi^2} + \frac{256}{r}\frac{\tau^{2}}{2^{\tau}}\biggr)\lceil r m_n \rceil \leq \biggl(\frac{2 c_H}{\chi^2} + \frac{256}{r}\frac{\tau^{2}}{2^{\tau}}\biggr)m_n= \biggl(\frac{c_H}{\chi^2} + \frac{128}{r}\frac{\tau^{2}}{2^{\tau}}\biggr)2^{k_n -1}. \end{equation*}$$

It is clear (see (3.30)) that

$$\begin{equation*} N_2\leq 2 \operatorname{Conn}(\widetilde{B}_{n})\leq \frac{8}{2^{\tau}}\cdot 2^{k_n} = \frac{16}{2^{\tau}}\cdot 2^{k_n - 1}. \end{equation*}$$

Therefore,

$$\begin{equation*} |\Lambda_{n}^{c}|\leq \biggl(\frac{c_H}{\chi^2} + \frac{128}{r}\frac{\tau^{2}}{2^{\tau}}+ \frac{16}{2^{\tau}}\biggr)2^{k_n -1} . \end{equation*}$$

Hence (see (3.15), (3.16)),

$$\begin{equation*} \begin{aligned} \, |\Lambda_n|&=2\lceil r m_n \rceil-|\Lambda_{n}^{c}|\geq 2\lceil r m_n \rceil - \biggl(\frac{c_H}{\chi^2} + \frac{128}{r}\frac{\tau^{2}}{2^{\tau}}+ \frac{16}{2^{\tau}}\biggr)2^{k_n -1} \\ &\geq 2 r m_n - \biggl(\frac{c_H}{\chi^2} + \frac{128}{r}\frac{\tau^{2}}{2^{\tau}}+ \frac{16}{2^{\tau}}\biggr)2^{k_n -1} \\ &=\biggl( 1- 0.5 \varepsilon - \biggl(\frac{c_H}{\chi^2} + \frac{128}{r}\frac{\tau^{2}}{2^{\tau}}+ \frac{16}{2^{\tau}}\biggr)\biggr)2^{k_n -1} \geq \big( 1- 0.6 \varepsilon \big)2^{k_n -1} \end{aligned} \end{equation*}$$

and (3.26) is proved.

We define $U$ as the linear hull (over $\mathbb{R}$) of

$$\begin{equation*} \begin{aligned} \, &\frac{1}{m_n} K_{m_n-1} \biggl(x-\frac{\pi s}{\lceil r m_n \rceil}\biggr)\cos (3 m_n x), \\ &\frac{1}{m_n} K_{m_n - 1} \biggl(x-\frac{\pi s}{\lceil r m_n \rceil}\biggr)\sin (3 m_n x),\qquad s\in \Lambda_n . \end{aligned} \end{equation*}$$

It is obvious that $U$ is a subspace of $E_{k_n}$. Since $k_n \geq n$, we have

$$\begin{equation*} m_n = 2^{k_n - 2}\geq 2^{n-2}\geq 2^{\tau + \eta_0 + 3}\geq \tau + \eta_0 +3 > \eta_0. \end{equation*}$$

It follows from Lemma 7 that $\dim U\,{=}\,2 |\Lambda_n|\,{\geq}\,(1\,{-}\,0.6 \varepsilon)2^{k_n}$. Hence,

$$\begin{equation*} \begin{aligned} \, \dim (U \cap L_n)&=\dim U + \dim L_n - \dim (U + L_n)\geq \dim U + \dim L_n - \dim E_{k_n} \\ &\geq (1-0.6 \varepsilon)2^{k_n} + \varepsilon 2^{k_n} - 2^{k_n}= 0.4 \varepsilon 2^{k_n}. \end{aligned} \end{equation*}$$

By Lemma 4, there is a $t\in U\cap L_n$ such that $\|t\|_{\infty} \leq 1$, $\|t\|_1 \geq 0.4 c_5 \varepsilon$. Put $t_n = t$. Since $t_n \in U$, we have

$$\begin{equation} \begin{aligned} \, t_{n}(x) &= \biggl(\,\sum_{s\in \Lambda_n} \alpha_{s}^{(n)}\frac{1}{m_{n}} K_{m_{n}- 1}\biggl(x-\frac{\pi s}{\lceil r m_{n}\rceil}\biggr)\biggr)\cos (3 m_n x) \notag \\ &\qquad+\biggl(\,\sum_{s\in \Lambda_n} \beta_{s}^{(n)}\frac{1}{m_{n}} K_{m_{n}- 1}\biggl(x-\frac{\pi s}{\lceil r m_{n}\rceil}\biggr)\biggr)\sin (3 m_n x), \end{aligned} \end{equation} \tag{ 3.31 }$$

where $\alpha_{s}^{(n)}$, $\beta_{s}^{(n)}$, $s\in \Lambda_n$, are real numbers. Since $m_n > \eta_0$ and $\|t_n\|_{\infty}\leq 1$, it follows from Lemma 6 that

$$\begin{equation} |\alpha_{s}^{(n)}|\leq \frac{A}{\varepsilon}=H(\varepsilon),\qquad |\beta_{s}^{(n)}|\leq \frac{A}{\varepsilon},\qquad s\in \Lambda_n . \end{equation} \tag{ 3.32 }$$

Thus we have constructed trigonometric polynomials $t_{n}(x)$, $n=1, 2, \dots$, satisfying the conditions in (3.20).

We now claim that these trigonometric polynomials satisfy (3.21). Let $n$ be a positive integer. If $1 \leq n \leq \tau + \eta_0 + 4$, then

$$\begin{equation*} \biggl\|\sum_{s=1}^{n} t_s\biggr\|_{\infty}\leq \sum_{s=1}^{n}\|t_s\|_{\infty}\leq n \leq \tau + \eta_0 + 4\leq \chi \sqrt{n} + \tau+ \eta_0 + 4 + \frac{8}{3} H(\varepsilon). \end{equation*}$$

Suppose that $n \geq \tau + \eta_0 + 5$. Put

$$\begin{equation*} \omega (x) := \max\{ 1\leq j \leq n-1\colon |S_{j}(x)| \leq \chi \sqrt{n}\}. \end{equation*}$$

Since

$$\begin{equation*} |S_{1}(x)| = |t_1 (x)| \leq 1 \leq \chi\leq \chi \sqrt{n}, \end{equation*}$$

the function $\omega (x)$ is well defined. Let $x\in (0,2\pi]$. Two cases may occur.

1) $\omega (x) \geq n - \tau - \eta_0 - 4$. Then

$$\begin{equation*} |S_{n}(x)| \leq |S_{\omega (x)} (x)| + \sum_{j=\omega (x) + 1}^{n}|t_{j}(x)|\leq \chi \sqrt{n} + \tau + \eta_0 + 4. \end{equation*}$$

2) $\omega (x) < n - \tau - \eta_0 - 4$. Then

$$\begin{equation*} \begin{aligned} \, |S_{n}(x)|&\leq |S_{\omega (x)} (x)|+\sum_{j=\omega (x) + 1}^{\omega (x)+\tau + \eta_0 + 4}|t_{j}(x)|+ \sum_{j=\omega (x) + \tau + \eta_0 + 5}^{n}|t_{j}(x)| \\ &\leq \chi \sqrt{n} + \tau + \eta_0 + 4 + \sum_{j=\omega (x) + \tau + \eta_0 + 5}^{n}|t_{j}(x)|. \end{aligned} \end{equation*}$$

We will estimate the last summand. Suppose that $\omega(x) + \tau + \eta_0 +5 \leq j \leq n$. We claim that

$$\begin{equation} x\in A_{j}^{\omega (x)+1}\subset \widetilde{B}_j. \end{equation} \tag{ 3.33 }$$

Indeed, since

$$\begin{equation*} |S_{\omega (x)+1}(x)|> \chi \sqrt{n}\geq \chi \sqrt{j} \end{equation*}$$

and $1\leq\omega (x)+1 \leq j -\tau$, the inclusion (3.33) is proved (see (3.22), (3.24)). Therefore (see (3.24), (3.25)), we obtain

$$\begin{equation} \operatorname{dist}\biggl(x,\frac{\pi s}{\lceil r m_j \rceil}\biggr)\geq \frac{\pi (j-\omega (x) - 1)^{2}}{\lceil r m_j \rceil} \end{equation} \tag{ 3.34 }$$

for any $s\in \Lambda_j$. We have (see (3.31), (3.32), (1.1), (3.34)) that

$$\begin{equation*} \begin{aligned} \, |t_j (x)| &\leq 2 H(\varepsilon) \sum_{s\in \Lambda_j} \frac{1}{m_j} K_{m_j - 1} \biggl(x-\frac{\pi s}{\lceil r m_j \rceil} \biggr) \\ &\leq2 H(\varepsilon) \sum_{s\in \Lambda_j} \min\biggl(1,\frac{{\pi}^{2}}{m_j^{2}\,\operatorname{dist}^{2}(x,\pi s/ \lceil r m_j \rceil)}\biggr) \\ &\leq 2 H(\varepsilon) \sum_{s\in \Lambda_j} \frac{{\pi}^{2}}{m_j^{2}\,\operatorname{dist}^{2}(x,\pi s/ \lceil r m_j \rceil)}\leq 4 H(\varepsilon) \sum_{s=(j-\omega (x) - 1)^{2}}^{\infty}\frac{\pi^2}{m_j^{2}(\pi s/ \lceil r m_j \rceil)^{2}} \\ &=4 H(\varepsilon) \frac{\lceil r m_j \rceil^{2}}{m_j^{2}} \sum_{s=(j-\omega (x) - 1)^{2}}^{\infty} \frac{1}{s^2}\leq 4 H(\varepsilon) \sum_{s=(j-\omega (x) - 1)^{2}}^{\infty} \frac{1}{s^2} \\ &\leq \frac{8 H(\varepsilon)}{(j-\omega (x)-1)^{2}}. \end{aligned} \end{equation*}$$

Hence,

$$\begin{equation*} \begin{aligned} \, \sum_{j= \omega (x) + \tau + \eta_0 +5}^{n} |t_j (x)|&\leq \sum_{j= \omega (x) + \tau + \eta_0 +5}^{n} \frac{8 H(\varepsilon)}{(j-\omega (x)-1)^{2}} \\ &\leq 8 H(\varepsilon) \sum_{s=\tau+\eta_0 +4}^{\infty}\frac{1}{s^2}\leq \frac{8 H(\varepsilon)}{\tau + \eta_0 +3}\leq \frac{8}{3} H(\varepsilon). \end{aligned} \end{equation*}$$

We finally have that in the second case

$$\begin{equation*} |S_n (x)|\leq \chi \sqrt{n} + \tau+ \eta_0 + 4 + \frac{8}{3} H(\varepsilon) \end{equation*}$$

and (3.21) is proved. Thus we have constructed trigonometric polynomials $t_n (x)$, $n=1, 2, \dots$, satisfying (3.20) and (3.21).

It is easy to see that $\log_{2} x \leq x$ when $x \geq 1$. Since $5440/ \varepsilon > 5440$, we have $\log_{2}(5440/\varepsilon) > 1$. Hence (see (3.14)),

$$\begin{equation*} \begin{gathered} \, \tau \leq 3 \log_{2}\frac{5440}{\varepsilon} + 1 \leq 4 \log_{2}\frac{5440}{\varepsilon}\leq 4\cdot \frac{5440}{\varepsilon}< \frac{22000}{\varepsilon}, \\ \eta_0\leq \frac{50}{\varepsilon^{3/2}} +1\leq\frac{50}{\varepsilon^{3/2}} +\frac{1}{\varepsilon^{3/2}}=\frac{51}{\varepsilon^{3/2}}. \end{gathered} \end{equation*}$$

Let $n\in \mathbb{N}$. We obtain (see (3.18))

$$\begin{equation*} \begin{aligned} \, \chi \sqrt{n} + \tau+ \eta_0 + 4 + \frac{8}{3} H(\varepsilon)&\leq \frac{5\cdot \sqrt{c_H}}{\sqrt{\varepsilon}}\sqrt{n}+\frac{22000}{\varepsilon} + \frac{51}{\varepsilon^{3/2}} + 4 + \frac{8}{3}\, \frac{A}{\varepsilon} \\ &\leq \frac{5\cdot \sqrt{c_H}}{\sqrt{\varepsilon}}\sqrt{n}+\frac{22000}{\varepsilon^{3/2}} + \frac{51}{\varepsilon^{3/2}} + \frac{4}{\varepsilon^{3/2}} + \frac{8}{3}\, \frac{A}{\varepsilon^{3/2}} \\ &=\frac{5\cdot \sqrt{c_H}}{\sqrt{\varepsilon}}\sqrt{n}+ \frac{22055+ (8/3)A}{\varepsilon^{3/2}}= \frac{\widetilde{a}}{\sqrt{\varepsilon}} \sqrt{n} + \frac{\widetilde{b}}{\varepsilon^{3/2}}. \end{aligned} \end{equation*}$$

Thus we have constructed a sequence $\{t_n(x)\}_{n=1}^{\infty}$ of trigonometric polynomials such that

$$\begin{equation*} \begin{gathered} \, t_n \in L_n,\qquad \|t_n\|_{\infty} \leq 1,\qquad \|t_n\|_{1} \geq \widetilde{c}\cdot \varepsilon,\qquad n=1, 2, \dots, \\ \biggl\|\sum_{s=1}^{n} t_s\biggr\|_{\infty}\leq \frac{\widetilde{a}}{\sqrt{\varepsilon}} \sqrt{n} + \frac{\widetilde{b}}{\varepsilon^{3/2}},\qquad n=1, 2, \dots\,. \end{gathered} \end{equation*}$$

It is evident that the constants $\widetilde{a}$, $\widetilde{b}$ and $\widetilde{c}$ can be replaced by $a$, $b$ and $c$, respectively (see (3.19)).

Now suppose that $0.03 \leq \varepsilon < 1$. Then

$$\begin{equation*} \dim L_n \geq \varepsilon \dim E_{k_n}\geq 0.015 \dim E_{k_n},\qquad n=1, 2, \dots\,. \end{equation*}$$

It follows from the argument above that there is a sequence $\{t_n (x)\}_{n=1}^{\infty}$ of trigonometric polynomials such that

$$\begin{equation*} \begin{gathered} \, t_n\in L_n,\qquad \|t_n\|_{\infty} \leq 1, \qquad \|t_n\|_{1} \geq \widetilde{c}\cdot0.015= c\geq c\cdot \varepsilon, \\ \biggl\|\sum_{s=1}^{n}t_s\biggr\|_{\infty}\leq \frac{\widetilde{a}}{\sqrt{0.015}} \sqrt{n} + \frac{\widetilde{b}}{0.015^{3/2}}= a\sqrt{n}+ b\leq \frac{a}{\sqrt{\varepsilon}}\sqrt{n} + \frac{b}{\varepsilon^{3/2}} \end{gathered} \end{equation*}$$

for any positive integer $n$. Theorem 2 is proved. $\Box$

Theorem 3.  Let $\varepsilon\,{\in}\,(0,1)$ be a real number, $1\,{\leq}\,k_1\,{<}\,\cdots\,{<}\, k_n\,{<}\,\cdots$ a sequence of positive integers, and $L_n$ a subspace of $E_{k_n}$ such that $\dim L_n \geq \varepsilon \dim E_{k_n}$, $n=1,2,\dots$ . Then

$$\begin{equation} \sup_{0\neq t\in L_1 \oplus\dots \oplus L_n}\frac{\|t\|_{\mathrm{QC}}}{\|t\|_{\infty}}\geq \frac{d \varepsilon n}{a \varepsilon^{-1/2}n^{1/2} + b \varepsilon^{-3/2}}\geq \gamma \varepsilon^{5/2} \sqrt{n},\qquad n=1, 2, \dots, \end{equation} \tag{ 3.35 }$$

where $a$, $b$, $d$ and $\gamma$ are positive absolute constants and, moreover, $a$ and $b$ are the constants in Theorem 2.

Proof.  By Theorem 2, there is a sequence $\{ t_{n}(x)\}_{n=1}^{\infty}$ of trigonometric polynomials such that

$$\begin{equation*} \begin{gathered} \, t_n \in L_n,\qquad \|t_n\|_{\infty} \leq 1,\quad \|t_n\|_{1}\geq c\cdot \varepsilon,\qquad n=1, 2, \dots, \\ \biggl\|\sum_{j=1}^{n} t_j\biggr\|_{\infty} \leq \frac{a}{\sqrt{\varepsilon}} \sqrt{n}+\frac{b}{\varepsilon^{3/2}},\qquad n=1, 2,\dots, \end{gathered} \end{equation*}$$

where $a$, $b$ and $c$ are positive absolute constants. Let $n$ be a positive integer. Put $t(x)=t_{1}(x)+\dots + t_{n}(x)$. Then $t\in L_1 \oplus \dots \oplus L_n$ and

$$\begin{equation*} \|t\|_{\infty} \leq \frac{a}{\sqrt{\varepsilon}} \sqrt{n}+\frac{b}{\varepsilon^{3/2}}. \end{equation*}$$

Since the spectra of the trigonometric polynomials $t_j (x)$, $j=1,\dots, n$, are pairwise disjoint and $\|t_1\|_1\geq c \varepsilon >0$, we obtain that $t(x)$ is a non-zero real trigonometric polynomial. By the inequality (3.2),

$$\begin{equation*} \|t\|_{\mathrm{QC}}\geq \frac{1}{48\pi}\sum_{j=1}^{n}\|t_j\|_{1}\geq \frac{1}{48\pi}c \varepsilon n = d \varepsilon n, \end{equation*}$$

where $d=c/(48\pi)$ is a positive absolute constant. Thus

$$\begin{equation*} \frac{\|t\|_{\mathrm{QC}}}{\|t\|_{\infty}}\geq \frac{d \varepsilon n}{a \varepsilon^{-1/2}n^{1/2} + b \varepsilon^{-3/2}} \end{equation*}$$

and the first inequality in (3.35) is proved. Further,

$$\begin{equation*} \begin{aligned} \, \frac{d \varepsilon n}{a \varepsilon^{-1/2}n^{1/2} + b \varepsilon^{-3/2}}&\geq \frac{d \varepsilon n}{a \varepsilon^{-1/2}n^{1/2} + b \varepsilon^{-3/2}n^{1/2}} \\ &\geq \frac{d \varepsilon n}{a \varepsilon^{-3/2}n^{1/2} + b \varepsilon^{-3/2}n^{1/2}}= \frac{d \varepsilon \sqrt{n}}{(a+b)\varepsilon^{-3/2}}=\gamma \varepsilon^{5/2} \sqrt{n}, \end{aligned} \end{equation*}$$

where $\gamma=d/ (a+b)$ is a positive absolute constant. Thus the second inequality in (3.35) is also proved. The proof of Theorem 3 is complete. $\Box$

Corollary 2.  Suppose that sequences $\{\tau_n\}_{n=1}^{\infty}$ and $\{\varepsilon_n\}_{n=1}^{\infty}$ of real numbers sasisfy the conditions

i) $\tau_n \geq 1$, $\tau_n \leq \tau_{n+1}$ ($n=1, 2, \dots$) and $\tau_n \to +\infty$ when $n\to +\infty$;

ii) $\varepsilon_n \in (0,1)$, $\varepsilon_n \geq \varepsilon_{n+1}$ ($n=1, 2, \dots$) and, starting at some $N$,

$$\begin{equation*} \varepsilon_n = \frac{\tau_n}{n^{1/3}}. \end{equation*}$$

Let $L_n$ be a subspace of $E_n$ such that

$$\begin{equation*} \dim L_n \geq \varepsilon_n \cdot 2^n = \varepsilon_n \dim E_n,\qquad n=1, 2, \dots\,. \end{equation*}$$

Then

$$\begin{equation*} \sup_{0\neq t\in L_1 \oplus\dots \oplus L_n}\frac{\|t\|_{\mathrm{QC}}}{\|t\|_{\infty}}\geq \rho\cdot \tau_n^{3/2},\qquad n\geq N, \end{equation*}$$

where $\rho$ is a positive absolute constant. In particular,

$$\begin{equation*} \lim_{n \to \infty} \sup_{0\neq t\in L_1 \oplus\dots \oplus L_n}\frac{\|t\|_{\mathrm{QC}}}{\|t\|_{\infty}}= +\infty . \end{equation*}$$

Proof.  Let $n \geq N$ be a positive integer. Write $\varepsilon = \varepsilon_n$. Then $\varepsilon \in (0,1)$ is a real number and, by condition ii), the inequality $\varepsilon_j \geq \varepsilon$ holds for any positive integer $j$, $1\leq j \leq n$. Put $\widetilde{L}_{j} = L_j$ when $1\leq j \leq n$ and $\widetilde{L}_j = E_j$ when $j>n$.

We claim that $\widetilde{L}_{j}$ is a subspace of $E_j$ and

$$\begin{equation} \dim \widetilde{L}_{j} \geq \varepsilon \dim E_j,\qquad j=1, 2, \dots\,. \end{equation} \tag{ 3.36 }$$

Indeed, if $1 \leq j \leq n$, then $\widetilde{L}_j = L_j$ is a subspace of $E_j$ and

$$\begin{equation*} \dim \widetilde{L}_j = \dim L_j \geq \varepsilon_j \dim E_j \geq \varepsilon \dim E_j. \end{equation*}$$

If $j>n$, then $\widetilde{L}_j = E_j$ is a subspace of $E_j$ and

$$\begin{equation*} \dim \widetilde{L}_j = \dim E_j\geq \varepsilon \dim E_j , \end{equation*}$$

and so (3.36) is proved.

Putting $k_j= j$, $j=1, 2, \dots$, in Theorem 3 we have

$$\begin{equation*} \sup_{0\neq t\in \widetilde{L}_1 \oplus\dots \oplus \widetilde{L}_j}\frac{\|t\|_{\mathrm{QC}}}{\|t\|_{\infty}}\geq \frac{d \varepsilon j}{a\varepsilon^{-1/2}\sqrt{j} + b\varepsilon^{-3/2}},\qquad j=1, 2,\dots\,. \end{equation*}$$

In particular,

$$\begin{equation*} \sup_{0\neq t\in L_1 \oplus\dots \oplus L_n}\frac{\|t\|_{\mathrm{QC}}}{\|t\|_{\infty}}\geq \frac{d \varepsilon n}{a\varepsilon^{-1/2}\sqrt{n} + b\varepsilon^{-3/2}} \end{equation*}$$

when $j=n$. Since $n\geq N$,

$$\begin{equation*} \varepsilon = \varepsilon_n = \frac{\tau_n}{n^{1/3}}. \end{equation*}$$

Thus,

$$\begin{equation*} \begin{aligned} \, \frac{d \varepsilon n}{a\varepsilon^{-1/2}\sqrt{n} + b\varepsilon^{-3/2}}&= \frac{d \tau_n n^{2/3}}{a \tau_{n}^{-1/2} n^{2/3} + b \tau_{n}^{-3/2} n^{1/2}}\geq \frac{d \tau_n n^{2/3}}{a \tau_{n}^{-1/2} n^{2/3} + b \tau_{n}^{-3/2} n^{2/3}} \\ &=\frac{d \tau_n }{a \tau_{n}^{-1/2} + b \tau_{n}^{-3/2} }\geq \frac{d \tau_n }{a \tau_{n}^{-1/2} + b \tau_{n}^{-1/2} } = \rho\cdot \tau_{n}^{3/2}, \end{aligned} \end{equation*}$$

where $\rho = d/ (a+b)$ is a positive absolute constant. Corollary 2 is proved. $\Box$

The author grateful to the referee for his careful reading of the paper.