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This article is cited in 2 scientific papers (total in 2 papers) On uniqueness for Franklin series with a convergent subsequence of partial sums G. G. Gevorkyan Yerevan State University, Yerevan, Republic of Armenia
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     § 1. IntroductionUniqueness problems for some classical orthogonal series play an important role in investigations of series in these systems. A central role in the study of uniqueness of trigonometric series is played by Cantor’s theorem (see [1], and also [2], Ch. 1, § 70), which asserts that if a trigonometric series converges to zero everywhere, then each coefficient of this series is zero. The uniqueness of trigonometric series has been studied extensively, and interest in this field is unabated (see, for example, [3]–[6]). The investigation of uniqueness for series in the Haar system began with [7]–[9], where a Cantor-type theorem for Haar series was established. Faber [10] showed that the singleton $\{1/2\}$ is not a uniqueness set for Haar series, that is, there exists a nontrivial series in the Haar system that converges to zero everywhere except this set. Moreover, it was found out later (see, for example, [11]) that no singleton is a uniqueness set for Haar series. Analogues of de la Vallée Poussin’s theorem for series in the Haar and Walsh systems whose coefficients satisfy certain necessary conditions were derived in [11]. Investigations in the field of uniqueness for series in the Haar and Walsh systems and their generalizations are still going on (see, for example, [12]–[15]). Investigations of uniqueness problems for series in the Franklin system started recently. The definition of the orthonormal Franklin system $\{f_n(x)\}_{n=1}^{\infty}$ will be given below. The following theorem (see [17]) was announced in [16]. Theorem 1. If the series in the Franklin system converges to zero everywhere, then each coefficient of this series is zero. Next, the following theorem was proved in [18]. Theorem 2. If the series (1.1) converges everywhere to a finite integrable function $f$, then this is the Fourier-Franklin series of this function. In [19] it was shown that no singleton is a uniqueness set for Franklin series, that is, for each $x_0\in[0,1]$ there exists a nontrivial series (1.1) that converges to zero everywhere except $x_0$. The coefficients of these series satisfy In this regard, we mention the following theorem (see Theorem 2.5 in [18]). Theorem 3. If the series (1.1) with coefficients converges in measure to an integrable function $f$ and everywhere, with the possible exception of a countable set, then this is the Fourier-Franklin series of $f$. Wronicz (see [20]) constructed a nontrivial Franklin series (1.1) satisfying
$$
\begin{equation}
\lim_{k\to\infty}\sum_{n=0}^{2^k}a_nf_n(x)=0 \quad\text{for each } x\in[0,1].
\end{equation}
\tag{1.3}
$$
The following problem was stated in [18]: does the series (1.1) have zero coefficients if the $2^k$th partial sums of this series converge to zero everywhere and condition (1.2) is met? A positive answer to this problem was given by Wronicz [21]. Theorem 4 (Wronicz). If conditions (1.2) and (1.3) are met, then each coefficient of series (1.1) is zero. In this paper we prove a more general theorem using methods different from those in [21]. Let $n_i$ be an increasing sequence of natural numbers satisfying $\sup_in_{i+1}/{n_{i}}< \infty$. Then, for some natural number $\kappa$, Let be the $n_i$th partial sum of the series (1.1). Theorem 5. Let the partial sums (1.5) of a series (1.1) with coefficients (1.2) converge in measure to zero and satisfy the condition As an immediate corollary to Theorem 5 we have the following. Theorem 6. Let the partial sums (1.5) of the series (1.1) with coefficients (1.2) converge to zero everywhere, with the possible exception of a countable set. Then each coefficient of (1.1) is zero. It is known that the Fourier-Franklin series of an integrable function converges almost everywhere to this function (see [22]) and that the coefficients of this series satisfy (1.2). Now from Theorem 5 we obtain the following theorem. Theorem 7. Let the partial sums (1.5) of a series (1.1) with coefficients (1.2) converge in measure to a bounded function $f$ and satisfy condition (1.6). Then (1.1) is the Fourier-Franklin series of $f$. In what follows we use the following notation:
§ 2. Definition of the Franklin system and some auxiliary lemmasGiven $n=2^{\mu}+\nu,$ where $\mu=0, 1, 2, \dots$, $1\leqslant \nu\leqslant 2^{\mu}$, we define
$$
\begin{equation*}
s_{n,i}=\begin{cases} \dfrac{i}{2^{\mu+1}}, & 0\leqslant i\leqslant 2\nu, \\ \dfrac{i-\nu}{2^{\mu}}, & 2\nu<i\leqslant n. \end{cases}
\end{equation*}
\notag
$$
We also set $s_{n,-1}=s_{n,0}=0$ and $s_{n,n+1}=s_{n,n}=1$. We let $\mathbf S_n$ denote the space of continuous piecewise linear functions on $[0,1]$ with knots $\{s_{n,i}\}_{i=0}^n$, that is, $f\in \mathbf S_n$ if $f$ is a continuous linear function on each closed interval $[s_{n,i-1}, s_{n,i}]$, $i=1, 2, \dots, n.$ It is clear that $\dim \mathbf S_n=n+1$ and that the set $\{s_{n,i}\}_{i=0}^n$ is obtained by adding the point $s_{n,2\nu-1}$ to the set $\{s_{n-1,i}\}_{i=0}^{n-1}.$ Therefore, there exists a unique (up to sign) function $f_n\in \mathbf S_n$, $\|f_n\|_2=1$, which is orthogonal to $\mathbf S_{n-1}$. Setting $f_0(x)=1$ and $f_1(x)=\sqrt{3}\,(2x-1),$ $x\in [0,1]$, we obtain an orthonormal system $\{f_n(x)\}_{n=0}^{\infty}$, which was defined by Franklin in [23] in an equivalent way. An important role in the study of the Franklin system is played by Ciesielski’s inequalities (see [22]). Namely, if $n=2^{\mu}+\nu,$ where $\mu=0, 1, 2, \dots$, $1\leqslant \nu\leqslant 2^{\mu}$, then
$$
\begin{equation}
|f_n(x)|\leqslant C_1 2^{\mu/2}q^{2^{\mu}|x-t_n|}, \quad\text{where } q=\sqrt{2-\sqrt{3}}<1, \quad t_n:=s_{n,2\nu-1}.
\end{equation}
\tag{2.1}
$$
It is easily checked that from (2.1) we obtain
$$
\begin{equation*}
\sum_{n=2^{\mu}+1}^{2^{\mu+1}}|f_n(x)|\leqslant C_2 2^{\mu/2}, \qquad x\in[0,1].
\end{equation*}
\notag
$$
Now, proceeding in the standard way we see that if the coefficients $a_n$ of the series (1.1) satisfy (1.2), then Let $\delta_{jk}$ be the Kronecker delta, that is, $\delta_{jk}=1$ if $j=k$, and $\delta_{jk}=0$ if $j\neq k$. For $n\geqslant 2$ we define the functions $\{N_{n,j}(t)\}_{j=0}^n$ by
$$
\begin{equation*}
N_{n,j}(x)=\begin{cases} \delta_{jk} & \text{if } x=s_{n,k}, \ k=0,\dots,n; \\ \text{is linear} & \text{on }[s_{n,k-1}, s_{n,k}], \ k=1,\dots, n. \end{cases}
\end{equation*}
\notag
$$
The functions $\{N_{n,j}(t)\}_{j=0}^n$ are normalized in $C[0,1]$. The system $\{N_{n,j}(t)\}_{i=0}^n$ is a basis of $\mathbf S_n$ since $N_{n,j}(s_{n,k})=\delta_{jk}$. Setting we obtain another basis of $\mathbf S_n$, which is now normalized in $L[0,1].$ We set It is clear that
$$
\begin{equation}
\Delta_{n,j}=\operatorname{supp}N_{n,j}=\operatorname{supp}M_{n,j}\quad\text{and} \quad \frac{1}{2n}\leqslant \operatorname{mes}(\Delta_{n,j})<\frac{2}{n}.
\end{equation}
\tag{2.3}
$$
Since the sequence $n_i$ satisfying (1.4) is fixed, we simplify the notation slightly by putting
$$
\begin{equation*}
\begin{gathered} \, \Delta^i_j:=\Delta_{n_i,j}, \qquad s^i_{j}:=s_{n_i,j}, \\ N_j^i(x):=N_{n_i,j}(x), \qquad M^i_j(x):=M_{n_i,j}(x)\quad\text{and} \qquad S^{(i)}(x):=S_{n_i}(x). \end{gathered}
\end{equation*}
\notag
$$
The inner product of a series (1.1) and a function $g\in \mathbf S_n$, as defined in [17], has proved to be useful in the study of uniqueness for series in the Franklin system. From the definition of the Franklin system for $m>n$ we have $f_m\bot \mathbf S_{m-1}$ and $\mathbf S_n\subset \mathbf S_{m-1}$. Therefore,
$$
\begin{equation*}
\int_0^1f_m(t)g(t)\,dt=0 \quad\text{for } m>n, \quad g\in \mathbf S_n.
\end{equation*}
\notag
$$
Hence, given a series (1.1) and a function $g\in \mathbf S_n$, their inner product can be defined by
$$
\begin{equation*}
(\mathcal S, g):=\sum_{m=0}^{\infty}a_m\int_0^1f_m(t)g(t)\,dt= \sum_{m=0}^{n}a_m\int_0^1f_m(t)g(t)\,dt=\int_0^1S_n(t)g(t)\,dt;
\end{equation*}
\notag
$$
here and in what follows $\mathcal S$ denotes the series (1.1). It is easily checked that, for all $\alpha, \beta \in \mathbb{R}$ and $g_1, g_2\in \mathbf S_n$,
$$
\begin{equation*}
(\mathcal S, \alpha g_1+\beta g_2)= \alpha(\mathcal S, g_1)+\beta(\mathcal S, g_2).
\end{equation*}
\notag
$$
Lemma 1. For each function $M^i_j$ and each $i_1>i$, where $0\leqslant \alpha_k\leqslant 4{n_{i_1}}/{n_i}$ and $\sum\alpha_k=1$. Indeed, to prove this it suffices to observe that
$$
\begin{equation*}
M^{i}_{j}(t)=\sum_{k\colon \operatorname{supp}M^{i_1}_k\subset \operatorname{supp}M^{i}_{j}}\alpha_kM^{i_1}_k(t), \quad\text{where} \ \alpha_k= \frac{M^{i}_{j}(s^{i_1}_k)}{M^{i_1}_k(s^{i_1}_k)},
\end{equation*}
\notag
$$
and then use the fact that the integral of $M^{i}_{j}$ is 1 and the integral of $M^{i_1}_k$ is 1. The next result is Lemma 2.3 in [24]. Lemma 3. Let the series (1.1) with coefficients (1.2) satisfy for some $M^{i_0}_{j_0}$. Then for each $x_0\in[0,1]$ there exist $i$ and $j$ such that Proof. By Lemma 1, for each $i>i_0$ we have
$$
\begin{equation*}
M^{i_0}_{j_0}(x)=\sum_{j\colon \operatorname{supp}M^i_j\subset\operatorname{supp}M^{i_0}_{j_0}}\alpha^i_jM^i_j(x)
\end{equation*}
\notag
$$
and (see (2.3)) Hence
Assume on the contrary that if $(\mathcal S, M^i_j)\neq 0$, then $x_0\in\Delta^i_j$. Note that, for each $i>i_0$, the inclusion $x_0\in\Delta^i_j$ can hold for at most three induces $j$. But this means that, for all $i>i_0$, the number of nonzero terms in the sum (2.6) is at most three. Now, using (2.5) and (2.2) we obtain
$$
\begin{equation*}
|(\mathcal S, M^{i_0}_{j_0})|\leqslant 12\,\frac{n_{i_0}}{n_i}\, o(n_i)=o(1), \qquad i\to\infty,
\end{equation*}
\notag
$$
which contradicts (2.4). This proves Lemma 3. Lemma 4. Let condition (2.4) be met. Then for each positive $M$ there exist $i$ and $j$ such that Proof. Assume on the contrary that
$$
\begin{equation*}
|(\mathcal S, M^i_j)|\leqslant M \quad\text{if } \Delta^i_j\subset\Delta^{i_0}_{j_0}.
\end{equation*}
\notag
$$
Let $\varepsilon$ be any positive number. We set
$$
\begin{equation}
G^i_1 :=\bigl\{j\colon \Delta^i_j\subset\Delta^{i_0}_{j_0},\ |(\mathcal S, M^i_j)|>\varepsilon,\ j\text{ is odd}\bigr\}
\end{equation}
\tag{2.7}
$$
and
$$
\begin{equation}
G^i_2 :=\bigl\{j\colon \Delta^i_j\subset\Delta^{i_0}_{j_0},\ |(\mathcal S, M^i_j)|>\varepsilon, \ j\text{ is even}\bigr\}.
\end{equation}
\tag{2.8}
$$
By Lemma 2, if $j\in G^i_1$, then
$$
\begin{equation}
\operatorname{mes}\biggl\{x\in\Delta^i_j\colon |S^{(i)}(x)|>\frac{\varepsilon}{2}\biggr\}\geqslant \frac{\operatorname{mes}(\Delta^i_j)}{9}.
\end{equation}
\tag{2.9}
$$
Since $\operatorname{mes}(\Delta^i_{j_1}\cap\Delta^i_{j_2})=0$ for $j_1\neq j_2$, $j_1,j_2\in G^i_1$, from (2.9) we obtain
$$
\begin{equation}
\operatorname{mes}\biggl\{x\in\Delta^{i_0}_{j_0}\colon |S^{(i)}(x)|>\frac{\varepsilon}{2}\biggr\} \geqslant\frac{\operatorname{mes}(\bigcup_{j\in G^i_1}\Delta^i_j)}{9}\geqslant\frac{1}{18n_i}\operatorname{card}(G^i_1).
\end{equation}
\tag{2.10}
$$
A similar argument shows that
$$
\begin{equation}
\operatorname{mes}\biggl\{x\in\Delta^{i_0}_{j_0}\colon |S^{(i)}(x)|>\frac{\varepsilon}{2}\biggr\} \geqslant\frac{1}{18n_i}\operatorname{card}(G^i_2).
\end{equation}
\tag{2.11}
$$
Using (2.10) and (2.11) we have
$$
\begin{equation}
\operatorname{card}(G^i_1\cup G^i_2)\leqslant 36n_i\operatorname{mes} \biggl\{x\in\Delta^{i_0}_{j_0}\colon |S^{(i)}(x)|>\frac{\varepsilon}{2}\biggr\}=o(n_i),
\end{equation}
\tag{2.12}
$$
since the series (1.1) converges to zero in measure.
Using Lemma 1 in combination with (2.7), (2.8) and (2.12) we find that
$$
\begin{equation*}
\begin{aligned} \, |(\mathcal S, M^{i_0}_{j_0})| &\leqslant\sum_{j\colon \operatorname{supp}M^{i}_j\subset \operatorname{supp}M^{i_0}_{j_0}}\alpha_j |(\mathcal S, M^{i}_j)|\leqslant \varepsilon+M\frac{n_{i_0}}{n_i}\operatorname{card}(G^i_1\cup G^i_2) \\ &=\varepsilon+M\frac{n_{i_0}}{n_i}\, o(n_i)=\varepsilon+Mn_{i_0}o(1), \qquad i\to\infty. \end{aligned}
\end{equation*}
\notag
$$
However, this contradicts condition (2.4) of the lemma. This proves Lemma 4. § 3. Proof of the main lemmaLemma 5 (main lemma). Let the series (1.1) with coefficients (1.2) converge to zero in measure on $\Delta^{i_0}_{j_0}$, and let Then there exist $i$ and $j$ such that and Proof. We set It is clear that $A_m$ is an open set and $A_m\subset A_{m+1}$. Let The set $A$ is also open and, as any open set, is a union of disjoint open intervals, that is,
Assume that there exist $i$ and $j$ such that $\Delta^i_j\subset A$ and $(\mathcal S, M^i_j)\neq 0$. Since each $A_m$ is an open set, the sets $A_m\subset A_{m+1}$ and $\Delta^i_j$ are closed, and therefore $\Delta^i_j\subset A_m$ for some $m$. It is clear that these $i$ and $j$ are as required. Now suppose that
$$
\begin{equation}
(\mathcal S, M^i_j)=0 \quad\text{if } \Delta^i_j\subset A.
\end{equation}
\tag{3.2}
$$
Let $q_0$ be such that (see (1.4))
$$
\begin{equation}
\operatorname{mes}\biggl(\bigcup_{q\geqslant q_0}I_q\biggr)< 2^{-\kappa-9}(n_{i_0+1})^{-1}.
\end{equation}
\tag{3.3}
$$
We set
$$
\begin{equation}
A^{q_0}:=\bigcup_{q<q_0}I_q, \qquad B:=\bigcup_{q\geqslant q_0}I_q\quad\text{and} \quad D:=\{x\in\Delta^{i_0}_{j_0}\colon \mathcal M(\chi_B,x)>2^{-\kappa-4}\},
\end{equation}
\tag{3.4}
$$
where $\mathcal M(\chi_B,x)$ is the maximal function of the characteristic function of the set $B$ with respect to the intervals $[s^i_j,s^i_{j+1}]$, that is,
$$
\begin{equation*}
\mathcal M(\chi_B,x):=\sup_{x\in[s^i_j,s^i_{j+1}]} \frac{\operatorname{mes}(B\cap[s^i_j,s^i_{j+1}])}{s^i_{j+1}-s^i_{j}}.
\end{equation*}
\notag
$$
Proceeding by induction, we arrive at the representations
$$
\begin{equation}
\begin{aligned} \, \notag M^{i_0}_{j_0}(x) &=\sum_{\nu=i_0}^i\sum_{\Delta^{\nu}_j\subset A^{q_0}}\alpha^{(\nu)}_jM^{\nu}_j(x)+ \sum_{\nu=i_0+1}^i\sum_{\Delta^{\nu}_j\subset D}\alpha^{(\nu)}_jM^{\nu}_j(x) \\ &\qquad +\sum_{\substack{\Delta^i_j\not\subset A^{q_0}\\ \Delta^i_j\not\subset D}} \alpha^{(i)}_jM^i_j(x)=:\Sigma^i_1(x)+\Sigma^i_2(x)+\Sigma^i_3(x), \qquad i>i_0, \end{aligned}
\end{equation}
\tag{3.5}
$$
with coefficients
$$
\begin{equation}
\alpha^{(\nu)}_j\geqslant 0, \qquad\sum_{\nu=i_0}^i\sum_j\alpha^{(\nu)}_j=1.
\end{equation}
\tag{3.6}
$$
For $i=i_0+1$, from Lemma 1 we obtain
$$
\begin{equation}
M^{i_0}_{j_0}(x)=\sum_{j\colon \Delta^{i_0+1}_j\subset\Delta^{i_0}_{j_0}}\alpha^{(i_0+1)}_jM^{i_0+1}_j(x).
\end{equation}
\tag{3.7}
$$
We also have $\Delta^{i_0+1}_j\not\subset D$ for each $j$. Indeed, supposing on the contrary that $\Delta^{i_0+1}_j\subset D$, we have
$$
\begin{equation*}
\operatorname{mes}(B)\geqslant \operatorname{mes}(\Delta^{i_0+1}_j\cap B)\geqslant 2^{-\kappa-4}\operatorname{mes}(\Delta^{i_0+1}_j)>\frac{2^{-\kappa-4}}{2n_{i_0+1}},
\end{equation*}
\notag
$$
which contradicts (3.3) (see also (3.4)).
Let $\Sigma^{i_0+1}_1(x)$ be the sum of the terms in (3.7) for which $\Delta^{i_0+1}_j\subset A^{q_0}$, and let $\Sigma^{i_0+1}_3(x)$ be the sum of the remaining terms. This completes the first induction step. Assume that representation (3.5) is true for some $i$. Let us verify (3.5) for $i+1$. Applying Lemma 1 to the terms $M^i_j$ involved in $\Sigma^i_3$ and collecting similar terms, we obtain the sum Let $\Sigma^{i+1}_1(x)$ be the sum of the terms from (3.8) for which $\Delta_j^{i+1}\subset A^{q_0}$, let $\Sigma^{i+1}_2(x)$ be the sum of the terms for which $\Delta_j^{i+1} \subset D$ and $\Delta_j^{i+1} \not\subset A^{q_0}$, and let $\Sigma^{i+1}_3(x)$ be the sum of the remaining terms. Thus, representation (3.5) holds for each $i$. That condition (3.6) is satisfied is clear. Now, for each $i$, using (3.5) we have
$$
\begin{equation}
\begin{aligned} \, \notag (\mathcal S, M^{i_0}_{j_0}) &=\sum_{\nu=i_0}^i\sum_{\Delta^{\nu}_j\subset A^{q_0}}\alpha^{(\nu)}_j(\mathcal S, M^{\nu}_j)+ \sum_{\nu=i_0+1}^i\sum_{\Delta^{\nu}_j\subset D}\alpha^{(\nu)}_j(\mathcal S, M^{\nu}_j) \\ &\qquad +\sum_{\substack{\Delta^i_j\not\subset A^{q_0}\\ \Delta^i_j\not\subset D}} \alpha^{(i)}_j(\mathcal S,M^i_j)=:\sigma^i_1+\sigma^i_2+\sigma^i_3. \end{aligned}
\end{equation}
\tag{3.9}
$$
The sum $\sigma_1^i$ is zero by assumption (see (3.2)). We claim that (see (3.5))
$$
\begin{equation}
|(\mathcal S, M^i_j)|<d \quad\text{if } M^i_j(x) \text{ is involved in the sum } \sigma^i_2.
\end{equation}
\tag{3.10}
$$
Suppose on the contrary that $|(\mathcal S, M^i_j)|\geqslant d$. Then by Lemma 2 we would have
$$
\begin{equation}
\operatorname{mes}\biggl\{x\in\Delta^i_j\colon |S^{(i)}(x)|\geqslant\frac{d}{2}\biggr\} \geqslant\frac{\operatorname{mes}(\Delta^i_j)}{9}.
\end{equation}
\tag{3.11}
$$
Let $\Delta^{i-1}_{\nu}$ be an interval such that $\Delta^i_j\subset\Delta^{i-1}_{\nu}$. Then (3.11) would imply that
$$
\begin{equation*}
\operatorname{mes}\biggl\{x\in\Delta^{i-1}_{\nu}\colon |S^{(i)}(x)|\geqslant\frac{d}{2}\biggr\} \geqslant\frac{\operatorname{mes}(\Delta^{i-1}_{\nu})}{9\cdot 2^{\kappa}}
\end{equation*}
\notag
$$
(also see (1.4)), which would give $\Delta^{i-1}_{\nu}\subset D$. But this is impossible by the construction of (3.5). This proves (3.10). As a result, we have (see (3.5) and (3.4))
$$
\begin{equation}
\begin{aligned} \, \notag |\sigma_2^i| &\leqslant d\sum_{\nu=i_0+1}^i\sum_{\Delta^{\nu}_j\subset D}\alpha^{(\nu)}_j=d \int\sum_{\nu=i_0+1}^i\sum_{\Delta^{\nu}_j\subset D}\alpha^{(\nu)}_jM^{\nu}_j(x)\,dx \\ \notag &\leqslant d\int_DM^{i_0}_{j_0}(x)\,dx\leqslant \frac{d}{2} \|M^{i_0}_{j_0}\|_{\infty}\operatorname{mes}(D)= \frac{d}{\operatorname{mes}(\Delta^{i_0}_{j_0})}\operatorname{mes}(D) \\ &\leqslant d\frac{10\cdot 2^{\kappa+4}}{\operatorname{mes}(\Delta^{i_0}_{j_0})}\operatorname{mes}(B)\leqslant d\frac{5\cdot 2^{-4}}{n_{i_0+1}\operatorname{mes}(\Delta^{i_0}_{j_0})}\leqslant\frac{d}{3}. \end{aligned}
\end{equation}
\tag{3.12}
$$
Now let us estimate $\sigma_3^i$. The sum $\Sigma^i_3(x)$ involves at most $4q_0$ terms satisfying $\Delta^i_j\cap A^{q_0}\neq\varnothing$. Next we set
$$
\begin{equation*}
\begin{aligned} \, J^i_4 &:=\{j\colon \Delta^i_j\not\subset D,\ \Delta^i_j\not\subset A^{q_0},\ \Delta^i_j\cap A^{q_0}\neq\varnothing\}, \\ J^i_5 &:=\{j\colon \Delta^i_j\not\subset D,\ \Delta^i_j\cap A^{q_0}=\varnothing\} \end{aligned}
\end{equation*}
\notag
$$
and define
$$
\begin{equation}
J^i_6:=\biggl\{j\in J^i_5\colon |(\mathcal S, M^i_j)|<\frac{d}{3}\biggr\}
\end{equation}
\tag{3.13}
$$
and
$$
\begin{equation}
J^i_7:=\biggl\{j\in J^i_5\colon |(\mathcal S, M^i_j)|\geqslant\frac{d}{3}\biggr\}.
\end{equation}
\tag{3.14}
$$
We have
$$
\begin{equation}
|\sigma^i_3|\leqslant \sum_{j\in J^{i}_4}\alpha^{i}_j|(\mathcal S, M^i_j)|+\sum_{j\in J^{i}_6}\alpha^{i}_j|(\mathcal S, M^i_j)|+ \sum_{j\in J^{i}_7}\alpha^{i}_j|(\mathcal S, M^i_j)|=:\sigma^i_4+\sigma^i_6+\sigma^i_7.
\end{equation}
\tag{3.15}
$$
Let us estimate $\sigma^i_4$, $\sigma^i_6$ and $\sigma^i_7$ separately. For $\sigma^i_4$, using (2.2) and (2.3), we have
$$
\begin{equation}
\begin{aligned} \, \notag |\sigma^i_4| &\leqslant o(n_i)\sum_{j\in J^i_4}\alpha^{(i)}_j=o(n_i)\sum_{j\in J^i_4}\alpha^{(i)}_j\int M^i_j(x)\,dx \\ &\leqslant o(n_i)\int_H M^{i_0}_{j_0}(x)\,dx\leqslant o(n_i)\frac{2}{\operatorname{mes}(\Delta^{i_0}_{j_0})}\frac{16q_0}{n_i}, \quad\text{where } H:=\bigcup_{j\in J^i_4}\Delta^i_j. \end{aligned}
\end{equation}
\tag{3.16}
$$
Therefore,
$$
\begin{equation}
|\sigma^i_4|\leqslant\frac{d}{10} \quad\text{for sufficiently large}\ i.
\end{equation}
\tag{3.17}
$$
For $\sigma^i_6$ we have (see (3.6) and (3.13)) Proceeding as the proof of (3.10), one can easily check that
$$
\begin{equation*}
|(\mathcal S, M^i_j)|\leqslant d \quad\text{for } j\in J^i_5.
\end{equation*}
\notag
$$
Hence (see (3.14))
$$
\begin{equation*}
|(\mathcal S, M^i_j)|\leqslant d\quad\text{for } j\in J^i_7.
\end{equation*}
\notag
$$
As a result,
$$
\begin{equation}
\begin{aligned} \, \notag |\sigma^i_7| &\leqslant d\sum_{j\in J^i_7}\alpha^{(i)}_j= d\sum_{j\in J^i_7}\alpha^{(i)}_j\int M^{i}_j(x)\,dx\leqslant d\int_EM^{i_0}_{j_0}(x)\,dx \\ &\leqslant d\|M^{i_0}_{j_0}\|_{\infty}\operatorname{mes}(E), \quad\text{where } E:=\bigcup_{j\in J^i_7}\Delta^i_j \end{aligned}
\end{equation}
\tag{3.19}
$$
(see (3.5)). Arguing as the proof of (2.12) and using (3.14), we find that
$$
\begin{equation}
\operatorname{card}(J^i_7)\leqslant 36n_i\operatorname{mes}\biggl\{x\in\Delta^{i_0}_{j_0}\colon |S^{(i)}(x)|>\frac{d}{6}\biggr\}.
\end{equation}
\tag{3.20}
$$
The series (1.1) converges to zero in measure, hence, for sufficiently large $i$,
$$
\begin{equation}
\operatorname{mes}\biggl\{x\in\Delta^{i_0}_{j_0}\colon |S^{(i)}(x)|>\frac{d}{6}\biggr\}<\frac{1}{2000n_{i_0}}.
\end{equation}
\tag{3.21}
$$
Using (3.19)–(3.21) we obtain
$$
\begin{equation}
\sigma^i_7<d4n_{i_0}\frac{4}{n_i}\operatorname{card}(J^i_7)\leqslant \frac{16n_{i_0}d}{n_i}\frac{36n_i}{2000n_{i_0}}<\frac{d}{3}.
\end{equation}
\tag{3.22}
$$
Now an appeal to (3.9), (3.12), (3.15)–(3.18) and (3.22) shows that
$$
\begin{equation*}
|(\mathcal S, M^{i_0}_{j_0})|<\frac{d}{3}+\frac{d}{10}+\frac{d}{10}+\frac{d}{3}<d,
\end{equation*}
\notag
$$
but this contradicts (3.1). Hence our assumption (3.2) is not true. Lemma 5 is proved.
§ 4. Proofs of the theorems Proof of Theorem 5. Let $B=\{y_p\}_{p=1}^{\infty}$, and let the series (1.1) satisfy the assumptions of Theorem 5. Assume that the series (1.1) is nontrivial, that is, there exists $k$ such that $a_k\neq 0$. For $i_0$ such that $n_{i_0}>k$ there exists $j_0$ such that $(\mathcal S, M^{i_0}_{j_0})\neq 0$ (recall that $\{M^{i}_j\}_{j=0}^{n_i}$ is a basis of $\mathbf S_{n_{i}})$.
Applying Lemma 3 we find a closed interval $\Delta^{i'_1}_{j'_1}$ such that
$$
\begin{equation*}
y_1\notin\Delta^{i'_1}_{j'_1}\subset \Delta^{i_0}_{j_0}\quad\text{and} \quad (\mathcal S, M^{i_1'}_{j_1'})\neq 0.
\end{equation*}
\notag
$$
Next, using Lemma 4 we find a closed interval $\Delta^{i''_1}_{j''_1}$ such that
$$
\begin{equation}
\Delta^{i''_1}_{j''_1}\subset \Delta^{i'_1}_{j'_1}, \qquad |(\mathcal S, M^{i_1''}_{j_1''})|>10.
\end{equation}
\tag{4.1}
$$
An application of Lemma 5 to (4.1) produces $i_1$ and $j_1$ such that
$$
\begin{equation}
y_1\notin\Delta^{i_1}_{j_1}\subset \Delta^{i''_1}_{j''_1}\subset \Delta^{i_0}_{j_0}\quad\text{and} \quad\max_{i_0\leqslant\nu\leqslant i_1} |S^{(\nu)}(x)|>1 \quad\text{for } x\in\Delta^{i_1}_{j_1}.
\end{equation}
\tag{4.2}
$$
Assume that we have already found $i_1,\dots,j_p$, $j_1,\dots, j_p$ such that
$$
\begin{equation*}
y_{\nu}\notin\Delta^{i_\nu}_{j_\nu}\subset \Delta^{i_{\nu-1}}_{j_{\nu-1}}, \qquad \nu=1,\dots,p,
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\max_{i_{\nu-1}\leqslant \mu\leqslant i_{\nu}}|S^{{\mu}}(x)|>\nu \quad\text{for } x\in\Delta^{i_{\nu}}_{j_{\nu}}, \quad\nu=1,\dots,p.
\end{equation*}
\notag
$$
Applying Lemma 3 we find a closed interval $\Delta^{i'_{p+1}}_{j'_{p+1}}$ such that
$$
\begin{equation*}
y_{p+1}\notin\Delta^{i'_{p+1}}_{j'_{p+1}}\subset \Delta^{i_p}_{j_p}, \qquad (\mathcal S, M^{i_{p+1}'}_{j_{p+1}'})\neq 0.
\end{equation*}
\notag
$$
Next, an appeal to Lemma 4 produces a closed interval $\Delta^{i''_{p+1}}_{j''_{p+1}}$ satisfying
$$
\begin{equation}
\Delta^{i''_{p+1}}_{j''_{p+1}}\subset \Delta^{i'_{p+1}}_{j'_{p+1}}, \qquad |(\mathcal S, M^{i_{p+1}''}_{j_{p+1}''})|>10(p+1).
\end{equation}
\tag{4.3}
$$
Applying Lemma 5 to (4.3) we find integers $i_{p+1}$ and $j_{p+1}$ such that
$$
\begin{equation*}
y_{p+1}\notin\Delta^{i_{p+1}}_{j_{p+1}}\subset \Delta^{i''_{p+1}}_{j''_{p+1}}\subset \Delta^{i_p}_{j_p}\quad\text{and} \quad \max_{i_p\leqslant\nu\leqslant i_{p+1}} |S^{(\nu)}(x)|>p+1 \quad\text{for } x\in\Delta^{i_{p+1}}_{j_{p+1}}.
\end{equation*}
\notag
$$
Proceeding in this way, we construct a nested sequence of closed intervals $\Delta^{i_p}_{j_p}$ such that
$$
\begin{equation}
y_p\notin\Delta^{i_p}_{j_p}\subset\Delta^{i_{p-1}}_{j_{p-1}}, \qquad p=1,2,\dots,
\end{equation}
\tag{4.4}
$$
$$
\begin{equation}
\max_{i_{p-1}\leqslant\nu\leqslant i_{p}}|S^{(\nu)}(x)|>p \quad\text{for } x\in\Delta^{i_p}_{j_p}.
\end{equation}
\tag{4.5}
$$
By (4.4) and (4.5) there exists $z$ such that
$$
\begin{equation*}
z\notin\{y_p\}_{p=1}^{\infty}=B\quad\text{and} \quad \sup_p|S^{(i_p)}(z)|=\infty,
\end{equation*}
\notag
$$
which contradicts (1.6). This proves Theorem 5. Proof of Theorem 7. Let $\sum_{n=0}^{\infty}b_nf_n(x)$ be the Fourier-Franklin series of a bounded function $f$. As already noted, in this case
$$
\begin{equation*}
\sup_i\biggl|\sum_{n=0}^{n_i}b_nf_n(x)\biggr|< C\|f\|_{\infty}, \qquad x\in[0,1],
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\lim_{i\to\infty}\sum_{n=0}^{n_i}b_nf_n(x)=f(x) \quad\text{in measure on}\ [0,1].
\end{equation*}
\notag
$$
Therefore, the series $\sum_{n=0}^{\infty}c_nf_n(x)$ with coefficients $c_n\,{:=}\,a_n-b_n$, $n\,{=}\,0, 1, 2, \dots$, satisfies the conditions of Theorem 5. Hence $c_n=0$, $n=0,1,2,\dots$, that is, $a_n=b_n$, $n=0,1,2,\dots$ . This proves Theorem 7. 
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\Bibitem{Gev23} |
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