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This article is cited in 2 scientific papers (total in 2 papers) Estimates for the Kolmogorov widths of an intersection of two balls in a mixed norm A. A. Vasil'evaab a Faculty of Mechanics and Mathematics, Lomonosov Moscow State University, Moscow, Russia b Moscow Center of Fundamental and Applied Mathematics, Moscow, Russia
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     § 1. IntroductionWe consider the order estimate problem for the Kolmogorov widths of intersections of two finite-dimensional balls in the mixed norm. First we give the requisite definitions and notation. Let $m, k\in \mathbb{N}$, $1\leqslant p<\infty$ and $1\leqslant \theta<\infty$. We let $l_{p,\theta}^{m,k}$ denote the space $\mathbb{R}^{mk}$ equipped with the norm
$$
\begin{equation*}
\|(x_{i,j})_{1\leqslant i\leqslant m,\,1\leqslant j\leqslant k}\|_{l_{p,\theta}^{m,k}} =\biggl(\sum _{j=1}^k\biggl(\sum _{i=1}^m |x_{i,j}|^p\biggr)^{\theta/p}\biggr)^{1/\theta}.
\end{equation*}
\notag
$$
For $p=\infty$ or $\theta=\infty$ the corresponding modifications are clear. The unit ball of the space $l_{p,\theta}^{m,k}$ is denoted by $B_{p,\theta}^{m,k}$. For $k=1$, the space and the unit ball are denoted, respectively, by $l_p^m$ and $B_p^m$. Let $X$ be a normed space, let $M\subset X$ and $n\in \mathbb{Z}_+$. Then the Kolmogorov $n$-width of the set $M$ in $X$ is defined by
$$
\begin{equation*}
d_n(M, X)=\inf _{L\in {\cal L}_n(X)} \sup _{x\in M} \inf _{y\in L} \|x-y\|;
\end{equation*}
\notag
$$
here ${\cal L}_n(X)$ is the set of all subspaces in $X$ of dimension $\leqslant n$. For more on widths, see, for example, [1]–[3]. The exact values of the widths $d_n(B_p^m, l_q^m)$ were found in [4], [5] (for $p\geqslant q$) and in [6], [7] (for $p=1$ and $q=2$). For $p\leqslant q<\infty$ order estimates were obtained in [8] and [9]. The problem of estimating $d_n(B_p^m, l_\infty^m)$ was considered in [10]–[12]: order estimates are known for $p\geqslant 2$; for $1\leqslant p<2$ the corresponding values are known up to a factor of $\log (em/n)$ in some power. Approximative characteristics of the balls $B^{m,k}_{p,\theta}$ in the space $l^{m,k}_{q,\sigma}$ are important for the study of Besov classes with dominating mixed smoothness (see [13]–[15]) and weighted Besov classes (see [16]). The Kolmogorov widths $d_n(B^{m,k}_{p,\theta}, l^{m,k}_{q,\sigma})$ for $n \leqslant mk/{2}$ were estimated in [14] and [16]–[21] (more precisely, the paper [14] was concerned with Gelfand widths, but if $p$, $\theta$, $q$, $\sigma \geqslant 1$, then this problem can be reformulated in terms of Kolmogorov widths; see [22]). Order estimates were obtained for the following values of the parameters:
In addition, Galeev [23] obtained a lower estimate in the case when $1\leqslant p\leqslant \infty$, $\theta=\infty$, $2\leqslant q<\infty$, $\sigma=q$ and $n\leqslant c(q)mk$ (where $c(q)$ is some positive number). The problem of estimates for the widths of the intersection of a family of Sobolev or Besov classes (see [13], [17], [24] and [25]) can be reduced, via discretization, to that of estimates for the widths $d_n(\bigcap _{\alpha \in A} \nu_\alpha B^m_{p_\alpha}, l_q^m)$ of intersections of balls. Galeev [24] found order estimates for this quantity for $n=m/{2}$; in [26] this result was extended to $n \leqslant {m}/{2}$. The problem of estimates for the widths of intersections of finite-dimensional balls in the mixed norm is natural. The results obtained in this way can be used to estimate the widths of intersections of weighed Besov classes or Besov classes with dominating mixed smoothness. In the present paper we consider the case of two balls $\nu_i B^{m,k}_{p_i,\theta_i}$, $i=1, 2$, where we assume that $2\leqslant q<\infty$, $2\leqslant \sigma <\infty$, $1\leqslant p_i\leqslant q$ and $1\leqslant \theta_i\leqslant \sigma$, $i=1, 2$. It turns out that for this range of parameters the above problem can be reduced to the evaluation of widths of a single ball in mixed norms; the orders of these widths were already found in [16] (see Theorem 1 below). Given two sets $X$ and $Y$ and functions $f_1$, $f_2\colon X\times Y\to \mathbb{R}_+$, we write
$$
\begin{equation*}
f_1(x, y)\underset{y}{\lesssim} f_2(x, y) \quad \Bigl(\text{or } f_2(x, y)\underset{y}{\gtrsim} f_1(x, y)\Bigr)
\end{equation*}
\notag
$$
if, for any $y\in Y$, there exists $c(y)>0$ such that $f_1(x, y)\leqslant c(y)f_2(x, y)$ for each $x\in X$; we also write if $f_1(x, y) \underset{y}{\lesssim} f_2(x, y)$ and $f_2(x,y)\underset{y}{\lesssim} f_1(x, y)$. Let $q>2$ and $1\leqslant p\leqslant q$. Then we set
$$
\begin{equation*}
\lambda_{p,q}=\min \biggl\{\frac{1/p-1/q}{1/2-1/q}, 1\biggr\},
\end{equation*}
\notag
$$
and for $q=2$ and $1\leqslant p\leqslant 2$ we set $\lambda_{p,2}=1$. Theorem 1 (see [16]). Let $m$, $k\in \mathbb{N}$, $n\in \mathbb{Z}_+$, $n\leqslant {mk}/{2}$, $2\leqslant q<\infty$, $2\leqslant \sigma <\infty$, $1\leqslant p\leqslant q$ and $1\leqslant \theta \leqslant \sigma$. Then In [16] this theorem was proved for $n \leqslant a(q, \sigma)mk$; in addition, in its statement in [16] the constants in order equalities depend on $p$, $\theta$, $q$ and $\sigma$. However, the proof shows that they are independent of $p$ and $\theta$. The upper estimate holds for all $n\leqslant mk$. For $a(q, \sigma)mk \leqslant n \leqslant mk/{2}$ the lower estimate is proved in § 2 (the corollary). Note that if $2\leqslant p\leqslant q$, $2\leqslant \theta\leqslant \sigma$ and $\lambda_{p,q}= \lambda_{\theta,\sigma}$, then
$$
\begin{equation}
\begin{aligned} \, \notag (n^{-1/2}m^{1/q}k^{1/\sigma})^{\lambda_{p,q}} &=m^{1/q-1/p}(n^{-1/2}m^{1/2}k^{1/\sigma})^{\lambda_{\theta,\sigma}} \\ &=(n^{-1/2}m^{1/q}k^{1/\sigma})^{\lambda_{\theta,\sigma}}= k^{1/\sigma-1/\theta}(n^{-1/2}k^{1/2}m^{1/q})^{\lambda_{p,q}}. \end{aligned}
\end{equation}
\tag{4}
$$
The main result of the paper is as follows. Theorem 2. Let $m$, $k\in \mathbb{N}$, $n\in \mathbb{Z}_+$, $n\leqslant {mk}/{2}$, ${2\leqslant q<\infty}$, $2\leqslant \sigma < \infty$, ${1\,{\leqslant}\, p_i\,{\leqslant}\, q}$, $1\leqslant \theta_i\leqslant \sigma$ and $\nu_i>0$, $i=1, 2$. Let $\Phi_j=\Phi_j(m,k,n,p_1,p_2,\theta_1,\theta_2,q,\sigma,\nu_1,\nu_2)$, $j=1, \dots, 5$, be defined by: 1) $\Phi_j=\nu_j d_n(B^{m,k}_{p_j,\theta_j}, l^{m,k}_{q,\sigma})$ for $j=1, 2$; 2) if $p_1\ne 2$ and there exists $\widetilde \lambda \in (0, 1)$ such that $\frac 12= \frac{1-\widetilde \lambda}{p_1}+\frac{\widetilde \lambda}{p_2}$, then $\Phi_3= \nu_1^{1-\widetilde \lambda}\nu_2^{\widetilde \lambda} d_n(B^{m,k}_{2,\widetilde\theta}, l^{m,k}_{q,\sigma})$, where $\widetilde \theta$ is defined by the equality $\frac{1}{\widetilde \theta}=\frac{1-\widetilde \lambda}{\theta_1}+\frac{\widetilde \lambda}{\theta_2}$; otherwise set $\Phi_3=+\infty$; 3) if $\theta_1\ne 2$ and there exists $\widetilde \mu \in (0, 1)$ such that $\frac 12= \frac{1-\widetilde \mu}{\theta_1}+\frac{\widetilde \mu}{\theta_2}$, then $\Phi_4=\nu_1^{1-\widetilde \mu}\nu_2^{\widetilde \mu} d_n(B^{m,k}_{\widetilde p,2}, l^{m,k}_{q,\sigma})$, where $\widetilde p$ is defined by the equality $\frac{1}{\widetilde p}=\frac{1-\widetilde \mu}{p_1}+ \frac{\widetilde \mu}{p_2}$; otherwise set $\Phi_4=+\infty$; 4) if $q>2$, $\sigma>2$ and $\frac{1/p_1-1/q}{1/2-1/q}\ne \frac{1/\theta_1-1/\sigma}{1/2-1/\sigma}$, and if there exist $\lambda \in (0, 1)$, $p\in (2, q]$ and $\theta\in (2, \sigma]$ such that $\frac 1p=\frac{1-\lambda}{p_1}+ \frac{\lambda}{p_2}$, $\frac{1}{\theta}=\frac{1-\lambda}{\theta_1}+ \frac{\lambda}{\theta_2}$ and $\lambda_{p,q}=\lambda_{\theta,\sigma}$, then $\Phi_5=\nu_1^{1-\lambda}\nu_2^{\lambda} d_n(B^{m,k}_{p,\theta}, l^{m,k}_{q,\sigma})$; otherwise set $\Phi_5=+\infty$. Then The above result was announced in [27] in an equivalent form. § 2. Auxiliary resultsLet $k,m,r,l\in \mathbb{N}$, $1\leqslant r\leqslant m$ and $1\leqslant l\leqslant k$. Set
$$
\begin{equation*}
G=\bigl\{(\tau_1, \tau_2, \varepsilon_1, \varepsilon_2)\colon \tau_1\in S_m,\, \tau_2\in S_k,\, \varepsilon_1\in \{1, -1\}^m,\, \varepsilon_2\in \{1, -1\}^k\bigr\},
\end{equation*}
\notag
$$
where $S_m$ and $S_k$ are the permutation groups on $m$ and $k$ elements, respectively. For $x=(x_{i,j})_{1\leqslant i\leqslant m, 1\leqslant j\leqslant k}\in \mathbb{R}^{mk}$, $\gamma=(\tau_1, \tau_2, \varepsilon_1, \varepsilon_2)\in G$, $\varepsilon_1=(\varepsilon_{1,i})_{1\leqslant i\leqslant m}$ and $\varepsilon_2=(\varepsilon_{2,j})_{1\leqslant j\leqslant k}$ we set
$$
\begin{equation}
\gamma(x)=(\varepsilon_{1,i}\varepsilon_{2,j}x_{\tau_1(i)\tau_2(j)})_{1\leqslant i\leqslant m,\,1\leqslant j\leqslant k}.
\end{equation}
\tag{5}
$$
We also define $e=(e_{i,j}^{m,k,r,l})_{1\leqslant i\leqslant m, 1\leqslant j\leqslant k}$ by
$$
\begin{equation}
e_{i,j}^{m,k,r,l}= \begin{cases} 1 &\text{if } 1\leqslant i\leqslant r,\, 1\leqslant j\leqslant l, \\ 0 &\text{otherwise}, \end{cases}
\end{equation}
\tag{6}
$$
and set
$$
\begin{equation}
V_{r,l}^{m,k}=\operatorname{conv}\{\gamma(e)\colon \gamma\in G\}.
\end{equation}
\tag{7}
$$
It was shown in [16], formula (34), that if $2\leqslant q<\infty$, $2\leqslant \sigma<\infty$, $n\in \mathbb{Z}_+$ and $n\leqslant a(q, \sigma) m^{2/q}k^{2/\sigma}r^{1-2/q} l^{1-2/\sigma}$, then
$$
\begin{equation}
d_n(V^{m,k}_{r,l}, l^{m,k}_{q,\sigma}) \geqslant b(q, \sigma) r^{1/q}l^{1/\sigma},
\end{equation}
\tag{8}
$$
where $a(q, \sigma)>0$ and $b(q, \sigma)>0$; in addition, the function $a(\,\cdot\,{,}\,\cdot\,) $ is nonincreasing in each argument and the function $b(\,\cdot\,{,}\,\cdot\,)$ is continuous. Here we obtain an estimate for all $n\leqslant mk/2$. The proof depends on Gluskin’s method presented in [8]. Proposition. Let $2\leqslant q<\infty$, $2\leqslant \sigma <\infty$, $n\in \mathbb{Z}_+$ and $n\leqslant mk/2$. Then Proof. For $n\leqslant a(q, \sigma)m^{2/q}k^{2/\sigma}r^{1-2/q} l^{1-2/\sigma}$ the required estimate follows from inequality (8).
Consider the case when $a(q, \sigma)m^{2/q}k^{2/\sigma}r^{1-2/q} l^{1-2/\sigma} \leqslant n\leqslant a(q, \sigma)mk$. Then there exist numbers $\widetilde q\in [2, q]$ and $\widetilde \sigma \in [2, \sigma]$ such that
$$
\begin{equation}
n=a(q, \sigma)m^{2/\widetilde q}k^{2/\widetilde\sigma}r^{1-2/\widetilde q} l^{1-2/\widetilde\sigma}.
\end{equation}
\tag{10}
$$
The function $a(\,\cdot\,{,}\,\cdot\,)$ is nonincreasing in each argument. As a result,
$$
\begin{equation*}
n\leqslant a(\widetilde q, \widetilde \sigma)m^{2/\widetilde q}k^{2/\widetilde\sigma}r^{1-2/\widetilde q} l^{1-2/\widetilde\sigma}.
\end{equation*}
\notag
$$
Hence by (8)
$$
\begin{equation*}
d_n(V^{m,k}_{r,l}, l^{m,k}_{\widetilde q,\widetilde \sigma}) \geqslant b(\widetilde q, \widetilde \sigma) r^{1/\widetilde q}l^{1/\widetilde\sigma} \underset{q,\sigma}{\gtrsim} r^{1/\widetilde q}l^{1/\widetilde\sigma}
\end{equation*}
\notag
$$
(here we have used the fact that $b$ is a continuous function). Therefore,
$$
\begin{equation*}
\begin{aligned} \, d_n(V^{m,k}_{r,l}, l^{m,k}_{q,\sigma}) &\geqslant m^{1/q-1/\widetilde q}k^{1/\sigma -1/\widetilde \sigma}d_n(V^{m,k}_{r,l}, l^{m,k}_{\widetilde q,\widetilde \sigma}) \\ &\underset{q,\sigma}{\gtrsim} m^{1/q-1/\widetilde q}k^{1/\sigma -1/\widetilde \sigma}r^{1/\widetilde q}l^{1/\widetilde\sigma} \stackrel{(10)}{\underset{q,\sigma}{\asymp}} m^{1/q}k^{1/\sigma}n^{-1/2}r^{1/2} l^{1/2}. \end{aligned}
\end{equation*}
\notag
$$
It remains to consider the case when $a(q, \sigma)mk \leqslant n \leqslant mk/2$. First we verify that
$$
\begin{equation}
d_n(V^{m,k}_{r,l}, l^{m,k}_{2,2}) \gtrsim r^{1/2}l^{1/2} \quad \text{for}\ n\leqslant \frac{mk}{2}.
\end{equation}
\tag{11}
$$
We follow the argument from [16], pp. 14–17, in this more simple case; instead of using inequality (35) from [16], we use the formula for the square of a sum, while Hölder’s and Young’s inequalities are not required at the end of the proof. As a result, for some $\xi \in \mathbb{R}$ we have
$$
\begin{equation*}
d_n(V^{m,k}_{r,l}, l^{m,k}_{2,2})\geqslant rl -2\biggl(\frac{nrl}{mk}\biggr)^{1/2}\xi+\xi^2 \geqslant rl \biggl(1-\frac{n}{mk}\biggr)\geqslant \frac{rl}{2}
\end{equation*}
\notag
$$
for $n\leqslant mk/2$.
Now let $q\in [2, \infty)$, $\sigma \in [2, \infty)$. Then Hence, for $a(q, \sigma)mk\leqslant n\leqslant mk/2$, which proves the proposition. Corollary. Let $2\leqslant q<\infty$, $2\leqslant \sigma <\infty$, $1\leqslant p\leqslant q$, $1\leqslant \theta \leqslant \sigma$, $m, k, n\in \mathbb{N}$ and $a(q, \sigma)mk \leqslant n \leqslant mk/2$. Then Proof. For $\min\{p, \theta\}\geqslant 2$ (respectively, for $\theta \leqslant 2\leqslant p$, $p\leqslant 2 \leqslant \theta$, or ${\max\{p, \theta\}\leqslant 2}$), we use the inclusion $m^{-1/p}k^{-1/\theta}V^{m,k}_{m,k} \subset B^{m,k}_{p,\theta}$ (respectively, the inclusion $m^{-1/p}V^{m,k}_{m,1}\subset B^{m,k}_{p,\theta}$, $k^{-1/\theta}V^{m,k}_{1,k} \subset B^{m,k}_{p,\theta}$, or $V^{m,k}_{1,1}\subset B^{m,k}_{p,\theta}$). Now the required estimate follows from (9). This proves the corollary. In what follows we assume that $m$, $k\in \mathbb{N}$, $n\in \mathbb{Z}_+$, $n\leqslant mk/2$ and $\nu_i>0$, $1=1, 2$. Lemma 1. Let $1\leqslant p_i\leqslant \infty$, $1\leqslant \theta_i\leqslant \infty$, $\lambda \in [0, 1]$, and let $p$, $\theta\in [1, \infty]$ be defined by Then Proof. It suffices to verify the inequality
$$
\begin{equation*}
\|(x_{i,j})_{1\leqslant i\leqslant m,\,1\leqslant j\leqslant k}\|_{l_{p,\theta}^{m,k}} \leqslant \|(x_{i,j})_{1\leqslant i\leqslant m,\, 1\leqslant j\leqslant k}\|^{1-\lambda}_{l_{p_1,\theta_1}^{m,k}} \|(x_{i,j})_{1\leqslant i\leqslant m,\, 1\leqslant j\leqslant k}\|_{l_{p_2,\theta_2}^{m,k}}^\lambda.
\end{equation*}
\notag
$$
Let $\beta$ be defined by the equality $\frac{\beta}{p}=\frac{\lambda}{p_2}$. From (12) we obtain $\frac{1-\beta}{p}=\frac{1-\lambda}{p_1}$, and so $\beta \in [0, 1]$. Applying Hölder’s inequality, for each $j\in \{1, \dots, k\}$ we have
$$
\begin{equation}
\begin{aligned} \, \notag \|(x_{i,j})_{1\leqslant i\leqslant m}\|_{l_p^m} &=\biggl(\sum _{i=1}^m |x_{i,j}|^{p(1-\lambda)} |x_{i,j}|^{p\lambda}\biggr)^{1/p} \\ & \leqslant \|(x_{i,j})_{1\leqslant i\leqslant m}\|_{l_{p_1}^m}^{1-\lambda}\|(x_{i,j})_{1\leqslant i\leqslant m}\|_{l_{p_2}^m}^{\lambda}. \end{aligned}
\end{equation}
\tag{13}
$$
Let $\gamma$ be defined by $\frac{\gamma}{\theta}=\frac{\lambda}{\theta_2}$. Then $\frac{1-\gamma}{\theta} \stackrel{(12)}{=} \frac{1-\lambda}{\theta_1}$, which shows that $\gamma \in [0, 1]$. Another appeal to Hölder’s inequality shows that
$$
\begin{equation*}
\begin{aligned} \, \|(x_{i,j})_{1\leqslant i\leqslant m, 1\leqslant j\leqslant k}\|_{l_{p,\theta}^{m,k}} &\stackrel{(13)}{\leqslant} \biggl(\sum _{j=1}^k \|(x_{i,j})_{1\leqslant i\leqslant m}\|_{l_{p_1}^m}^{(1-\lambda)\theta}\|(x_{i,j})_{1\leqslant i\leqslant m}\|_{l_{p_2}^m}^{\lambda\theta}\biggr)^{1/\theta} \\ &\leqslant \|(x_{i,j})_{1\leqslant i\leqslant m,\,1\leqslant j\leqslant k}\|^{1-\lambda}_{l_{p_1,\theta_1}^{m,k}} \|(x_{i,j})_{1\leqslant i\leqslant m,\,1\leqslant j\leqslant k}\|_{l_{p_2,\theta_2}^{m,k}}^\lambda. \end{aligned}
\end{equation*}
\notag
$$
This proves Lemma 1. Lemma 2. Let $\lambda \in [0, 1]$, $\frac 1p=\frac{1-\lambda}{p_1}+\frac{\lambda}{p_2}$, $\frac{1}{\theta}=\frac{1-\lambda}{\theta_1}+\frac{\lambda}{\theta_2}$, $\widetilde r \in [1, m]$, $\widetilde l=[1, k]$, $r=\lfloor \widetilde r \rfloor$ or $r=\lceil \widetilde r \,\rceil$, $l=\lfloor \widetilde l\, \rfloor$ or $l=\lceil \,\widetilde l\, \rceil$, and Then Proof. In view of (5)–(7) it suffices to show that
$$
\begin{equation*}
\nu_1^{1-\lambda}\nu_2^{\lambda} \widetilde r^{\,1/p_1 -1/p} \widetilde l^{\,1/\theta_1-1/\theta}\leqslant \nu_1\quad\text{and}\quad \nu_1^{1-\lambda}\nu_2^{\lambda} \widetilde r^{\,1/p_2-1/p} \widetilde l^{\,1/\theta_2-1/\theta}\leqslant \nu_2.
\end{equation*}
\notag
$$
However, this result follows from (14). Lemma 2 is proved. Proof. By the inclusion $\min\{\nu_1, \nu_2\} V^{m,k}_{1,1} \subset \nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}$ and the above proposition we have
$$
\begin{equation*}
\begin{aligned} \, &d_n(\nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}, l^{m,k}_{q,\sigma}) \\ &\qquad \geqslant d_n(\min\{\nu_1, \nu_2\} V^{m,k}_{1,1}, l^{m,k}_{q,\sigma}) \underset{q,\sigma}{\gtrsim} \min\{\nu_1, \nu_2\}\min\{1, n^{-1/2}m^{1/q}k^{1/\sigma}\}, \end{aligned}
\end{equation*}
\notag
$$
which proves Lemma 3. Lemma 4. Let $2\leqslant q<\infty$, $2\leqslant \sigma <\infty$ and $m^{2/q} k^{2/\sigma}< n \leqslant mk/2$. 1. Let $2\leqslant p_1\leqslant q$, $\lambda_{p_1,q}\leqslant \lambda_{\theta_1,\sigma}$ and
$$
\begin{equation}
\frac{\nu_1}{\nu_2} \leqslant \begin{cases} (n^{1/2}m^{-1/q} k^{-1/\sigma})^{\frac{1/p_1-1/p_2}{1/2-1/q}}, & m^{2/q} k^{2/\sigma}< n \leqslant mk^{2/\sigma}, \\ m^{1/p_1 -1/p_2} (n^{1/2}m^{-1/2} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}}, & mk^{2/\sigma}< n \leqslant \dfrac{mk}{2}, \, \theta_1\geqslant 2, \\ m^{1/p_1 -1/p_2}, & mk^{2/\sigma}< n \leqslant \dfrac{mk}{2}, \, \theta_1< 2. \end{cases}
\end{equation}
\tag{15}
$$
Then
$$
\begin{equation}
\begin{aligned} \, \notag &d_n(\nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}, l^{m,k}_{q,\sigma}) \\ &\qquad \underset{q,\sigma}{\gtrsim} \nu_1\min\{(n^{-1/2} m^{1/q} k^{1/\sigma})^{\lambda_{p_1,q}}, m^{1/q-1/p_1} (n^{-1/2}m^{1/2} k^{1/\sigma})^{\lambda_{\theta_1,\sigma}}\}. \end{aligned}
\end{equation}
\tag{16}
$$
2. Let $2\leqslant \theta_1\leqslant \sigma$, $\lambda_{\theta_1,\sigma}\leqslant \lambda_{p_1,q}$ and
$$
\begin{equation*}
\frac{\nu_1}{\nu_2} \leqslant \begin{cases} (n^{1/2}m^{-1/q} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}}, & m^{2/q}k^{2/\sigma}< n \leqslant m^{2/q}k, \\ k^{1/\theta_1 -1/\theta_2} (n^{1/2}m^{-1/q} k^{-1/2})^{\frac{1/p_1-1/p_2}{1/2-1/q}}, & m^{2/q}k < n\leqslant \dfrac{mk}{2}, \, p_1\geqslant 2, \\ k^{1/\theta_1 -1/\theta_2}, & m^{2/q}k < n\leqslant \dfrac{mk}{2}, \, p_1< 2. \end{cases}
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
\begin{aligned} \, &d_n(\nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}, l^{m,k}_{q,\sigma}) \\ &\qquad\underset{q,\sigma}{\gtrsim} \nu_1\min \bigl\{(n^{-1/2} m^{1/q} k^{1/\sigma})^{\lambda_{\theta_1,\sigma}}, k^{1/\sigma-1/\theta_1} (n^{-1/2}m^{1/q} k^{1/2})^{\lambda_{p_1,q}}\bigr\}. \end{aligned}
\end{equation*}
\notag
$$
Proof. We prove assertion 1 (the proof of assertion 2 is similar).
Let $m^{2/q}k^{2/\sigma}< n \leqslant mk^{2/\sigma}$. Then $q>2$, and the right-hand side of (16) is
$$
\begin{equation*}
\nu_1 (n^{-1/2} m^{1/q} k^{1/\sigma})^{\frac{1/p_1-1/q}{1/2-1/q}}.
\end{equation*}
\notag
$$
We set
$$
\begin{equation}
r=\bigl\lceil(n^{1/2}m^{-1/q}k^{-1/\sigma})^{\frac{1}{1/2-1/q}}\bigr\rceil.
\end{equation}
\tag{17}
$$
Since $m^{2/q} k^{2/\sigma}\leqslant n\leqslant mk^{2/\sigma}$, we have $1\leqslant r\leqslant m$. We claim that
$$
\begin{equation}
\nu_1 r^{-1/p_1} V^{m,k}_{r,1} \subset 2(\nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}).
\end{equation}
\tag{18}
$$
It suffices to show that (see (6) and (7)). This result follows from (15) and (17).
By (17) we have $n\leqslant m^{2/q}k^{2/\sigma} r^{1-2/q}$. Hence
$$
\begin{equation*}
d_n(\nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}, l^{m,k}_{q,\sigma}) \stackrel{(18)}{\gtrsim} \nu_1 r^{-1/p_1}d_n(V_{r,1}^{m,k}, l_{q,\sigma}^{m,k}) \stackrel{(9)}{\underset{q,\sigma}{\gtrsim}} \nu_1 r^{1/q-1/p_1},
\end{equation*}
\notag
$$
and now (16) follows from (17).
Let $mk^{2/\sigma}< n \leqslant mk/2$, $\theta_1\geqslant 2$. Then $\sigma > 2$, and the right-hand side of (16) is
$$
\begin{equation*}
\nu_1m^{1/q-1/p_1}(n^{-1/2} m^{1/2} k^{1/\sigma})^{\frac{1/\theta_1-1/\sigma}{1/2-1/\sigma}}.
\end{equation*}
\notag
$$
Let
$$
\begin{equation}
l=\bigl\lceil (n^{1/2}m^{-1/2}k^{-1/\sigma})^{\frac{1}{1/2-1/\sigma}}\bigr\rceil.
\end{equation}
\tag{19}
$$
Since $mk^{2/\sigma}\leqslant n \leqslant mk/2$, we have $1\leqslant l\leqslant k$.
We claim that
$$
\begin{equation}
\nu_1 m^{-1/p_1} l^{-1/\theta_1} V^{m,k}_{m,l} \subset 2(\nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}).
\end{equation}
\tag{20}
$$
It suffices to show that
$$
\begin{equation*}
\nu_1 m^{1/p_2-1/p_1} l^{1/\theta_2-1/\theta_1} \leqslant 2\nu_2.
\end{equation*}
\notag
$$
This result follows from (15) and (19).
Next, by (19) we have $n\leqslant m^{2/q}k^{2/\sigma} m^{1-2/q}l^{1-2/\sigma}$. As a result,
$$
\begin{equation*}
\begin{aligned} \, &d_n(\nu_1B^{m,k}_{p_1,\theta_1}\cap \nu_2B^{m,k}_{p_2,\theta_2}, l^{m,k}_{q,\sigma}) \\ &\qquad\stackrel{(20)}{\gtrsim} \nu_1 m^{-1/p_1} l^{-1/\theta_1}d_n(V_{m,l}^{m,k}, l^{m,k}_{q,\sigma})\stackrel{(9)}{\underset{q,\sigma}{\gtrsim}} \nu_1m^{1/q-1/p_1}l^{1/\sigma-1/\theta_1}. \end{aligned}
\end{equation*}
\notag
$$
Now (16) follows from (19).
Let $mk^{2/\sigma}< n \leqslant mk/2$, $\theta_1< 2$. As above, we have $\sigma > 2$, and the right-hand side of (16) is Since $\nu_1 m^{1/p_2-1/p_1}\,{\stackrel{(15)}{\leqslant}}\, \nu_2$ , we have $\nu_1 m^{-1/p_1}V^{m,k}_{m,1} \subset \nu_1B_{p_1,\theta_1}^{m,k}\cap \nu_2 B_{p_2,\theta_2}^{m,k}$. Next, $n\geqslant m^{2/q}k^{2/\sigma} m^{1-2/q}$. Hence
$$
\begin{equation*}
\begin{aligned} \, &d_n(\nu_1B_{p_1,\theta_1}^{m,k}\cap \nu_2 B_{p_2,\theta_2}^{m,k}, l_{q,\sigma}^{m,k}) \\ &\qquad\geqslant d_n(\nu_1 m^{-1/p_1}V_{m,1}^{m,k}, l_{q,\sigma}^{m,k}) \stackrel{(9)}{\underset{q,\sigma}{\gtrsim}} \nu_1 m^{-1/p_1} n^{-1/2} m^{1/q} k^{1/\sigma}m^{1/2}. \end{aligned}
\end{equation*}
\notag
$$
This proves Lemma 4. Lemma 5. Let $q>2$ and $\sigma>2$. Assume that the following holds: 1) if $m^{2/q}k^{2/\sigma} \leqslant n\leqslant \min\{mk^{2/\sigma}, m^{2/q}k\}$, then
$$
\begin{equation*}
(n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/p_1-1/p_2} {1/2-1/q}} \leqslant \frac{\nu_1}{\nu_2} \leqslant (n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}}
\end{equation*}
\notag
$$
or
$$
\begin{equation}
(n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}} \leqslant \frac{\nu_1}{\nu_2} \leqslant (n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/p_1-1/p_2} {1/2-1/q}};
\end{equation}
\tag{21}
$$
2) if $km^{2/q}\leqslant n \leqslant mk^{2/\sigma}$, then
$$
\begin{equation*}
(n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/p_1-1/p_2}{1/2-1/q}}\leqslant \frac{\nu_1}{\nu_2} \leqslant k^{1/\theta_1-1/\theta_2} (n^{1/2} m^{-1/q} k^{-1/2})^{\frac{1/p_1-1/p_2}{1/2-1/q}}
\end{equation*}
\notag
$$
or
$$
\begin{equation}
k^{1/\theta_1-1/\theta_2} (n^{1/2} m^{-1/q} k^{-1/2})^{\frac{1/p_1-1/p_2}{1/2-1/q}} \leqslant \frac{\nu_1}{\nu_2} \leqslant (n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/p_1-1/p_2}{1/2-1/q}};
\end{equation}
\tag{22}
$$
3) if $mk^{2/\sigma}\leqslant n \leqslant km^{2/q}$, then
$$
\begin{equation*}
(n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}}\leqslant \frac{\nu_1}{\nu_2} \leqslant m^{1/p_1-1/p_2} (n^{1/2} m^{-1/2} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}}
\end{equation*}
\notag
$$
or
$$
\begin{equation}
m^{1/p_1-1/p_2} (n^{1/2} m^{-1/2} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}} \leqslant \frac{\nu_1}{\nu_2} \leqslant (n^{1/2} m^{-1/q} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}};
\end{equation}
\tag{23}
$$
4) if $\max\{mk^{2/\sigma}, m^{2/q}k\}\leqslant n \leqslant mk/2$, then
$$
\begin{equation*}
m^{1/p_1-1/p_2} (n^{1/2} m^{-1/2} k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}} \,{\leqslant}\, \frac{\nu_1}{\nu_2} \,{\leqslant}\, k^{1/\theta_1 -1/\theta_2}(n^{1/2} m^{-1/q} k^{-1/2})^{\frac{1/p_1-1/p_2}{1/2-1/q}}
\end{equation*}
\notag
$$
or
$$
\begin{equation}
k^{1/\theta_1 -1/\theta_2}(n^{1/2} m^{-1/q} k^{-1/2})^{\frac{1/p_1-1/p_2}{1/2-1/q}} \,{\leqslant}\, \frac{\nu_1}{\nu_2} \,{\leqslant}\, m^{1/p_1-1/p_2} (n^{1/2} m^{-1/2} k^{-1/\sigma})^{\frac{1/\theta_1\mkern-1.5mu-\mkern-0.5mu 1/\theta_2}{1/2-1/\sigma}}.
\end{equation}
\tag{24}
$$
Let $\lambda \in [0, 1]$, $\frac 1p=\frac{1-\lambda}{p_1}+\frac{\lambda}{p_2} \in [\frac1{q}, \frac12]$, $\frac{1}{\theta}=\frac{1-\lambda}{\theta_1}+\frac{\lambda}{\theta_2}\in [\frac1{\sigma}, \frac12]$ and Then Proof. Let $1\leqslant \widetilde r\leqslant m$ and $1\leqslant \widetilde l\leqslant k$ satisfy (14). We set $r=\lceil \widetilde r\rceil$, $l=\lceil \widetilde l \rceil$ and $W=\nu_1^{1-\lambda} \nu_2^\lambda r^{-1/p}l^{-1/\theta} V_{r,l}^{m,k}$. By Lemma 2 The numbers $\widetilde r$ and $\widetilde l$ are defined as follows.
In case 1) we set
$$
\begin{equation}
\widetilde r=(n^{1/2}m^{-1/q} k^{-1/\sigma})^{\frac{1-\alpha}{1/2-1/q}}\quad\text{and} \quad \widetilde l=(n^{1/2}m^{-1/q} k^{-1/\sigma})^{\frac{\alpha}{1/2-1/\sigma}},
\end{equation}
\tag{28}
$$
in case 2) we set
$$
\begin{equation}
\widetilde r=(n^{1/2} m^{-1/q}k^{-1/\sigma})^{\frac{1-\alpha}{1/2-1/q}}(n^{1/2} m^{-1/q}k^{-1/2})^{\frac{\alpha}{1/2-1/q}}, \qquad \widetilde l=k^\alpha,
\end{equation}
\tag{29}
$$
in case 3) we put
$$
\begin{equation}
\widetilde r=m^\alpha\quad\text{and} \quad \widetilde l=(n^{1/2} m^{-1/q}k^{-1/\sigma})^{\frac{1-\alpha}{1/2-1/\sigma}}(n^{1/2} m^{-1/2}k^{-1/\sigma})^{\frac{\alpha}{1/2-1/\sigma}},
\end{equation}
\tag{30}
$$
and in case 4)
$$
\begin{equation}
\widetilde r=m^{1-\alpha}(n^{1/2}m^{-1/q}k^{-1/2})^{\frac{\alpha}{1/2-1/q}}\quad\text{and} \quad \widetilde l=(n^{1/2} m^{-1/2} k^{-1/\sigma})^{\frac{1-\alpha}{1/2-1/\sigma}}k^\alpha,
\end{equation}
\tag{31}
$$
where $\alpha \in [0, 1]$ is chosen to satisfy (14); such $\alpha $ exists by (21), (22), (23), and (24), respectively. In each case, from the corresponding restriction on $n$ we obtain ${1\leqslant \widetilde r \leqslant m}$ and $1\leqslant \widetilde l\leqslant k$.
By (28)–(31) we have $n\leqslant m^{2/q} k^{2/\sigma} r^{1-2/q} l^{1-2/\sigma}$. Hence
$$
\begin{equation*}
d_n(\nu_1B^{m,k}_{p_1,\theta_1} \cap \nu_2 B_{p_2,\theta_2}^{m,k}, l^{m,k}_{q,\sigma}) \stackrel{(27)}{\gtrsim} d_n(W, l^{m,k}_{q,\sigma}) \stackrel{(9)}{\underset{q,\sigma}{\gtrsim}} \nu_1^{1-\lambda} \nu_2^\lambda r^{1/q-1/p} l^{1/\sigma -1/\theta}.
\end{equation*}
\notag
$$
Now (26) follows from the last inequality and from (4), (25) and (28)–(31).
This proves Lemma 5. Lemma 6. Let $q\geqslant 2$ and $\sigma\geqslant 2$. 1. Let $mk^{2/\sigma}\leqslant n \leqslant \frac{mk}2$, $\widetilde \mu\in [0, 1]$, $\frac12= \frac{1-\widetilde \mu}{\theta_1}+\frac{\widetilde \mu}{\theta_2}$, $\frac{1}{\widetilde p}=\frac{1-\widetilde \mu}{p_1}+\frac{\widetilde \mu}{p_2}$ and
$$
\begin{equation*}
m^{1/p_1-1/p_2} (n^{1/2}m^{-1/2}k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}} \leqslant \frac{\nu_1}{\nu_2} \leqslant m^{1/p_1-1/p_2}
\end{equation*}
\notag
$$
or
$$
\begin{equation*}
m^{1/p_1-1/p_2} \leqslant \frac{\nu_1}{\nu_2} \leqslant m^{1/p_1-1/p_2} (n^{1/2}m^{-1/2}k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}}.
\end{equation*}
\notag
$$
Then
$$
\begin{equation*}
d_n(\nu_1B_{p_1,\theta_1}^{m,k}\cap \nu_2 B_{p_2,\theta_2}^{m,k}, l_{q,\sigma}^{m,k}) \underset{q,\sigma}{\gtrsim} \nu_1^{1-\widetilde \mu}\nu_2^{\widetilde \mu} m^{1/q-1/\widetilde p+1/2} k^{1/\sigma} n^{-1/2}.
\end{equation*}
\notag
$$
2. Let $m^{2/q}k\leqslant n \leqslant \frac{mk}2$, $\widetilde \lambda\in [0, 1]$, $\frac12= \frac{1-\widetilde \lambda}{p_1}+\frac{\widetilde \lambda}{p_2}$, $\frac{1}{\widetilde \theta}=\frac{1-\widetilde \lambda}{\theta_1}+\frac{\widetilde \lambda}{\theta_2}$ and
$$
\begin{equation*}
k^{1/\theta_1-1/\theta_2} (n^{1/2}m^{-1/q}k^{-1/2})^{\frac{1/p_1-1/p_2}{1/2-1/q}} \leqslant \frac{\nu_1}{\nu_2} \leqslant k^{1/\theta_1-1/\theta_2}
\end{equation*}
\notag
$$
or Then Proof. We prove assertion 1 (the proof of assertion 2 is similar).
We have $mk^{2/\sigma}\leqslant n\leqslant mk/2$, and so $\sigma >2$. Set
$$
\begin{equation*}
W=\nu_1^{1-\widetilde \mu}\nu_2^{\widetilde \mu} m^{-1/\widetilde p} l^{-1/2}V_{m,l}^{m,k},\quad \text{where } l=\lfloor \widetilde l\rfloor\text{ and } \widetilde l=(n^{1/2}m^{-1/2}k^{-1/\sigma})^{\frac{1-\alpha}{1/2-1/\sigma}};
\end{equation*}
\notag
$$
the number $\alpha \in [0, 1]$ is chosen so as to have ${\nu_1}/{\nu_2}=m^{1/p_1-1/p_2} \widetilde l^{\,1/\theta_1 -1/\theta_2}$. Since $mk^{2/\sigma} \leqslant n \leqslant mk$, we have $1\leqslant l\leqslant k$. By Lemma 2, $W \subset 4(\nu_1B_{p_1,\theta_1}^{m,k}\cap \nu_2 B_{p_2,\theta_2}^{m,k})$. We also note that $n \geqslant m^{2/q} k^{2/\sigma} m^{1-2/q} l^{1-2/\sigma}$. Hence
$$
\begin{equation*}
\begin{aligned} \, d_n(\nu_1B_{p_1,\theta_1}^{m,k}\cap \nu_2 B_{p_2,\theta_2}^{m,k}, l_{q,\sigma}^{m,k}) &\gtrsim d_n(\nu_1^{1-\widetilde \mu}\nu_2^{\widetilde \mu} m^{-1/\widetilde p} l^{-1/2}V_{m,l}^{m,k}, l_{q,\sigma}^{m,k}) \\ &\!\stackrel{(9)}{\underset{q,\sigma}{\gtrsim}} \nu_1^{1-\widetilde \mu}\nu_2^{\widetilde \mu} m^{-1/\widetilde p} l^{-1/2} n^{-1/2} m^{1/q} k^{1/\sigma} m^{1/2}l^{1/2} \\ &= \nu_1^{1-\widetilde \mu}\nu_2^{\widetilde \mu} m^{1/q-1/\widetilde p} n^{-1/2} m^{1/2}k^{1/\sigma}. \end{aligned}
\end{equation*}
\notag
$$
This proves Lemma 6. Lemma 7. Let $q\geqslant 2$ and $\sigma\geqslant 2$. 1. Let
$$
\begin{equation*}
1\leqslant \frac{\nu_1}{\nu_2} \leqslant r_0^{1/p_1-1/p_2} \quad\textit{or}\quad r_0^{1/p_1-1/p_2} \leqslant \frac{\nu_1}{\nu_2} \leqslant 1,
\end{equation*}
\notag
$$
where
$$
\begin{equation*}
r_0= \begin{cases} (n^{1/2}m^{-1/q}k^{-1/\sigma})^{\frac{1}{1/2-1/q}} & \textit{if }m^{2/q}k^{2/\sigma} < n \leqslant mk^{2/\sigma}, \\ m& \textit{if }mk^{2/\sigma}< n \leqslant \dfrac{mk}{2}. \end{cases}
\end{equation*}
\notag
$$
Suppose that there exists $\widetilde \lambda\in [0, 1]$ such that $\frac{1}{2}=\frac{1-\widetilde \lambda}{p_1}+\frac{\widetilde \lambda}{p_2}$. Then
$$
\begin{equation*}
d_n(\nu_1B_{p_1,\theta_1}^{m,k} \cap \nu_2B_{p_2,\theta_2}^{m,k}, l_{q,\sigma}^{m,k}) \underset{q,\sigma}{\gtrsim} \nu_1^{1-\widetilde \lambda} \nu_2^{\widetilde \lambda} n^{-1/2}m^{1/q} k^{1/\sigma}.
\end{equation*}
\notag
$$
2. Let
$$
\begin{equation*}
1\leqslant \frac{\nu_1}{\nu_2} \leqslant l_0^{1/\theta_1-1/\theta_2} \quad\textit{or}\quad l_0^{1/\theta_1-1/\theta_2}\leqslant \frac{\nu_1}{\nu_2} \leqslant 1,
\end{equation*}
\notag
$$
where Suppose that there exists $\widetilde \mu\in [0, 1]$ such that $\frac{1}{2}=\frac{1-\widetilde \mu}{\theta_1}+\frac{\widetilde \mu}{\theta_2}$. Then Proof. We prove assertion 1 (the proof of assertion 2 is similar).
If $m^{2/q}k^{2/\sigma} < n \leqslant mk^{2/\sigma}$, then $q>2$ and the formula for $r_0$ is correct. We set $r=\lfloor r_0^\alpha \rfloor$, where $\alpha \in [0, 1]$ is chosen so as to have ${\nu_1}/{\nu_2}=r_0^{\alpha(1/p_1-1/p_2)}$. By the definition of $r_0$ we have $1\leqslant r\leqslant m$. By Lemma 2,
$$
\begin{equation*}
\nu_1^{1-\widetilde \lambda} \nu_2^{\widetilde \lambda} r^{-1/2}V^{m,k}_{r,1} \subset 4(\nu_1B_{p_1,\theta_1}^{m,k} \cap \nu_2B_{p_2,\theta_2}^{m,k}).
\end{equation*}
\notag
$$
In addition, $n\geqslant m^{2/q}k^{2/\sigma} r^{1-2/q}$. Hence
$$
\begin{equation*}
\begin{aligned} \, &d_n(\nu_1B_{p_1,\theta_1}^{m,k} \cap \nu_2B_{p_2,\theta_2}^{m,k}, l_{q,\sigma}^{m,k}) \\ &\qquad\gtrsim d_n(\nu_1^{1-\widetilde \lambda} \nu_2^{\widetilde \lambda} r^{-1/2}V^{m,k}_{r,1}, l_{q,\sigma}^{m,k})\stackrel{(9)}{\underset{q,\sigma}{\gtrsim}} \nu_1^{1-\widetilde \lambda} \nu_2^{\widetilde \lambda} r^{-1/2} n^{-1/2}m^{1/q} k^{1/\sigma} r^{1/2} \\ &\qquad=\nu_1^{1-\widetilde \lambda} \nu_2^{\widetilde \lambda} n^{-1/2}m^{1/q} k^{1/\sigma}. \end{aligned}
\end{equation*}
\notag
$$
This proves Lemma 7. § 3. Estimates for the widths of intersections of finite-dimensional balls Proof of Theorem 2. The upper estimate for widths is secured by Lemma 1. Let us prove the lower estimate.
For $n\leqslant m^{2/q}k^{2/\sigma}$ we use Lemma 3 and (1)–(3). Next, we consider the case $m^{2/q}k^{2/\sigma} < n \leqslant mk/2$. We set
$$
\begin{equation*}
\widetilde \Phi_j=\nu_jD_{m,k,n,q,\sigma}(p_j, \theta_j),\qquad j=1, 2
\end{equation*}
\notag
$$
(see Theorem 1). For $j=3, 4, 5$ we set
$$
\begin{equation*}
\widetilde \Phi_j=+\infty \quad\text{if } \Phi_j=+\infty,
\end{equation*}
\notag
$$
otherwise
$$
\begin{equation*}
\begin{gathered} \, \widetilde \Phi_3 :=\nu_1^{1-\widetilde \lambda} \nu_2^{\widetilde \lambda}D_{m,k,n,q,\sigma}(2, \widetilde\theta),\qquad \widetilde \Phi_4 :=\nu_1^{1-\widetilde \mu} \nu_2^{\widetilde \mu}D_{m,k,n,q,\sigma}(\widetilde p, 2) \\\text{and}\quad \widetilde \Phi_5 := \nu_1^{1-\lambda} \nu_2^{\lambda}D_{m,k,n,q,\sigma}(p, \theta). \end{gathered}
\end{equation*}
\notag
$$
In view of Theorem 1 it suffices to show that
$$
\begin{equation*}
d_n(\nu_1 B_{p_1,\theta_1}^{m,k}\cap \nu_2 B_{p_2,\theta_2}^{m,k}, l_{q,\sigma}^{m,k}) \underset{q,\sigma}{\gtrsim} \min _{1\leqslant j\leqslant 5} \widetilde\Phi_j=:\Psi.
\end{equation*}
\notag
$$
Note that it suffices to consider the case when
$$
\begin{equation*}
\biggl(\frac12, \frac12\biggr) \notin \biggl[\biggl(\frac1{p_1}, \frac1{\theta_1}\biggr), \biggl(\frac1{p_2}, \frac1{\theta_2}\biggr)\biggr],\qquad p_i\ne 2,\quad \theta_i\ne 2\quad (i=1, 2),
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\frac{1/p_i-1/q}{1/2-1/q}\ne \frac{1/\theta_i-1/\sigma}{1/2-1/\sigma}\quad (i=1, 2)\quad\text{for } q>2,\quad \sigma>2,
\end{equation*}
\notag
$$
for otherwise we can replace $p_i$ and $\theta_i$ by sufficiently close quantities for which the above conditions are satisfied (here we use the fact that the functions $D_{m,k,n,q,\sigma}$ are continuous; if $\widetilde \Phi_j=+\infty$, then we can shift $p_i$ and $\theta_i$ so that this condition still holds). As a result, $(1/p_i, 1/\theta_i)$ lies in one of the following domains:
$$
\begin{equation*}
\begin{aligned} \, G_1 &=\biggl\{(t, s)\colon \frac12<t\leqslant 1,\, \frac12< s\leqslant 1\biggr\}, \\ G_2&=\biggl\{(t, s)\colon \frac12<t\leqslant 1,\, \frac 1\sigma\leqslant s<\frac12\biggr\}, \\ G_3&=\biggl\{(t, s)\colon \frac1q\leqslant t<\frac12,\, \frac12<s\leqslant 1\biggr\}, \\ G_4&=\biggl\{(t, s)\colon \frac1q\leqslant t<\frac12, \frac1\sigma\leqslant s<\frac12,\, \lambda_{1/t,q}<\lambda_{1/s,\sigma}\biggr\}, \\ G_5&=\biggl\{(t, s)\colon \frac1q\leqslant t<\frac12, \frac1\sigma\leqslant s<\frac12,\, \lambda_{1/t,q}>\lambda_{1/s,\sigma}\biggr\}. \end{aligned}
\end{equation*}
\notag
$$
Case 1: $\Psi=\widetilde \Phi_j$, where $j\in \{1, 2\}$. We can assume without loss of generality that $j=1$. Let $(1/p_1, 1/\theta_1)\in G_i$ for some $i\in \{1, \dots, 5\}$. We set
$$
\begin{equation*}
\begin{gathered} \, \lambda_*=\sup\biggl\{\mu \in [0, 1]\colon \biggl(\frac{1-\mu}{p_1}+\frac{\mu}{p_2},\, \frac{1-\mu}{\theta_1}+\frac{\mu}{\theta_2}\biggr) \in G_i\biggr\}, \\ \frac{1}{p_*}=\frac{1-\lambda_*}{p_1}+\frac{\lambda_*}{p_2}\quad\text{and} \quad \frac{1}{\theta_*}=\frac{1-\lambda_*}{\theta_1}+\frac{\lambda_*}{\theta_2}. \end{gathered}
\end{equation*}
\notag
$$
Hence $\lambda_*>0$. If $\lambda_*=1$, then $(p_*, \theta_*)=(p_2, \theta_2)$; in other cases $(p_*, \theta_*)=(2, \widetilde \theta)$, $(p_*, \theta_*)=(\widetilde p, 2)$, or $(p_*, \theta_*)=(p, \theta)$. Now using the condition $\Psi=\widetilde \Phi_1$ we find that
$$
\begin{equation}
\nu_1 D_{m,k,n,q,\sigma}(p_1, \theta_1) \leqslant \nu_1^{1-\lambda_*} \nu_2^{\lambda_*} D_{m,k,n,q,\sigma}(p_*, \theta_*).
\end{equation}
\tag{32}
$$
Let $p_1>2$, $\theta_1>2$ and $\lambda_{p_1,q}< \lambda_{\theta_1,\sigma}$. Using (32) and (2) we have
$$
\begin{equation*}
\nu_1(n^{-1/2}m^{1/q}k^{1/\sigma})^{\frac{1/p_1-1/q}{1/2-1/q}} \leqslant \nu_1^{1-\lambda_*} \nu_2^{\lambda_*} (n^{-1/2}m^{1/q}k^{1/\sigma})^{\frac{(1-\lambda_*)/p_1+\lambda_*/p_2-1/q}{1/2-1/q}}
\end{equation*}
\notag
$$
for $m^{2/q}k^{2/\sigma}< n\leqslant mk^{2/\sigma}$ and
$$
\begin{equation*}
\begin{aligned} \, &\nu_1m^{1/q-1/p_1}(n^{-1/2}m^{1/2}k^{1/\sigma})^{\frac{1/\theta_1-1/\sigma}{1/2-1/\sigma}} \\ &\qquad \leqslant \nu_1^{1-\lambda_*} \nu_2^{\lambda_*} m^{1/q-(1-\lambda_*)/p_1-\lambda_*/p_2} (n^{-1/2}m^{1/2}k^{1/\sigma})^{\frac{(1-\lambda_*)/\theta_1+\lambda_*/\theta_2-1/\sigma} {1/2-1/\sigma}} \end{aligned}
\end{equation*}
\notag
$$
for $mk^{2/\sigma}< n\leqslant mk/2$. This implies (15). It remains to invoke Lemma 4. The case when $p_1>2$, $\theta_1>2$ and $\lambda_{p_1,q}> \lambda_{\theta_1,\sigma}$ is dealt with similarly.
Let $p_1>2$ and $\theta_1<2$. From (32) and (2) we arrive at (15) again, and now the estimate for widths follows from Lemma 4. The case when $p_1<2$ and $\theta_1>2$ is similar. Let $p_1<2$ and $\theta_1<2$. Then by (32) and (1) we have $\nu_1\leqslant \nu_2$. It remains to employ Lemma 3. Before we turn to the remaining cases, we note that the condition $(1/2, 1/2) \notin [(1/p_1, 1/\theta_1), (1/p_2, 1/\theta_2)]$ implies that $\widetilde \theta \ne 2$ for $\Psi=\widetilde \Phi_3$ and $\widetilde p\ne 2$ for $\Psi=\widetilde \Phi_4$. Case 2a: $\Psi=\widetilde \Phi_3$ and $\widetilde \theta<2$. We can assume without loss of generality that $p_1>p_2$. Note that on the line segment $[(1/p_1, 1/\theta_1), (1/p_2, 1/\theta_2)]$ sufficiently small half-neighbourhoods of the point $(1/2, 1/\widetilde \theta)$ lie in $G_1$ and $G_3$, respectively. We set
$$
\begin{equation}
\nonumber \lambda_*=\inf \biggl\{\mu\in [0, \widetilde \lambda]\colon \biggl(\frac{1-\mu}{p_1}+\frac{\mu}{p_2}, \frac{1-\mu}{\theta_1}+\frac{\mu}{\theta_2}\biggr)\in G_3\biggr\},
\end{equation}
\notag
$$
$$
\begin{equation}
\nonumber \lambda_{**}=\sup \biggl\{\mu\in [\widetilde \lambda, 1]\colon \biggl(\frac{1-\mu}{p_1}+\frac{\mu}{p_2}, \frac{1-\mu}{\theta_1}+\frac{\mu}{\theta_2}\biggr)\in G_1\biggr\},
\end{equation}
\notag
$$
$$
\begin{equation}
\frac{1}{p_*}=\frac{1-\lambda_*}{p_1}+\frac{\lambda_*}{p_2}, \qquad \frac{1}{\theta_*}= \frac{1-\lambda_*}{\theta_1}+\frac{\lambda_*}{\theta_2},
\end{equation}
\tag{33}
$$
$$
\begin{equation}
\frac{1}{p_{**}}=\frac{1-\lambda_{**}}{p_1}+\frac{\lambda_{**}}{p_2}\quad\text{and} \quad \frac{1}{\theta_{**}}=\frac{1-\lambda_{**}}{\theta_1}+\frac{\lambda_{**}}{\theta_2}.
\end{equation}
\tag{34}
$$
From the condition $\Psi=\widetilde \Phi_3$ we obtain
$$
\begin{equation}
\begin{gathered} \, \nu_1^{1-\widetilde \lambda}\nu_2^{\widetilde \lambda} D_{m,k,n,q,\sigma}(2, \widetilde \theta) \leqslant \nu_1^{1-\lambda_*}\nu_2^{\lambda_*} D_{m,k,n,q,\sigma}(p_*, \theta_*), \\ \nu_1^{1-\widetilde \lambda}\nu_2^{\widetilde \lambda} D_{m,k,n,q,\sigma}(2, \widetilde \theta) \leqslant \nu_1^{1-\lambda_{**}}\nu_2^{\lambda_{**}} D_{m,k,n,q,\sigma}(p_{**}, \theta_{**}). \end{gathered}
\end{equation}
\tag{35}
$$
We have $(1/p_{**}, 1/\theta_{**})\in G_1$, and now from (1) and the second inequality in (35) we find that ${\nu_1}/{\nu_2}\leqslant 1$. Next, from (2) and the first inequality in (35) we see that if $n\leqslant mk^{2/\sigma}$, then
$$
\begin{equation*}
\frac{\nu_1}{\nu_2} \geqslant (n^{1/2}m^{-1/q}k^{-1/\sigma})^{\frac{1/p_1-1/p_2}{1/2-1/q}},
\end{equation*}
\notag
$$
and if $n> mk^{2/\sigma}$, then ${\nu_1}/{\nu_2} \geqslant m^{1/p_1-1/p_2}$. Now it remains to use Lemma 7.
Case 2b: $\Psi=\widetilde \Phi_4$ and $\widetilde p<2$. We argue as in Case 2a. Case 3a: $\Psi=\widetilde \Phi_3$ and $\widetilde \theta>2$. We can assume without loss of generality that $p_1>p_2$. Note that on the line segment $[(1/p_1, 1/\theta_1), (1/p_2, 1/\theta_2)]$ sufficiently small half-neighbourhoods of the point $(1/2, 1/\widetilde \theta)$ lie in $G_2$ and $G_5$, respectively. We set
$$
\begin{equation*}
\lambda_* =\inf \biggl\{\mu\in [0, \widetilde \lambda]\colon \biggl(\frac{1-\mu}{p_1}+\frac{\mu}{p_2}, \frac{1-\mu}{\theta_1}+\frac{\mu}{\theta_2}\biggr)\in G_5\biggr\}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\lambda_{**} =\sup \biggl\{\mu\in [\widetilde \lambda, 1]\colon \biggl(\frac{1-\mu}{p_1}+\frac{\mu}{p_2}, \frac{1-\mu}{\theta_1}+\frac{\mu}{\theta_2}\biggr)\in G_2\biggr\},
\end{equation*}
\notag
$$
and define $p_*$, $\theta_*$, $p_{**}$ and $\theta_{**}$ by (33) and (34). Now (35) follows.
In the case when $n\leqslant m^{2/q}k$, from (35) and (3) we obtain
$$
\begin{equation*}
\frac{\nu_1}{\nu_2}= (n^{1/2}m^{-1/q}k^{-1/\sigma})^{\frac{1/\theta_1-1/\theta_2}{1/2-1/\sigma}}.
\end{equation*}
\notag
$$
Hence $\nu_1^{1-\widetilde \lambda}\nu_2^{\widetilde \lambda}D_{m,k,n,q,\sigma}(2, \widetilde \theta)=\nu_1^{1-\lambda_{**}}\nu_2^{\lambda_{**}}D_{m,k,n,q,\sigma}(p_{**}, \theta_{**})$, which gives us ${\Psi=\widetilde\Phi_2}$ or ${\Psi=\widetilde \Phi_4}$ and $\widetilde p<2$. These cases have already been taken care of.
In the case when $n> m^{2/q}k$, from (35) and (3) we obtain
$$
\begin{equation*}
k^{1/\theta_1-1/\theta_2}(n^{1/2}m^{-1/q}k^{-1/2})^{\frac{1/p_1-1/p_2}{1/2-1/q}}\leqslant \frac{\nu_1}{\nu_2}\leqslant k^{1/\theta_1-1/\theta_2};
\end{equation*}
\notag
$$
and then one applies Lemma 6.
Case 3b: $\Psi=\widetilde \Phi_4$ and $\widetilde p>2$. Then we argue as in Case 3a. Case 4: $\Psi=\widetilde \Phi_5$. We can assume without loss of generality that
$$
\begin{equation*}
\frac{1/p_1-1/q}{1/2-1/q}< \frac{1/\theta_1-1/\sigma}{1/2-1/\sigma}\quad\text{and} \quad \frac{1/p_2-1/q}{1/2-1/q}> \frac{1/\theta_2-1/\sigma}{1/2-1/\sigma}.
\end{equation*}
\notag
$$
Note that on the interval $[(1/p_1, 1/\theta_1), (1/p_2, 1/\theta_2)]$ sufficiently small half-neighbourhoods of the point $(1/p, 1/\theta)$ lie in $G_4$ and $G_5$. We set
$$
\begin{equation*}
\begin{aligned} \, \lambda_* &=\inf \biggl\{\mu\in [0, \widetilde \lambda]\colon \biggl(\frac{1-\mu}{p_1}+\frac{\mu}{p_2}, \frac{1-\mu}{\theta_1}+\frac{\mu}{\theta_2}\biggr)\in G_4\biggr\}, \\ \lambda_{**} &=\sup \biggl\{\mu\in [\widetilde \lambda, 1]\colon \biggl(\frac{1-\mu}{p_1}+\frac{\mu}{p_2}, \frac{1-\mu}{\theta_1}+\frac{\mu}{\theta_2}\biggr)\in G_5\biggr\}, \end{aligned}
\end{equation*}
\notag
$$
and define $p_*$, $\theta_*$, $p_{**}$ and $\theta_{**}$ by (33) and (34). Hence
$$
\begin{equation*}
\nu_1^{1-\lambda}\nu_2^{\lambda} D_{m,k,n,q,\sigma}(p, \theta) \leqslant \nu_1^{1-\lambda_*}\nu_2^{\lambda_*} D_{m,k,n,q,\sigma}(p_*, \theta_*)
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\nu_1^{1- \lambda}\nu_2^{\lambda} D_{m,k,n,q,\sigma}(p, \theta) \leqslant \nu_1^{1-\lambda_{**}}\nu_2^{\lambda_{**}} D_{m,k,n,q,\sigma}(p_{**}, \theta_{**}).
\end{equation*}
\notag
$$
In view of (2)–(4) the conditions of Lemma 5 are met. Consequently, we have estimate (26).
This proves Theorem 2. 
Citation in format AMSBIB
\Bibitem{Vas24} |
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