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This article is cited in 3 scientific papers (total in 3 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∈N, 1⩽p<∞ and 1⩽θ<∞. We let lm,kp,θ denote the space Rmk equipped with the norm For p=∞ or θ=∞ the corresponding modifications are clear.The unit ball of the space lm,kp,θ is denoted by Bm,kp,θ. For k=1, the space and the unit ball are denoted, respectively, by lmp and Bmp. Let X be a normed space, let M⊂X and n∈Z+. Then the Kolmogorov n-width of the set M in X is defined by 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
\begin{equation*}
d_n(B^{m,k}_{p,\theta}, l_{q,\sigma}^{m,k}) \underset{q,\sigma}{\asymp} D_{m,k,n,q,\sigma}(p, \theta),
\end{equation*}
\notag
where D_{m,k,n,q,\sigma}(p, \theta) is defined as follows:
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
\begin{equation}
d_n(V^{m,k}_{r,l}, l^{m,k}_{q,\sigma}) \underset{q,\sigma}{\gtrsim} \begin{cases} r^{1/q}l^{1/\sigma} & \textit{if }n\leqslant m^{2/q}k^{2/\sigma}r^{1-2/q} l^{1-2/\sigma}, \\ n^{-1/2}m^{1/q}k^{1/\sigma} r^{1/2} l^{1/2} & \textit{if }n\geqslant m^{2/q}k^{2/\sigma}r^{1-2/q} l^{1-2/\sigma}. \end{cases}
\end{equation}
\tag{9}
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
\begin{equation*}
d_n(V^{m,k}_{r,l}, l^{m,k}_{q,\sigma}) \stackrel{(11)}{\gtrsim} m^{1/q-1/2} k^{1/\sigma-1/2}r^{1/2} l^{1/2}, \qquad n\leqslant \frac{mk}{2}.
\end{equation*}
\notag
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
\begin{equation*}
d_n(B^{m,k}_{p,\theta}, l^{m,k}_{q,\sigma}) \underset{q,\sigma}{\gtrsim} \begin{cases} m^{1/q-1/p} k^{1/\sigma-1/\theta} & \textit{if } \min\{p, \theta\}\geqslant 2, \\ m^{1/q -1/p+1/2}n^{-1/2}k^{1/\sigma} & \textit{if } p\geqslant 2,\, \theta \leqslant 2, \\ k^{1/\sigma-1/\theta+1/2}n^{-1/2}m^{1/q} & \textit{if } \theta\geqslant 2,\, p \leqslant 2, \\ n^{-1/2}m^{1/q}k^{1/\sigma} & \textit{if }\max\{p, \theta\}\leqslant 2. \end{cases}
\end{equation*}
\notag
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
\begin{equation}
W \subset 4(\nu_1 B^{m,k}_{p_1,\theta_1} \cap \nu_2 B^{m,k}_{p_2,\theta_2}).
\end{equation}
\tag{27}
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
\begin{equation*}
l_0=\begin{cases} (n^{1/2}m^{-1/q}k^{-1/\sigma})^{\frac{1}{1/2-1/\sigma}} & \textit{if }m^{2/q}k^{2/\sigma} < n \leqslant km^{2/q}, \\ k & \textit{if }km^{2/q}< n \leqslant \dfrac{mk}{2}. \end{cases}
\end{equation*}
\notag
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 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}
This publication is cited in the following 3 articles:
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