Estimates of probabilistic widths of the diagonal operator of finite-dimensional sets with the Gaussian measure

Abstract

In this paper, we estimate the asymptotic orders of probabilistic and average widths of the compact embedding operators from the Sobolev space $W_2^r(\mathbb{T})$ into $L_q(\mathbb{T})$ ($1\le q\le \mathrm{\infty }$) with the Gaussian measure.

MSC:41A10, 41A46, 42A61, 46C99.

1 Introduction and main results

Problems of n-widths in the approximation theory have by now been studied in depth. A great deal of classical problems have been solved, and interesting new developments have appeared. For example, the problems of probabilistic, average and stochastic widths, which can reflect the behavior of function on the whole class and give information about the measure of the elements in the class that can be approximated to this or that degree, are the problems of this kind. For the results related to the probabilistic, average and stochastic widths, the reader may be referred to Sul’din [1, 2], Traub et al. [3], Maiorov [4–7], Mathé [8–12], Sun [13, 14], and Ritter [15]. The new developments in this direction can be found in Fang’s papers [16–21]. Moreover, Carl and Pajor [22] proved an inequality with respect to the Gelfand numbers of an operator u from ${\ell }_1^N$ into a Hilbert space, from which one can immediately derive the inequality related to the Kolmogorov numbers by the known duality. In this article we continue the previous works and prove the estimates of probabilistic widths of the diagonal operators from ${\mathbb{R}}^m$ onto ${\ell }_q^m$.

First, we recall some useful concepts. Let W be a bounded subset of a normed linear space X with the norm ${\parallel \cdot \parallel }_X$, and $F_N$ be an N-dimensional subspace of X. The quantity

$$e(W,F_N,X)=\underset{x\in W}{sup}e(x,F_N,X),$$

where

$$e(x,F_N,X)=\underset{y\in F_N}{inf}{\parallel x-y\parallel }_X$$

is called the deviation of W from $F_N$. It shows how well the ‘worst’ elements of W can be approximated by $F_N$. The number

$$d_N(W,X)=\underset{F_N}{inf}e(W,F_N,X)=\underset{F_N}{inf}\underset{x\in W}{sup}\underset{y\in F_N}{inf}{\parallel x-y\parallel }_X,$$

where $F_N$ runs through all possible linear subspaces of X of dimension at most N, is called the Kolmogorov’s N-width of W in X. Assume that W contains a Borel field ℬ consisting of open subsets of W and equipped with a probabilistic measure μ defined on ℬ. That is, μ is a σ-additive nonnegative function on ℬ, and $\mu (W)=1$. Let $\delta \in [0,1]$ be an arbitrary number. The corresponding probabilistic Kolmogorov’s $(N,\delta )$-width of a set W with a measure μ in the space X is defined by

$$d_{N,\delta }(W,\mu ,X)=\underset{G_{\delta }}{inf}{d}_N(W\mathrm{\setminus }G,X),$$

(1)

where $G_{\delta }$ runs through all possible subsets in ℬ with measure $\mu (G_{\delta })\le \delta$. The p-average Kolmogorov’s N-width is defined by

$$d_N^{(a)}{(W,\mu ,X)}_p=\underset{F_N}{inf}{({\int }_{W}e{(x,F_N,X)}^{p}d\mu (x))}^{1/p},0<p<\mathrm{\infty },$$

(2)

where $F_N$ in (2) runs over all linear subspaces of X of dimension at most N. Let ${\ell }_p^m$ be an m-dimensional normed space of vectors $x=(x_1,\dots ,x_m)\in {\mathbb{R}}^m$, with a norm

$${\parallel x\parallel }_{{\ell }_p^m}=\{\begin{array}{cc}{({\sum }_{i=1}^m{|x_i|}^p)}^{1/p},\hfill & 1\le p<\mathrm{\infty },\hfill \\ {max}_{1\le i\le m}|x_i|,\hfill & p=\mathrm{\infty }.\hfill \end{array}$$

Consider in ${\mathbb{R}}^m$ the standard Gaussian measure $v=v_m$, which is defined as

$$v(G)={(2\pi )}^{-m/2}{\int }_{G}exp(-\frac{1}{2}{\parallel x\parallel }_2^2)dx,$$

where G is any Borel subset in ${\mathbb{R}}^m$. Obviously, $v({\mathbb{R}}^m)=1$.

Denote by $B_p^m(\rho )=\{x\in {\ell }_p^m:{\parallel x\parallel }_p\le \rho \}$ the ball of radius ρ in ${\ell }_p^m$. Let $B_p^m=B_p^m(1)$.

Let $N=0,1,\dots$ , $\delta \in [0,1)$ be arbitrary and $T_m$ be a linear invertible operator from ${\mathbb{R}}^m$ onto ${\ell }_q^m$. We define the probabilistic $(N,\delta )$-width of the operator acting in space ${\mathbb{R}}^m$ equipped with the Gaussian measure v in ${\ell }_q^m$-norm:

$$d_{N,\delta }(T_m:{\mathbb{R}}^m\to {\ell }_q^m,v)=\underset{G}{inf}\underset{{\mathcal{L}}_N}{inf}e(T_m\left({\mathbb{R}}^m\mathrm{\setminus }G\right),{\mathcal{L}}_N,{\ell }_q^m),$$

where $v(G)<\delta$, $dim{\mathcal{L}}_N\le N$.

Maiorov in [6] proved the following result.

Theorem A [6]

For $m>2N$, $\delta \in (0,1/2]$, $1\le q\le 2$, then

$$d_{N,\delta }({\mathbb{R}}^m,{\ell }_q^m,v)\asymp m^{\frac{1}{q}-\frac{1}{2}}\sqrt{m+ln(1/\delta )}.$$

In [22], Carl and Pajor proved the following result with respect to Gelfand numbers of an operator with values in a Hilbert space.

Theorem B [22]

Let T be an operator from ${\ell }_1^m$ into a Hilbert space H. Then

$$d^N(T)\le C{\left(\frac{log(\frac{m}{N}+1)}{N}\right)}^{1/2}\parallel T\parallel$$

for $1\le N\le m$, $m=1,2,\dots$ , where $C>0$ is a universal constant.

Detailed facts about the usual widths, such as the Kolmogorov’s N-widths and N th Gelfand numbers (or Gelfand N-widths) of T were given in the books [23–26].

Remark 1

• (a) Theorem A shows the asymptotic expression of the probabilistic widths of the identity embedding from ${\mathbb{R}}^m$ into ${\ell }_q^m$, $1\le q\le 2$.

• (b) Theorem B gives the upper estimate of Gelfand numbers of operators from ${\ell }_1^m$ into a Hilbert space, and some of its striking applications in the geometry of Banach spaces and Rademacher processes can be found in [22]. By the dual relation, it is easy to obtain the similar upper estimate of the Kolmogorov’s N-widths $d_N(T)$ of operators from ${\ell }_2^m$ into ${\ell }_{\mathrm{\infty }}^m$, i.e.,

  $$d_N(T)\le C{\left(\frac{log(\frac{m}{N}+1)}{N}\right)}^{1/2}\parallel T\parallel .$$

• (c) Motivated by Theorems A and B, in general cases, here we investigate the asymptotic estimate of probabilistic widths for diagonal operators from ${\mathbb{R}}^m$ onto ${\ell }_q^m$, $1\le q\le \mathrm{\infty }$.

Now we are in a position to formulate our main results.

Theorem 1 For $m>N$, $\delta \in (0,1/2]$, then

$$d_{N,\delta }(T_m:{\mathbb{R}}^m\to {\ell }_{\mathrm{\infty }}^m,v)\le C\parallel T_m\parallel \sqrt{(1+(1/N)ln(1/\delta ))ln(em/N)}.$$

Theorem 2 For $m>2N$, $\delta \in (0,1/2]$, then

$$d_{N,\delta }(T_m:{\mathbb{R}}^m\to {\ell }_1^m,v)\ge C^{\prime }\parallel T_m\parallel \sqrt{m+ln(1/\delta )}.$$

2 Proof of main results

In order to prove Theorems 1 and 2, we also need some auxiliary assertions.

Lemma 1 Let $\delta \in (0,\sqrt{2/e\pi }]$, and let $T_m$ be a bounded linear invertible operator from ${\mathbb{R}}^m$ onto ${\ell }_{\mathrm{\infty }}^m$. Then, for any vector $z\in {\mathbb{R}}^m$,

$$v(x:|(T_{m}x,z)|\ge 2\parallel T_m\parallel \sqrt{ln(1/\delta )}{\parallel z\parallel }_2)\le \delta .$$

Proof First, assume that $T_m$ is a diagonal operator of ${\mathbb{R}}^m$, i.e., $T_{m}x={({\lambda }_i{x}_i)}_{i=1}^m$, for any $x\in {\mathbb{R}}^m$. Without loss of generality, assume that the sequence of the absolute of eigenvalues ${\lambda }_i$, $i=1,\dots ,m$, is arranged non-increasingly, i.e., $|{\lambda }_1|\ge |{\lambda }_2|\ge \cdots \ge |{\lambda }_m|>0$. It is known that

$$\parallel T_m\parallel =\underset{1\le i\le m}{max}|{\lambda }_i|=|{\lambda }_1|.$$

Since v is invariant with respect to orthogonal transformation of ${\mathbb{R}}^m$, it suffices to prove the lemma for the vector $z^{\ast }=({\parallel z\parallel }_2,0,\dots ,0)$. Let $\epsilon \in (0,e^{-1}]$ be arbitrary. We have

$$\begin{array}{c}v(x:\left|(T_{m}x,z^{\ast })\right|\ge \parallel T_m\parallel \sqrt{ln(1/\epsilon )}{\parallel z\parallel }_2)\hfill \\ =v(x:\left|{\lambda }_1{x}_1\right|{\parallel z\parallel }_2\ge |{\lambda }_1|\sqrt{ln(1/\epsilon )}{\parallel z\parallel }_2)\hfill \\ =v(x:|x_1|\ge \sqrt{ln(1/\epsilon )})\hfill \\ =\frac{2}{\sqrt{\pi }}{\int }_{\sqrt{(1/2)ln(1/\epsilon )}}^{\mathrm{\infty }}exp(-t^2)dt\le {\left(\frac{2\epsilon }{\pi }\right)}^{1/2}.\hfill \end{array}$$

(3)

Here we use the inequality

$${\int }_u^{\mathrm{\infty }}exp(-t^2)dt<\frac{1}{2u}exp(-u^2),u\ge \frac{1}{\sqrt{2}}.$$

From (3) we obtain the assertion of the lemma by $\delta =\sqrt{2\epsilon /\pi }$.

Next, assume that $T_m$ is a symmetric transformation of ${\mathbb{R}}^m$, then there is an orthogonal matrix U of order m such that the matrix $U{T}_m{U}^T$ is a diagonal matrix. Since the Gaussian measure is invariant for orthogonal transformation, the result holds for symmetric transformation $T_m$.

Finally, assume that $T_m$ is a general invertible linear transformation from ${\mathbb{R}}^m$ onto ${\mathbb{R}}^m$, then there are two matrices U and S such that $T_m=US$, where U is an orthogonal matrix and S is a positive definite symmetric matrix. As the same reason above, the result holds for the transformation $T_m$.

Thus Lemma 1 is proved. □

The following inequality will be used (see [27]). For any integers N and m with $m>N\ge 1$, there exists a subspace H of ${\mathbb{R}}^m$ of dimension $dimH\ge m-N$ such that for any $x\in H$,

$${\parallel x\parallel }_2\le c_0{\left(\frac{ln(em/N)}{N}\right)}^{1/2}{\parallel x\parallel }_1,$$

(4)

where $c_0$ is an absolute constant.

Let $G\subset {\mathbb{R}}^m$ be a set. We introduce in ${\mathbb{R}}^m$ another norm for the operator $T_m:{\mathbb{R}}^m\to {\ell }_{\mathrm{\infty }}^m$:

$${\parallel x\parallel }_G=\underset{y\in {\mathbb{R}}^m\mathrm{\setminus }G}{sup}|(T_{m}y,x)|.$$

Lemma 2 For any $\delta \in (0,1/2]$ and an arbitrary operator $T_m$ from ${\mathbb{R}}^m$ onto ${\ell }_{\mathrm{\infty }}^m$, there exists a subset $G=G_{\delta }$ of ${\mathbb{R}}^m$ with measure $v(G)\le \delta$ such that

$$\underset{z\in B_1^m\cap H}{sup}{\parallel z\parallel }_G\le c_1\parallel T_m\parallel {(\frac{1}{N}ln\frac{exp(aN)}{\delta }ln\frac{em}{N})}^{1/2},$$

where a and $c_1$ are absolute constants.

Proof Let $k=[8(N+1)/ln(em/N)]$ and consider the $k^{-1}$-net

$$S=\{(s_1/k,\dots ,s_m/k):s_1,\dots ,s_m\in \mathbb{Z},|s_1|+\cdots +|s_m|\le k\}$$

for the $B_1^m$ in ${\ell }_{\mathrm{\infty }}^m$-norm. Using the inequality $\left(\begin{array}{c}m\\ \ell \end{array}\right)\le {(\frac{em}{\ell })}^{\ell }$, we estimate the cardinality of S:

$$cardS\le \sum _{\ell =1}^k{2}^{\ell }\left(\begin{array}{c}m+\ell \\ \ell \end{array}\right)\le \sum _{\ell =1}^k{2}^{\ell }{\left(\frac{4em}{\ell }\right)}^{\ell }\le {\left(\frac{8em}{k}\right)}^k\le e^{aN},$$

(5)

where a is some absolute constant.

Consider the polyhedron $Q=B_1^m\cap k^{-1}{B}_{\mathrm{\infty }}^m$. Let $Q^{\prime }$ be the set of extremal points of Q. The set $Q^{\prime }$ consists of vectors with k coordinates equal to $\pm k^{-1}$ and the remaining coordinates zero. This implies that $Q^{\prime }\subset S$, and hence $card{Q}^{\prime }\le exp(aN)$.

Let $\epsilon =\delta /exp(aN)$. In ${\mathbb{R}}^m$ we consider the set $G={\bigcup }_{s\in S}{G}_s$, where

$$G_s=\{y\in {\mathbb{R}}^m:|(T_{m}y,s)|\ge 2\parallel T_m\parallel \sqrt{ln(1/\epsilon )}{\parallel s\parallel }_2\}.$$

Let $z\in B_1^m\cap H$ be any point, and $s\in S$ be a point closest to z in ${\ell }_{\mathrm{\infty }}^m$-norm. Then $z=s+t$ for some $t\in Q$. From Lemma 1,

$${\parallel z\parallel }_G\le {\parallel s\parallel }_G+{\parallel t\parallel }_G\le 2\parallel T_m\parallel \sqrt{ln(1/\epsilon )}{\parallel s\parallel }_2+{\parallel t\parallel }_G.$$

(6)

Using the definition of Q, we have ${\parallel t\parallel }_2^2\le {\parallel t\parallel }_1{\parallel t\parallel }_{\mathrm{\infty }}\le k^{-1}$. From this and the definition of $Q^{\prime }$ and G,

$$\begin{array}{rcl}{\parallel t\parallel }_G& \le & \underset{t\in Q^{\prime }}{max}{\parallel t\parallel }_G\le \underset{t\in S\cap Q}{max}{\parallel t\parallel }_G\\ \le & 2\parallel T_m\parallel \sqrt{ln(1/\epsilon )}\underset{t\in S\cap Q}{max}{\parallel t\parallel }_2\le 2\parallel T_m\parallel \sqrt{k^{-1}ln(1/\epsilon )}.\end{array}$$

Therefore from (6),

$$\begin{array}{rcl}{\parallel z\parallel }_G& \le & 2\parallel T_m\parallel \sqrt{ln(1/\epsilon )}({\parallel z\parallel }_2+{\parallel t\parallel }_2)+{\parallel t\parallel }_G\\ \le & 2\parallel T_m\parallel \sqrt{ln(1/\epsilon )}({\parallel z\parallel }_2+2k^{-1/2}).\end{array}$$

(7)

Since $z\in H$, it follows from the inequalities (4) and (7) that

$${\parallel z\parallel }_G\le c_1\parallel T_m\parallel \sqrt{k^{-1}ln(1/\epsilon )}\le c_2\parallel T_m\parallel \sqrt{(1/N)ln(1/\epsilon )ln(em/N)}.$$

Using Lemma 1 and the inequality (5), we can estimate the measure of G:

$$v(G)\le \sum _{s\in S}v(G_s)\le \epsilon cardS\le \epsilon exp(aN)=\delta .$$

Thus, Lemma 2 is proved. □

Proof of Theorem 1 Using the duality in ${\mathbb{R}}^m$ and Lemma 2, we have

$$\begin{array}{rcl}\underset{x\in {\mathbb{R}}^m\mathrm{\setminus }G}{sup}\underset{y\in H^{\perp }}{inf}{\parallel T_{m}x-y\parallel }_{\mathrm{\infty }}& =& \underset{x\in {\mathbb{R}}^m\mathrm{\setminus }G}{sup}\underset{z\in H\cap B_1^m}{sup}|(T_{m}x,z)|\\ =& \underset{z\in H\cap B_1^m}{sup}{\parallel z\parallel }_G\le c\parallel T_m\parallel \sqrt{(1/N)ln(1/\epsilon )ln(em/N)}\\ =& c\parallel T_m\parallel \sqrt{(1/N)ln(exp(aN)/\delta )ln(em/N)},\end{array}$$

where $H^{\perp }$ is the orthogonal complement of H and $dim{H}^{\perp }\le N$. The proof of Theorem 1 is completed. □

Let us proceed to the proof of Theorem 2. For this, we first prove four lemmas. We introduce a definition. For arbitrary $\epsilon >0$, the ε-cardinality of a subset K of ${\ell }_1^m$ is defined to be

$${\mathcal{N}}_{\epsilon }(K)=min\{\mathcal{N}:z_1,\dots ,z_{\mathcal{N}}\in {\mathbb{R}}^m,e(K,\{z_1,\dots ,z_{\mathcal{N}}\})\le \epsilon \},$$

where

$$e(K,\{z_1,\dots ,z_{\mathcal{N}}\})=\underset{x\in K}{sup}\underset{i=1,\dots ,\mathcal{N}}{min}{\parallel x-z_i\parallel }_1$$

is the deviation of K from the set $\{z_1,\dots ,z_{\mathcal{N}}\}$ in ${\ell }_1^m$.

Let λ be a Lebesgue measure in ℝ, normalized by the condition $\lambda (B)=1$, where $B=B_2^m$. We consider Kolmogorov’s $(N,\delta )$-width of the ball B with measure λ in the ${\ell }_1^m$-norm:

$$d_{N,\delta }\equiv d_{N,\delta }(T_m:B\to {\ell }_1^m,\lambda )=\underset{G}{inf}\underset{\mathcal{L}}{inf}e(T_m(B\mathrm{\setminus }G),\mathcal{L},{\ell }_1^m),$$

where the $T_m$ is as above, the infima are over all possible subsets $G\subset B$ of measure $\lambda (G)\le \delta$ and all subspaces $\mathcal{L}\subset {\mathbb{R}}^m$ with $dim\mathcal{L}\le N$.

Lemma 3 Suppose that $D\subset B$ is an arbitrary subset with measure $\lambda (D)\le \delta$. Then, for any $\epsilon >d_{N,\delta }$,

$${\mathcal{N}}_{\epsilon }(T_m(B\mathrm{\setminus }D))\le {(1+\frac{4\parallel T_m\parallel }{\epsilon -d_{N,\delta }})}^m.$$

Proof Let $h>0$ be any number, and let H be any subspace of ℝ with $dimH\le N$ such that

$$e(T_m(B\mathrm{\setminus }D),H,{\ell }_1^m)-h\le d_{N,\delta }.$$

(8)

Let ${\epsilon }^{\prime }=\epsilon -d_{N,\delta }$. We consider the set $Q=2(T_m(B\mathrm{\setminus }D))\cap H$. Let $Q_{{\epsilon }^{\prime }}=\{z_1,\dots ,z_{\mathcal{N}}\}$ be the maximal subset of Q such that ${\parallel z_i-z_j\parallel }_1\ge {\epsilon }^{\prime }$ for all $i\ne j$. Clearly, by maximality $Q_{{\epsilon }^{\prime }}$ is a ${\epsilon }^{\prime }$-net of Q for ${\parallel \cdot \parallel }_1$. The balls $z_i+({\epsilon }^{\prime }/2)B_1^m$ are disjoint and all contained in $Q+({\epsilon }^{\prime }/2)B_1^m$. Therefore, taking volumes we can obtain

$$\sum _{i=1}^{\mathcal{N}}vol(z_i+({\epsilon }^{\prime }/2)B_1^m)\le vol(Q+({\epsilon }^{\prime }/2)B_1^m).$$

Hence, we have

$$\mathcal{N}\cdot ({\epsilon }^{\prime }/2)vol\left(B_1^m\right)\le vol(Q+({\epsilon }^{\prime }/2)B_1^m).$$

(9)

By $Q=2(T_m(B\mathrm{\setminus }D))\cap H\subset 2T_m(B)\subset 2T_m(B_1^m)\subset 2\parallel T_m\parallel (B_1^m)$ and (9), we have

$$\mathcal{N}{({\epsilon }^{\prime }/2)}^{m}vol\left(B_1^m\right)\le {(2\parallel T_m\parallel +({\epsilon }^{\prime }/2))}^{m}vol\left(B_1^m\right),$$

that is,

$$\mathcal{N}\le {(1+\frac{4\parallel T_m\parallel }{{\epsilon }^{\prime }})}^m.$$

(10)

Now, we need to establish $e(T_m(B\mathrm{\setminus }D),\{z_1,\dots ,z_{\mathcal{N}}\})\le \epsilon$. Since $t\in Q\subset H$, we have from (8)

$$\begin{array}{rcl}\underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{t\in Q}{min}{\parallel x-t\parallel }_1& \le & \underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{t\in Q}{min}\underset{h_0\in H}{min}({\parallel x-h_0\parallel }_1+{\parallel h_0-t\parallel }_1)\\ \le & \underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{t\in Q}{min}{\parallel x-h_0\parallel }_1+\underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{t\in Q}{min}\underset{h_0\in H}{min}{\parallel h_0-t\parallel }_1\\ \le & d_{N,\delta }+h.\end{array}$$

(11)

From the inequality (11), the definition of $Q_{{\epsilon }^{\prime }}$ and (8), it follows that

$$\begin{array}{rcl}e(T_m(B\mathrm{\setminus }D),\{z_1,\dots ,z_{\mathcal{N}}\})& =& \underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{i=1,\dots ,\mathcal{N}}{min}{\parallel x-z_i\parallel }_1\\ \le & \underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{i=1,\dots ,\mathcal{N}}{min}\underset{t\in Q}{min}({\parallel x-t\parallel }_1+{\parallel t-z_i\parallel }_1)\\ \le & \underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{t\in Q}{min}({\parallel x-t\parallel }_1+\underset{i=1,\dots ,\mathcal{N}}{min}{\parallel t-z_i\parallel }_1)\\ =& \underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{t\in Q}{min}{\parallel x-t\parallel }_1+{\epsilon }^{\prime }\\ \le & d_{N,\delta }+h+{\epsilon }^{\prime }\\ =& \epsilon +h.\end{array}$$

Consequently, letting $h\to 0$, we get that $e(T_m(B\mathrm{\setminus }D),\{z_1,\dots ,z_{\mathcal{N}}\})\le \epsilon$, which together with (10) completes the proof of Lemma 3. □

From the relation (see [28])

$$vol\left(B_p^m\right)={[2\mathrm{\Gamma }(1/p+1)]}^m/\mathrm{\Gamma }(m/p+1),1\le p\le \mathrm{\infty },$$

the balls $B_p^m$ satisfy the inequalities

$${\left(c_p^{\prime }m\right)}^{-m/p}<vol\left(B_p^m\right)<{\left(c_p^{″}m\right)}^{-m/p},$$

(12)

where Γ is the Euler Γ-function, and $c_p^{\prime }$, $c_p^{″}$ depend only on p.

To estimate ${\mathcal{N}}_{\epsilon }(T_m(B\mathrm{\setminus }D))$ from below, we now need another auxiliary result.

Lemma 4 If $T_m$ is a diagonal operator from ${\mathbb{R}}^m$ onto ${\ell }_1^m$, then

$$|det(T_m)|=|\prod _{i=1}^m{\lambda }_i(T_m)|\asymp {\left(\frac{\parallel T_m\parallel }{\sqrt{m}}\right)}^m,$$

where ${\lambda }_i(T_m)$, $i=1,\dots ,m$, are non-zero eigenvalues of the operator $T_m$ rearranged as usual so that $|{\lambda }_i(T_m)|$ is non-increasing and each eigenvalue is repeated according to its multiplicity.

Proof It is known that

$$\parallel T_m\parallel ={\left(\sum _{i=1}^m{({\lambda }_i(T_m))}^2\right)}^{1/2}.$$

Accordingly,

$$\sqrt{m}|{\lambda }_m(T_m)|\le \parallel T_m\parallel \le \sqrt{m}|{\lambda }_1(T_m)|.$$

Obviously,

$${|{\lambda }_m(T_m)|}^m\le |\prod _{i=1}^m{\lambda }_i(T_m)|\le {|{\lambda }_m(T_1)|}^m,$$

from which the result of Lemma 4 follows immediately. □

Lemma 5 If $\delta \in [0,1]$ and $\lambda (D)\le \delta$, then

$${\mathcal{N}}_{\epsilon }(T_m(B\mathrm{\setminus }D))\ge \frac{1}{3}(1-\delta ){(c_0\parallel T_m\parallel /\epsilon )}^m.$$

Proof We first establish the inequality

$${\mathcal{N}}_{\epsilon }(T_m(B\mathrm{\setminus }D))\ge \frac{1}{3}\frac{\lambda (T_m(B\mathrm{\setminus }D))}{\lambda (B_1^m(2\epsilon ))}.$$

(13)

Indeed, suppose that (13) does not hold. Then, for $\mathcal{N}={\mathcal{N}}_{\epsilon }(T_m(B\mathrm{\setminus }D))$ and some set of points $z_1^{\ast },\dots ,z_{\mathcal{N}}^{\ast }$, by

$$\begin{array}{rcl}\epsilon & \ge & \underset{x\in T_m(B\mathrm{\setminus }D)}{sup}\underset{i=1,\dots ,\mathcal{N}}{min}{\parallel x-z_i^{\ast }\parallel }_1\\ \ge & \frac{1}{\lambda (T_m(B\mathrm{\setminus }D))}{\int }_{T_m(B\mathrm{\setminus }D)}\underset{i}{min}{\parallel x-z_i^{\ast }\parallel }_1\lambda (dx)\\ \ge & \frac{1}{\lambda (T_m(B\mathrm{\setminus }D))}{\int }_{T_m(B\mathrm{\setminus }D)\mathrm{\setminus }{\bigcup }_{i=1}^{\mathcal{N}}(z_i^{\ast }+B_1^m(2\epsilon ))}\underset{i}{min}{\parallel x-z_i^{\ast }\parallel }_1\lambda (dx)\\ \ge & \frac{1}{\lambda (T_m(B\mathrm{\setminus }D))}(2\epsilon )\lambda (T_m(B\mathrm{\setminus }D)\setminus \bigcup _{i=1}^{\mathcal{N}}(z_i^{\ast }+B_1^m(2\epsilon )))\\ =& 2\epsilon \frac{\lambda (T_m(B\mathrm{\setminus }D))-\mathcal{N}\cdot \lambda (B_1^m(2\epsilon ))}{\lambda (T_m(B\mathrm{\setminus }D))}\\ =& 2\epsilon (1-\mathcal{N}\cdot \frac{\lambda (B_1^m(2\epsilon ))}{\lambda (T_m(B\mathrm{\setminus }D))})\\ \ge & 2\epsilon \cdot \frac{2}{3}=\frac{4}{3}\epsilon ,\end{array}$$

we have obtained a contradiction.

In the sequel, we may as well assume that $T_m$ is a diagonal operator from ${\mathbb{R}}^m$ onto ${\ell }_1^m$. Using the inequality (13), (12) and Lemma 4, we have

$$\begin{array}{rcl}{\mathcal{N}}_{\epsilon }(T_m(B\mathrm{\setminus }D))& \ge & \frac{1}{3}\frac{\lambda (T_m(B\mathrm{\setminus }D))}{\lambda (B_1^m(2\epsilon ))}\\ =& \frac{1}{3}\frac{|det(T_m)|\lambda (B\mathrm{\setminus }D)}{\lambda (B_1^m(2\epsilon ))}\\ \ge & \frac{1}{3}c{\left(\frac{\parallel T_m\parallel }{\sqrt{m}}\right)}^m(1-\delta )\frac{vol(B)}{vol(B_1^m(2\epsilon ))}\\ \ge & \frac{1}{3}{\left(\frac{c_1\parallel T_m\parallel }{\sqrt{m}}\right)}^m(1-\delta ){\left(\frac{c_2\sqrt{m}}{2\epsilon }\right)}^m\\ =& \frac{1}{3}(1-\delta ){\left(\frac{c_0\parallel T_m\parallel }{\epsilon }\right)}^m.\end{array}$$

Next, assume that $T_m$ is a symmetric transformation of ${\mathbb{R}}^m$, then there is an orthogonal matrix U of order m such that the matrix $U{T}_m{U}^T$ is a diagonal matrix. Since the Lebesgue measure is invariant for orthogonal transformation, the result holds for symmetric transformation $T_m$.

Finally, in the general case, $T_m$ is a general invertible linear transformation from ${\mathbb{R}}^m$ onto ${\ell }_1^m$, then there are two matrices U and S such that $T_m=US$, where U is an orthogonal matrix and S is a positive definite symmetric matrix. As the same reason above, the result holds for the transformation $T_m$.

Thus, we complete the proof of Lemma 5. □

Lemma 6 If $\delta \in [0,1/2]$, then

$$d_{N,\delta }(T_m:B\to {\ell }_1^m,\lambda )\ge c_1\parallel T_m\parallel .$$

Proof From Lemma 3 and Lemma 5, we get

$$c{(1+\frac{4\parallel T_m\parallel }{\epsilon -d_{N,\delta }})}^m\ge N_{\epsilon }(T_m(B\mathrm{\setminus }D))\ge \frac{1}{3}(1-\delta ){\left(\frac{c_0\parallel T_m\parallel }{\epsilon }\right)}^m\ge \frac{1}{6}{\left(\frac{c_0\parallel T_m\parallel }{\epsilon }\right)}^m.$$

(14)

Let $\epsilon =5d_{N,\delta }$. Taking the logarithm of the inequality (14), we get the inequality

$$d_{N,\delta }(T_m:B\to {\ell }_1^m,\lambda )\ge c_1\parallel T_m\parallel$$

for some constants $c_1$ and $c_2$ and N with $m\ge c_{2}N$. Lemma 6 is proved. □

Proof of Theorem 2 According to Lemma 6, for any $\delta \in [0,1/2]$ and any subspace $\mathcal{L}\subset {\mathbb{R}}^m$ with $dim\mathcal{L}\le N$, there is a set $K\subset B$ with Lebesgue measure $\lambda (K)>\delta$ such that

$$e(T_{m}x,\mathcal{L},{\ell }_1^m)\ge c_1\parallel T_m\parallel$$

(15)

for any element $x\in K$. On the unit sphere $S^{m-1}=\{x\in {\mathbb{R}}^m:{\parallel x\parallel }_2=1\}$, we consider the subset $K^{\prime }=\{x/{\parallel x\parallel }_2:x\in K\}$.

Let ${\lambda }_{S^{m-1}}$ be a Lebesgue measure on the sphere $S^{m-1}$. We prove that

$${\lambda }_{S^{m-1}}\left(K^{\prime }\right)>\delta {\lambda }_{S^{m-1}}\left(S^{m-1}\right).$$

(16)

Indeed, assume that

$${\lambda }_{S^{m-1}}\left(K^{\prime }\right)\le \delta {\lambda }_{S^{m-1}}\left(S^{m-1}\right).$$

We introduce in ${\mathbb{R}}^m$ a polar system of coordinates $(r,s)$, where $r\ge 0$ and $s\in S^{m-1}$, and consider in B the cone

$$C=\{(r,s):0\le r\le 1,s\in K^{\prime }\}.$$

Then

$$\begin{array}{rcl}\lambda (K)& \le & \lambda (C)=\frac{1}{vol(B)}{\int }_0^1{r}^{m-1}dr{\int }_{K^{\prime }}{\lambda }_{S^{m-1}}(ds)\\ \le & \frac{\delta }{vol(B)}{\int }_0^1{r}^{m-1}dr{\int }_{S^{m-1}}{\lambda }_{S^{m-1}}(ds)=\delta .\end{array}$$

We have obtained a contradiction.

Consider the set $K_t=\{(r,s):r\ge t,s\in K^{\prime }\}$, $t\ge 0$. Using the inequality (16), we estimate the Gaussian measure of $K_t$:

$$\begin{array}{rcl}v(K_t)& =& {(2\pi )}^{-m/2}{\int }_{K_t}exp(-\frac{1}{2}{\parallel u\parallel }_2^2)(du)\\ =& {(2\pi )}^{-m/2}{\int }_t^{\mathrm{\infty }}{r}^{m-1}exp(-\frac{r^2}{2})dr{\int }_{K^{\prime }}{\lambda }_{S^{m-1}}(ds)\\ \ge & {(2\pi )}^{-m/2}\delta {\int }_t^{\mathrm{\infty }}{r}^{m-1}exp(-\frac{r^2}{2})dr{\int }_{S^{m-1}}{\lambda }_{S^{m-1}}(ds)\\ =& \delta v(x:{\parallel x\parallel }_2\ge t).\end{array}$$

(17)

A direct computation shows that for $t\ge \sqrt{m}$,

$$v({\parallel x\parallel }_2\ge t)\ge exp(-t^2)\text{and}v({\parallel x\parallel }_2\ge \sqrt{m})\ge c_0>0,$$

where $c_0$ is some absolute constant. It follows from this and (17) that for $t_0=max\{\sqrt{m},\sqrt{ln(1/\delta )}\}$ and for any $\delta \in (0,c_0]$,

$$v(K_{t_0})\ge \delta \cdot v({\parallel x\parallel }_2>max\{\sqrt{m},\sqrt{ln(1/\delta )}\})>{\delta }^2.$$

(18)

For any element $y=rs\in K_{t_0}$, we have from (15)

$$\begin{array}{rcl}e(T_{m}y,\mathcal{L},{\ell }_1^m)& =& re(T_{m}s,\mathcal{L},{\ell }_1^m)\ge c_{1}r\parallel T_m\parallel \ge c_2\parallel T_m\parallel max\{\sqrt{m},\sqrt{ln(1/\delta )}\}\\ \ge & \frac{c_2}{2}\parallel T_m\parallel \sqrt{m+ln(1/\delta )}.\end{array}$$

(19)

Since ℒ is an arbitrary subspace with $dim\mathcal{L}\le N$, it follows from (18) and (19) that

$$d_{N,{\delta }^2}(T_m:{\mathbb{R}}^m\to {\ell }_1^m,v)\ge \frac{c_2}{2}\parallel T_m\parallel \sqrt{m+ln(1/\delta )}.$$

Theorem 2 is a direct consequence of this. □