Asymptotic estimates for n-width of fuzzy numbers

Abstract

n-widths in approximation theory characterize how well one can approximate a subset by some “good” subsets of a normed linear space. Especially, n-widths of sets of $\mathbb{R}^{N}$ have been studied deeply. Now the following problem is posed: we know that $\mathbb{R} ^{N}$ can be embedded in the fuzzy number space $E^{N}$. Is it then possible to define n-widths of set A in $E^{N}$ and obtain asymptotic estimates for these n-widths?

In this paper, we shall introduce four n-widths of A in $E^{N}$ and determine these n-widths of Zadeh’s extension of diagonal matrices.

1 Introduction

Let A and B be two subsets of a normed linear space X. One may ask: how well A can be approximated by B? In the theory of n-widths of A in X, B will be a simple subspace of X. We will consider the possibility of allowing the simple subspaces to vary within X and find the one best adjusted to A. In many cases very simple sets may approximate A in an asymptotically optimal manner. It is then possible to judge whether it is worthwhile or not to spend additional time and money in using better but more complicated subspaces. The results of n-widths of A in $\mathbb{R}^{N}$ may be found in [1,2,3,4,5,6,7,8].

It is well known that $\mathbb{R}^{N}$ can be embedded in $E^{N}$. Thus if we restrict $d_{s}^{p}$-metric (see Sect. 4) convergence and level convergence on $\mathbb{R}^{N}$, then these types of convergence both become $l_{p}$-metric (induced by $\Vert \cdot \Vert _{p}$) convergence. Motivated by the study of n-widths of A in $\mathbb{R}^{N}$, we introduce definitions of four n-widths of A in $E^{N}$. Moreover, asymptotic estimates of these n-widths of Zadeh’s extension of diagonal matrices are obtained.

2 Preliminaries

2.1 Fuzzy numbers

For a fuzzy set $u:\mathbb{R}^{N}\rightarrow [0,1]$, suppose that:

• (1) u is normal, i.e., there exists $x\in \mathbb{R}^{N}$ such that $u(x)=1$;

• (2) u is upper semi-continuous;

• (3) $\operatorname{supp} u=\operatorname{cl}\{x\in \mathbb{R}^{N}: u(x)>0\}$ is compact;

• (4) u is fuzzy convex, i.e.,

  $$ u\bigl(\lambda x+(1-\lambda )y\bigr)\geq \min \bigl\{ u(x),u(y)\bigr\} , \quad 0 \leq \lambda \leq 1, $$

  for all $x,y\in \mathbb{R}^{N}$. Then u is called a fuzzy number. Let $E^{N}$ be the family of all fuzzy numbers. $\mathbb{R}^{N}$ can be embedded in $\mathbb{E}^{N}$, as any $u\in \mathbb{R}^{N}$ can be viewed as the fuzzy number

  $$ \hat{u}(x)= \textstyle\begin{cases} 1, & u=x; \\ 0, & u\neq x. \end{cases} $$

For $u, v\in E^{n}$, $\alpha \in [0,1]$, the α-cut of u is defined as follows:

$$ [u]^{\alpha }= \textstyle\begin{cases} \{x\in \mathbb{R}^{N}:u(x)\geq \alpha \}, & \text{if }0< \alpha \leq 1; \\ \operatorname{supp} u, & \text{if }\alpha =0; \end{cases} $$

and the algebraic operations on $E^{N}$ are defined as

$$ [u+v]^{\alpha }=[u]^{\alpha }+[v]^{\alpha }, \quad\quad [ku]^{\alpha }=k[u]^{ \alpha },\quad k\in \mathbb{R}, \alpha \in [0,1]. $$

If $f: \mathbb{R}^{N}\rightarrow \mathbb{R}^{N}$ is a function, we define Zadeh’s extension of f by

$$\begin{aligned}& \widetilde{f}:E^{N}\rightarrow E^{N}, \\& \widetilde{f}(u) (x)= \textstyle\begin{cases} \sup_{z\in f^{-1}(x)} u(z), & \text{if }f^{-1}(x)\neq \emptyset ; \\ 0, & \text{if }f^{-1}(x)= \emptyset . \end{cases}\displaystyle \end{aligned}$$

Lemma 1

([9])

If $f: \mathbb{R}^{N}\rightarrow \mathbb{R}^{N}$ is continuous, then f̃ is a well-defined function and

$$ \bigl[\widetilde{f}(u)\bigr]^{\alpha }=f\bigl([u]^{\alpha }\bigr),\quad \forall \alpha \in [0,1], \forall u\in E^{N}. $$

2.2 n-Widths of diagonal matrix

Definition 1

([10])

Let $(X,\Vert \cdot \Vert )$ be a normed linear space, and $A\subseteq X$.

• (1) The Kolmogorov n-width of A in X is defined by

  $$ d_{n}(A;X)=\inf_{{X_{n}}} \sup_{x\in A} \inf_{y\in X_{n}} \Vert x-y \Vert , $$

  where the left-most infimum is taken over all n-dimensional subspace $X_{n}$ of X.

• (2) The Bernstein n-width of A in X is defined by

  $$\begin{aligned} b_{n}(A;X) =& \sup_{{X_{n+1}}}\sup \bigl\{ \lambda : \lambda S({X_{n+1}}) \subseteq A\bigr\} \\ =& \sup_{{X_{n+1}}}\inf_{x\in \partial (A\cap { X_{n+1}})} \Vert x \Vert , \end{aligned}$$

  where $X_{n+1}$ is any $(n+1)$-dimensional subspace of X, and $S({X_{n+1}})$ is the unit ball of $X_{n+1}$.

• (3) The Gelfand n-width of A in X is defined as

  $$ d^{n}(A;X)=\inf_{{L^{n}}}\sup_{x\in A\cap {L^{n}}} \Vert x \Vert , $$

  where the infimum is taken over all subspaces ${L^{n}}$ of X of codimension n.

• (4) The linear n-width is given by

  $$ \delta _{n}(A;X)=\inf_{{P_{n}}(A)}\sup_{x\in A} \bigl\Vert x-P_{n}(x) \bigr\Vert , $$

  where the infimum is taken over all continuous linear operators ${P_{n}}$ of X into X of rank n.

Lemma 2

([11])

Let $X_{n+1}$ be any $(n+1)$-dimensional subspace of a normed linear space $(X,\Vert \cdot \Vert )$, and let $S(X_{n+1})$ denote the unit ball of $X_{n+1}$. Then

$$ d_{k}\bigl(S(X_{n+1});X\bigr)=1, \quad k=1,2,\ldots,n. $$

Let $l_{p}^{N}$ be the N-dimensional normed spaces of $x=(x_{1},\ldots,x_{N})\in \mathbb{R}^{N}$, with the normed

$$ \Vert x \Vert _{p}= \textstyle\begin{cases} (\sum_{i=1}^{N} \vert x_{i} \vert ^{p} )^{1/p}, & 1\leq p< \infty , \\ \max_{1\leq i\leq N} \vert x_{i} \vert , & p=\infty . \end{cases} $$

Let $D=\operatorname{diag}\{D_{1},\ldots,D_{N}\}$ be an $N\times N$ diagonal matrix. Without loss of generality, we assume that

$$ D_{1}\geq D_{2}\geq \cdots \geq D_{N}>0. $$

n-widths of $\mathfrak{D}_{p}=\{Dx:\Vert x\Vert _{p}\leq 1\}$ can be found in [1, 2, 10].

Theorem A

([2, 10])

For $1\leq p\leq \infty$,

$$ d_{n}\bigl(\mathfrak{D}_{p};l_{p}^{N} \bigr)=d^{n}\bigl(\mathfrak{D}_{p};l_{p}^{N} \bigr)=b _{n}\bigl(\mathfrak{ D}_{p};l_{p}^{N} \bigr)=\delta _{n}\bigl(\mathfrak{D}_{p};l_{p} ^{N}\bigr)=D_{n+1}. $$

Theorem B

([1])

Given $1\leq q\leq p\leq \infty$. Let $1/r=1/q-1/p$. Then

$$ d_{n}\bigl(\mathfrak{D}_{p};l_{q}^{N} \bigr)=d^{n}\bigl(\mathfrak{D}_{p};l_{q}^{N} \bigr)=b _{n}\bigl(\mathfrak{ D}_{p};l_{q}^{N} \bigr)=\delta _{n}\bigl(\mathfrak{D}_{p};l_{q} ^{N}\bigr)= \Biggl(\sum_{k=n+1}^{N}D_{k}^{r} \Biggr)^{1/r}. $$

3 n-Widths of fuzzy numbers

The following notation will be used throughout this paper. Let $X_{n}$ be an n-dimensional subspace of $\mathbb{R}^{N}$, ${L^{n}}$ be subspaces of $\mathbb{R}^{N}$ of codimension n. $S({X_{n}})$ denotes the unit ball of $X_{n}$. Set

$$\begin{aligned}& \widetilde{X_{n}}=\bigl\{ u:u\in E^{N}, [u]^{0} \subseteq X_{n}\bigr\} , \\& \widetilde{L^{n}}=\bigl\{ u:u\in E^{N},[u]^{0} \subseteq L^{n}\bigr\} , \\& S(\widetilde{X_{n+1}})=\bigl\{ u: d(u,\hat{0})\leq 1, u\in \widetilde{X_{n+1}}\bigr\} . \end{aligned}$$

Let $\widetilde{P_{n}}$ be Zadeh’s extension of the continuous linear operators ${P_{n}}$ of $\mathbb{R}^{N}$ into $\mathbb{R}^{N}$ of rank n.

Definition 2

Let $(E^{N},d)$ be a metric space, and $A\subseteq E^{N}$.

• (1) The Kolmogorov n-width of A in $E^{N}$ is defined by

  $$ d_{n}\bigl(A;E^{N}\bigr)=\inf_{\widetilde{X_{n}}} \sup _{u\in A} \inf_{v\in \widetilde{X_{n}}}d(u,v), $$

  where the left-most infimum is taken over all $\widetilde{X_{n}} \subseteq \mathbb{E}^{N}$.

• (2) The Bernstein n-width of A in $E^{N}$ is defined by

  $$\begin{aligned} b_{n}\bigl(A;E^{N}\bigr) =& \sup_{\widetilde{X_{n+1}}} \sup \bigl\{ \lambda :\lambda \geq 0, \lambda S(\widetilde{X_{n+1}}) \subseteq A\bigr\} . \end{aligned}$$

• (3) The Gelfand n-width of A in $E^{N}$ is defined as

  $$ d^{n}\bigl(A;E^{N}\bigr)=\inf_{\widetilde{L^{n}}}\sup _{u\in A\cap \widetilde{L^{n}}}d(u,\hat{0}), $$

  where the infimum is taken over all subspaces $\widetilde{L^{n}}$ of $E^{N}$.

• (4) The linear n-width of A in $E^{N}$ is given by

  $$ \delta _{n}\bigl(A;E^{N}\bigr)=\inf_{ \widetilde{P_{n}}} \sup_{u\in A} d\bigl(u, \widetilde{P_{n}}(u)\bigr), $$

  where the infimum is taken over all $\widetilde{P_{n}}$.

Proposition 1

Let $(E^{N},d)$ be a metric space, and $A\subseteq E^{N}$.

• (i) $\delta _{n}(A;E^{N})\geq d_{n}(A;E^{N})$.

• (ii) $\delta _{n}(A;E^{N})\geq d^{n}(A;E^{N})$.

Proof

Let $\widetilde{P_{n}}$ be Zadeh’s extension of the continuous linear operators ${P_{n}}$ of $\mathbb{R}^{N}$ into $\mathbb{R}^{N}$ of rank n.

(i) From Lemma 1 and $\operatorname{rank}P_{n}=n$, we know that there exists an n-dimensional subspace $X_{n}$ of $\mathbb{R}^{N}$ subject to the following relation:

$$ \bigl[\widetilde{P_{n}}(u)\bigr]^{0}=P_{n} \bigl([u]^{0}\bigr)\subseteq X_{n}, \quad u\in A\subseteq E^{N}. $$

Then $\widetilde{P_{n}}(u)\in \widetilde{X_{n}}$, i.e., $\widetilde{P_{n}}(A)\subseteq \widetilde{X_{n}}$. By the definitions of $d_{n}(A;E^{N})$ and $\delta _{n}(A;E^{N})$

$$ \delta _{n}\bigl(A;E^{N}\bigr)\geq d_{n} \bigl(A;E^{N}\bigr). $$

(ii) If $\widetilde{P_{n}}(u)=\hat{0}$, then

$$ \bigl[\widetilde{P_{n}}(u)\bigr]^{0}=P_{n} \bigl([u]^{0}\bigr)=\{0\}, $$

and $[u]^{0}\subseteq L^{n}$, i.e., $u\in \widetilde{L^{n}}$. Therefore

$$ \sup_{u\in A} d\bigl(u,\widetilde{P_{n}}(u)\bigr)\geq \sup_{\substack{u\in A\\ \widetilde{P_{n}}(u)=0}}d(u,\hat{0}), $$

whence it follows that $\delta _{n}(A;E^{N})\geq d^{n}(A;E^{N})$. □

4 n-Widths of D̃

We first choose a suitable metric $d(\cdot ,\cdot )$ on $E^{N}$ to establish a relation between $\Vert x-y\Vert _{p}$ and $d(u,v)$, $x,y\in l_{p}^{N}$, $u,v \subset E^{N}$. Then the distance between subsets of $E^{N}$ can be estimated by $\Vert x-y\Vert _{p}$.

Let $\mathcal{K}(\mathbb{R}^{N})$ be the family of nonempty compact subsets of $l_{p}^{N}$. If $A, B\in \mathcal{K}(\mathbb{R}^{N})$, $1\leq p<\infty$, the Hausdorff distance between A and B is defined by

$$ d_{H}^{p}(A,B)=\max \Bigl\{ \sup_{a\in A}\inf _{b\in B} \Vert a-b \Vert _{p}, \sup _{b\in B}\inf_{a\in A} \Vert a-b \Vert _{p} \Bigr\} . $$

For $u,v\in E^{N}$, $1\leq s<\infty$, $\alpha \in [0,1]$, we define

$$ d_{s}^{p}(u,v)= \biggl( \int _{0}^{1}d_{H}^{p} \bigl([u]^{\alpha },[v]^{\alpha }\bigr)^{s}\,d \alpha \biggr)^{1/s}, $$

then $d_{s}^{p}$ is called the $L_{s}$-metric on $E^{N}$ [12]. Let $L_{s,p}^{N}:=(E^{N},d_{s}^{p})$.

Proposition 2

([12])

$(E^{N},d_{s}^{p})$ is a metric space for $1\leq s, p < \infty$.

Let $D=\operatorname{diag}\{D_{1},\ldots,D_{N}\}$, $D_{1}\geq D_{2}\geq\cdots \geq D_{N}>0$, be an $N\times N$ diagonal matrix, and D̃ be Zadeh’s extension of D.

Lemma 3

Let $u\in E^{N}$, $k\in \mathbb{R}$, $1\leq s<\infty$, $\alpha \in [0,1]$. Then

$$ d_{s}^{p}(\widetilde{D}u,k\hat{0})= \biggl( \int _{0}^{1}\Bigl( \sup_{a\in [u]^{\alpha }} \Vert Da \Vert _{p}\Bigr)^{s}\,d\alpha \biggr)^{1/s}, $$

and

$$ d_{s}^{p}\bigl(\widetilde{D}(ku),\hat{0}\bigr)= \vert k \vert d_{s}^{p}\bigl(\widetilde{D}(u), \hat{0}\bigr). $$

Proof

Since $\operatorname{supp}\hat{0}=\{0\}$, it follows that

$$\begin{aligned} d_{H}^{p}\bigl([\widetilde{D}u]^{\alpha },k[ \hat{0}]^{\alpha }\bigr) =& d_{H} ^{p}\bigl({D} \bigl([u]^{\alpha }\bigr),k[\hat{0}]^{\alpha }\bigr) \\ =&\max \Bigl\{ \sup_{a\in [u]^{\alpha }}\inf_{b\in k[\hat{0}]^{\alpha }} \Vert Da-b \Vert _{p},\sup_{b\in k[\hat{0}]^{\alpha }}\inf_{a\in [u]^{\alpha }} \Vert Da-b \Vert _{p} \Bigr\} \\ =& \sup_{a\in [u]^{\alpha }} \Vert Da \Vert _{p}. \end{aligned}$$

Hence

$$ d_{s}^{p}(\widetilde{D}u,k\hat{0})= \biggl( \int _{0}^{1}\Bigl( \sup_{a\in [u]^{\alpha }} \Vert Da \Vert _{p}\Bigr)^{s}\,d\alpha \biggr)^{1/s}. $$

Similarly, we can get the second equation. □

In this paper we are concerned with the estimate of n-widths of

$$ \widetilde{\mathfrak{D}_{s,p}}=\bigl\{ \widetilde{D}u:u\in E^{N}, d_{s}^{p}(u, \hat{0})\leq 1\bigr\} $$

and

$$ \widetilde{\mathfrak{D}_{s,p}^{-1}}=\bigl\{ \widetilde{D^{-1}}u:u\in E^{N}, d_{s}^{p}(u, \hat{0})\leq 1\bigr\} . $$

It is often the case, see examples in [10], that a very simple form of $b_{n}(A;X) = \sup_{{X_{n+1}}} \inf_{x\in \partial (A\cap { X_{n+1}})}\Vert x\Vert$ is used. We introduce a similar definition of $b_{n}(\widetilde{\mathfrak{ D}_{s,p}};L_{s,p} ^{N})$ for easy computation.

Definition 3

For $1\leq s,p<\infty$,

$$ b_{n}\bigl(\widetilde{\mathfrak{ D}_{s,p}};L_{s,p}^{N} \bigr)= \sup_{\widetilde{X_{n+1}}} \inf_{\substack{u\in \widetilde{ X_{n+1}}\\ d_{s}^{p}(u,\hat{0})=1}}d( \widetilde{D}u,\hat{0}). $$

Now we state our main results.

Theorem 1

For $1\leq s,p<\infty$,

$$ d_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)=d^{n}\bigl( \widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)=b_{n}\bigl( \widetilde{\mathfrak{ D}_{s,p}};L_{s,p}^{N} \bigr)=\delta _{n}\bigl( \widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)=D_{n+1}. $$

Theorem 2

Given $1\leq s<\infty$, $1\leq q\leq p<\infty$. Let $1/r=1/q-1/p$. Then

$$ d_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N} \bigr)=d^{n}\bigl( \widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N} \bigr)=\delta _{n}\bigl( \widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N} \bigr)= \Biggl(\sum_{k=n+1}^{N}D _{k}^{r} \Biggr)^{1/r}. $$

Remark

For $1\leq q\leq p<\infty$, Theorem 1 and 2 are obvious generalizations of Theorems A and B.

Before proving these two theorems, we need some lemmas.

Lemma 4

For $1\leq s,p<\infty$,

• (i) $\delta _{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p} ^{N})\geq d_{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N})\geq b_{n}( \widetilde{\mathfrak{ D}_{s,p}};L_{s,p}^{N})$.

• (ii) $\delta _{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p} ^{N})\geq d^{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N})\geq b_{n}( \widetilde{\mathfrak{ D}_{s,p}};L_{s,p}^{N})$.

Proof

From Proposition 1 the following results are known:

$$ \delta _{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)\geq d_{n}\bigl( \widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr), \quad\quad \delta _{n}\bigl( \widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)\geq d^{n}\bigl( \widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr). $$

Now we prove that $d_{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N}) \geq b_{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N})$. If $\lambda S(\widetilde{X_{n+1}})\subseteq \widetilde{\mathfrak{D}_{s,p}}$, then from the definition of $d_{n}(A;E^{N})$

$$ d_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)\geq d_{n}\bigl(\lambda S( \widetilde{X_{n+1}});L_{s,p}^{N} \bigr)\geq d_{n}\bigl(\lambda S({X_{n+1}});L _{s,p}^{N}\bigr). $$

(1)

For $\hat{a}\in \lambda S({X_{n+1}})$, $v\in \widetilde{X_{n}}$, a direct computation shows that

$$ d_{H}^{p}\bigl(\{a\},[v]^{\alpha }\bigr)=\sup _{b\in [v]^{\alpha }} \Vert a-b \Vert _{p} \geq \inf _{b\in X_{n}} \Vert a-b \Vert _{p} $$

and

$$ d_{s}^{p}(\hat{a},v)= \biggl( \int _{0}^{1}d_{H}^{p}\bigl( \{a\},[v]^{\alpha }\bigr)^{s}\,d \alpha \biggr)^{1/s}\geq \inf_{b\in X_{n}} \Vert a-b \Vert _{p}. $$

Therefore

$$\begin{aligned} \sup_{\hat{a}\in \lambda S({X_{n+1}})}\inf_{v\in \widetilde{X_{n+1}}}d _{s}^{p}( \hat{a},v) \geq & \sup_{a\in \lambda S({X_{n+1}})} \inf_{b\in X_{n}} \Vert a-b \Vert _{p}, \end{aligned}$$

which implies that

$$ d_{n}\bigl(\lambda S({X_{n+1}});L_{s,p}^{N} \bigr)\geq d_{n}\bigl(\lambda S({X_{n+1}});l _{p}^{N}\bigr). $$

From (1) and Lemma 2

$$ d_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)\geq \lambda . $$

By the definition of $b_{n}(A;E^{N})$ in Definition 2 (2), we have $d_{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N})\geq b_{n}( \widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N})$.

The proof of $d^{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N})\geq b _{n}(\widetilde{\mathfrak{ D}_{s,p}};L_{s,p}^{N})$ is totally analogous to the proof of $d_{n}(\widetilde{\mathfrak{D}_{s,p}}; L_{s,p}^{N}) \geq b_{n}(\widetilde{\mathfrak{ D}_{s,p}};L_{s,p}^{N})$, here we omit it. The theorem is proved. □

Let $a=(a_{1},\ldots,a_{N})\in \mathbb{R}^{N}$, $A,B\subseteq \mathbb{R}^{N}$, we write $a\perp A$ if $\sum_{i=1}^{N}a_{i}x_{i}=0$ for all $x=(x_{1}, \ldots,x_{N})\in A$ and $A\perp B$ if $\sum_{i=1}^{N}a _{i}b_{i}=0$ for $\forall a=(a_{1},\ldots,a_{N})\in A$, $b=(b_{1},\ldots,b_{N})\in B$. Set

$$ A^{\perp }= \Biggl\{ y\in \mathbb{R}^{N}: \sum _{i=1}^{N}x_{i}y_{i}=0, \forall x=(x_{1},\ldots,x_{N})\in A \Biggr\} . $$

Let $u\in E^{N}$, $C\subseteq E^{N}$, we write $u\perp C$ if $[u]^{0}\perp [v]^{0}$, $\forall v\in C$.

Lemma 5

For $u\in E^{N}$, the following properties are equivalent.

• (i) $u\perp \widetilde{X_{N-n}}$.

• (ii) $u\in \widetilde{X_{n}}$.

Proof

(i) ⇒ (ii). Since $u\perp \widetilde{X_{N-n}}$, we know that for $\forall v\in \widetilde{X_{N-n}}$, i.e., $[v]^{0}\subseteq X_{N-n}$, we have $[u]^{0}\perp [v]^{0}$.

$$ [u]^{0}\subset X_{N-n}^{\perp }. $$

Note that $X_{N-n}^{\perp }=X_{n}$. Then (ii) follows.

(ii) ⇒ (i). If $u\in \widetilde{X_{n}}$, i.e., $[u]^{0} \subseteq X_{n}$, then for all $x\in [u]^{0}$ there is an $(N-n)$-dimensional subspace $X_{N-n}$ such that $x\perp X_{N-n}$. Consequently, we have $[u]^{0}\perp {X_{N-n}}$, i.e., $u\perp \widetilde{X_{N-n}}$. □

Lemma 6

For $1\leq s,p<\infty$, $n< N$,

$$ b_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}},L_{s,q}^{N} \bigr)d^{N-n-1}\bigl( \widetilde{\mathfrak{D}_{s,q}^{-1}},L_{s,p}^{N} \bigr)=1. $$

Proof

By the definition of $d^{N-n-1}(\widetilde{\mathfrak{D}_{s,q}^{-1}},L _{s,p}^{N})$ and Lemma 5, we have

$$\begin{aligned} d^{N-n-1}\bigl(\widetilde{\mathfrak{D}_{s,p}^{-1}},L_{s,q}^{N} \bigr) =& \min_{\widetilde{X_{N-n-1}}} \max_{\substack{u\in \widetilde{X_{n+1}}\\ {d_{s}^{p}(u,\hat{0})}\leq 1}} {d_{s}^{q}\bigl(\widetilde{D^{-1}}u,\hat{0}\bigr)} \\ =& \min_{\widetilde{X_{N-n-1}}} \max_{\substack{u\perp \widetilde{X_{N-n-1}}\\ u\neq \hat{0}}}\frac{d_{s} ^{q}(\widetilde{D^{-1}}u,\hat{0})}{d_{s}^{p}(u,\hat{0})} \\ =& \min_{\widetilde{X_{N-n-1}}} \max_{\substack{u\perp \widetilde{X_{N-n-1}}\\ u\neq \hat{0}}} \biggl[ \frac{d _{s}^{p}(u,\hat{0})}{d_{s}^{q}(\widetilde{D^{-1}}u,\hat{0})} \biggr]^{-1}. \end{aligned}$$

Setting $v=\widetilde{D^{-1}}u$, by Lemma 1 we have

$$ [v]^{\alpha }=D^{-1}[u]^{\alpha }, \quad \forall \alpha \in [0,1], $$

and

$$ [u]^{\alpha }=DD^{-1}[u]^{\alpha }=D\bigl[\widetilde{D^{-1}}u \bigr]^{\alpha }=D[v]^{ \alpha }=[\widetilde{D}v]^{\alpha }, \quad \forall \alpha \in [0,1]. $$

Since D is invertible, hence $v\perp \widetilde{X_{N-n-1}}$. Now

$$\begin{aligned} d^{N-n-1}\bigl(\widetilde{\mathfrak{D}_{s,q}^{-1}},L_{s,p}^{N} \bigr) =& \min_{\widetilde{X_{N-n-1}}} \max_{\substack{v\perp \widetilde{X_{N-n-1}}\\ v\neq \hat{0}}} \biggl[ \frac{d _{s,p}(\widetilde{D}v,\hat{0})}{d_{s}^{q}(v,\hat{0})} \biggr]^{-1} \\ =& \biggl[\max_{\widetilde{X_{n+1}}} \min_{\substack{v\in \widetilde{X_{n+1}}\\ v\neq \hat{0}}} \frac{d_{s}^{p}( \widetilde{D}v,\hat{0})}{d_{s}^{q}(v,\hat{0})} \biggr]^{-1} \\ =& \bigl[b_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}},L_{s,q}^{N} \bigr) \bigr]^{-1}. \end{aligned}$$

 □

Proof of Theorem 1

Let $P_{n}=\operatorname{diag}(D_{1},\ldots,D_{n},0,\ldots,0)$. For any $u\in E^{N}$,

$$\begin{aligned} d_{H}^{p}\bigl([\widetilde{D}u]^{\alpha }\bigr),[ \widetilde{P_{n}} u]^{\alpha }) =& d_{H}^{p} \bigl(D\bigl([u]^{\alpha }\bigr),P_{n}\bigl([u]^{\alpha }\bigr) \bigr) \\ =& \max \Bigl\{ \sup_{a\in [u]^{\alpha }}\inf_{b\in [u]^{\alpha }} \Vert Da-P _{n}b \Vert _{p},\sup_{b\in [u]^{\alpha }}\inf _{a\in [u]^{\alpha }} \Vert Da-P _{n}b \Vert _{p} \Bigr\} \\ =& \max_{a\in [u]^{\alpha }} \bigl\Vert (D-P_{n})a \bigr\Vert _{p}. \end{aligned}$$

(2)

Then

$$\begin{aligned} \delta _{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr) \leq & \max_{d_{s}^{p}(u,\hat{0})\leq 1}d_{s}^{p}( \widetilde{D}u, \widetilde{P_{n}}u) \\ =& \max_{u\neq \hat{0}}\frac{d_{s}^{p}(\widetilde{D}u, \widetilde{P_{n}}u)}{d_{s}^{p}(u,\hat{0})} \\ =& \max_{u\neq \hat{0}}\frac{ (\int _{0}^{1}(\max_{a\in [u]^{ \alpha }} \Vert (D-P_{n})a \Vert _{p})^{s}\,d\alpha )^{1/s}}{ (\int _{0} ^{1}(\max_{a\in [u]^{\alpha }} \Vert a \Vert _{p})^{s}\,d\alpha )^{1/s}} \\ =& \max_{u\neq \hat{0}}\frac{ (\int _{0}^{1}(\max_{a\in [u]^{ \alpha }}(\sum^{N}_{i=n+1} \vert D_{i}a_{i} \vert ^{p})^{1/p})^{s}\,d\alpha )^{1/s}}{ (\int _{0}^{1}(\max_{a\in [u]^{\alpha }} \Vert a \Vert _{p})^{s}\,d\alpha )^{1/s}} \\ \leq & \max_{u\neq \hat{0}}\frac{D_{n+1} (\int _{0}^{1}( \max_{a\in [u]^{\alpha }}(\sum^{N}_{i=n+1} \vert a_{i} \vert ^{p})^{1/p})^{s}\,d \alpha )^{1/s}}{ (\int _{0}^{1}(\max_{a\in [u]^{\alpha }} \Vert a \Vert _{p})^{s}\,d\alpha )^{1/s}} \\ \leq & D_{n+1}. \end{aligned}$$

In a similar way,

$$ \delta _{n}\bigl(\widetilde{\mathfrak{D}^{-1}_{s,p}};L_{s,p}^{N} \bigr) \leq 1/D _{n+1}. $$

From Lemma 6

$$\begin{aligned} b_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr) =& \bigl(d^{M-n-1}\bigl( \widetilde{\mathfrak{D}_{s,p}^{-1}};L_{s,p}^{N} \bigr) \bigr)^{-1} \\ \geq & \bigl(\delta _{n}\bigl(\widetilde{\mathfrak{D}_{s,p}^{-1}};L_{s,p} ^{N}\bigr) \bigr)^{-1}. \end{aligned}$$

Thus

$$ b_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,p}^{N} \bigr)\geq D_{n+1}. $$

By Lemma 4, we prove that these four n-widths equal $D_{n+1}$. □

Let $1\leq p,q<\infty$, and $1/p+1/p'=1/q+1/q'=1$. We use the notation $x'(x)=\langle x,x'\rangle$ for $x\in l_{p}^{N}$, $x'\in l_{p'}^{N}$, and similarly for $l_{q}^{N}$. Diagonal matrix $D:l_{p}^{N}\rightarrow l _{q}^{N}$ has an adjoint $D':l_{q'}^{N}\rightarrow l_{p'}^{N}$, defined by $\langle Dx,y'\rangle =\langle x,D'y'\rangle$ for $x\in l_{p}^{N}$ and $y'\in l_{q'}^{N}$. It is well known that $D=D'$. Let ${L^{n}}$ be subspaces of $l_{p}^{N}$ of codimension n and $\widetilde{L^{n}}=\{u \in E^{N}:[u]^{0}\subseteq L^{n}\}$, set

$$\begin{aligned}& L^{n}_{\perp }= \bigl\{ x': x'\in l_{p'}^{N}, \bigl\langle x,x'\bigr\rangle =0, \text{all } x\in L^{n} \bigr\} , \\& \widetilde{L^{n}_{\perp }}= \bigl\{ v: v\in E^{N}, [v]^{0}\subset L_{ \perp }^{n} \bigr\} . \end{aligned}$$

It is well known that $\operatorname{dim}L^{n}_{\perp }=n$.

Lemma 7

Let $1\leq s, p,q<\infty$, and $1/p+1/p'=1/q+1/q'=1$. Then

$$ d_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N} \bigr)\geq d^{n}\bigl( \widetilde{\mathfrak{D}_{s,q'}};L_{s,p'}^{N} \bigr). $$

Proof

Let $\widetilde{A_{p}}=\{u\in E^{N}:d_{s}^{p}(u,\hat{0})\leq 1\}$. Then

$$\begin{aligned} \sup_{u\in \widetilde{A_{p}}\cap \widetilde{L^{n}}}\sup_{a\in [u]^{0}} \Vert Da \Vert _{q} =&\sup_{u\in \widetilde{A_{p}}\cap \widetilde{L^{n}}} \sup_{a\in [u]^{0}}\sup _{ \Vert b' \Vert _{q'}\leq 1}\bigl\langle Da,b'\bigr\rangle \\ \leq &\sup_{u\in \widetilde{A_{p}}\cap \widetilde{L^{n}}} \sup_{a\in [u]^{0}}\sup _{v\in \widetilde{A_{q'}}} \sup_{b'\in [v]^{\alpha }}\bigl\langle Da,b'\bigr\rangle \\ =&\sup_{v\in \widetilde{A_{q'}}} \sup_{u\in \widetilde{A_{p}}\cap \widetilde{L^{n}}} \sup _{b'\in [v]^{\alpha }}\sup_{a\in [u]^{0}}\bigl\langle a,D'b'\bigr\rangle . \end{aligned}$$

(3)

Let $u\in \widetilde{A_{p}}\cap \widetilde{L^{n}}$ and $a'\in {L_{ \perp }^{n}}$. Then $\langle a,a'\rangle =0$, $\forall a\in [u]^{0}$. Hence

$$ \bigl\langle a,D'b'\bigr\rangle =\bigl\langle a,D'b'-a'\bigr\rangle \leq \Vert a \Vert _{p} \bigl\Vert D'b'-a' \bigr\Vert _{p'}\leq \bigl\Vert D'b'-a' \bigr\Vert _{p'}. $$

For $u'\in \widetilde{{L_{\perp }^{n}}}$, we have

$$\begin{aligned} \sup_{b'\in [v]^{\alpha }}\sup_{a\in [u]^{0}}\bigl\langle a,D'b'\bigr\rangle \leq& \sup _{b'\in [v]^{\alpha }} \inf_{a'\in [u']^{\alpha }} \bigl\Vert Db'-a' \bigr\Vert _{p'}. \end{aligned}$$

(4)

Following (3), (4), and $D=D'$

$$\begin{aligned} \sup_{v\in \widetilde{A_{q'}}} \sup_{u\in \widetilde{A_{p}}\cap \widetilde{L^{n}}} \sup _{b'\in [v]^{\alpha }}\sup_{a\in [u]^{0}}\bigl\langle a,D'b'\bigr\rangle \leq & \sup_{v\in \widetilde{A_{q'}}} \inf_{u'\in \widetilde{L_{\perp }^{n}}}\sup_{b'\in [v]^{\alpha }} \inf _{a'\in [u']^{\alpha }} \bigl\Vert Db'-a' \bigr\Vert _{p'} \\ \leq &\sup_{v\in \widetilde{A_{q'}}} \inf_{u'\in \widetilde{L_{\perp }^{n}}}d_{s}^{p'} \bigl(u',\widetilde{D}v\bigr). \end{aligned}$$

(5)

Combining Lemma 3, (3), and (5)

$$ \sup_{u\in \widetilde{A_{p}}\cap \widetilde{L^{n}}}d_{s}^{q}( \widetilde{D}u, \hat{0})\leq \sup_{v\in \widetilde{A_{q'}}} \inf_{u'\in \widetilde{L_{\perp }^{n}}}d_{s}^{p'} \bigl(u',\widetilde{D}v\bigr), $$

then taking the infimum over $\widetilde{L^{n}}$, we have the result. □

Proof of Theorem 2

We first prove that $\delta _{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,q} ^{N})\leq (\sum_{k=n+1}^{N}D_{k}^{r} )^{1/r}$.

Let $P_{n}=\operatorname{diag}(D_{1},\ldots,D_{n},0,\ldots,0)$. For any $u\in E^{N}$, as (2) in the proof of Theorem 1

$$\begin{aligned} d_{H}^{q}\bigl([\widetilde{D}u]^{\alpha }\bigr),[ \widetilde{P_{n}} u]^{\alpha }) =& \sup_{a\in [u]^{\alpha }} \bigl\Vert (D-P_{n})a \bigr\Vert _{q} \\ =&\sup_{a\in [u]^{\alpha }} \Biggl(\sum_{k=n+1}^{N} \vert D_{k}a_{k} \vert ^{q} \Biggr)^{1/q}. \end{aligned}$$

From $1/r=1/q-1/p$ and Hölder’s inequality

$$ \Biggl(\sum_{k=n+1}^{N} \vert D_{k}a_{k} \vert ^{q} \Biggr)^{1/q} \leq \Biggl( \sum_{k=n+1}^{N} \vert D_{k} \vert ^{r} \Biggr)^{1/r} \Biggl(\sum _{k=n+1} ^{N} \vert a_{k} \vert ^{p} \Biggr)^{1/p}, $$

we have

$$\begin{aligned} d_{H}^{q}\bigl([\widetilde{D}u]^{\alpha }\bigr),[ \widetilde{P_{n}} u]^{\alpha }) \leq & \sup_{a\in [u]^{\alpha }} \Biggl(\sum_{k=n+1}^{N} \vert D_{k} \vert ^{r} \Biggr)^{1/r} \Biggl(\sum _{k=n+1}^{N} \vert a_{k} \vert ^{p} \Biggr)^{1/p} \\ \leq & \Biggl(\sum_{k=n+1}^{N} \vert D_{k} \vert ^{r} \Biggr)^{1/r} \sup _{a\in [u]^{\alpha }} \Biggl(\sum_{k=1}^{N} \vert a_{k} \vert ^{p} \Biggr)^{1/p} \\ =& \Biggl(\sum_{k=n+1}^{N} \vert D_{k} \vert ^{r} \Biggr)^{1/r}d^{p}_{H} \bigl([u]^{ \alpha },[\hat{0}]^{\alpha }\bigr). \end{aligned}$$

Then

$$\begin{aligned} \delta _{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N} \bigr) \leq & \max_{d_{s}^{p}(u,\hat{0})\leq 1}d_{s}^{q}\bigl([ \widetilde{D}u]^{\alpha }\bigr),[ \widetilde{P_{n}} u]^{\alpha }) \\ \leq & \max_{d_{s}^{p}(u,\hat{0})\leq 1} \biggl( \int _{0}^{1}d_{H}^{q}\bigl(D \bigl([u]^{ \alpha }\bigr),P_{n}\bigl([u]^{\alpha }\bigr) \bigr)^{s}\,d\alpha \biggr)^{1/s} \\ \leq & \max_{d_{s}^{p}(u,\hat{0})\leq 1} \Biggl(\sum_{k=n+1} ^{N} \vert D_{k} \vert ^{r} \Biggr)^{1/r} d_{s}^{p}(u,\hat{0}) \\ \leq & \Biggl(\sum_{k=n+1}^{N} \vert D_{k} \vert ^{r} \Biggr)^{1/r}. \end{aligned}$$

(6)

Now we are to prove $d^{n}(\widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N}) \geq (\sum_{k=n+1}^{N}\vert D_{k}\vert ^{r} )^{1/r}$. From Lemma 3

$$\begin{aligned} d_{H}^{q}\bigl([\widetilde{D}u]^{\alpha }\bigr),[ \hat{0}]^{\alpha }) =& d_{H} ^{q}\bigl(D \bigl([u]^{\alpha }\bigr),[\hat{0}]^{\alpha }\bigr) \\ =& \max_{a\in [u]^{\alpha }} \Vert Da \Vert _{q} \\ =&\max_{a\in [u]^{\alpha }} \Biggl(\sum_{k=1}^{N} \vert D_{k}a_{k} \vert ^{q} \Biggr)^{1/q} \end{aligned}$$

and

$$\begin{aligned} d_{H}^{p}\bigl([u]^{\alpha }\bigr),[ \hat{0}]^{\alpha }) =&\max_{a\in [u]^{ \alpha }} \Biggl(\sum _{k=1}^{N} \vert a_{k} \vert ^{p} \Biggr)^{1/p}. \end{aligned}$$

By the definition of $d^{n}(\widetilde{\mathfrak{D}_{s,p}},L_{s,q} ^{N})$ and Theorem B

$$\begin{aligned} d^{n}\bigl(\widetilde{\mathfrak{D}_{s,p}},L_{s,q}^{N} \bigr) \geq d^{n}\bigl({\mathfrak{D} _{p}},L_{s,q}^{N} \bigr) = d^{n}\bigl({\mathfrak{D}_{p}},l_{q}^{N} \bigr) = \Biggl( \sum_{k=n+1}^{N}D_{k}^{r} \Biggr)^{1/r}. \end{aligned}$$

(7)

Combine (6), (7), and Proposition 1(ii),

$$ \delta _{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N} \bigr)=d^{n}\bigl( \widetilde{\mathfrak{D}_{s,p}},L_{s,q}^{N} \bigr)= \Biggl(\sum_{k=n+1} ^{N}D_{k}^{r} \Biggr)^{1/r}. $$

Similarly, we can have

$$ \delta _{n}\bigl(\widetilde{\mathfrak{D}_{s,q'}};L_{s,p'}^{N} \bigr)= \Biggl(\sum_{k=n+1}^{N}D_{k}^{r} \Biggr)^{1/r} $$

and

$$ d^{n}\bigl(\widetilde{\mathfrak{D}_{s,q'}},L_{s,p'}^{N} \bigr)\geq \Biggl(\sum_{k=n+1}^{N}D_{k}^{r} \Biggr)^{1/r}. $$

By Proposition 1(i) and Lemma 7

$$ d_{n}\bigl(\widetilde{\mathfrak{D}_{s,p}};L_{s,q}^{N} \bigr)= \Biggl(\sum_{k=n+1}^{N}D_{k}^{r} \Biggr)^{1/r}. $$

The theorem is proved. □