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Asymptotic estimates for n-width of fuzzy numbers
Journal of Inequalities and Applications volume 2019, Article number: 86 (2019)
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:
and the algebraic operations on \(E^{N}\) are defined as
If \(f: \mathbb{R}^{N}\rightarrow \mathbb{R}^{N}\) is a function, we define Zadeh’s extension of f by
Lemma 1
([9])
If \(f: \mathbb{R}^{N}\rightarrow \mathbb{R}^{N}\) is continuous, then f̃ is a well-defined function and
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
Let \(l_{p}^{N}\) be the N-dimensional normed spaces of \(x=(x_{1},\ldots,x_{N})\in \mathbb{R}^{N}\), with the normed
Let \(D=\operatorname{diag}\{D_{1},\ldots,D_{N}\}\) be an \(N\times N\) diagonal matrix. Without loss of generality, we assume that
n-widths of \(\mathfrak{D}_{p}=\{Dx:\Vert x\Vert _{p}\leq 1\}\) can be found in [1, 2, 10].
Theorem A
For \(1\leq p\leq \infty \),
Theorem B
([1])
Given \(1\leq q\leq p\leq \infty \). Let \(1/r=1/q-1/p\). Then
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
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:
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 \(\delta _{n}(A;E^{N})\)
(ii) If \(\widetilde{P_{n}}(u)=\hat{0}\), then
and \([u]^{0}\subseteq L^{n}\), i.e., \(u\in \widetilde{L^{n}}\). Therefore
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
For \(u,v\in E^{N}\), \(1\leq s<\infty \), \(\alpha \in [0,1]\), we define
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
and
Proof
Since \(\operatorname{supp}\hat{0}=\{0\}\), it follows that
Hence
Similarly, we can get the second equation. □
In this paper we are concerned with the estimate of n-widths of
and
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 \(1\leq s,p<\infty \),
Now we state our main results.
Theorem 1
For \(1\leq s,p<\infty \),
Theorem 2
Given \(1\leq s<\infty \), \(1\leq q\leq p<\infty \). Let \(1/r=1/q-1/p\). Then
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:
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 \(d_{n}(A;E^{N})\)
For \(\hat{a}\in \lambda S({X_{n+1}})\), \(v\in \widetilde{X_{n}}\), a direct computation shows that
and
Therefore
which implies that
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
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 \([u]^{0}\perp [v]^{0}\).
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 \(1\leq s,p<\infty \), \(n< N\),
Proof
By the definition of \(d^{N-n-1}(\widetilde{\mathfrak{D}_{s,q}^{-1}},L _{s,p}^{N})\) and Lemma 5, we have
Setting \(v=\widetilde{D^{-1}}u\), by Lemma 1 we have
and
Since D is invertible, hence \(v\perp \widetilde{X_{N-n-1}}\). Now
□
Proof of Theorem 1
Let \(P_{n}=\operatorname{diag}(D_{1},\ldots,D_{n},0,\ldots,0)\). For any \(u\in E^{N}\),
Then
In a similar way,
From Lemma 6
Thus
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
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
Proof
Let \(\widetilde{A_{p}}=\{u\in E^{N}:d_{s}^{p}(u,\hat{0})\leq 1\}\). Then
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
For \(u'\in \widetilde{{L_{\perp }^{n}}}\), we have
Following (3), (4), and \(D=D'\)
Combining Lemma 3, (3), and (5)
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
From \(1/r=1/q-1/p\) and Hölder’s inequality
we have
Then
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
and
By the definition of \(d^{n}(\widetilde{\mathfrak{D}_{s,p}},L_{s,q} ^{N})\) and Theorem B
Combine (6), (7), and Proposition 1(ii),
Similarly, we can have
and
By Proposition 1(i) and Lemma 7
The theorem is proved. □
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This work was supported by the Key Scientific Research Fund of Xihua University (Grant No. z1412621) and the National Natural Science Foundation of China (Grant No. 15233593).
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Han, Y.J., Liang, L. & Chen, G.G. Asymptotic estimates for n-width of fuzzy numbers. J Inequal Appl 2019, 86 (2019). https://doi.org/10.1186/s13660-019-2041-7
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DOI: https://doi.org/10.1186/s13660-019-2041-7