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

Journal of Inequalities and Applications20132013:277

https://doi.org/10.1186/1029-242X-2013-277

Received: 20 October 2012

Accepted: 15 May 2013

Published: 3 June 2013

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 ( T ) into L q ( T ) ( 1 q ) with the Gaussian measure.

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

Keywords

probabilistic widthaverage widthsSobolev spaceGaussian measure

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 [47], Mathé [812], Sun [13, 14], and Ritter [15]. The new developments in this direction can be found in Fang’s papers [1621]. Moreover, Carl and Pajor [22] proved an inequality with respect to the Gelfand numbers of an operator u from 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 R m onto  q m .

First, we recall some useful concepts. Let W be a bounded subset of a normed linear space X with the norm X , and F N be an N-dimensional subspace of X. The quantity
e ( W , F N , X ) = sup x W e ( x , F N , X ) ,
where
e ( x , F N , X ) = inf y F N x y 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 ) = inf F N e ( W , F N , X ) = inf F N sup x W inf y F N x y 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 consisting of open subsets of W and equipped with a probabilistic measure μ defined on . That is, μ is a σ-additive nonnegative function on , and μ ( W ) = 1 . Let δ [ 0 , 1 ] be an arbitrary number. The corresponding probabilistic Kolmogorov’s ( N , δ ) -width of a set W with a measure μ in the space X is defined by
d N , δ ( W , μ , X ) = inf G δ d N ( W G , X ) ,
(1)
where G δ runs through all possible subsets in with measure μ ( G δ ) δ . The p-average Kolmogorov’s N-width is defined by
d N ( a ) ( W , μ , X ) p = inf F N ( W e ( x , F N , X ) p d μ ( x ) ) 1 / p , 0 < p < ,
(2)
where F N in (2) runs over all linear subspaces of X of dimension at most N. Let p m be an m-dimensional normed space of vectors x = ( x 1 , , x m ) R m , with a norm
x p m = { ( i = 1 m | x i | p ) 1 / p , 1 p < , max 1 i m | x i | , p = .
Consider in R m the standard Gaussian measure v = v m , which is defined as
v ( G ) = ( 2 π ) m / 2 G exp ( 1 2 x 2 2 ) d x ,

where G is any Borel subset in R m . Obviously, v ( R m ) = 1 .

Denote by B p m ( ρ ) = { x p m : x p ρ } the ball of radius ρ in p m . Let B p m = B p m ( 1 ) .

Let N = 0 , 1 ,  , δ [ 0 , 1 ) be arbitrary and T m be a linear invertible operator from R m onto q m . We define the probabilistic ( N , δ ) -width of the operator acting in space R m equipped with the Gaussian measure v in q m -norm:
d N , δ ( T m : R m q m , v ) = inf G inf L N e ( T m ( R m G ) , L N , q m ) ,

where v ( G ) < δ , dim L N N .

Maiorov in [6] proved the following result.

Theorem A [6]

For m > 2 N , δ ( 0 , 1 / 2 ] , 1 q 2 , then
d N , δ ( R m , q m , v ) m 1 q 1 2 m + ln ( 1 / δ ) .

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 1 m into a Hilbert space H. Then
d N ( T ) C ( log ( m N + 1 ) N ) 1 / 2 T

for 1 N m , m = 1 , 2 ,  , 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 [2326].

Remark 1
  1. (a)

    Theorem A shows the asymptotic expression of the probabilistic widths of the identity embedding from R m into q m , 1 q 2 .

     
  2. (b)
    Theorem B gives the upper estimate of Gelfand numbers of operators from 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 2 m into m , i.e.,
    d N ( T ) C ( log ( m N + 1 ) N ) 1 / 2 T .
     
  3. (c)

    Motivated by Theorems A and B, in general cases, here we investigate the asymptotic estimate of probabilistic widths for diagonal operators from R m onto q m , 1 q .

     

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

Theorem 1 For m > N , δ ( 0 , 1 / 2 ] , then
d N , δ ( T m : R m m , v ) C T m ( 1 + ( 1 / N ) ln ( 1 / δ ) ) ln ( e m / N ) .
Theorem 2 For m > 2 N , δ ( 0 , 1 / 2 ] , then
d N , δ ( T m : R m 1 m , v ) C T m m + ln ( 1 / δ ) .

2 Proof of main results

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

Lemma 1 Let δ ( 0 , 2 / e π ] , and let T m be a bounded linear invertible operator from R m onto m . Then, for any vector z R m ,
v ( x : | ( T m x , z ) | 2 T m ln ( 1 / δ ) z 2 ) δ .
Proof First, assume that T m is a diagonal operator of R m , i.e., T m x = ( λ i x i ) i = 1 m , for any x R m . Without loss of generality, assume that the sequence of the absolute of eigenvalues  λ i , i = 1 , , m , is arranged non-increasingly, i.e., | λ 1 | | λ 2 | | λ m | > 0 . It is known that
T m = max 1 i m | λ i | = | λ 1 | .
Since v is invariant with respect to orthogonal transformation of R m , it suffices to prove the lemma for the vector z = ( z 2 , 0 , , 0 ) . Let ε ( 0 , e 1 ] be arbitrary. We have
v ( x : | ( T m x , z ) | T m ln ( 1 / ε ) z 2 ) = v ( x : | λ 1 x 1 | z 2 | λ 1 | ln ( 1 / ε ) z 2 ) = v ( x : | x 1 | ln ( 1 / ε ) ) = 2 π ( 1 / 2 ) ln ( 1 / ε ) exp ( t 2 ) d t ( 2 ε π ) 1 / 2 .
(3)
Here we use the inequality
u exp ( t 2 ) d t < 1 2 u exp ( u 2 ) , u 1 2 .

From (3) we obtain the assertion of the lemma by δ = 2 ε / π .

Next, assume that T m is a symmetric transformation of 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 R m onto R m , then there are two matrices U and S such that T m = U S , 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 1 , there exists a subspace H of R m of dimension dim H m N such that for any x H ,
x 2 c 0 ( ln ( e m / N ) N ) 1 / 2 x 1 ,
(4)

where c 0 is an absolute constant.

Let G R m be a set. We introduce in R m another norm for the operator T m : R m m :
x G = sup y R m G | ( T m y , x ) | .
Lemma 2 For any δ ( 0 , 1 / 2 ] and an arbitrary operator T m from R m onto m , there exists a subset G = G δ of R m with measure v ( G ) δ such that
sup z B 1 m H z G c 1 T m ( 1 N ln exp ( a N ) δ ln e m N ) 1 / 2 ,

where a and c 1 are absolute constants.

Proof Let k = [ 8 ( N + 1 ) / ln ( e m / N ) ] and consider the k 1 -net
S = { ( s 1 / k , , s m / k ) : s 1 , , s m Z , | s 1 | + + | s m | k }
for the B 1 m in m -norm. Using the inequality ( m ) ( e m ) , we estimate the cardinality of S:
card S = 1 k 2 ( m + ) = 1 k 2 ( 4 e m ) ( 8 e m k ) k e a N ,
(5)

where a is some absolute constant.

Consider the polyhedron Q = B 1 m k 1 B m . Let Q be the set of extremal points of Q. The set Q consists of vectors with k coordinates equal to ± k 1 and the remaining coordinates zero. This implies that Q S , and hence card Q exp ( a N ) .

Let ε = δ / exp ( a N ) . In R m we consider the set G = s S G s , where
G s = { y R m : | ( T m y , s ) | 2 T m ln ( 1 / ε ) s 2 } .
Let z B 1 m H be any point, and s S be a point closest to z in m -norm. Then z = s + t for some t Q . From Lemma 1,
z G s G + t G 2 T m ln ( 1 / ε ) s 2 + t G .
(6)
Using the definition of Q, we have t 2 2 t 1 t k 1 . From this and the definition of Q and G,
t G max t Q t G max t S Q t G 2 T m ln ( 1 / ε ) max t S Q t 2 2 T m k 1 ln ( 1 / ε ) .
Therefore from (6),
z G 2 T m ln ( 1 / ε ) ( z 2 + t 2 ) + t G 2 T m ln ( 1 / ε ) ( z 2 + 2 k 1 / 2 ) .
(7)
Since z H , it follows from the inequalities (4) and (7) that
z G c 1 T m k 1 ln ( 1 / ε ) c 2 T m ( 1 / N ) ln ( 1 / ε ) ln ( e m / N ) .
Using Lemma 1 and the inequality (5), we can estimate the measure of G:
v ( G ) s S v ( G s ) ε card S ε exp ( a N ) = δ .

Thus, Lemma 2 is proved. □

Proof of Theorem 1 Using the duality in R m and Lemma 2, we have
sup x R m G inf y H T m x y = sup x R m G sup z H B 1 m | ( T m x , z ) | = sup z H B 1 m z G c T m ( 1 / N ) ln ( 1 / ε ) ln ( e m / N ) = c T m ( 1 / N ) ln ( exp ( a N ) / δ ) ln ( e m / N ) ,

where H is the orthogonal complement of H and dim H 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 ε > 0 , the ε-cardinality of a subset K of 1 m is defined to be
N ε ( K ) = min { N : z 1 , , z N R m , e ( K , { z 1 , , z N } ) ε } ,
where
e ( K , { z 1 , , z N } ) = sup x K min i = 1 , , N x z i 1

is the deviation of K from the set { z 1 , , z N } in 1 m .

Let λ be a Lebesgue measure in , normalized by the condition λ ( B ) = 1 , where B = B 2 m . We consider Kolmogorov’s ( N , δ ) -width of the ball B with measure λ in the 1 m -norm:
d N , δ d N , δ ( T m : B 1 m , λ ) = inf G inf L e ( T m ( B G ) , L , 1 m ) ,

where the T m is as above, the infima are over all possible subsets G B of measure λ ( G ) δ and all subspaces L R m with dim L N .

Lemma 3 Suppose that D B is an arbitrary subset with measure λ ( D ) δ . Then, for any ε > d N , δ ,
N ε ( T m ( B D ) ) ( 1 + 4 T m ε d N , δ ) m .
Proof Let h > 0 be any number, and let H be any subspace of with dim H N such that
e ( T m ( B D ) , H , 1 m ) h d N , δ .
(8)
Let ε = ε d N , δ . We consider the set Q = 2 ( T m ( B D ) ) H . Let Q ε = { z 1 , , z N } be the maximal subset of Q such that z i z j 1 ε for all i j . Clearly, by maximality Q ε is a ε -net of Q for 1 . The balls z i + ( ε / 2 ) B 1 m are disjoint and all contained in Q + ( ε / 2 ) B 1 m . Therefore, taking volumes we can obtain
i = 1 N vol ( z i + ( ε / 2 ) B 1 m ) vol ( Q + ( ε / 2 ) B 1 m ) .
Hence, we have
N ( ε / 2 ) vol ( B 1 m ) vol ( Q + ( ε / 2 ) B 1 m ) .
(9)
By Q = 2 ( T m ( B D ) ) H 2 T m ( B ) 2 T m ( B 1 m ) 2 T m ( B 1 m ) and (9), we have
N ( ε / 2 ) m vol ( B 1 m ) ( 2 T m + ( ε / 2 ) ) m vol ( B 1 m ) ,
that is,
N ( 1 + 4 T m ε ) m .
(10)
Now, we need to establish e ( T m ( B D ) , { z 1 , , z N } ) ε . Since t Q H , we have from (8)
sup x T m ( B D ) min t Q x t 1 sup x T m ( B D ) min t Q min h 0 H ( x h 0 1 + h 0 t 1 ) sup x T m ( B D ) min t Q x h 0 1 + sup x T m ( B D ) min t Q min h 0 H h 0 t 1 d N , δ + h .
(11)
From the inequality (11), the definition of Q ε and (8), it follows that
e ( T m ( B D ) , { z 1 , , z N } ) = sup x T m ( B D ) min i = 1 , , N x z i 1 sup x T m ( B D ) min i = 1 , , N min t Q ( x t 1 + t z i 1 ) sup x T m ( B D ) min t Q ( x t 1 + min i = 1 , , N t z i 1 ) = sup x T m ( B D ) min t Q x t 1 + ε d N , δ + h + ε = ε + h .

Consequently, letting h 0 , we get that e ( T m ( B D ) , { z 1 , , z N } ) ε , which together with (10) completes the proof of Lemma 3. □

From the relation (see [28])
vol ( B p m ) = [ 2 Γ ( 1 / p + 1 ) ] m / Γ ( m / p + 1 ) , 1 p ,
the balls B p m satisfy the inequalities
( c p m ) m / p < vol ( B p m ) < ( c p m ) m / p ,
(12)

where Γ is the Euler Γ-function, and c p , c p depend only on p.

To estimate N ε ( T m ( B D ) ) from below, we now need another auxiliary result.

Lemma 4 If T m is a diagonal operator from R m onto 1 m , then
| det ( T m ) | = | i = 1 m λ i ( T m ) | ( T m m ) m ,

where λ i ( T m ) , i = 1 , , m , are non-zero eigenvalues of the operator T m rearranged as usual so that | λ i ( T m ) | is non-increasing and each eigenvalue is repeated according to its multiplicity.

Proof It is known that
T m = ( i = 1 m ( λ i ( T m ) ) 2 ) 1 / 2 .
Accordingly,
m | λ m ( T m ) | T m m | λ 1 ( T m ) | .
Obviously,
| λ m ( T m ) | m | i = 1 m λ i ( T m ) | | λ m ( T 1 ) | m ,

from which the result of Lemma 4 follows immediately. □

Lemma 5 If δ [ 0 , 1 ] and λ ( D ) δ , then
N ε ( T m ( B D ) ) 1 3 ( 1 δ ) ( c 0 T m / ε ) m .
Proof We first establish the inequality
N ε ( T m ( B D ) ) 1 3 λ ( T m ( B D ) ) λ ( B 1 m ( 2 ε ) ) .
(13)
Indeed, suppose that (13) does not hold. Then, for N = N ε ( T m ( B D ) ) and some set of points z 1 , , z N , by
ε sup x T m ( B D ) min i = 1 , , N x z i 1 1 λ ( T m ( B D ) ) T m ( B D ) min i x z i 1 λ ( d x ) 1 λ ( T m ( B D ) ) T m ( B D ) i = 1 N ( z i + B 1 m ( 2 ε ) ) min i x z i 1 λ ( d x ) 1 λ ( T m ( B D ) ) ( 2 ε ) λ ( T m ( B D ) i = 1 N ( z i + B 1 m ( 2 ε ) ) ) = 2 ε λ ( T m ( B D ) ) N λ ( B 1 m ( 2 ε ) ) λ ( T m ( B D ) ) = 2 ε ( 1 N λ ( B 1 m ( 2 ε ) ) λ ( T m ( B D ) ) ) 2 ε 2 3 = 4 3 ε ,

we have obtained a contradiction.

In the sequel, we may as well assume that T m is a diagonal operator from R m onto 1 m . Using the inequality (13), (12) and Lemma 4, we have
N ε ( T m ( B D ) ) 1 3 λ ( T m ( B D ) ) λ ( B 1 m ( 2 ε ) ) = 1 3 | det ( T m ) | λ ( B D ) λ ( B 1 m ( 2 ε ) ) 1 3 c ( T m m ) m ( 1 δ ) vol ( B ) vol ( B 1 m ( 2 ε ) ) 1 3 ( c 1 T m m ) m ( 1 δ ) ( c 2 m 2 ε ) m = 1 3 ( 1 δ ) ( c 0 T m ε ) m .

Next, assume that T m is a symmetric transformation of 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 R m onto 1 m , then there are two matrices U and S such that T m = U S , 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 δ [ 0 , 1 / 2 ] , then
d N , δ ( T m : B 1 m , λ ) c 1 T m .
Proof From Lemma 3 and Lemma 5, we get
c ( 1 + 4 T m ε d N , δ ) m N ε ( T m ( B D ) ) 1 3 ( 1 δ ) ( c 0 T m ε ) m 1 6 ( c 0 T m ε ) m .
(14)
Let ε = 5 d N , δ . Taking the logarithm of the inequality (14), we get the inequality
d N , δ ( T m : B 1 m , λ ) c 1 T m

for some constants c 1 and c 2 and N with m c 2 N . Lemma 6 is proved. □

Proof of Theorem 2 According to Lemma 6, for any δ [ 0 , 1 / 2 ] and any subspace L R m with dim L N , there is a set K B with Lebesgue measure λ ( K ) > δ such that
e ( T m x , L , 1 m ) c 1 T m
(15)

for any element x K . On the unit sphere S m 1 = { x R m : x 2 = 1 } , we consider the subset K = { x / x 2 : x K } .

Let λ S m 1 be a Lebesgue measure on the sphere S m 1 . We prove that
λ S m 1 ( K ) > δ λ S m 1 ( S m 1 ) .
(16)
Indeed, assume that
λ S m 1 ( K ) δ λ S m 1 ( S m 1 ) .
We introduce in R m a polar system of coordinates ( r , s ) , where r 0 and s S m 1 , and consider in B the cone
C = { ( r , s ) : 0 r 1 , s K } .
Then
λ ( K ) λ ( C ) = 1 vol ( B ) 0 1 r m 1 d r K λ S m 1 ( d s ) δ vol ( B ) 0 1 r m 1 d r S m 1 λ S m 1 ( d s ) = δ .

We have obtained a contradiction.

Consider the set K t = { ( r , s ) : r t , s K } , t 0 . Using the inequality (16), we estimate the Gaussian measure of K t :
v ( K t ) = ( 2 π ) m / 2 K t exp ( 1 2 u 2 2 ) ( d u ) = ( 2 π ) m / 2 t r m 1 exp ( r 2 2 ) d r K λ S m 1 ( d s ) ( 2 π ) m / 2 δ t r m 1 exp ( r 2 2 ) d r S m 1 λ S m 1 ( d s ) = δ v ( x : x 2 t ) .
(17)
A direct computation shows that for t m ,
v ( x 2 t ) exp ( t 2 ) and v ( x 2 m ) c 0 > 0 ,
where c 0 is some absolute constant. It follows from this and (17) that for t 0 = max { m , ln ( 1 / δ ) } and for any δ ( 0 , c 0 ] ,
v ( K t 0 ) δ v ( x 2 > max { m , ln ( 1 / δ ) } ) > δ 2 .
(18)
For any element y = r s K t 0 , we have from (15)
e ( T m y , L , 1 m ) = r e ( T m s , L , 1 m ) c 1 r T m c 2 T m max { m , ln ( 1 / δ ) } c 2 2 T m m + ln ( 1 / δ ) .
(19)
Since is an arbitrary subspace with dim L N , it follows from (18) and (19) that
d N , δ 2 ( T m : R m 1 m , v ) c 2 2 T m m + ln ( 1 / δ ) .

Theorem 2 is a direct consequence of this. □

Declarations

Acknowledgements

The authors thank the editor and the referees for their valuable suggestions to improve the quality of this paper. The present investigations was supported by the Natural Science Foundation of Inner Mongolia Province of China under Grant 2011MS0103.

Authors’ Affiliations

(1)
Hulunbeier Vocational and Technological College, Hulunbeier, Inner Mongolia, China
(2)
School of Mathematical Sciences, Hulunbeier University, Hulunbeier, Inner Mongolia, China

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© Zhou and Li; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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