Sharp sufficient condition for the convergence of greedy expansions with errors in coefficient computation

Artur R. Valiullin; Albert R. Valiullin; Alexei P. Solodov

doi:10.1515/dema-2022-0019

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Sharp sufficient condition for the convergence of greedy expansions with errors in coefficient computation

Artur R. Valiullin , Albert R. Valiullin and Alexei P. Solodov

Published/Copyright: July 13, 2022

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From the journal Demonstratio Mathematica Volume 55 Issue 1

Abstract

Generalized approximate weak greedy algorithms (gAWGAs) were introduced by Galatenko and Livshits as a generalization of approximate weak greedy algorithms, which, in turn, generalize weak greedy algorithm and thus pure greedy algorithm. We consider a narrower case of gAWGA in which only a sequence of absolute errors { ξ n } n = 1 ∞ is nonzero. In this case sufficient condition for a convergence of a gAWGA expansion to an expanded element obtained by Galatenko and Livshits can be written as ∑ n = 1 ∞ ξ n 2 < ∞ . In the present article, we relax this condition and show that the convergence is guaranteed for ξ n = o 1 n . This result is sharp because the convergence may fail to hold for ξ n ≍ 1 n .

Keywords: greedy expansion; prescribed coefficients; Hilbert space; greedy approximation; convergence

MSC 2010: 41-xx; 41A58

1 Introduction

In this article, we consider generalized approximate weak greedy algorithms (gAWGAs), which were introduced by Galatenko and Livshits [1]. Let us recall the definition of gAWGA.

Definition 1.1

Let H be a Hilbert space over R , D be a symmetric unit-normed dictionary in H (i.e., span D ¯ = H , all elements in D have a unit norm, and if g ∈ D , then − g ∈ D ). In addition, let { t n } n = 1 ∞ ⊂ ( 0 , 1 ] , { q n } n = 1 ∞ ⊂ [ 0 , ∞ ) be weakness sequences and { ( ε n , ξ n ) } n = 1 ∞ ⊂ R 2 be an error sequence. For an expanded element f ∈ H , coefficients { c n } n = 1 ∞ ⊂ R , remainders { r n } n = 0 ∞ ⊂ H and expanding elements { e n } n = 1 ∞ ⊂ D are defined as follows.

Initially, r 0 is set to f . Next, if r n − 1 ∈ H ( n ∈ N ) has already been defined, then an (arbitrary) element satisfying ( r n − 1 , e n ) ⩾ t n sup e ∈ D ( r n − 1 , e ) − q n is selected as e n . We set c n = ( r n − 1 , e n ) ( 1 + ε n ) + ξ n and define r n = r n − 1 − c n e n .

The process described earlier is called a gAWGA. The series ∑ n = 1 ∞ c n e n is called a gAWGA expansion of f in the dictionary D with the weakening sequences { t n } n = 1 ∞ , { q n } n = 1 ∞ and the error sequence { ( ε n , ξ n ) } n = 1 ∞ .

It immediately follows from the definition of gAWGA that

r N = f − ∑ n = 1 N c n e n ( N ∈ N ) ,

and hence, the convergence of the expansion to an expanded element is equivalent to that of the remainders r n to zero.

As a selection of an expanding element e n is potentially not unique, there may exist different realizations of gAWGA expansion for a given expanded element f and a given dictionary D . Furthermore, if t n = 1 and q n = 0 for at least one n ∈ N , gAWGA expansion may turn out to be nonrealizable due to the absence of an element e ∈ D which provides sup e ∈ D ( r n − 1 , e ) .

If q n = ξ n = 0 for every n ∈ N , then gAWGA coincides with the approximate weak greedy algorithm (AWGA) proposed by Gribonval and Nielsen [2]. If q n = ξ n = ε n = 0 for every n ∈ N , then gAWGA coincides with the weak greedy algorithm (WGA), which was introduced by Temlyakov in [3]. If q n = ξ n = ε n = 0 and t n = 1 for every n ∈ N , then gAWGA coincides with the pure greedy algorithms [3], also known as “projection pursuit regression” or “matching pursuit” [4,5].

The error sequence { ( ε n , ξ n ) } n = 1 ∞ can be separated into two sequences, i.e., into a relative error sequence { ε n } n = 1 ∞ and an absolute error sequence { ξ n } n = 1 ∞ . As each computational error can be quantified by its absolute value, here we consider the error sequences { ( 0 , ξ n ) } n = 1 ∞ . Furthermore, we assume that all t n and q n have their default values – 1 and 0, respectively.

Let us note that if c n > 0 for all n , then we can interpret this realization of a greedy expansion as a special case of a greedy algorithm with prescribed coefficients (GAPCs), which was initially introduced by Temlyakov [6,7]. Indeed, we can take the sequence of coefficients { c n } n = 1 ∞ from the realization of gAWGA, and consider GAPC for the same expanded element with this sequence of coefficients (treated as “predefined”). GAPC in this case has a realization with the expanding elements coinciding with element selected in gAWGA, and these realizations of gAWGA and GAPC are identical. Thus, if { c n } n = 1 ∞ in gAWGA satisfy conditions sufficient for the convergence of GAPC, it guarantees convergence of gAWGA as well.

Galatenko and Livshits found sufficient conditions on weakening sequences and error sequences for a convergence of gAWGA expansion [1, Theorem 2]. For the considered case, these conditions take the form ∑ n = 1 ∞ ξ n 2 < ∞ . In the same article, they also showed [1, Theorem 3] that if ξ n ≍ 1 n , then the convergence can be violated. Thus, there remained a gray zone between l 2 and 1 n .

Similar results for GAPC were obtained in [8]. More precisely, it was shown that if the sequence of coefficients satisfies conditions ∑ n = 1 ∞ c n = ∞ and ∑ n = 1 ∞ c n 2 < ∞ , then GAPC expansion converges to an expanded element, but for c n ≍ 1 n , the convergence may fail to hold. However, for GAPC, the specified gray zone was eliminated. Specifically, in [9, Theorem 2.2], the authors showed that the convergence is guaranteed for { c n } n = 1 ∞ satisfying conditions ∑ n = 1 ∞ c n = ∞ and c n = o 1 n .

2 Main result

In this article, we present a result for gAWGA, similar to the one proved in [9] for GAPC, which removes the gray zone between l 2 and 1 n for gAWGA. This result can be stated as follows.

Theorem 1

Let H be a Hilbert space, D be a symmetric unit-normed dictionary in H , weakening sequences { t n } n = 1 ∞ and { q n } n = 1 ∞ be identically equal to 1 and 0, respectively, and a relative error sequence { ε n } n = 1 ∞ be identically equal to 0. Let an absolute error sequence { ξ n } n = 1 ∞ satisfy the condition ξ n = o 1 n . Then, for every element f ∈ H , its gAWGA-expansion in dictionary D with the weakening sequences { t n } n = 1 ∞ , { q n } n = 1 ∞ and the error sequence { ( ε n , ξ n ) } n = 1 ∞ converges to f .

The proof of Theorem 1 uses certain methods and technique that were used in the proof of the similar result for GAPC [9, Theorem 2.2]. However, a straightforward adaptation of the proof of [9, Theorem 2.2] is insufficient for proving Theorem 1: to obtain the result for gAWGA, we introduce new ideas in steps 2 and, especially, 4 and 5, as well as new Lemmas (specifically, Lemmas 3, 5, 6, and 9).

3 Proof of Theorem 1

In the following text, we write “gAWGA expansion” as a short form of “gAWGA expansion in dictionary D with the weakening sequences { t n } n = 1 ∞ , { q n } n = 1 ∞ and the error sequence { ( ε n , ξ n ) } n = 1 ∞ .”

We note that for proving Theorem 1, it is sufficient to show that

lim n → ∞ ‖ r n ‖ = 0 .

We split the proof into nine steps. Steps 1–5 are the preparation for the main part of the proof. Step 6 is the proof for one simple case, and steps 7–9 constitute the proof for the more difficult case.

1. We begin the proof by showing that there exists the limit

lim n → ∞ ‖ r n ‖ < ∞ .

We split the proof of this fact into two lemmas.

Lemma 1

Let α n = arccos ( r n − 1 , e n ) ‖ r n − 1 ‖ (i.e., α n ∈ [ 0 , π ] is the angle between r n − 1 and e n ), and let h n = ‖ r n − 1 ‖ sin α n . Then { h n } n = 1 ∞ is a nonincreasing sequence.

Proof

Let β n = min { ( r n , e n ) ^ , ( r n , − e n ) ^ } , where ( a , b ) ^ denotes the angle between vectors a and b . Then h n = ‖ r n ‖ sin β n . Since the expansion is greedy with t n ≡ 1 and q n ≡ 0 , we have α n + 1 ⩽ β n , and so

h n + 1 = ‖ r n ‖ sin α n + 1 ⩽ ‖ r n ‖ sin β n = h n . □

Lemma 2

The limit

lim n → ∞ ‖ r n ‖ < ∞

exists.

Proof

Similar to Lemma 1, we set h n = ‖ r n − 1 ‖ sin α n . In view of Lemma 1, the sequence { h n } n = 1 ∞ is nonincreasing. Hence, there exists a limit lim n → ∞ h n = h .

For gAWGA, we have

(1) ‖ r n ‖ 2 = h n 2 + ξ n 2 .

Hence, ‖ r n ‖ ⟶ h as n ⟶ ∞ .□

2. We prove Theorem 1 by contradiction. Let us assume that lim n → ∞ ‖ r n ‖ > 0 . If ‖ r k ‖ = 0 for some k > 0 , and then it is obvious that the expansion converges to the expanded element. Otherwise, there exists a number r > 0 , such that for every k ∈ N

(2) ‖ r k ‖ ⩾ r .

Let us note that in the considered case of gAWGA expansion, we can always assume that c n ⩾ 0 . Indeed, assume that an expanding element e n + 1 ∈ D was selected at step n + 1 . If c n + 1 < 0 , then the same element e n + 1 will be selected as e n + j until ( r n , e n + 1 ) ⩾ ( r n + j − 1 , e n + 1 ) for some j > 1 . It follows from the fact that if r n + j − 1 = r n + α e n + 1 with α > 0 for j > 1 , then

( r n + j − 1 , e n + 1 ) = ( r n , e n + 1 ) + α > ( r n + α e n + 1 , e ) = ( r n + j − 1 , e )

for every e ∈ D ⧹ { e n + 1 } .

But if the sequence { ξ n } n = 1 ∞ satisfies the condition ξ n = o 1 n , then it is obvious that every subsequence { ξ n k } k = 1 ∞ satisfies the condition ξ n k = o 1 k . Therefore, we can combine several consecutive steps (namely, consecutive steps with identical selection of an expanding element) into one and consider the sequence of errors { ξ n k } k = 1 ∞ instead of { ξ n } n = 1 ∞ . As a result, we ensure that all c n ⩾ 0 . If c n = 0 for some n , then at step n , the remainder stays unchanged, and we can simply exclude such steps from consideration. Therefore, for the same reason, we can assume that c n > 0 for every n ∈ N .

Note that due to Lemma 2, the assertion of the theorem, i.e., the convergence of r n to zero, follows from the convergence of a subsequence r n k to zero.

We need the following lemma.

Lemma 3

If ‖ r k ‖ ⩾ r > 0 for all k > 0 , then

∑ n = 1 ∞ c n = ∞ .

Proof

We prove this lemma by contradiction. Assume on the contrary that

∑ n = 1 ∞ c n < ∞ .

Then the series ∑ n = 1 ∞ c n e n converges. Therefore, since

r n = f − ∑ k = 1 n c k e k ,

there exists the limit r n ⟶ a ≠ 0 . As span D ¯ = H and since D is symmetric, there exists d ∈ D such that ( a , d ) = b > 0 . Consequently, there exists a number n ∈ N such that b 2 < ( r n , d ) ⩽ ( r n , e n + 1 ) , ∣ ξ n + 1 ∣ < b 4 , and c n + 1 < b 4 simultaneously. Therefore, for this n , we have

b 4 > c n + 1 = ( r n , e n + 1 ) + ξ n + 1 > b 2 − b 4 = b 4 .

This contradiction completes the proof of Lemma 3.□

3. For every nonzero element f ∈ H , we set

F f ( g ) ≔ ( f , g ) ‖ f ‖ , r D ( f ) ≔ sup g ∈ D F f ( g ) = sup g ∈ D ( f , g ) ‖ f ‖ .

Let S k be the k th partial sum of the sequence { c n } n = 1 ∞ , i.e., S k = ∑ j = 1 k c j .

We need the following lemma.

Lemma 4

If ‖ r k ‖ ⩾ r > 0 for all k > 0 , then

liminf n → ∞ S n r D ( r n − 1 ) > 0 .

Proof

As mentioned earlier, assume on the contrary that

liminf n → ∞ S n r D ( r n − 1 ) = 0 .

The monotonicity of S n implies that

liminf n → ∞ S n r D ( r n ) = 0 .

Therefore, there exists a subsequence { n k } k = 1 ∞ , such that

lim k → ∞ S n k r D ( r n k ) = 0 .

Let us consider a sequence of functionals { F r n k } k = 1 ∞ of norm 1. By the Banach-Alaoglu theorem, the unit sphere is weakly^✱ compact. Hence, there exists a weakly^✱-converging subsequence { F r n k i } i = 1 ∞ . For simplicity, we set F r n k i = F i . As noted earlier, there exists the weak^✱-limit

F ≔ lim i → ∞ F i .

The dictionary D is symmetric, and hence,

F i ( f ) = F i r n k i + ∑ j = 1 n k i c j e j = ‖ r n k i ‖ + ∑ j = 1 n k i c j F i ( e j ) ⩾ r − S n k i r D ( r n k i ) .

Passage to the limit, we have F ( f ) ⩾ r , which implies that F ≠ 0 .

Conversely, for every g from the dictionary D , we have

F ( g ) = lim i → ∞ F i ( g ) ⩽ lim i → ∞ r D ( r n k i ) = 0 , F ( − g ) = lim i → ∞ F i ( − g ) ⩽ lim i → ∞ r D ( r n k i ) = 0 .

Hence, F ( g ) = 0 for all g ∈ D and, since D is complete, we obtain F = 0 . The contradiction completes the proof of the lemma.□

4. We split the set of indices k ∈ N into two parts.

Let M ˜ be the set of indices k such that ∣ ξ k ∣ < ( r k − 1 , e k ) 10 , and let N ˜ = N ⧹ M ˜ .

If k ∈ N ˜ , then

(3) c k = ( r k − 1 , e k ) + ξ k ⩽ 11 ∣ ξ k ∣ .

If k ∈ M ˜ , we have

(4) c k = ( r k − 1 , e k ) + ξ k > 9 ( r k − 1 , e k ) 10 , c k = ( r k − 1 , e k ) + ξ k < 11 ( r k − 1 , e k ) 10 .

We further split M ˜ into the disjoint sets M ˜ 1 , M ˜ 2 , … , which satisfy the following conditions:

Every set consists of sequential indices.
The union of any two consecutive sets does not consist of sequential indices.

In other words, M ˜ j are maximum blocks of consequent elements in M ˜ .

Let l j be the first element of M ˜ j . For every l j , we find the index p j ∈ M ˜ j (if it exists) such that c i + 1 ⩽ c i 2 for every i ∈ { l j , l j + 1 , … , p j − 1 } , but not for i = p j . Let M j ¯ be defined as M ˜ j if such an index does not exist, and as { l j , l j + 1 , … , p j − 1 , p j } otherwise; M j = M ˜ j ⧹ M j ¯ (some of these M j may be empty).

Let N be defined as the union of N ˜ and all M j ¯ , and let M = N ⧹ N . We note that M = ⋃ j M j .

Thus, we have split the set of indices k ∈ N into two sets N and M with the aforementioned properties.

5. Now, consider the sequences

x k = c k , if k ∈ M , 0 , otherwise; y k = c k , if k ∈ N , 0 , otherwise.

Let also S n M and S n N be the n th partial sums of the sequences { x n } n = 1 ∞ and { y n } n = 1 ∞ , respectively.

We note that

(5) ‖ r n ‖ 2 = ( r n , r n ) = ( r n − 1 − c n e n , r n − 1 − c n e n ) = ‖ r n − 1 ‖ 2 − 2 c n ( r n − 1 , e n ) + c n 2 = … = ‖ r 0 ‖ 2 − 2 ∑ k = 1 n c k ( r k − 1 , e k ) + ∑ k = 1 n c k 2 = ‖ r 0 ‖ 2 − 2 ∑ k ⩽ n , k ∈ M c k ( r k − 1 , e k ) + ∑ k ⩽ n , k ∈ M c k 2 − 2 ∑ k ⩽ n , k ∈ N c k ( r k − 1 , e k ) + ∑ k ⩽ n , k ∈ N c k 2 .

We also note that

(6) 2 ∑ k ⩽ n , k ∈ M c k ( r k − 1 , e k ) − ∑ k ⩽ n , k ∈ M c k 2 = 2 ∑ k ⩽ n , k ∈ M ( ( r k − 1 , e k ) + ξ k ) ( r k − 1 , e k ) − ∑ k ⩽ n , k ∈ M ( ( r k − 1 , e k ) + ξ k ) 2 = ∑ k ⩽ n , k ∈ M ( r k − 1 , e k ) 2 − ∑ k ⩽ n , k ∈ M ξ k 2 .

If k ∈ M , then by (4) and since M ⊂ M ˜ , we have

( r k − 1 , e k ) 2 − ξ k 2 > ( r k − 1 , e k ) 2 − ( r k − 1 , e k ) 2 100 = 99 ( r k − 1 , e k ) 2 100 > 99 100 ⋅ 10 c k 11 ( r k − 1 , e k ) > c k ( r k − 1 , e k ) 2 .

Combining this estimate with (5) and (6), we find that

(7) ‖ r n ‖ 2 = ‖ r 0 ‖ 2 − ∑ k ⩽ n , k ∈ M ( r k − 1 , e k ) 2 + ∑ k ⩽ n , k ∈ M ξ k 2 − 2 ∑ k ⩽ n , k ∈ N c k ( r k − 1 , e k ) + ∑ k ⩽ n , k ∈ N c k 2 ⩽ ‖ r 0 ‖ 2 − 1 2 ∑ k ⩽ n , k ∈ M c k ( r k − 1 , e k ) − 2 ∑ k ⩽ n , k ∈ N c k ( r k − 1 , e k ) + ∑ k ⩽ n , k ∈ N c k 2 .

Rewriting inequality (7) in terms of the sequences { x k } k = 1 ∞ and { y k } k = 1 ∞ , we have

(8) ‖ r n ‖ 2 ⩽ ‖ r 0 ‖ 2 − 1 2 ∑ k = 1 n x k ( r k − 1 , e k ) − 2 ∑ k = 1 n y k ( r k − 1 , e k ) + ∑ k = 1 n y k 2 .

Now, using the inequality

( r k − 1 , e k ) 2 − ξ k 2 ⩽ ( r k − 1 , e k ) 2 ⩽ 10 9 2 c k 2 < 2 c k 2 ,

which holds for k ∈ M , we additionally obtain the complementary estimate:

(9) ‖ r n ‖ 2 ⩾ ‖ r 0 ‖ 2 − 2 ∑ k = 1 n x k 2 − 2 ∑ k = 1 n y k ( r k − 1 , e k ) + ∑ k = 1 n y k 2 .

6. There are two possible cases for the sequence { y k } k = 1 ∞ : the series ∑ k = 1 ∞ y k 2 either converges or diverges.

In the first case, using (4) and passing to the limit in (8), we obtain that

∑ k = 1 ∞ x k 2 < ∞ .

It implies that

∑ k = 1 ∞ ( x k + y k ) 2 = ∑ k = 1 ∞ c k 2 < ∞ .

At the same time, taking into account Lemma 3, we have ∑ k = 1 ∞ c k = ∞ .

But due to [8, Theorem 2], which gives sufficient condition for the convergence for GAPC expansion, we have r n ⟶ 0 . This is a contradiction to (2). Therefore, the first case ∑ k = 1 ∞ y k 2 < ∞ is not possible.

7. It remains to show that the case ∑ k = 1 ∞ y k 2 = ∞ is also impossible.

We first prove two lemmas, which describe the properties of the sequences { x k } k = 1 ∞ and { y k } k = 1 ∞ that follow from the convergence of ∑ k = 1 ∞ y k 2 .

Lemma 5

Under the aforementioned conditions,

(10) y n = o 1 n .

Proof

From the construction of the set N , it follows that

N = ⋃ j M j ¯ ∪ N ˜ .

Let n ∈ M j ¯ for some index j . Similar to the fourth step of the proof, let l j be the first element of the set M ˜ j . Then l j − 1 ∈ N ˜ , and so, by (3) and (4), we have:

y l j = c l j < 11 10 ( r l j − 1 , e l j ) < 2 ( r l j − 2 − c l j − 1 e l j − 1 , e l j ) ⩽ 2 ( r l j − 2 , e l j − 1 ) + 2 c l j − 1 ⩽ 20 ∣ ξ l j − 1 ∣ + 22 ∣ ξ l j − 1 ∣ = 42 ∣ ξ l j − 1 ∣

(if j = 1 and l j = 1 , then we omit this estimate).

It follows from the selection of the index p j that for i ∈ { l j + 1 , l j + 2 , … , p j } (or for all i ∈ M ˜ j ⧹ l j if such an index p j does not exist for this j ), then y i ⩽ y i − 1 2 holds. Thus, the sequence { y i } i ∈ M ˜ j ⧹ M is decreasing at least as fast as the geometric progression with the common ratio 1 2 .

If n ∈ N ˜ , then by (3), we have:

y n = c n = ( r n − 1 , e n ) + ξ n ⩽ 11 ∣ ξ n ∣ .

Now Lemma 5 follows from the aforementioned estimates and the fact that ξ n = o 1 n .□

Lemma 6

Under the aforementioned conditions,

(11) ∑ n = 1 ∞ x n 2 < ∞ .

Proof

Let us consider an arbitrary nonempty set M j . The set M j is finite, since otherwise ∑ n = 1 ∞ y n < ∞ , which contradicts the condition of the considered case. Combining (1) and (4) for n ∈ M j , we obtain that

(12) h n 2 + ξ n 2 = ‖ r n ‖ 2 = ( r n , r n ) = ( r n − 1 − c n e n , r n − 1 − c n e n ) = ‖ r n − 1 ‖ 2 − 2 c n ( r n − 1 , e n ) + c n 2 = h n − 1 2 + ξ n − 1 2 − 2 c n ( r n − 1 , e n ) + c n 2 ⩽ h n − 1 2 + ξ n − 1 2 − 20 11 c n 2 + c n 2 = h n − 1 2 + ξ n − 1 2 − 9 11 c n 2 .

Let q j and s j = p j + 1 be the last and the first elements of the set M j , respectively. Summing (12) over all n ∈ M j , we obtain

(13) h q j 2 + ξ q j 2 ⩽ h s j − 1 2 + ξ s j − 1 2 − 9 11 ∑ i = s j q j c i 2 .

Note that s j − 1 = p j ∈ M ˜ j , and so, c s j − 1 < 2 c s j due to the selection of the index p j . Therefore,

(14) ∣ ξ s j − 1 ∣ ⩽ ( r s j − 2 , e s j − 1 ) 10 < c s j − 1 9 < 2 c s j 9 .

From (13) and (14), we have:

9 11 ∑ i = s j q j c i 2 ⩽ h s j − 1 2 + ξ s j − 1 2 − h q j 2 − ξ q j 2 < h s j − 1 2 − h q j 2 + 4 c s j 2 81 .

As a result,

(15) 1 2 ∑ i = s j q j c i 2 < h q j 2 − h s j − 1 2 .

Summing (15) over all j (with nonempty M j ) and using the fact that { h n } n = 1 ∞ is monotone (Lemma 1), we find that

∑ j ∈ M c j 2 = ∑ n = 1 ∞ x n 2 < ∞ ,

which completes the proof of Lemma 6.□

8. In this step, we need to prove two more auxiliary lemmas.

Lemma 7

Under the conditions of this case,

(16) liminf n → ∞ S n N r D ( r n − 1 ) = 0 .

Proof

By Lemma 5, we have ∑ n = 1 ∞ y n = ∞ and y n = o 1 n .

In view of these properties, the proof of the lemma can be carried out so as in [9, Lemma 1]. For the sake of completeness, we provide all the details below.

Assume on the contrary that there exists a number c > 0 and p ∈ N such that

r D ( r k − 1 ) S k N ⩾ c ( k ⩾ p ) .

Without the loss of generality, let us assume that p = 1 (we can achieve that by shifting the sequence of the remainders; if S k N ≡ 0 , then the conclusion of the lemma is obvious).

From (8), we obtain

(17) ‖ r n ‖ 2 ⩽ ‖ r 0 ‖ 2 − 1 2 ∑ k = 1 n x k ( r k − 1 , e k ) − 2 ∑ k = 1 n y k ( r k − 1 , e k ) + ∑ k = 1 n y k 2 ⩽ ‖ r n ‖ 2 − 2 ∑ k = 1 n y k S k N ‖ r k − 1 ‖ r D ( r k − 1 ) S k N + ∑ k = 1 n y k 2 ⩽ ‖ r n ‖ 2 − 2 c r ∑ k = 1 n y k S k N + ∑ k = 1 n y k 2 .

It is known (see Abel-Dini theorem [10]) that if ∑ k = 1 ∞ y k = ∞ , then ∑ k = 1 ∞ y k S k N = ∞ . Also, in our case, we have ∑ k = 1 ∞ y k 2 = ∞ .

Next, since there exists a number a > 0 such that y k < a k for all k ∈ N , there exists a number A > 0 such that S k N < A k for all k . As y k = o 1 k , there exists a function f ( k ) (here, we assume that f ( k ) might be equal to ∞ ) such that y k = 1 k f ( k ) and f ( k ) → ∞ as k → ∞ . Therefore,

(18) ∑ k = 1 n y k S k N = ∑ k = 1 n 1 k f ( k ) S k N ⩾ ∑ k = 1 n 1 A k f ( k ) .

We note that in the considered case

∑ k = 1 ∞ y k 2 = ∑ k = 1 ∞ 1 k f 2 ( k ) = ∞ ,

and so

(19) ∑ k = 1 ∞ 1 k f ( k ) = ∞ .

Combining (17)–(19), we obtain the estimate

0 < ‖ r n ‖ 2 ⩽ ‖ r 0 ‖ 2 − 2 c r ∑ k = 1 n 1 A k f ( k ) + ∑ k = 1 n 1 k f 2 ( k ) = ‖ r 0 ‖ 2 − ∑ k = 1 n 1 k f ( k ) 2 c r A − 1 f ( k ) ⟶ − ∞ ,

as n ⟶ ∞ . This contradiction completes the proof of Lemma 7.□

Lemma 8

In the considered case,

(20) liminf n → ∞ S n M r D ( r n − 1 ) = 0 .

Proof

We argue by contradiction. Assume, on the contrary, that there exists a number β > 0 such that S n M r D ( r n − 1 ) > β for every n ∈ N (similarly to the previous lemma).

Let us note that if ∑ k = 1 ∞ x k < ∞ , then the inequality S n N > S n M holds for all sufficiently large indices. Therefore, using Lemma 7, we obtain

liminf n → ∞ S n r D ( r n − 1 ) = 0 ,

which contradicts the assertion of Lemma 4. Thus,

(21) ∑ k = 1 ∞ x k = ∞ .

Now, we note that from Lemma 2 and inequalities (9) and (11), we have

(22) ∑ k = 1 ∞ ( 2 y k ( r k − 1 , e k ) − y k 2 ) < ∞ .

Now by using (2) and applying the Abel-Dini theorem from [10] to the sequence { x k } k = 1 ∞ , we find that

∑ k = 1 n x k ( r k − 1 , e k ) = ∑ k = 1 n x k S k M S k M r D ( r k − 1 ) ‖ r k − 1 ‖ > r β ∑ k = 1 n x k S k M ⟶ ∞ , n ⟶ ∞ ,

which together with (8) contradicts (22) . This contradiction proves Lemma 8.□

9. In this step, we finalize the proof of Theorem 1 by proving the following lemma.

Lemma 9

In the considered case,

(23) liminf n → ∞ S n r D ( r n − 1 ) = 0 .

Proof

Assume on the contrary, that there exists a number α > 0 such that for all n ∈ N , the inequality

(24) S n r D ( r n − 1 ) ⩾ α .

In view of (2) and (20), there exists a number l 1 such that

S l 1 M ( r l 1 − 1 , e l 1 ) < α 4 .

By (2) and (16), there exists a number k 2 > l 1 such that

S k 2 N ( r k 2 − 1 , e k 2 ) < α 4 .

Let k 1 be the largest number such that k 1 < k 2 and

S k 1 M ( r k 1 − 1 , e k 1 ) < α 4 .

Combining these estimates with (24), we obtain

S k 1 M ( r k 1 − 1 , e k 1 ) < α 4 , S k 2 N ( r k 2 − 1 , e k 2 ) < α 4 , S k 2 M ( r k 2 − 1 , e k 2 ) > 3 α 4 , S k 1 N ( r k 1 − 1 , e k 1 ) > 3 α 4 .

As a result, we have

3 α 4 < S k 2 M ( r k 2 − 1 , e k 2 ) < α 4 S k 2 M S k 2 N , 3 α 4 < S k 1 N ( r k 1 − 1 , e k 1 ) < α 4 S k 1 N S k 1 M ,

and hence,

(25) S k 1 M < 1 3 S k 1 N ⩽ 1 3 S k 2 N < 1 9 S k 2 M .

Lemma 6 implies that in the considered case we have x k → 0 , ( k → ∞ ) , and so by selecting a sufficiently large l 1 , we can guarantee that 2 x k < S k M for every k ∈ M exceeding k 1 .

It is easy to see that, for x ∈ 0 , 1 2 , the following inequality

4 x ⩾ − ln ( 1 − x )

holds.

From (25), we have

(26) ∑ k = k 1 + 1 k 2 x k S k M ⩾ − 1 4 ∑ k = k 1 + 1 k 2 ln 1 − x k S k M = − 1 4 ∑ k = k 1 + 1 k 2 ln S k − 1 M S k M = − 1 4 ln S k 1 M S k 2 M > ln 3 2 .

Using (26) and the fact that for k 1 < k ⩽ k 2 , we have

S k M ( r k − 1 , e k ) ⩾ α 4 ,

and then we obtain that

∑ k = k 1 + 1 k 2 x k ( r k − 1 , e k ) = ∑ k = k 1 + 1 k 2 x k S k M S k M r D ( r k − 1 ) ‖ r k − 1 ‖ ⩾ r α 4 ∑ k = k 1 + 1 k 2 x k S k M > r α ln 3 8 .

Now, we find a number l 2 > k 2 such that

S l 1 M ( r l 1 − 1 , e l 1 ) < α 4 .

Similar to the selection of k 1 and k 2 , we select k 3 and k 4 ( k 2 < k 3 < k 4 ) to satisfy

∑ k = k 3 + 1 k 4 x k ( r k − 1 , e k ) > r α ln 3 8 .

Continuing this procedure, we obtain that ∑ k = 1 ∞ x k ( r k − 1 , e k ) = ∞ , which together with (22) contradicts (8). This competes the proof of Lemma 9.□

The assertions of Lemmas 4 and 9 contradict each other, and this contradiction comes from the assumption that r n ↛ 0 , ( n → ∞ ) . Thus, r n → 0 and the proof of Theorem 1 is complete.

Acknowledgements

The authors thank Dr. Vladimir V. Galatenko and Dr. Alexey R. Alimov for valuable discussions, comments and suggestions.

Funding information: The work of Al. R. Valiullin and A. P. Solodov on Lemmas 16 was supported by the Russian Science Foundation (project no. 21-11-00131) at Lomonosov Moscow State University. The work of Ar. R. Valiullin on Lemmas 79 was supported by the Government of the Russian Federation (grant no. 14.W03.31.0031). This work was also supported by the company Intelligent Solutions.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and approved its submission.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analysed during the current study.

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Received: 2021-10-26

Revised: 2022-04-14

Accepted: 2022-05-10

Published Online: 2022-07-13

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/dema-2022-0019

Keywords for this article

greedy expansion; prescribed coefficients; Hilbert space; greedy approximation; convergence

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