Sharp conditions for the convergence of greedy expansions with prescribed coefficients

1 Introduction

We consider greedy expansions with prescribed coefficients in Hilbert spaces. This type of greedy expansion was initially introduced by V. N. Temlyakov [1,2] (see also [3]) in a more general case of Banach spaces. In case of Hilbert spaces, the definition of greedy expansions with prescribed coefficients (further — just greedy expansions) takes the following form.

It immediately follows from the definition of a greedy expansion that

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

As a selection of an expanding element e _n is potentially not unique, there may exist different realizations of a greedy expansion for a given expanded element f and a given dictionary D. Furthermore, if t _n  = 1 for at least one n ∈ N , greedy 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 ) .

In this paper, we consider only symmetric unit-normed dictionaries and the weakness sequence t n = 1 ( n ∈ N ) .

Earlier we have shown [4, Theorem 2] (see also [1] or [2] for the case of Banach spaces) that if ∑ n = 1 ∞ c n = ∞ and ∑ n = 1 ∞ c n 2 < ∞ , then a greedy expansion of f converges to f (in other words, all realizations of this expansion converge to f).

Also, we have constructed an example [4, Theorem 3] which shows that a convergence can be violated for a coefficient sequence { c n } n = 1 ∞ with c n ≤ 1 n and ∑ n = 1 ∞ c n = ∞ . However, the absence of convergence has been shown only for one of the possible realizations of this expansion.

In this paper, we present improvements for both the negative and the positive results.

2 Main results

An improvement of the negative result can be stated as follows.

An improvement of the positive result can be stated as follows.

3 Proof of Theorem 2.1

Let r ₀ ∈ H be an arbitrary vector with ∥ r 0 ∥ = 1 2 .

Let us consider the inequality

For c n = 1 n the left part of (1) equals

and the right part equals

Hence, as ∥ r 0 ∥ = 1 2 , there exists a number N, that for all n > N the inequality (1) holds.

Now, let us consider the inequality

For c n = 1 n the left part of (2) equals

and the right part equals

Hence, there exists a number M, that for all n > M the inequality (2) holds. Thus, we can reenumerate (shift) a sequence:

and as a result obtain a sequence for which the inequalities (1) and (2) hold for all n ∈ N .

Let α _n denote arccos ( r n − 1 , e n ) ∥ r n − 1 ∥ (i.e., an angle between r _n−1 and e _n ), and h _n denote ∥ r n − 1 ∥ sin α n . Let e ₋₁, e ₀ be vectors (which will be included in the dictionary D), such that r ₀, e ₋₁, e ₀ lie on the same plane, (e ₀, r ₀) =(e ₋₁, r ₀) and the angle between r ₀ and e ₀ is greater than arccos c 1 − c 2 2 ∥ r 0 ∥ (the inequality c n + 1 − c n + 2 2 ∥ r n ∥ < 1 for every n = 0,1, 2,… will be proven further).

We construct the example inductively. First, as e ₁ we take an (arbitrary) element of H which satisfies the following conditions:

1. the angle between e ₁ and r ₀ equals α 1 = arccos c 1 − c 2 2 ∥ r 0 ∥ ;

2. the projection of e ₁ on the plane ⟨e ₋₁, e ₀⟩ is collinear to r ₀.

Let vectors e ₋₁, e ₀, e ₁, …, e _n and r ₀, r ₁, r ₂, …, r _n have already been constructed. Then as e _n+1 we take an (arbitrary) element of H which satisfies the following conditions:

1. the angle between e _n+1 and r _n equals α n + 1 = arccos c n + 1 − c n + 2 2 ∥ r n ∥ ;

2. the projection of e _n+1 on the subspace ⟨e ₋₁, e ₀, …, e _n ⟩ is collinear to r _n ;

We note that in this construction the angle α _n+1 is equal to the angle between e _n+1 and the subspace ⟨e ₋₁, e ₀, …, e _n ⟩.

Let D be defined as { e n } n = − 1 ∞ ∪ { − e n } n = − 1 ∞ . It is sufficient to show that for each n = 0, 1, 2, … the element e _n+1 is the only vector from D that can be selected as an expanding element at the step n, and that ∥r _n ∥ ↛ 0 as n → ∞.

We split the proof of these assertions into the following steps. First, we show that ∥r _n ∥ ↛ 0. Second, we show that at the step (n + 1) the vector  −e _n cannot be selected as an expanding element. Third, we show the same for vectors e _k and  −e _k , k > n + 1. As ( − e n + 1 , r n ) ^ and ( e n , r n ) ^ are obtuse (where ( a , b ) ^ is the angle between vectors a and b), vectors  −e _n+1 and e _n also cannot be selected as expanding elements at the step (n + 1). And finally, we show that it also holds for vectors e _k and −e _k , k < n.

1. Due to the law of cosines

(the last inequality holds due to monotonicity of the sequence { c n } n = 1 ∞ ). Hence, using (3), which directly implies that c 1 2 < 1 9 , we get the inequality

We see that c n + 1 − c n + 2 2 ∥ r n ∥ < c 1 ∥ r n ∥ < 1 3 ⋅ 6 5 < 1 and, additionally, that the sequence { ∥ r n ∥ } n = 0 ∞ is monotonically decreasing. Thus, ∥r _n ∥ ↛ 0, which completes the first part of the proof.

2. Next, we note that α _n+1 < γ _n where γ n = arccos ( r n , − e n ) ∥ r n ∥ (i.e., an angle between r _n and −e _n , see Figure 1), which implies that a vector −e _n is not selected at the step (n + 1) of the algorithm. Indeed, it immediately follows from the inequality

which is a direct corollary of the monotonicity of the { c n } n = 1 ∞ .

Figure 1

Selection of e _n .

3. Here we show that vectors e _n+2, −e _n+2 are not selected at the step (n + 1). As α _n+1 is the angle between the vector e _n+1 and the subspace ⟨e ₋₁, e ₀, …, e _n ⟩, it is sufficient to show that α _n+1 < α _n+2, i.e.,

Using (2), (3) and (4) we can note that the following inequalities hold:

which is equivalent to (5). Similarly, vectors e _k and  −e _k (k > n + 2) are not selected at this step.

4. Now, it remains to prove that if k < n, then the angles ( − e k , r n ) ^ and ( e k , r n ) ^ exceed α _n . We prove it inductively by n. The induction base holds as the angle between e ₀ and r ₀ exceeds α ₁. Let β n k be the min { ( r n , e k ) ^ , ( r n , − e k ) ^ } , and let x _n be the angle between r _n and r _n−1 (Figure 1). Without loss of generality let β n + 1 k = ( r n + 1 , − e k ) ^ . Due to the spherical law of cosines [5, Chapter 12] we get

where γ n k = ( r n , − e k ) ^ . As ∥ r n ∥ 2 > 5 36 , ∥ r n + 1 ∥ 2 > 5 36 and c n 2 < c 1 2 < 1 9 , then ∥r _n ∥ > c _n and ∥r _n+1∥ > c _n . Therefore, the angle x _n is acute. Thus, the sign of cos β n + 1 k coincides with the sign of cos γ n k , therefore, γ n k = β n k .

Due to the induction assumption, the inequality cos α n > cos β n k holds (in the case of k = n − 1 this inequality holds due to the point 2 of the proof). Hence, it is sufficient to show that

Let us prove the inequality (6). The following series of inequalities holds:

Combining (7) and (6), we see that it is sufficient to prove that

which is equivalent to the inequality

We have already shown (see step 1 of the proof) that the sequence { ∥ r n ∥ } n = 0 ∞ is monotonically decreasing. Therefore, it is sufficient to show that

or, equivalently

And due to the monotonicity of the sequence { ∥ r n ∥ } n = 0 ∞ , it is sufficient to prove that

which coincides with (1). It completes the proof of this step and the theorem in general.

4 Proof of Theorem 2.2

For every non-zero element f ∈ H, let F f ( g ) = ( f , g ) ∥ f ∥ and let r _D (f) =  sup g ∈ D F f ( g ) = sup g ∈ D ( f , g ) ∥ f ∥ .

We split the proof of Theorem 2.2 into two parts.

1. First, we show that

Let us assume the contrary, i.e.,

If there exists a number k > 0 such that ∥r _k ∥ = 0, then it is obvious that a greedy expansion converges to an expanded element. Otherwise, there exists a number r > 0, such that for every k ∈ N

Let S _k be the kth partial sum of the sequence { c n } n = 1 ∞ , i.e., S k = ∑ j = 1 k c j . Then the following lemma holds.

Now, let us proceed to the proof of Theorem 2.2.

Lemma 4.1 and monotonicity of S _n imply that

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

Let us consider a sequence of functionals { F r n k } k = 1 ∞ . The norm of each functional equals 1. As a unit sphere is a weak compact (according to the Banach–Alaoglu theorem), there exists a weakly converging subsequence { F r n k i } i = 1 ∞ . For simplicity we denote F r n k i as F _i . As noted above, there exists a weak limit

Due to the fact that the dictionary D is symmetric, for all sufficiently large i the following inequality holds:

Passage to the limit results in an estimate F(f) ≥ r, which implies that F ≠ 0. 

On the other hand, for every g from the dictionary D

Hence, F(g) equals 0 for all g ∈ D and, due to completeness of D, we get that F = 0. We have come to a contradiction, which proved the equality

Thus, we completed the first part of the proof of Theorem 2.2.

2. Now it remains to prove the following two lemmas.

Thus, we have proved both Lemmas 4.2 and 4.3, and therefore, we completed the proof of Theorem 2.2.

5 Generalization

One of the disadvantages of a greedy expansion with the prescribed coefficients in case of t _n  ≡ 1 is that there might be no greedy expansion for an expanded element. Another disadvantage is that (irrespective of t _n ) in a general case selection of expanding element is not constructive.

We consider the following generalization of the greedy algorithm that eliminates these undesirable properties. Let symmetric sets { D n } n = 1 ∞ be an exhaustion of a dictionary D, i.e., D _n  ⊂ D _n+1 for all n ∈ N , and ⋃ n = 1 ∞ D n = D .

At the step n of a greedy algorithm we select an element e _n from D _n such that ( r n − 1 , e n ) = sup e ∈ D n ( r n − 1 , e ) . In other words, at the step n we consider D _n instead of D. If D _n is finite, then e _n always exists and can be constructively selected by a simple exhaustive search. Let us call this modification of a greedy expansion “a greedy expansion with prescribed coefficients in the exhaustion { D n } n = 1 ∞ .”

Let us note that for this modification the analogue of Theorem 2.2 holds. It can be stated in the following way.

6 Conclusion

In this paper, we presented improvements for both negative and positive results on the convergence of greedy expansions with prescribed coefficients. These improvements, in particular, allowed to remove completely a gap between positive and negative results.

We expect that the technique we used in this research is applicable also to greedy expansions with errors in coefficient calculation [7], where a gap between negative and positive results still exists, and we are going to present results for greedy expansions with errors in coefficient calculation in our subsequent publications. We are also going to study a possibility of generalizing positive results to the case of t _n  ≡ t ∈ (0, 1).