# Greedy Expansions with Prescribed Coefficients in Hilbert Spaces

Authors: Artur R. Valiullin; Albert R. Valiullin; Vladimir V. Galatenko
Journal: International Journal of Mathematics and Mathematical Sciences
 (2018)

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## Abstract

Greedy expansions with prescribed coefficients, which have been studied by V. N. Temlyakov in Banach spaces, are considered here in a narrower case of Hilbert spaces. We show that in this case the positive result on the convergence does not require monotonicity of coefficient sequenceC. Furthermore, we show that the condition sufficient for the convergence, namely, the inclusion C∈l2∖l1, can not be relaxed at least in the power scale. At the same time, in finite-dimensional spaces, the condition C∈l2 can be replaced by convergence of C to zero.


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## Body

## 1. Introduction

Expansion in Fourier series [1] is a classical and comprehensively studied tool of theoretical and applied mathematics which takes an expanded function as an input and constructs a sequence of its expansion coefficients. Greedy expansions [2, 3], which are equivalent in the simplest case to Fourier series reordered by decreasing norms of terms and known in statistics and signal processing as Projection Pursuit Regression [4, 5] and Matching Pursuit [6], respectively, perform parallel computation of expansion coefficients and selection of expansion elements from a predefined dictionary. V. N. Temlyakov [3, 7] (see also [8]) proposed a type of a greedy expansion that performs only selection of expansion elements, while coefficients are prescribed in advance. The definition proposed by V. N. Temlyakov for the case of Banach spaces, in the case of Hilbert spaces, takes the following form.Definition 1.
LetH be a Hilbert space over R with a scalar product (·,·), D be a symmetric unit-normed dictionary in H (i.e., spanD¯=H, all elements in D have a unit norm, and if g∈D, then -g also belongs to D). In addition, let f∈H, t∈(0;1], and C=cnn=1∞ be a sequence of positive numbers. We define inductively a sequence of remainders {rn}n=0∞⊂H and a sequence of expanding elements enn=1∞⊂D. First, we set r0=f. Then, if rn-1∈H(n∈1,2,3,…) has already been defined, we select en as an (arbitrary) element which satisfies the condition (1)rn-1,en⩾tsupe∈D⁡rn-1,e,and set rn=rn-1-cnen.
The series∑n=1∞cnen(f) is called a greedy expansion of f in the dictionary D with the prescribed coefficients C and the weakness parameter t.It immediately follows from the definition of a greedy expansion thatrN=f-∑n=1Ncnen(f)(N∈N), and hence the convergence of the expansion to an expanded element is equivalent to the convergence of remainders rn to zero as n→∞.As a selection of an expanding elementen is potentially not unique, there may exist different realizations of a greedy expansion for a given dictionary D, weakness parameter t and sequence of coefficients C. Furthermore, for t=1 greedy expansion may turn out to be nonrealizable due to the absence of an element e∈D which provides supe∈D⁡(rn-1,e).V. N. Temlyakov showed [3, Theorem 2.1] that if a number of conditions hold which are equivalent in the case of Hilbert spaces to the divergence of the series ∑n=1∞cn and the convergence of the series ∑n=1∞cn2, a greedy expansion with prescribed coefficients C=cnn=1∞ converges to an expanded element at least for a subsequence of indexes, i.e., liminfn→∞⁡rn=0. Later V. N. Temlyakov proved the standard convergence (i.e., limn→∞⁡rn=0) under the additional condition of monotonicity of C [7, Theorem 4]. Yet, it remained unknown whether the condition C∈l2 and the monotonicity condition could be essentially relaxed without violating the guaranteed convergence to an expanded element.


## 2. Main Results

We start with a positive result which states that in Hilbert spaces the monotonicity is not required for the standard convergence. Namely, the following theorem holds.Theorem 2.
LetH be a Hilbert space, D be a symmetric unit-normed dictionary in H, t∈(0,1], C=cnn=1∞ be a sequence of positive numbers which satisfies the conditions (2)∑n=1∞cn=∞;∑n=1∞cn2<∞(i.e., C∈l2∖l1). Then for every element f∈H all realizations of its greedy expansion in the dictionary D with the prescribed coefficients C and the weakness parameter t converge to f.It is clear that if the first condition onC is violated, then there is no convergence to an expanded element for all f with the norm exceeding the sum ∑n=1∞cn. The significance of the second condition on C follows from the following theorem.Theorem 3.
There exist a Hilbert spaceH, a symmetric unit-normed dictionary D⊂H, an element f∈H and a sequence of positive numbers C=cnn=1∞ such that (3)∑n=1∞cn=∞,cn⩽1n∀n∈1,2,3.…,but a greedy expansion of f in the dictionary D with the prescribed coefficients C and the weakness parameter t=1 does not converge to f.As for the second condition of Theorem2 a boundary in the power scale is 1/n, Theorem 3 in fact shows that this condition in Theorem 2 can not be relaxed at least in the power scale.However, the question about a possibility of a more delicate relaxation of the conditionC∈l2 remains open. This question can be stated as follows: is it true that for every sequence C of positive numbers that converges to zero but does not belong to l2 there exist a Hilbert space H, a symmetric unit-normed dictionary D⊂H and an expanded element f∈H such that at least one realization of greedy expansion of f in D with the prescribed coefficients C (and, e.g., the weakness parameter t=1) does not converge to f?We note that assertions similar to Theorems2 and 3 have been announced by O. Rassudova in her conference talk [9], but the proofs have not been published. To the best of our knowledge, in her proof of an analogue of Theorem 3 O. Rassudova used a modification of the construction [10, Theorem 3] which is based on analytical estimates and does not have a clear geometric interpretation. The construction presented in our work is geometrically demonstrative.We also note that at least fort=1 for the natural class of monotonic coefficients in case of finite-dimensional Hilbert spaces the condition ∑n=1∞cn2<∞ in Theorem 2 can be replaced by an essentially weaker condition cn→0(n→∞). The proof of this fact is presented in sectionThe case of finite-dimensional spaces.


## 3. Proof of Theorem2

The theorem can be easily derived from the equalityliminfn→∞⁡rn=0, which holds due to the aforementioned result by V. N. Temlyakov [3, Theorem 2.1]. From the definition of the greedy expansion it immediately follows that(4)rm2=rm-1-cmem,rm-1-cmem=rm-12-2cmrm-1,em+cm2.As coefficients cn are positive and the dictionary is symmetric, 2cm(rm-1,em)⩾0. Hence (5)rm2⩽rm-12+cm2,and thus(6)rm+k2⩽rm2+∑j=m+1m+kcj2.The condition C=cnn=1∞∈l2 implies that (7)∀ε>0∃N1>0:∀m>N1∑j=m+1∞cj2<ε2.Due to the equality liminfn→∞⁡rn=0 we have that(8)∀ε>0∀N>0∃m>N:rm2<ε2.From two last assertions we obtain that for every ε>0 there exists m>0 such that the following two conditions simultaneously hold: (9)∑j=m+1∞cj2<ε2,rm2<ε2.Thus using estimate (6) we get that rm+k2<ε for all k∈N. But according to the definition of the limit it directly means that limn→∞⁡rn=0. The proof of Theorem 2 is complete.We note that for monotonic coefficients and the weakness parametert=1 the statement of Theorem 2, which is covered in this case by [7, Theorem 4], can be also derived as a corollary of the same result by V. N. Temlyakov about the convergence for a subsequence of indexes [3, Theorem 2.1] and the following lemma. We find this lemma to be interesting on its own.Lemma 4.
Lett=1 and cn monotonically converge to zero as n→∞. Then for every realization of greedy expansion with the prescribed coefficients C=cnn=1∞ the sequence of norms of its remainders rnn=0∞ converges.We begin our proof of this lemma with an estimate of a possible increase of the remainder norms. LetM denote the set of all indexes n for which rn+1>rn. If the set M is finite, then starting from a certain index the sequence rnn=0∞ is monotonic and thus convergent. Hence it remains to consider the case of countably infinite M.For the sake of brevity we denote the scalar products(rn,en+1) by xn+1(n=0,1,2,…). It follows from the definition of the greedy expansion and the symmetric property of the dictionary that all xn are nonnegative.The Pythagorean theorem implies that for all indexesn the equality(10)rn+12=rn2-xn+12+cn+1-xn+12holds. If n∈M, then cn+1>2xn+1, rn+12⩽rn2+cn+12 and thus(11)xn+2=supe∈D⁡rn+1,e⩾-rn+1,en+1=cn+1-xn+1.Consequently(12)rn+22=rn+12-xn+22+cn+2-xn+22=rn2-xn+12-xn+22+cn+1-xn+12+cn+2-xn+22=rn2+cn+12+cn+22-2cn+1xn+1-2cn+2xn+2⩽rn2+cn+12+cn+22-2cn+1xn+1-2cn+2cn+1-xn+1=rn2+cn+1-cn+22-2xn+1cn+1-cn+2⩽rn2+cn+1-cn+22.At the same time(13)rn+22=rn+12+cn+22-2cn+2xn+2=rn+12+cn+2cn+2-2xn+2⩽rn+12+cn+2cn+2-2cn+1-xn+1=rn+12+cn+2cn+2-cn+1+2xn+1-cn+1⩽rn+12.It means that if the remainder norm increased at the transition from rn to rn+1, then the increase of the norm square does not exceed cn+12, for the next expansion step the increase is impossible, and the joint increase of the square of remainder norm for two steps of the expansion does not exceed (cn+1-cn+2)2 and hence does not exceed C(cn+1-cn+2), where C can be set to 2c1.Having this estimate, let us complete the proof. We note that the series∑n∈M(cn+1-cn+2) converges: it can be easily derived either from the Leibniz’s alternating series test or from the inequality(14)∑n∈M,n⩽Kcn+1-cn+2⩽∑n=0Kcn+1-cn+2=c1-cK+2⩽c1.Let us fix an arbitrary positive ε and find an index N0 such that ∑n∈M,n>N0(cn+1-cn+2)<ε/4C and simultaneously supn>N0⁡cn2<ε/4. Next we find an index N1>N0 such that rN12<r2+ε/4, where r denotes the infimum of the remainder norms rnn>N0. Then for every n>N1 we have that (15)r2⩽rn2⩽rN12+∑n∈M,n⩾N1Ccn+1-cn+2+cn2<r2+ε.Hence rn2→r2 (n→∞) and consequently rn→r. The proof of Lemma 4 is complete.


## 4. Proof of Theorem3

Our proof of Theorem3 includes the following blocks: construction of a dictionary with simultaneous construction of coefficients C=cnn=1∞; description of realization of greedy expansion; proof of the absence of convergence to the expanded element; obtaining the required estimate of cn. As a Hilbert space H we take an arbitrary infinite-dimensional separable space, e.g., l2.Figure1 illustrates certain steps of the proof.Figure 1
An illustration for certain steps of the proof of Theorem3. All vectors except e lie in the vertical plane; e (as well as rn-1) lies in the horizontal plane; e represents any vector from the set e-1,e0,-e1,-e2,…,-en-1. The spherical law of cosines is applied to angles formed by the vectors rn-1, rn, e. The law of sines is applied to the triangle formed by vectors rn-1, rn, -cnen.


### 4.1. Description of the Construction

We first present the structure of the example; i.e., we describe the construction of dictionary elementsen and coefficients cn. As a part of this construction we also define the sequence of vectors (remainders) rn, including the expanded element f=r0.Letf be an arbitrary non-zero element of H with f⩽1/2, r0=f. We define dictionary elements e-1 and e0 as arbitrary (unequal) unit vectors such that e-1, e0 and r0 lie in one plane and the angle α0 between e0 and r0 equals the angle between e-1 and r0 and belongs to the interval (π/4,π/2).Next, we set dictionary elemente1 to an arbitrary unit vector with the following two properties: its orthogonal projection on the plane <e-1,e0> lies on the line with directing vector r0, and the angle between this vector and r0 also equals α0. We also find the coefficient c1 such that for r1=r0-c1e1 scalar products (r1,-e1),(r1,e0),(r1,e-1) are equal or, equivalently, angles between r1 and the vectors e-1, e0, -e1 are equal. We denote this angle by α1. Note that α1∈α0,π/2: the formal justification of this fact can be based, e.g., on the equality cos⁡α1=cos⁡α0cos⁡(π-α0-α1), which directly follows from the spherical law of cosines.Similarly, we sete2 to an arbitrary unit vector with an orthogonal projection on the subspace <e-1,e0,e1> lying on the line with directing vector r1 and the angle between this vector and r1 equal α1, and select the coefficient c2 in such a way that for r2=r1-c2e2 scalar products (r2,-e2), (r2,-e1), (r2,e0), (r2,e-1) are equal or, equivalently, angles between r2 and the vectors e-1, e0, -e1, -e2 are equal. We denote this angle by α2. Again it is easy to see that α2∈α1,π/2, as due to the spherical law of cosines cos⁡α2=cos⁡α1cos⁡(π-α1-α2).We continue the construction inductively. Namely, after constructingrjj=0n-1, ejj=-1n-1, cjj=1n-1 and αjj=0n-1 we set en to an arbitrary unit vector with an orthogonal projection on the subspace <e-1,e0,…,en-1> lying on the line with directing vector rn-1 and angle between this vector and rn-1 equal αn-1, and select the coefficient cn in such a way that for rn=rn-1-cnen scalar products (rn,-en), (rn,-en-1), …, (rn,-e1), (rn,e0), (rn,e-1) are equal or, equivalently, angles between rn and the vectors e-1, e0, -e1, -e2, …, -en-1, -en are equal. We denote this angle by αn and note that as cos⁡αn=cos⁡αn-1cos⁡(π-αn-1-αn) due to the spherical law of cosines, αn∈αn-1,π/2.Let us justify formally that it is possible to finden and cn with the required properties. Let θn denote an arbitrary unit vector orthogonal to the subspace <e-1,e0,…,en-1> (which also contains vector rn-1), and En denote the set of vectors e-1,e0,-e1,-e2,…,-en-1. As en we can take any of two unit vectors from the plane <rn-1,θn> that have an angle with rn-1 equal to αn-1. Let en=anrn-1+bnθn; note that an>0 as the angle αn-1 is acute. By construction for all e∈En the scalar product (rn-1,e) equals rn-1cos⁡αn-1. For the sake of brevity we denote rn-1cos⁡αn-1 by βn. Hence for all e∈E, as θn⊥e, scalar products of en and e are the same and equal anβn. Consequently, for an arbitrary positive c and every e∈En we have equalities (rn-1-cen,-en)=-βn+c, (rn-1-cen,e)=βn-canβn. Thus, it remains to set cn to the solution of the linear equation -βn+c=βn-canβn, i.e., to 2βn/(1+anβn).


### 4.2. Realization of Greedy Expansion

We note that a possible realization of greedy expansion in the dictionary(16)D=±e-1,±e0,±e1,±e2,…with the prescribed coefficients C=cnn=1∞ (and the weakness parameter t=1) is a realization in which en is selected as an expanding element at the n-th step, and hence n-th remainder coincides with rn. Indeed, an angle between a vector and its orthogonal projection on a subspace does not exceed an angle between this vector and any non-zero vector from the subspace. Thus while for every n∈1,2,3,… and every e∈±e-1,±e0,±e1,±e2,…,±en we have an equality (17)rn-1,e=rn-1cos⁡αn-1and for e=en the scalar product (rn-1,e) is positive, for e=±en+j(j∈1,2,3,…) we have an inequality(18)rn-1,e≤rn-1cos⁡αn+j-1≤rn-1cos⁡αn-1.


### 4.3. Absence of Convergence

In this subsection we show that the greedy expansion∑n=1∞cnen does not converge to f or, equivalently, remainders rn do not converge to zero.First we find the limit ofαn. It exists as αnn=1∞ is a nondecreasing sequence with all values belonging to the interval π/4,π/2. As noted above, for all n∈0,1,2,… the equality (19)cos⁡αn+1=cos⁡αncos⁡π-αn-αn+1holds. Consequently, if α denotes the limit limn→∞⁡αn, then cos⁡(π-2α)cos⁡α=cos⁡α. It implies that α=π/2.Now we show that∑n=0∞(1/tan⁡αn)=∞. Indeed,(20)cos⁡π-αn-αn+1cos⁡αn=cos⁡αn+1⇒cos⁡αn+αn+1cos⁡αn+cos⁡αn+1=0⇒cos⁡αncos⁡αn+1-sin⁡αnsin⁡αn+1cos⁡αn+cos⁡αn+1=0⇒cos⁡αn-sin⁡αntan⁡αn+1cos⁡αn+1=0⇒cos2⁡αn-sin⁡αntan⁡αn+1cos⁡αn+1=0⇒tan⁡αn+1=cos2⁡αn+1sin⁡αncos⁡αn=cos⁡2αn+1/2+1sin⁡2αn/2=cos⁡2αn+3sin⁡2αn=1-tan2⁡αn/1+tan2⁡αn+32tan⁡αn/1+tan2⁡αn=2tan2⁡αn+42tan⁡αn=tan⁡αn+2tan⁡αn.Thus(21)tan⁡αn+1=tan⁡α0+21tan⁡α0+1tan⁡α1+…+1tan⁡αn.As αn→π/2, tan⁡αn→∞ and hence ∑n=0∞(1/tan⁡αn)=∞.The law of sines gives the equalities(22)r0sin⁡α1=r1sin⁡α0⇒r1=sin⁡α0sin⁡α1r0;r1sin⁡α2=r2sin⁡α1⇒r2=sin⁡α1sin⁡α2r1;⋮rn-1sin⁡αn=rnsin⁡αn-1⇒rn=sin⁡αn-1sin⁡αnrn-1.Consequently(23)rn=sin⁡αn-1sin⁡αnsin⁡αn-2sin⁡αn-1…sin⁡α0sin⁡α1r0=r0sin⁡α0sin⁡αn.Taking into consideration the convergence αn→π/2(n→∞), we derive from this equality that(24)rn↛n→∞0,so the absence of convergence to the expanded element is proved.


### 4.4. Estimate ofcn

Here we first show that∑n=1∞cn=∞. Applying again the law of sines, but this time for the other pairs of angles, we get equalities(25)r0sin⁡α1=c1sin⁡π-α0-α1⇒c1=r0sin⁡α0+α1sin⁡α1;⋮rnsin⁡αn+1=cn+1sin⁡π-αn-αn+1⇒cn+1=rnsin⁡αn+αn+1sin⁡αn+1.Let us use equality (23):(26)cn+1=r0sin⁡α0sin⁡αn+αn+1sin⁡αn+1sin⁡αn=r0sin⁡α0sin⁡αncos⁡αn+1+cos⁡αnsin⁡αn+1sin⁡αn+1sin⁡αn=r0sin⁡α01tan⁡αn+1tan⁡αn+1.Taking into account the monotonicity of αn, it implies the inequality (27)cn+1>2r0sin⁡α0tan⁡αn+1.Consequently, as the series ∑n=0∞(1/tan⁡αn) diverges, the series ∑n=1∞cn also diverges.Then we show thatcn⩽1/n. The equality (26) and monotonicity of αn imply that (28)cn+1<2r0sin⁡α0tan⁡αnn∈0,1,2,….It remains to establish the inequality tan⁡αn>n+1: as r0⩽1/2, it directly gives the required upper estimate of cn. For n=0 this inequality holds because α0∈π/4,π/2. Next, from equality (20), taking into account that the minimum value of the function f(x)=x+2/x on the positive semi-axis equals 22, we obtain that tan⁡α1⩾22>2. Besides, f(x) is an increasing function on [2,+∞), and the justification of the inequality can be completed by induction: (29)tan⁡αn+1=ftan⁡αn⩾fn+1=n+1+2n+1>n+2.Justification of this inequality completes the proof of the upper estimate of cn and in total the proof Theorem 3.We note that the construction described in the proof of Theorem3 can be straightforwardly adapted from the case of greedy expansions with prescribed coefficients to the generalized Approximate Weak Greedy Algorithm (see [10, Theorem 3]).


## 4.1. Description of the Construction

We first present the structure of the example; i.e., we describe the construction of dictionary elementsen and coefficients cn. As a part of this construction we also define the sequence of vectors (remainders) rn, including the expanded element f=r0.Letf be an arbitrary non-zero element of H with f⩽1/2, r0=f. We define dictionary elements e-1 and e0 as arbitrary (unequal) unit vectors such that e-1, e0 and r0 lie in one plane and the angle α0 between e0 and r0 equals the angle between e-1 and r0 and belongs to the interval (π/4,π/2).Next, we set dictionary elemente1 to an arbitrary unit vector with the following two properties: its orthogonal projection on the plane <e-1,e0> lies on the line with directing vector r0, and the angle between this vector and r0 also equals α0. We also find the coefficient c1 such that for r1=r0-c1e1 scalar products (r1,-e1),(r1,e0),(r1,e-1) are equal or, equivalently, angles between r1 and the vectors e-1, e0, -e1 are equal. We denote this angle by α1. Note that α1∈α0,π/2: the formal justification of this fact can be based, e.g., on the equality cos⁡α1=cos⁡α0cos⁡(π-α0-α1), which directly follows from the spherical law of cosines.Similarly, we sete2 to an arbitrary unit vector with an orthogonal projection on the subspace <e-1,e0,e1> lying on the line with directing vector r1 and the angle between this vector and r1 equal α1, and select the coefficient c2 in such a way that for r2=r1-c2e2 scalar products (r2,-e2), (r2,-e1), (r2,e0), (r2,e-1) are equal or, equivalently, angles between r2 and the vectors e-1, e0, -e1, -e2 are equal. We denote this angle by α2. Again it is easy to see that α2∈α1,π/2, as due to the spherical law of cosines cos⁡α2=cos⁡α1cos⁡(π-α1-α2).We continue the construction inductively. Namely, after constructingrjj=0n-1, ejj=-1n-1, cjj=1n-1 and αjj=0n-1 we set en to an arbitrary unit vector with an orthogonal projection on the subspace <e-1,e0,…,en-1> lying on the line with directing vector rn-1 and angle between this vector and rn-1 equal αn-1, and select the coefficient cn in such a way that for rn=rn-1-cnen scalar products (rn,-en), (rn,-en-1), …, (rn,-e1), (rn,e0), (rn,e-1) are equal or, equivalently, angles between rn and the vectors e-1, e0, -e1, -e2, …, -en-1, -en are equal. We denote this angle by αn and note that as cos⁡αn=cos⁡αn-1cos⁡(π-αn-1-αn) due to the spherical law of cosines, αn∈αn-1,π/2.Let us justify formally that it is possible to finden and cn with the required properties. Let θn denote an arbitrary unit vector orthogonal to the subspace <e-1,e0,…,en-1> (which also contains vector rn-1), and En denote the set of vectors e-1,e0,-e1,-e2,…,-en-1. As en we can take any of two unit vectors from the plane <rn-1,θn> that have an angle with rn-1 equal to αn-1. Let en=anrn-1+bnθn; note that an>0 as the angle αn-1 is acute. By construction for all e∈En the scalar product (rn-1,e) equals rn-1cos⁡αn-1. For the sake of brevity we denote rn-1cos⁡αn-1 by βn. Hence for all e∈E, as θn⊥e, scalar products of en and e are the same and equal anβn. Consequently, for an arbitrary positive c and every e∈En we have equalities (rn-1-cen,-en)=-βn+c, (rn-1-cen,e)=βn-canβn. Thus, it remains to set cn to the solution of the linear equation -βn+c=βn-canβn, i.e., to 2βn/(1+anβn).


## 4.2. Realization of Greedy Expansion

We note that a possible realization of greedy expansion in the dictionary(16)D=±e-1,±e0,±e1,±e2,…with the prescribed coefficients C=cnn=1∞ (and the weakness parameter t=1) is a realization in which en is selected as an expanding element at the n-th step, and hence n-th remainder coincides with rn. Indeed, an angle between a vector and its orthogonal projection on a subspace does not exceed an angle between this vector and any non-zero vector from the subspace. Thus while for every n∈1,2,3,… and every e∈±e-1,±e0,±e1,±e2,…,±en we have an equality (17)rn-1,e=rn-1cos⁡αn-1and for e=en the scalar product (rn-1,e) is positive, for e=±en+j(j∈1,2,3,…) we have an inequality(18)rn-1,e≤rn-1cos⁡αn+j-1≤rn-1cos⁡αn-1.


## 4.3. Absence of Convergence

In this subsection we show that the greedy expansion∑n=1∞cnen does not converge to f or, equivalently, remainders rn do not converge to zero.First we find the limit ofαn. It exists as αnn=1∞ is a nondecreasing sequence with all values belonging to the interval π/4,π/2. As noted above, for all n∈0,1,2,… the equality (19)cos⁡αn+1=cos⁡αncos⁡π-αn-αn+1holds. Consequently, if α denotes the limit limn→∞⁡αn, then cos⁡(π-2α)cos⁡α=cos⁡α. It implies that α=π/2.Now we show that∑n=0∞(1/tan⁡αn)=∞. Indeed,(20)cos⁡π-αn-αn+1cos⁡αn=cos⁡αn+1⇒cos⁡αn+αn+1cos⁡αn+cos⁡αn+1=0⇒cos⁡αncos⁡αn+1-sin⁡αnsin⁡αn+1cos⁡αn+cos⁡αn+1=0⇒cos⁡αn-sin⁡αntan⁡αn+1cos⁡αn+1=0⇒cos2⁡αn-sin⁡αntan⁡αn+1cos⁡αn+1=0⇒tan⁡αn+1=cos2⁡αn+1sin⁡αncos⁡αn=cos⁡2αn+1/2+1sin⁡2αn/2=cos⁡2αn+3sin⁡2αn=1-tan2⁡αn/1+tan2⁡αn+32tan⁡αn/1+tan2⁡αn=2tan2⁡αn+42tan⁡αn=tan⁡αn+2tan⁡αn.Thus(21)tan⁡αn+1=tan⁡α0+21tan⁡α0+1tan⁡α1+…+1tan⁡αn.As αn→π/2, tan⁡αn→∞ and hence ∑n=0∞(1/tan⁡αn)=∞.The law of sines gives the equalities(22)r0sin⁡α1=r1sin⁡α0⇒r1=sin⁡α0sin⁡α1r0;r1sin⁡α2=r2sin⁡α1⇒r2=sin⁡α1sin⁡α2r1;⋮rn-1sin⁡αn=rnsin⁡αn-1⇒rn=sin⁡αn-1sin⁡αnrn-1.Consequently(23)rn=sin⁡αn-1sin⁡αnsin⁡αn-2sin⁡αn-1…sin⁡α0sin⁡α1r0=r0sin⁡α0sin⁡αn.Taking into consideration the convergence αn→π/2(n→∞), we derive from this equality that(24)rn↛n→∞0,so the absence of convergence to the expanded element is proved.


## 4.4. Estimate ofcn

Here we first show that∑n=1∞cn=∞. Applying again the law of sines, but this time for the other pairs of angles, we get equalities(25)r0sin⁡α1=c1sin⁡π-α0-α1⇒c1=r0sin⁡α0+α1sin⁡α1;⋮rnsin⁡αn+1=cn+1sin⁡π-αn-αn+1⇒cn+1=rnsin⁡αn+αn+1sin⁡αn+1.Let us use equality (23):(26)cn+1=r0sin⁡α0sin⁡αn+αn+1sin⁡αn+1sin⁡αn=r0sin⁡α0sin⁡αncos⁡αn+1+cos⁡αnsin⁡αn+1sin⁡αn+1sin⁡αn=r0sin⁡α01tan⁡αn+1tan⁡αn+1.Taking into account the monotonicity of αn, it implies the inequality (27)cn+1>2r0sin⁡α0tan⁡αn+1.Consequently, as the series ∑n=0∞(1/tan⁡αn) diverges, the series ∑n=1∞cn also diverges.Then we show thatcn⩽1/n. The equality (26) and monotonicity of αn imply that (28)cn+1<2r0sin⁡α0tan⁡αnn∈0,1,2,….It remains to establish the inequality tan⁡αn>n+1: as r0⩽1/2, it directly gives the required upper estimate of cn. For n=0 this inequality holds because α0∈π/4,π/2. Next, from equality (20), taking into account that the minimum value of the function f(x)=x+2/x on the positive semi-axis equals 22, we obtain that tan⁡α1⩾22>2. Besides, f(x) is an increasing function on [2,+∞), and the justification of the inequality can be completed by induction: (29)tan⁡αn+1=ftan⁡αn⩾fn+1=n+1+2n+1>n+2.Justification of this inequality completes the proof of the upper estimate of cn and in total the proof Theorem 3.We note that the construction described in the proof of Theorem3 can be straightforwardly adapted from the case of greedy expansions with prescribed coefficients to the generalized Approximate Weak Greedy Algorithm (see [10, Theorem 3]).


## 5. The Case of Finite-Dimensional Spaces

In this section we prove the following theorem.Theorem 5.
Let the spaceH be finite-dimensional, and let the coefficients C=cnn=1∞ converge to zero monotonically and satisfy the condition ∑n=1∞cn=∞. Then for every symmetric unit-normed dictionary D⊂H and any f∈H all realizations of greedy expansion of f in D with the prescribed coefficients C and the weakness parameter t=1 converge to f.Thus in the finite-dimensional case at least fort=1 in the natural class of monotonic coefficient sequences conditions sufficient for the convergence of a greedy expansion to an expanded element include only the convergence of coefficients to zero and infinity of their sum. Clearly, convergence of coefficients to zero is also a necessary condition for the convergence of greedy expansion.As the first step of the proof we note that due to Lemma4 there exists a limit of norms of remainders of a greedy expansion with prescribed coefficients limn→∞⁡rn. Let us suppose that this limit is non-zero. Then starting from a certain index remainder norms are separated from zero. Due to the compactness of a sphere in a finite-dimensional space and the completeness and the symmetric property of the dictionary it implies that scalar products(30)xn+1=rn,en+1=supe∈D⁡rn,eare also separated from zero. In other words, there exist a positive number γ and an index N0 such that for all n>N0 the inequality xn+1>γ holds. In addition there exists such an index N1>N0 that for all n>N1 coefficient cn does not exceed γ. But consequently for n⩾N1 it follows from equality (10) that (31)rn+12=rn2-xn+12+cn+1-xn+12=rn2-cn+12xn+1-cn+1⩽rn2-γcn+1.Hence(32)rN1+K2⩽rN12-γ∑n=N1+1N1+Kcn→-∞K→∞.This contradiction completes the proof of Theorem 5.


## 6. Conclusion

The main results of the paper state that in Hilbert spaces a greedy expansion with prescribed coefficients converges to an expanded element if coefficients satisfy certain relatively weak conditions that do not include monotonicity, and these conditions can not be essentially relaxed. At the same time we showed that for the finite-dimensional case the relaxation is possible.



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