# On Algebraic Basis of the Algebra of Symmetric Polynomials onlp(Cn)

Authors: Victoriia Kravtsiv; Taras Vasylyshyn; Andriy Zagorodnyuk
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We consider polynomials on spaceslpCn,1≤p<+∞, of p-summing sequences of n-dimensional complex vectors, which are symmetric with respect to permutations of elements of the sequences, and describe algebraic bases of algebras of continuous symmetric polynomials on lpCn.


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## Body

## 1. Introduction

Algebras of polynomials and analytic functions on a Banach spaceX which are invariant (symmetric) with respect to a group of linear operators GX acting on X were studied by a number of authors [1–10] (see also a survey [11]). If X has a symmetric structure, then it is natural to consider the case when GX is a group of operators which preserve this structure. In particular, if X is a rearrangement-invariant sequence space, then GX is used to be the group of permutations of positive integers. In [8] Nemirovskii and Semenov described algebraic bases of algebras of continuous symmetric polynomials on real spaces lp, where 1≤p<+∞. Their results were generalized by González et al. [7] to real separable rearrangement-invariant sequence spaces.Algebraic basis plays a crucial role in the problem of description of spectra of algebras generated by polynomials [1–4]. For example, each complex homomorphism on the algebra of symmetric polynomials on lp is completely defined by its values on the basis elements.Note that an algebra of symmetric functions essentially depends on a representation of a given groupG on X. In particular, in [12–14] the group of permutations of positive integers was considered which acts on the complex space l1 permutating “blocks” of coordinates. Polynomials which are invariant with respect to the action are called block-symmetric. It is natural to consider such polynomials as symmetric polynomials on l1Cn.In this work we get an explicit description of algebraic bases of algebras of symmetric polynomials onlpCn, where 1≤p<+∞.


## 2. Materials and Methods

We denote byN the set of all positive integers and by Z+ the set of all nonnegative integers.A mappingP:X→C, where X is a complex Banach space, is called an N-homogeneous polynomial if there exists an N-linear form AP:XN→C such that P is the restriction to the diagonal of AP, that is, P(x)=AP(x,…,x︸N) for every x∈X. By [15, Corollary 2.3], N-homogeneous polynomial P is continuous if and only if its norm P=supx≤1⁡Px is finite. Definition of N-homogeneous polynomial implies the inequality Px≤PxN for every x∈X. A mapping P=P0+P1+⋯+Pm, where P0∈C and Pj is a j-homogeneous polynomial for every j∈{1,…,m}, is called a polynomial of degree at most m.Letn∈N and p∈[1,+∞). Let us denote lp(Cn) the vector space of all sequences (1)x=x1,x2,…,where xj=xj1,…,xjn∈Cn for j∈N, such that the series ∑j=1∞∑s=1nxjsp is convergent. The space lpCn with norm (2)xp=∑j=1∞∑s=1nxjsp1/pis a Banach space.Definition 1.
A functionf:lpCn→C is called symmetric if fx∘σ=fx for every x∈lpCn and for every bijection σ:N→N, where x∘σ=xσ1,xσ2,….Let us denotePslpCn the algebra of all symmetric continuous polynomials on lpCn.


## 3. Results and Discussion

### 3.1. Power Sum Symmetric Polynomials onlp(Cn)

For a multi-indexk=k1,…,kn∈Z+n let |k|=k1+⋯+kn. For every k∈Z+n such that k≥p, where p is a ceiling of p, let us define a mapping Hk:lpCn→C by (3)Hkx=∑j=1∞∏s=1ks>0nxjsks.Also we set H0,…,0x≡1. Note that Hk is a symmetric k-homogeneous polynomial. Polynomials Hk are generalizations of so-calledpower sum symmetric polynomials on finite-dimensional spaces (see, e.g., [16, page 23] or [17, page 297]).Proposition 2.
Forp∈1,+∞ and for every k∈Z+n such that k≥p, polynomial Hk on lpCn is continuous and Hk≤1.Proof.
Letx∈lpCn such that xp≤1. Note that (4)Hkx≤∑j=1∞∏s=1ks>0nxjsks.Since xjs≤max1≤m≤n⁡xjm for every s∈1,…,n and j∈N, it follows that (5)∏s=1ks>0nxjsks≤max1≤m≤nxjmkfor every j∈N. Note that (6)max1≤m≤nxjmk=max1≤m≤nxjmk≤∑m=1nxjmk.Therefore, (7)Hkx≤∑j=1∞∑m=1nxjmk.Since xp≤1, it follows that xjm≤1 for every m∈1,…,n and j∈N. Therefore, xjmk≤xjmp. Thus, (8)Hkx≤∑j=1∞∑m=1nxjmp=xpp≤1.Therefore, Hk=supxp≤1⁡Hkx≤1. Hence, Hk is bounded and, consequently, it is continuous.Form∈N, let c00mCn be the space of all sequences x=x1,…,xm,0,…, where x1,…,xm∈Cn and 0=0,…,0∈Cn. Note that c00mCn is isomorphic to Cnm. Let c00(Cn)=⋃m=1∞c00mCn. Note that c00Cn is a dense subspace in lpCn. Also note that Hk is well-defined on c00Cn for every k∈Z+n.For arbitraryx=x1,…,xm,0,…,y=y1,…,ys,0,…∈c00Cn, we set (9)x⊕y=x1,…,xm,y1,…,ys,0,….For x1,…,xr∈c00Cn, let (10)⨁j=1rxj=x1⊕⋯⊕xr.Note that(11)⨁j=1rxjpp=∑j=1rxjpp.Also note that for every k∈Z+n, such that |k|≥1,(12)Hk⨁j=1rxj=∑j=1rHkxj.For everym∈N and j∈{1,…,m}, we set(13)αmj=1m1/mexp⁡2πijm.Also we set α01=0. For l=l1,…,ln∈Z+n, let(14)al=⨁j1=1l1^⋯⨁jn=1ln^αl1j1,…,αlnjn,0,…,0,…,where lj^=max⁡1,lj for j∈1,…,n.Let us define a partial order onZ+n by the following way. For k,l∈Z+n we set k⪰l if and only if there exists m∈Z+n such that ks=msls for every s∈1,…,n. We write k≻l, if k⪰l and k≠l.Proposition 3.
Fork∈Z+n such that k≥1 and for arbitrary l∈Z+n(15)Hkal=∏s=1ks>0n1lsks/ls-1∏s=1ks=0nls^,ifk⪰l0,otherwise,where, by the definition, product of an empty set of multipliers is equal to 1. In particular, Hkak=1.Proof.
By (12) and (14), (16)Hkal=∑j1=1l1^⋯∑jn=1ln^Hkαl1j1,…,αlnjn,0,…,0,….By the definition of Hk, (17)Hkαl1j1,…,αlnjn,0,…,0,…=∏s=1ks>0nαlsjsks.Therefore,(18)Hkal=∑j1=1l1^⋯∑jn=1ln^∏s=1ks>0nαlsjsks=∏s=1ks>0n∑js=1ls^αlsjsks∏s=1ks=0n∑js=1ls^1=∏s=1ks>0n∑js=1ls^αlsjsks∏s=1ks=0nls^.
Letk⪰l. Then there exists m∈Z+n such that ks=msls for every s∈1,…,n. For s∈1,…,n such that ks>0, we have that ls>0 too. Consequently, for such s we have ls^=ls, and, by (13),(19)∑js=1ls^αlsjsks=∑js=1ls1ls1/lsexp⁡2πijslsmsls=1lsms∑js=1lsexp⁡2πijsms=1lsms∑js=1ls1=1lsms-1=1lsks/ls-1.Therefore, by (18), (20)Hkal=∏s=1ks>0n1lsks/ls-1∏s=1ks=0nls^.In the case k=l we have (21)Hkak=∏s=1ks>0n1ksks/ks-1∏s=1ks=0nks^=1.
Letk⪰l. Then we have two cases.Case1. There exists s∈1,…,n such that ks>ls=0. Then (22)∑js=1ls^αlsjsks=α01ks=0;therefore, Hkal=0.Case2. There exists s∈1,…,n such that ls>ks>0. Then (23)∑js=1ls^αlsjsks=∑js=1ls1ls1/lsexp⁡2πijslsks=1lsks/ls∑js=1lsexp⁡2πijslsks.It is known that(24)∑j=1qexp⁡2πijqr=0for every q∈2,3,… and r∈1,…,q-1. Therefore, (25)∑js=1lsexp⁡2πijslsks=0and, consequently, Hkal=0.Let us prove the following auxiliary proposition.Proposition 4.
A functiong:0,+∞→R,gx=c1x+⋯+cmx1/x, where m∈N and c1,…,cm>0, is strictly decreasing.Proof.
Let us prove thatg′x<0 for every x∈0,+∞. Note that gx=1/xln⁡c1x+⋯+cmx. Therefore,(26)g′x=gx-1x2ln⁡c1x+⋯+cmx+1xc1xln⁡c1+⋯+cmxln⁡cmc1x+⋯+cmx=-gxx2c1x+⋯+cmxc1x+⋯+cmxln⁡c1x+⋯+cmx-xc1xln⁡c1+⋯+cmxln⁡cm=-gxx2c1x+⋯+cmxc1xln⁡c1x+⋯+cmx-ln⁡c1x+⋯+cmxln⁡c1x+⋯+cmx-ln⁡cmx.Since gx/x2c1x+⋯+cmx>0 and lnc1x+⋯+cmx>ln⁡cjx for every j∈1,…,m, it follows that g′(x)<0.Corollary 5.
For everyx∈lpCn and for every q≥p(27)xp≥xq.For an arbitrary nonempty finite setM⊂Z+n let us define a mapping πM:c00Cn→C|M|, where M is the cardinality of M, by(28)πMx=Hkxk∈M,where Hkxk∈M is an M-dimensional vector of values of Hk on x, indexed by k∈M.We endow the space CM with norm ξ∞=maxk∈M|ξk|, where ξ=ξkk∈M∈CM.Theorem 6.
LetM be a finite nonempty subset of Z+n such that k≥1 for every k∈M. Then(i)
there existsm∈N such that for every ξ=ξkk∈M∈CM there exists xξ∈c00mCn such that πMxξ=ξ;(ii)
there exists a constantρM>0 such that if ξ∞<1, then xξp<ρM for every p∈1,+∞.Proof.
(i) Letξ=ξkk∈M∈CM. For every k∈M, let us define ηk∈C and bk∈c00Cn by the following way. For minimal elements k of the partially ordered set M,⪯, let ηk=ξk and bk=ηkkak, where ak is defined by (14) and (29)ηkk=ηkkeiarg⁡ηk/k,ifηk≠00,ifηk=0.For k∈M, which are not minimal elements of M,⪯, we define ηk and bk inductively by(30)ηk=ξk-∑l∈Ml≺kHkbl,(31)bk=ηkkak.We set xξ=⨁l∈Mbl. Note that xξ∈c00(m)(Cn), where (32)m=∑k∈Mmin⁡j∈N:ak∈c00jCn.For k∈M, by (12), Hkxξ=∑l∈MHkbl. Since Hk is a k-homogeneous polynomial,(33)Hkbl=ηllkHkal.By Proposition 3, Hkal is not equal to zero only for l∈M such that l⪯k. Therefore, (34)Hkxξ=Hkbk+∑l∈Ml≺kHkbl.By Proposition 3, Hkak=1, and therefore, by (33), Hkbk=ηk. Hence, (35)Hkxξ=ηk+∑l∈Ml≺kHkbl.Taking into account (30), we have Hkxξ=ξk. Hence, πMxξ=ξ.
(ii) Letξ=ξkk∈M∈CM be such that ξ∞<1. For k∈M let (36)k=max⁡s∈N:∃l1,…,ls∈Msuch thatl1≺⋯≺ls=k.Note that for minimal elements k∈M we have 〈k〉=1.
Let(37)C=max⁡1,maxk∈M⁡ak1.Let(38)r=maxk∈M⁡k,and for every j∈1,…,r let (39)μj=∏s=1j1+ms,where(40)ms=k∈M:k=s.Also we set μ0=1.
Note that for everyj∈1,…,r(41)μj=μj-11+mj=μj-1+μj-1mj=μj-2+μj-2mj-1+μj-1mj=⋯=μ0+μ0m1+μ1m2+⋯+μj-1mj.
Let us prove that for everyk∈M(42)bk1<μk-1Ck.We proceed by induction on k. In the case k=1, we have ηk=ξk, and therefore, bk1=ξk|k|ak1. Since ξk<1, it follows that bk1<ak1≤C=μ0C. If r=1, then (42) is proved. Let r≥2 and j∈2,…,r. Suppose that inequality (42) holds for every k∈M such that k∈1,…,j-1. Let us prove (42) for k∈M such that k=j. By (31) and (37),(43)bk1≤ηkkak1≤ηkkC.By (30), (44)ηk≤ξk+∑l∈Ml≺kHkbl.Since Hk is a k-homogeneous polynomial on the space l1Cn and Hk≤1, (45)Hkbl≤Hkbl1k≤bl1k.Therefore, taking into account ξk<1, we have (46)ξk+∑l∈Ml≺kHkbl<1+∑l∈Ml≺kbl1k.Therefore,(47)ηkk<1+∑l∈Ml≺kbl1k1/k.By Proposition 4,(48)1+∑l∈Ml≺kbl1k1/k≤1+∑l∈Ml≺kbl1.Note that if l≺k, then l<k. Therefore,(49)∑l∈Ml≺kbl1≤∑l∈Ml<kbl1.Since k=j,(50)∑l∈Ml<kbl1=∑s=1j-1∑l∈Ml=sbl1.By the induction hypothesis, if l=s, where s∈1,…,j-1, then bl1<μs-1Cs. Therefore,(51)∑l∈Ml=sbl1<∑l∈Ml=sμs-1Cs=μs-1Cs∑l∈Ml=s1=μs-1msCs.Since C≥1, it follows that Cs≤Cj-1 for every s∈1,…,j-1, and therefore,(52)1+∑s=1j-1μs-1msCs≤1+Cj-1∑s=1j-1μs-1ms≤1+∑s=1j-1μs-1msCj-1.Since μ0=1, by (41),(53)1+∑s=1j-1μs-1ms=μj-1.By (47)–(53),(54)ηkk<μj-1Cj-1.By (43) and (54), bk1≤μj-1Cj. Hence, inequality (42) holds for every k∈M.
By (11) and by Proposition 4, (55)xξ1≤∑l∈Mbl1.By (42),(56)∑l∈Mbl1=∑j=1r∑l∈Ml=jbl1<∑j=1r∑l∈Ml=jμj-1Cj=∑j=1rμj-1Cj∑l∈Ml=j1=∑j=1rμj-1mjCj≤∑j=1rμj-1mjCr<μ0+∑j=1rμj-1mjCr=μrCr.Set ρM=μrCr. We have that xξ1<ρM if ξ∞<1. By Corollary 5, xξp≤xξ1≤ρM for every p∈1,+∞.Corollary 7.
LetM=k1,…,ks⊂Z+n such that kj≥1 for every j∈1,…,s. Then there exists m∈N such that for every m′≥m polynomials Hk1,…,Hks are algebraically independent on c00m′Cn.Proof.
By Theorem6, there exists m∈N such that for every ξ=ξ1,…,ξs∈Cs there exists xξ∈c00mCn such that(57)Hkjxξ=ξjfor every j∈1,…,s. Let us show that Hk1,…,Hks are algebraically independent on c00m′Cn for every m′≥m. Let Q:Cs→C be a polynomial such that (58)QHk1x,…,Hksx=0for every x∈c00m′Cn. Set x=xξ. Taking into account (57), we have Qξ1,…,ξs=0 for arbitrary ξ1,…,ξs∈C, that is, Q≡0. Hence, Hk1,…,Hks are algebraically independent.


### 3.2. Algebraic Basis of the AlgebraPsl1Cn

Theorem 8.
EveryN-homogeneous polynomial P∈Psc00mCn, where m is an arbitrary positive integer, can be represented as an algebraic combination of polynomials Hk, where k∈Z+n such that 1≤k≤N.Proof.
We proceed by induction onm. In the case m=1 for x=x1,0,…∈c001Cn, we have (59)Px=∑k∈Z+nk=Nαkx11k1⋯x1nkn=∑k∈Z+nk=NαkHkx,where αk∈C. Suppose the statement holds for m-1 and prove it for m. Let P∈Psc00mCn and x=x1,…,xm,0,…∈c00mCn. Then Px can be represented as a sum of terms (60)βkxm1k1⋯xmnknfkx1,…,xm-1,0,…,where βk∈C,k=k1,…,kn∈Z+n such that 1≤k≤N, and fk is an N-k-homogeneous polynomial. Note that fk∈Psc00m-1Cn, and therefore, by the induction hypothesis, fkx1,…,xm-1,0,… can be represented as an algebraic combination of Hlx1,…,xm-1,0,…, where l∈Z+n such that 1≤l≤N-k. Note that (61)Hlx1,…,xm-1,0,…=Hlx-xm1l1⋯xmnln.Therefore, P(x) can be represented as an algebraic combination of Hlx and xm1,…,xmn. Since P and Hl are symmetric, it follows that together with term (62)γr1,…,rn,t1,…,tsxm1r1⋯xmnrnHl1t1x⋯Hlstsx,where γr1,…,rn,t1,…,ts∈C,l1,…,ls∈Z+n and r1,…,rn,t1,…,ts∈Z+, the sum must contain terms (63)γr1,…,rn,t1,…,tsxj1r1⋯xjnrnHl1t1x⋯Hlstsx,where j∈1,…,m-1. Therefore, Px can be represented as a sum of terms (64)γr1,…,rn,t1,…,ts1m∑j=1mxj1r1⋯xjnrnHl1t1x⋯Hlstsx.Since ∑j=1mxj1r1⋯xjnrn=Hrx, where r=r1,…,rn, it follows that P is an algebraic combination of polynomials Hk, where k∈Z+n such that 1≤k≤N.Theorem 9.
LetP:c00Cn→C be a symmetric N-homogeneous polynomial. Let MN=k∈Z+n:1≤k≤N. There exists a polynomial q:CMN→C such that P=q∘πMN, where the mapping πMN is defined by (28).Proof.
By Corollary7, there exists m∈N such that for every m′≥m polynomials Hk, where k∈M, are algebraically independent. Therefore, the representation, given by Theorem 8 for the restriction of P to c00m′Cn, is unique. Thus, for every m′≥m there exists a unique polynomial qm′:CMN→C such that Px=(qm′∘πMN)(x) for every x∈c00m′Cn. Since c00m′Cn⊃c00(m)(Cn), it follows that qm is the restriction of qm′ to πMN(c00(m)(Cn)). By Theorem 6, πMNc00mCn=CMN, and therefore, qm′≡qm. Let q=qm. Then Px=q∘πMNx for every x∈c00Cn.Theorem 10.
PolynomialsHk, where k∈Z+n, form an algebraic basis of the algebra Psl1Cn.Proof.
Let us prove that every symmetric continuous polynomial onl1Cn can be uniquely represented as an algebraic combination of polynomials Hk. It suffices to prove the statement only for homogeneous polynomials. Let P:l1Cn→C be a symmetric continuous N-homogeneous polynomial. By Theorem 9, the restriction of P to c00Cn can be uniquely represented as an algebraic combination of polynomials Hk, where k∈Z+n such that 1≤k≤N. Since c00Cn is dense in l1Cn and polynomials Hk are well-defined and continuous on l1Cn, it follows that given representation can be extended to l1Cn.


### 3.3. Algebraic Basis of the AlgebraPslpCn

Letp∈1,+∞. In this section, we describe an algebraic basis of the algebra PslpCn.Let us prove a complex analog of [8, Lemma 2].Lemma 11.
LetK⊂Cm and ϰ:K→Cm-1 be an orthogonal projection: ϰx1,x2,…,xm=x2,…,xm. Let K1=ϰ(K),intK1≠∅ and for every open set U⊂K1 a set ϰ-1U is unbounded. If polynomial Qx1,…,xm is bounded on K, then Q does not depend on x1.Proof.
Suppose thatQ depends on x1. Then (65)Qx1,…,xm=∑j=0kqjx2,…,xmx1j,where 1≤k≤degQ and qk≡0. Note that qk≡0 on intK1, and therefore, there exists point a∈intK1 such that qk(a)≠0. Since intK1 is open and qk is continuous, there exists r>0 such that Ba,r⊂intK1 and infb∈Ba,rqkb>0, where Ba,r is an open ball with center a and radius r in the space Cm-1. Note that, for x1,…,xm∈ϰ-1Ba,r,(66)Qx1,…,xm≥qkx2,…,xmx1k-∑j=0k-1qjx2,…,xmx1j≥cx1k-∑j=0k-1djx1j,where c=infb∈Ba,rqkb and dj=supb∈Ba,rqj(b) for j∈0,…,k-1. Note that for the polynomial cx1k+∑j=0k-1djx1j there exists R>0 such that if x1>R, then cx1k>2∑j=0k-1djx1j, that is, ∑j=0k-1djx1j<1/2c|x1|k. Therefore, if x1>R, then(67)cx1k-∑j=0k-1djx1j>cx1k-12cx1k=12cx1k.Since ϰ-1Ba,r is unbounded, there exists a sequence x1n,…,xmnn∈N⊂ϰ-1Ba,r such that x1n→∞ as n→+∞. Taking into account (66) and (67), we have (68)Qx1n,…,xmn>12cx1nk⟶+∞as n→+∞, which contradicts the boundedness of Q on K.Fork=k1,…,kn∈Z+n, let V(k)={s∈1,…,n:ks≠0} and νk=Vk.Lemma 12.
Fork,l∈Z+n if l≻k and νl≥νk, then l>k.Proof.
Sincel≻k, there exists m∈Z+n such that l1,…,ln=m1k1,…,mnkn, and l≠k. Therefore, if ks=0 for some s∈1,…,n, then ls=0 too. It means that Vl⊂Vk. On the other hand, νl≥νk. Therefore, Vl=Vk; that is, for s∈1,…,n, we have that ls≠0 if and only if ks≠0. Therefore, for every s∈Vl we have that ms≥1. Since l≠k, there exists s0∈Vl such that ms0≥2. Therefore, (69)l=m1k1+⋯+mnkn>k1+⋯+kn=k.ForN∈N and J∈1,…,n let(70)MNJ=l∈Z+n:1≤l<p,νl≥J∪l∈Z+n:p≤l≤N.By Theorem 6, for M=MN1 there exists ρ=ρM>0 such that πMVρ contains the open unit ball of the space CM with norm ·∞, where(71)Vρ=x∈c00Cn:xp<ρ.Proposition 13.
ForJ∈1,…,N, let qξll∈MNJ be a polynomial on CMNJ. If q is bounded on πMNJVρ, then q does not depend on ξk such that νk=J and 1≤k<p.Proof.
Letk∈Z+n such that νk=J and 1≤k<p. Let K=πMNJVρ,K1=πMN(J)∖{k}(Vρ) and ϰ:K→K1 be an orthogonal projection, defined by (72)ϰ:ξll∈MNJ⟼ξll∈MNJ∖k.Let us show that, for every ball (73)Bu,r=ξ∈CMNJ∖k:ξ-u∞<rwith center u=(ul)l∈MN(J)∖{k}∈C|MN(J)∖{k}| and radius r>0 such that Bu,r⊂πMNJ∖kVρ, a set ϰ-1Bu,r is unbounded. Since u∈πMNJ∖kVρ, there exists xu∈Vρ such that πMNJ∖kxu=u. For m∈N, we set xm=⨁j=1m1/j1/|k|ak, where ak is defined by (14). Choose ε such that (74)0<ε<min⁡1,ρ-xupakpζp/k1/p,rak1Nζ1+1/k,where ζ· is a Riemann zeta-function. Let xm,ε=εxm⊕xu. Let us show that xm,ε∈Vρ. By (11), (75)xmpp=∑j=1m1j1/kakpp=∑j=1m1jp/kakpp=akpp∑j=1m1jp/k<akppζpk.Therefore, xmp<akpζp/k1/p. By the triangle inequality, (76)xm,εp≤εxmp+xup<εakpζpk1/p+xup.Since ε<ρ-xup/akpζp/k1/p, it follows that xm,εp<ρ. Hence, xm,ε∈Vρ.
Note that for arbitraryl∈Z+n such that l≥1, by (12), (77)Hlxm=∑j=1m1jl/kHlak=Hlak∑j=1m1jl/k,(78)Hlxm,ε=εlHlxm+Hlxu=εlHlak∑j=1m1jl/k+Hlxu.Let us show that πMNJ∖kxm,ε∈Bu,r. For l∈MNJ∖k such that l≻k, by Proposition 3, Hlak=0, and therefore, by (78), (79)Hlxm,ε=Hlxu=ul.Let l∈MNJ∖k be such that l≻k. If p≤l≤N, then l>k, since k<p. If 1≤l<p and νl≥J, then l>k by Lemma 12. Hence, l>k in both cases. By (78), (80)Hlxm,ε-ul≤εlHlak∑j=1m1jl/k.Since ε<1, it follows that εl≤ε. Since Hl≤1, it follows that Hlak≤ak1l. Taking into account akp≥1 and l≤N, we have that |Hl(ak)|≤ak1N. Since l and k are integer numbers and l>k, it follows that l≥k+1, and therefore, (81)∑j=1m1jl/k≤∑j=1m1j1+1/k<ζ1+1k.Hence,(82)Hlxm,ε-ul<εak1Nζ1+1k.Since ε<r/ak1Nζ(1+1/|k|), it follows that Hlxm,ε-ul<r, and therefore, πMNJ∖kxm,ε∈Bu,r.
By Proposition3, Hk(ak)=1, and therefore, by (78), (83)Hkxm,ε=εl∑j=1m1j+Hkxu⟶∞as m→+∞. Hence, ϰ-1Bu,r is unbounded. By Lemma 11, q does not depend on ξk.Theorem 14.
LetP∈PslpCn be an N-homogeneous polynomial. If N<p, then P≡0. Otherwise, there exists a unique polynomial q:CMp,N→C such that P=q∘πMp,Np, where Mp,N=k∈Z+n:p≤k≤N and πMp,Np:lpCn→C|Mp,N| is defined by πMp,Npx=Hkxk∈Mp,N.Proof.
LetP0 be the restriction of P to c00Cn. Note that P0 is a continuous symmetric N-homogeneous polynomial. By Theorem 9, there exists a unique polynomial q:CMN→C, where MN=MN1 such that P0=q∘πMN. Since P0 is continuous, P0 is bounded on Vρ, defined by (71). Therefore, q is bounded on πMNVρ.
Let us prove thatq does not depend on arguments ξk such that 1≤k<p by induction on νk. By Proposition 13, for J=1 we have that qξkk∈MN does not depend on arguments ξk such that νk=1 and 1≤k<p. Suppose that the statement holds for νk∈1,…,J-1, where J∈{2,…,n}, that is, qξkk∈MN does not depend on arguments ξk such that 1≤νk≤J-1 and 1≤|k|<⌈p⌉. Then the restriction of q to CMNJ, by Proposition 13, does not depend on ξk such that ν(k)=J and 1≤k<p. Hence, q does not depend on ξk such that 1≤k<p.
Since polynomialsHk, where k∈Mp,N, are well-defined and continuous on lpCn and c00Cn is dense in lpCn, it follows that P=q∘πMp,Np. Note that in the case N<p we have Mp,N=∅ and, therefore, P≡0.Corollary 15.
PolynomialsHk, where k∈l∈Z+n:l≥p∪0, form an algebraic basis of the algebra PslpCn.


## 3.1. Power Sum Symmetric Polynomials onlp(Cn)

For a multi-indexk=k1,…,kn∈Z+n let |k|=k1+⋯+kn. For every k∈Z+n such that k≥p, where p is a ceiling of p, let us define a mapping Hk:lpCn→C by (3)Hkx=∑j=1∞∏s=1ks>0nxjsks.Also we set H0,…,0x≡1. Note that Hk is a symmetric k-homogeneous polynomial. Polynomials Hk are generalizations of so-calledpower sum symmetric polynomials on finite-dimensional spaces (see, e.g., [16, page 23] or [17, page 297]).Proposition 2.
Forp∈1,+∞ and for every k∈Z+n such that k≥p, polynomial Hk on lpCn is continuous and Hk≤1.Proof.
Letx∈lpCn such that xp≤1. Note that (4)Hkx≤∑j=1∞∏s=1ks>0nxjsks.Since xjs≤max1≤m≤n⁡xjm for every s∈1,…,n and j∈N, it follows that (5)∏s=1ks>0nxjsks≤max1≤m≤nxjmkfor every j∈N. Note that (6)max1≤m≤nxjmk=max1≤m≤nxjmk≤∑m=1nxjmk.Therefore, (7)Hkx≤∑j=1∞∑m=1nxjmk.Since xp≤1, it follows that xjm≤1 for every m∈1,…,n and j∈N. Therefore, xjmk≤xjmp. Thus, (8)Hkx≤∑j=1∞∑m=1nxjmp=xpp≤1.Therefore, Hk=supxp≤1⁡Hkx≤1. Hence, Hk is bounded and, consequently, it is continuous.Form∈N, let c00mCn be the space of all sequences x=x1,…,xm,0,…, where x1,…,xm∈Cn and 0=0,…,0∈Cn. Note that c00mCn is isomorphic to Cnm. Let c00(Cn)=⋃m=1∞c00mCn. Note that c00Cn is a dense subspace in lpCn. Also note that Hk is well-defined on c00Cn for every k∈Z+n.For arbitraryx=x1,…,xm,0,…,y=y1,…,ys,0,…∈c00Cn, we set (9)x⊕y=x1,…,xm,y1,…,ys,0,….For x1,…,xr∈c00Cn, let (10)⨁j=1rxj=x1⊕⋯⊕xr.Note that(11)⨁j=1rxjpp=∑j=1rxjpp.Also note that for every k∈Z+n, such that |k|≥1,(12)Hk⨁j=1rxj=∑j=1rHkxj.For everym∈N and j∈{1,…,m}, we set(13)αmj=1m1/mexp⁡2πijm.Also we set α01=0. For l=l1,…,ln∈Z+n, let(14)al=⨁j1=1l1^⋯⨁jn=1ln^αl1j1,…,αlnjn,0,…,0,…,where lj^=max⁡1,lj for j∈1,…,n.Let us define a partial order onZ+n by the following way. For k,l∈Z+n we set k⪰l if and only if there exists m∈Z+n such that ks=msls for every s∈1,…,n. We write k≻l, if k⪰l and k≠l.Proposition 3.
Fork∈Z+n such that k≥1 and for arbitrary l∈Z+n(15)Hkal=∏s=1ks>0n1lsks/ls-1∏s=1ks=0nls^,ifk⪰l0,otherwise,where, by the definition, product of an empty set of multipliers is equal to 1. In particular, Hkak=1.Proof.
By (12) and (14), (16)Hkal=∑j1=1l1^⋯∑jn=1ln^Hkαl1j1,…,αlnjn,0,…,0,….By the definition of Hk, (17)Hkαl1j1,…,αlnjn,0,…,0,…=∏s=1ks>0nαlsjsks.Therefore,(18)Hkal=∑j1=1l1^⋯∑jn=1ln^∏s=1ks>0nαlsjsks=∏s=1ks>0n∑js=1ls^αlsjsks∏s=1ks=0n∑js=1ls^1=∏s=1ks>0n∑js=1ls^αlsjsks∏s=1ks=0nls^.
Letk⪰l. Then there exists m∈Z+n such that ks=msls for every s∈1,…,n. For s∈1,…,n such that ks>0, we have that ls>0 too. Consequently, for such s we have ls^=ls, and, by (13),(19)∑js=1ls^αlsjsks=∑js=1ls1ls1/lsexp⁡2πijslsmsls=1lsms∑js=1lsexp⁡2πijsms=1lsms∑js=1ls1=1lsms-1=1lsks/ls-1.Therefore, by (18), (20)Hkal=∏s=1ks>0n1lsks/ls-1∏s=1ks=0nls^.In the case k=l we have (21)Hkak=∏s=1ks>0n1ksks/ks-1∏s=1ks=0nks^=1.
Letk⪰l. Then we have two cases.Case1. There exists s∈1,…,n such that ks>ls=0. Then (22)∑js=1ls^αlsjsks=α01ks=0;therefore, Hkal=0.Case2. There exists s∈1,…,n such that ls>ks>0. Then (23)∑js=1ls^αlsjsks=∑js=1ls1ls1/lsexp⁡2πijslsks=1lsks/ls∑js=1lsexp⁡2πijslsks.It is known that(24)∑j=1qexp⁡2πijqr=0for every q∈2,3,… and r∈1,…,q-1. Therefore, (25)∑js=1lsexp⁡2πijslsks=0and, consequently, Hkal=0.Let us prove the following auxiliary proposition.Proposition 4.
A functiong:0,+∞→R,gx=c1x+⋯+cmx1/x, where m∈N and c1,…,cm>0, is strictly decreasing.Proof.
Let us prove thatg′x<0 for every x∈0,+∞. Note that gx=1/xln⁡c1x+⋯+cmx. Therefore,(26)g′x=gx-1x2ln⁡c1x+⋯+cmx+1xc1xln⁡c1+⋯+cmxln⁡cmc1x+⋯+cmx=-gxx2c1x+⋯+cmxc1x+⋯+cmxln⁡c1x+⋯+cmx-xc1xln⁡c1+⋯+cmxln⁡cm=-gxx2c1x+⋯+cmxc1xln⁡c1x+⋯+cmx-ln⁡c1x+⋯+cmxln⁡c1x+⋯+cmx-ln⁡cmx.Since gx/x2c1x+⋯+cmx>0 and lnc1x+⋯+cmx>ln⁡cjx for every j∈1,…,m, it follows that g′(x)<0.Corollary 5.
For everyx∈lpCn and for every q≥p(27)xp≥xq.For an arbitrary nonempty finite setM⊂Z+n let us define a mapping πM:c00Cn→C|M|, where M is the cardinality of M, by(28)πMx=Hkxk∈M,where Hkxk∈M is an M-dimensional vector of values of Hk on x, indexed by k∈M.We endow the space CM with norm ξ∞=maxk∈M|ξk|, where ξ=ξkk∈M∈CM.Theorem 6.
LetM be a finite nonempty subset of Z+n such that k≥1 for every k∈M. Then(i)
there existsm∈N such that for every ξ=ξkk∈M∈CM there exists xξ∈c00mCn such that πMxξ=ξ;(ii)
there exists a constantρM>0 such that if ξ∞<1, then xξp<ρM for every p∈1,+∞.Proof.
(i) Letξ=ξkk∈M∈CM. For every k∈M, let us define ηk∈C and bk∈c00Cn by the following way. For minimal elements k of the partially ordered set M,⪯, let ηk=ξk and bk=ηkkak, where ak is defined by (14) and (29)ηkk=ηkkeiarg⁡ηk/k,ifηk≠00,ifηk=0.For k∈M, which are not minimal elements of M,⪯, we define ηk and bk inductively by(30)ηk=ξk-∑l∈Ml≺kHkbl,(31)bk=ηkkak.We set xξ=⨁l∈Mbl. Note that xξ∈c00(m)(Cn), where (32)m=∑k∈Mmin⁡j∈N:ak∈c00jCn.For k∈M, by (12), Hkxξ=∑l∈MHkbl. Since Hk is a k-homogeneous polynomial,(33)Hkbl=ηllkHkal.By Proposition 3, Hkal is not equal to zero only for l∈M such that l⪯k. Therefore, (34)Hkxξ=Hkbk+∑l∈Ml≺kHkbl.By Proposition 3, Hkak=1, and therefore, by (33), Hkbk=ηk. Hence, (35)Hkxξ=ηk+∑l∈Ml≺kHkbl.Taking into account (30), we have Hkxξ=ξk. Hence, πMxξ=ξ.
(ii) Letξ=ξkk∈M∈CM be such that ξ∞<1. For k∈M let (36)k=max⁡s∈N:∃l1,…,ls∈Msuch thatl1≺⋯≺ls=k.Note that for minimal elements k∈M we have 〈k〉=1.
Let(37)C=max⁡1,maxk∈M⁡ak1.Let(38)r=maxk∈M⁡k,and for every j∈1,…,r let (39)μj=∏s=1j1+ms,where(40)ms=k∈M:k=s.Also we set μ0=1.
Note that for everyj∈1,…,r(41)μj=μj-11+mj=μj-1+μj-1mj=μj-2+μj-2mj-1+μj-1mj=⋯=μ0+μ0m1+μ1m2+⋯+μj-1mj.
Let us prove that for everyk∈M(42)bk1<μk-1Ck.We proceed by induction on k. In the case k=1, we have ηk=ξk, and therefore, bk1=ξk|k|ak1. Since ξk<1, it follows that bk1<ak1≤C=μ0C. If r=1, then (42) is proved. Let r≥2 and j∈2,…,r. Suppose that inequality (42) holds for every k∈M such that k∈1,…,j-1. Let us prove (42) for k∈M such that k=j. By (31) and (37),(43)bk1≤ηkkak1≤ηkkC.By (30), (44)ηk≤ξk+∑l∈Ml≺kHkbl.Since Hk is a k-homogeneous polynomial on the space l1Cn and Hk≤1, (45)Hkbl≤Hkbl1k≤bl1k.Therefore, taking into account ξk<1, we have (46)ξk+∑l∈Ml≺kHkbl<1+∑l∈Ml≺kbl1k.Therefore,(47)ηkk<1+∑l∈Ml≺kbl1k1/k.By Proposition 4,(48)1+∑l∈Ml≺kbl1k1/k≤1+∑l∈Ml≺kbl1.Note that if l≺k, then l<k. Therefore,(49)∑l∈Ml≺kbl1≤∑l∈Ml<kbl1.Since k=j,(50)∑l∈Ml<kbl1=∑s=1j-1∑l∈Ml=sbl1.By the induction hypothesis, if l=s, where s∈1,…,j-1, then bl1<μs-1Cs. Therefore,(51)∑l∈Ml=sbl1<∑l∈Ml=sμs-1Cs=μs-1Cs∑l∈Ml=s1=μs-1msCs.Since C≥1, it follows that Cs≤Cj-1 for every s∈1,…,j-1, and therefore,(52)1+∑s=1j-1μs-1msCs≤1+Cj-1∑s=1j-1μs-1ms≤1+∑s=1j-1μs-1msCj-1.Since μ0=1, by (41),(53)1+∑s=1j-1μs-1ms=μj-1.By (47)–(53),(54)ηkk<μj-1Cj-1.By (43) and (54), bk1≤μj-1Cj. Hence, inequality (42) holds for every k∈M.
By (11) and by Proposition 4, (55)xξ1≤∑l∈Mbl1.By (42),(56)∑l∈Mbl1=∑j=1r∑l∈Ml=jbl1<∑j=1r∑l∈Ml=jμj-1Cj=∑j=1rμj-1Cj∑l∈Ml=j1=∑j=1rμj-1mjCj≤∑j=1rμj-1mjCr<μ0+∑j=1rμj-1mjCr=μrCr.Set ρM=μrCr. We have that xξ1<ρM if ξ∞<1. By Corollary 5, xξp≤xξ1≤ρM for every p∈1,+∞.Corollary 7.
LetM=k1,…,ks⊂Z+n such that kj≥1 for every j∈1,…,s. Then there exists m∈N such that for every m′≥m polynomials Hk1,…,Hks are algebraically independent on c00m′Cn.Proof.
By Theorem6, there exists m∈N such that for every ξ=ξ1,…,ξs∈Cs there exists xξ∈c00mCn such that(57)Hkjxξ=ξjfor every j∈1,…,s. Let us show that Hk1,…,Hks are algebraically independent on c00m′Cn for every m′≥m. Let Q:Cs→C be a polynomial such that (58)QHk1x,…,Hksx=0for every x∈c00m′Cn. Set x=xξ. Taking into account (57), we have Qξ1,…,ξs=0 for arbitrary ξ1,…,ξs∈C, that is, Q≡0. Hence, Hk1,…,Hks are algebraically independent.


## 3.2. Algebraic Basis of the AlgebraPsl1Cn

Theorem 8.
EveryN-homogeneous polynomial P∈Psc00mCn, where m is an arbitrary positive integer, can be represented as an algebraic combination of polynomials Hk, where k∈Z+n such that 1≤k≤N.Proof.
We proceed by induction onm. In the case m=1 for x=x1,0,…∈c001Cn, we have (59)Px=∑k∈Z+nk=Nαkx11k1⋯x1nkn=∑k∈Z+nk=NαkHkx,where αk∈C. Suppose the statement holds for m-1 and prove it for m. Let P∈Psc00mCn and x=x1,…,xm,0,…∈c00mCn. Then Px can be represented as a sum of terms (60)βkxm1k1⋯xmnknfkx1,…,xm-1,0,…,where βk∈C,k=k1,…,kn∈Z+n such that 1≤k≤N, and fk is an N-k-homogeneous polynomial. Note that fk∈Psc00m-1Cn, and therefore, by the induction hypothesis, fkx1,…,xm-1,0,… can be represented as an algebraic combination of Hlx1,…,xm-1,0,…, where l∈Z+n such that 1≤l≤N-k. Note that (61)Hlx1,…,xm-1,0,…=Hlx-xm1l1⋯xmnln.Therefore, P(x) can be represented as an algebraic combination of Hlx and xm1,…,xmn. Since P and Hl are symmetric, it follows that together with term (62)γr1,…,rn,t1,…,tsxm1r1⋯xmnrnHl1t1x⋯Hlstsx,where γr1,…,rn,t1,…,ts∈C,l1,…,ls∈Z+n and r1,…,rn,t1,…,ts∈Z+, the sum must contain terms (63)γr1,…,rn,t1,…,tsxj1r1⋯xjnrnHl1t1x⋯Hlstsx,where j∈1,…,m-1. Therefore, Px can be represented as a sum of terms (64)γr1,…,rn,t1,…,ts1m∑j=1mxj1r1⋯xjnrnHl1t1x⋯Hlstsx.Since ∑j=1mxj1r1⋯xjnrn=Hrx, where r=r1,…,rn, it follows that P is an algebraic combination of polynomials Hk, where k∈Z+n such that 1≤k≤N.Theorem 9.
LetP:c00Cn→C be a symmetric N-homogeneous polynomial. Let MN=k∈Z+n:1≤k≤N. There exists a polynomial q:CMN→C such that P=q∘πMN, where the mapping πMN is defined by (28).Proof.
By Corollary7, there exists m∈N such that for every m′≥m polynomials Hk, where k∈M, are algebraically independent. Therefore, the representation, given by Theorem 8 for the restriction of P to c00m′Cn, is unique. Thus, for every m′≥m there exists a unique polynomial qm′:CMN→C such that Px=(qm′∘πMN)(x) for every x∈c00m′Cn. Since c00m′Cn⊃c00(m)(Cn), it follows that qm is the restriction of qm′ to πMN(c00(m)(Cn)). By Theorem 6, πMNc00mCn=CMN, and therefore, qm′≡qm. Let q=qm. Then Px=q∘πMNx for every x∈c00Cn.Theorem 10.
PolynomialsHk, where k∈Z+n, form an algebraic basis of the algebra Psl1Cn.Proof.
Let us prove that every symmetric continuous polynomial onl1Cn can be uniquely represented as an algebraic combination of polynomials Hk. It suffices to prove the statement only for homogeneous polynomials. Let P:l1Cn→C be a symmetric continuous N-homogeneous polynomial. By Theorem 9, the restriction of P to c00Cn can be uniquely represented as an algebraic combination of polynomials Hk, where k∈Z+n such that 1≤k≤N. Since c00Cn is dense in l1Cn and polynomials Hk are well-defined and continuous on l1Cn, it follows that given representation can be extended to l1Cn.


## 3.3. Algebraic Basis of the AlgebraPslpCn

Letp∈1,+∞. In this section, we describe an algebraic basis of the algebra PslpCn.Let us prove a complex analog of [8, Lemma 2].Lemma 11.
LetK⊂Cm and ϰ:K→Cm-1 be an orthogonal projection: ϰx1,x2,…,xm=x2,…,xm. Let K1=ϰ(K),intK1≠∅ and for every open set U⊂K1 a set ϰ-1U is unbounded. If polynomial Qx1,…,xm is bounded on K, then Q does not depend on x1.Proof.
Suppose thatQ depends on x1. Then (65)Qx1,…,xm=∑j=0kqjx2,…,xmx1j,where 1≤k≤degQ and qk≡0. Note that qk≡0 on intK1, and therefore, there exists point a∈intK1 such that qk(a)≠0. Since intK1 is open and qk is continuous, there exists r>0 such that Ba,r⊂intK1 and infb∈Ba,rqkb>0, where Ba,r is an open ball with center a and radius r in the space Cm-1. Note that, for x1,…,xm∈ϰ-1Ba,r,(66)Qx1,…,xm≥qkx2,…,xmx1k-∑j=0k-1qjx2,…,xmx1j≥cx1k-∑j=0k-1djx1j,where c=infb∈Ba,rqkb and dj=supb∈Ba,rqj(b) for j∈0,…,k-1. Note that for the polynomial cx1k+∑j=0k-1djx1j there exists R>0 such that if x1>R, then cx1k>2∑j=0k-1djx1j, that is, ∑j=0k-1djx1j<1/2c|x1|k. Therefore, if x1>R, then(67)cx1k-∑j=0k-1djx1j>cx1k-12cx1k=12cx1k.Since ϰ-1Ba,r is unbounded, there exists a sequence x1n,…,xmnn∈N⊂ϰ-1Ba,r such that x1n→∞ as n→+∞. Taking into account (66) and (67), we have (68)Qx1n,…,xmn>12cx1nk⟶+∞as n→+∞, which contradicts the boundedness of Q on K.Fork=k1,…,kn∈Z+n, let V(k)={s∈1,…,n:ks≠0} and νk=Vk.Lemma 12.
Fork,l∈Z+n if l≻k and νl≥νk, then l>k.Proof.
Sincel≻k, there exists m∈Z+n such that l1,…,ln=m1k1,…,mnkn, and l≠k. Therefore, if ks=0 for some s∈1,…,n, then ls=0 too. It means that Vl⊂Vk. On the other hand, νl≥νk. Therefore, Vl=Vk; that is, for s∈1,…,n, we have that ls≠0 if and only if ks≠0. Therefore, for every s∈Vl we have that ms≥1. Since l≠k, there exists s0∈Vl such that ms0≥2. Therefore, (69)l=m1k1+⋯+mnkn>k1+⋯+kn=k.ForN∈N and J∈1,…,n let(70)MNJ=l∈Z+n:1≤l<p,νl≥J∪l∈Z+n:p≤l≤N.By Theorem 6, for M=MN1 there exists ρ=ρM>0 such that πMVρ contains the open unit ball of the space CM with norm ·∞, where(71)Vρ=x∈c00Cn:xp<ρ.Proposition 13.
ForJ∈1,…,N, let qξll∈MNJ be a polynomial on CMNJ. If q is bounded on πMNJVρ, then q does not depend on ξk such that νk=J and 1≤k<p.Proof.
Letk∈Z+n such that νk=J and 1≤k<p. Let K=πMNJVρ,K1=πMN(J)∖{k}(Vρ) and ϰ:K→K1 be an orthogonal projection, defined by (72)ϰ:ξll∈MNJ⟼ξll∈MNJ∖k.Let us show that, for every ball (73)Bu,r=ξ∈CMNJ∖k:ξ-u∞<rwith center u=(ul)l∈MN(J)∖{k}∈C|MN(J)∖{k}| and radius r>0 such that Bu,r⊂πMNJ∖kVρ, a set ϰ-1Bu,r is unbounded. Since u∈πMNJ∖kVρ, there exists xu∈Vρ such that πMNJ∖kxu=u. For m∈N, we set xm=⨁j=1m1/j1/|k|ak, where ak is defined by (14). Choose ε such that (74)0<ε<min⁡1,ρ-xupakpζp/k1/p,rak1Nζ1+1/k,where ζ· is a Riemann zeta-function. Let xm,ε=εxm⊕xu. Let us show that xm,ε∈Vρ. By (11), (75)xmpp=∑j=1m1j1/kakpp=∑j=1m1jp/kakpp=akpp∑j=1m1jp/k<akppζpk.Therefore, xmp<akpζp/k1/p. By the triangle inequality, (76)xm,εp≤εxmp+xup<εakpζpk1/p+xup.Since ε<ρ-xup/akpζp/k1/p, it follows that xm,εp<ρ. Hence, xm,ε∈Vρ.
Note that for arbitraryl∈Z+n such that l≥1, by (12), (77)Hlxm=∑j=1m1jl/kHlak=Hlak∑j=1m1jl/k,(78)Hlxm,ε=εlHlxm+Hlxu=εlHlak∑j=1m1jl/k+Hlxu.Let us show that πMNJ∖kxm,ε∈Bu,r. For l∈MNJ∖k such that l≻k, by Proposition 3, Hlak=0, and therefore, by (78), (79)Hlxm,ε=Hlxu=ul.Let l∈MNJ∖k be such that l≻k. If p≤l≤N, then l>k, since k<p. If 1≤l<p and νl≥J, then l>k by Lemma 12. Hence, l>k in both cases. By (78), (80)Hlxm,ε-ul≤εlHlak∑j=1m1jl/k.Since ε<1, it follows that εl≤ε. Since Hl≤1, it follows that Hlak≤ak1l. Taking into account akp≥1 and l≤N, we have that |Hl(ak)|≤ak1N. Since l and k are integer numbers and l>k, it follows that l≥k+1, and therefore, (81)∑j=1m1jl/k≤∑j=1m1j1+1/k<ζ1+1k.Hence,(82)Hlxm,ε-ul<εak1Nζ1+1k.Since ε<r/ak1Nζ(1+1/|k|), it follows that Hlxm,ε-ul<r, and therefore, πMNJ∖kxm,ε∈Bu,r.
By Proposition3, Hk(ak)=1, and therefore, by (78), (83)Hkxm,ε=εl∑j=1m1j+Hkxu⟶∞as m→+∞. Hence, ϰ-1Bu,r is unbounded. By Lemma 11, q does not depend on ξk.Theorem 14.
LetP∈PslpCn be an N-homogeneous polynomial. If N<p, then P≡0. Otherwise, there exists a unique polynomial q:CMp,N→C such that P=q∘πMp,Np, where Mp,N=k∈Z+n:p≤k≤N and πMp,Np:lpCn→C|Mp,N| is defined by πMp,Npx=Hkxk∈Mp,N.Proof.
LetP0 be the restriction of P to c00Cn. Note that P0 is a continuous symmetric N-homogeneous polynomial. By Theorem 9, there exists a unique polynomial q:CMN→C, where MN=MN1 such that P0=q∘πMN. Since P0 is continuous, P0 is bounded on Vρ, defined by (71). Therefore, q is bounded on πMNVρ.
Let us prove thatq does not depend on arguments ξk such that 1≤k<p by induction on νk. By Proposition 13, for J=1 we have that qξkk∈MN does not depend on arguments ξk such that νk=1 and 1≤k<p. Suppose that the statement holds for νk∈1,…,J-1, where J∈{2,…,n}, that is, qξkk∈MN does not depend on arguments ξk such that 1≤νk≤J-1 and 1≤|k|<⌈p⌉. Then the restriction of q to CMNJ, by Proposition 13, does not depend on ξk such that ν(k)=J and 1≤k<p. Hence, q does not depend on ξk such that 1≤k<p.
Since polynomialsHk, where k∈Mp,N, are well-defined and continuous on lpCn and c00Cn is dense in lpCn, it follows that P=q∘πMp,Np. Note that in the case N<p we have Mp,N=∅ and, therefore, P≡0.Corollary 15.
PolynomialsHk, where k∈l∈Z+n:l≥p∪0, form an algebraic basis of the algebra PslpCn.


## 4. Conclusions

Power sum symmetric polynomials onlpCn are algebraically independent and form an algebraic basis of the algebra of all continuous symmetric polynomials on lpCn.Results of this work generalize results of works [7, 8, 14].



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