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Axioms
, Volume 11 (2) – Jan 21, 2022

/lp/multidisciplinary-digital-publishing-institute/algebraic-basis-of-the-algebra-of-all-symmetric-continuous-polynomials-EBVDkpyCth

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axioms Article Algebraic Basis of the Algebra of All Symmetric Continuous Polynomials on the Cartesian Product of ` -Spaces 1, 2 2 Andriy Bandura * , Viktoriia Kravtsiv and Taras Vasylyshyn Department of Advanced Mathematics, Ivano-Frankivsk National Technical Unuversity of Oil and Gas, 76019 Ivano-Frankivsk, Ukraine Faculty of Mathematics and Computer Sciences, Vasyl Stefanyk Precarpathian National University, 76018 Ivano-Frankivsk, Ukraine; viktoriia.kravtsiv@pnu.edu.ua (V.K.); taras.vasylyshyn@pnu.edu.ua (T.V.) * Correspondence: andriykopanytsia@gmail.com Abstract: We construct a countable algebraic basis of the algebra of all symmetric continuous poly- nomials on the Cartesian product ` . . . ` , where p , . . . , p 2 [1, +¥), and ` is the complex p p 1 n p 1 n Banach space of all p-power summable sequences of complex numbers for p 2 [1, +¥). Keywords: symmetric polynomial on a Banach space; continuous polynomial on a Banach space; algebraic basis; space of p-summable sequences MSC: 46G25; 47H60; 46B45; 46G20 1. Introduction For classical results on symmetric polynomials on ﬁnite dimensional spaces, we refer to [1–3]. Symmetric polynomials on inﬁnite dimensional Banach spaces were studied, Citation: Bandura, A.; Kravtsiv, V.; ﬁrstly, by Nemirovski and Semenov in [4]. In particular, in [4] the authors constructed a Vasylyshyn, T. Algebraic Basis of the countable algebraic basis (see deﬁnition below) of the algebra of symmetric continuous Algebra of All Symmetric real-valued polynomials on the real Banach space ` and a ﬁnite algebraic basis of the Continuous Polynomials on the algebra of symmetric continuous real-valued polynomials on the real Banach space L [0, 1], Cartesian Product of ` -Spaces. where 1 p < +¥. Axioms 2022, 11, 41. https://doi.org/ In [5], these results were generalized to separable sequence real Banach spaces with 10.3390/axioms11020041 symmetric basis (see, e.g., ([6], Deﬁnition 3.a.1, p. 113) for the deﬁnition of a Banach space Academic Editor: Clemente Cesarano with symmetric basis) and to separable rearrangement invariant function real Banach spaces (see, e.g., ([7], Deﬁnition 2.a.1, p. 117) for the deﬁnition of a rearrangement invariant Received: 27 November 2021 function Banach space) resp. In [8], it was shown that there are only trivial symmetric con- Accepted: 19 January 2022 tinuous polynomials on the space ` . Consequently, the results of [5] cannot be generalized Published: 21 January 2022 to nonseparable sequence Banach spaces. The most general approach to the studying of Publisher’s Note: MDPI stays neutral symmetric functions on Banach spaces was introduced in [9–13]. with regard to jurisdictional claims in Note that the existence of a ﬁnite or countable algebraic basis in some algebra of published maps and institutional afﬁl- symmetric continuous polynomials gives us the opportunity to obtain some information iations. or, even, to describe spectra of topological algebras of symmetric holomorphic functions, which contain the algebra of symmetric continuous polynomials as a dense subalgebra. For example, in [14], the authors constructed an algebraic basis of the algebra of symmetric continuous complex-valued polynomials on the complex Banach space L [0, 1] of complex- Copyright: © 2022 by the authors. valued Lebesgue measurable essentially bounded functions on [0, 1]. Licensee MDPI, Basel, Switzerland. This result gave us the opportunity to describe the spectrum of the Fréchet algebra This article is an open access article H (L [0, 1]) of symmetric analytic entire functions, which are bounded on bounded sets distributed under the terms and ¥ bs on the complex Banach space L [0, 1] (see [14]) and to show that the algebra H (L [0, 1]) conditions of the Creative Commons ¥ ¥ bs Attribution (CC BY) license (https:// is isomorphic to the algebra of all analytic functions on the strong dual of the topological creativecommons.org/licenses/by/ vector space of entire functions on the complex plane C (see [15]). 4.0/). Axioms 2022, 11, 41. https://doi.org/10.3390/axioms11020041 https://www.mdpi.com/journal/axioms Axioms 2022, 11, 41 2 of 14 In [16,17], there were constructed algebraic bases of algebras of symmetric continuous polynomials on Cartesian powers of complex Banach spaces L [0, 1] and L [0, +¥) of all p p complex-valued Lebesgue integrable in a power p functions on [0, 1] and [0, +¥) resp., where 1 p < +¥. These results gave us the opportunity to represent Fréchet algebras of symmetric entire analytic functions of bounded type on these Cartesian powers as Fréchet algebras of entire analytic functions on their spectra (see [18]). The spectra of algebras with countable algebraic bases and completions of such alge- bras also were studied in [19–21]. Symmetric analytic functions of unbounded type were studied in [22–25]. Applications of symmetric analytic functions to the spectra of linear operators were introduced in [26]. Symmetric polynomials and symmetric holomorphic functions on spaces ` were studied by a number of authors [22,27–41] (see also the survey [42]). Symmetric polynomi- als and symmetric holomorphic functions on Cartesian powers of spaces ` were studied in [43–47]. In particular, in [46] there was constructed a countable algebraic basis of the algebra of all symmetric continuous complex-valued polynomials on the Cartesian power of the complex Banach space ` . This result was generalized to the real case in [47]. In this work, we generalize the results of the work [46] to the algebra of symmetric continuous polynomials on the arbitrary Cartesian product ` . . . ` . p p 2. Preliminaries We denote by N and Z the set of all positive integers and the set of all nonnegative integers resp. 2.1. Polynomials Let X be a complex Banach space with norm kk . A function P : X ! C is called an N-homogeneous polynomial if there exist N 2 N and an N-linear form A : X ! C such that P is the restriction of A to the diagonal, i.e., P(x) = A x, . . . , x | {z } for all x 2 X. A function P : X ! C, which can be represented in the form P = P + P + . . . + P , (1) 0 1 N where P is a constant function on X and P : X ! C is a j-homogeneous polynomial for 0 j every j 2 f1, . . . , Ng, which is called a polynomial of degree at most N. It is known that a polynomial P : X ! C is continuous if and only if its norm kPk = sup jP(x)j kxk 1 is ﬁnite. Consequently, if P : X ! C is a continuous N-homogeneous polynomial, then we have jP(x)j kPkkxk (2) for every x 2 X. For details on polynomials on Banach spaces, we refer the reader to [48] or [49,50]. 2.2. Algebraic Combinations and Algebraic Basis Let functions f , f , . . . , f act from T to C, where T is an arbitrary nonempty set. If 1 m there exists a polynomial Q : C ! C such that f (x) = Q( f (x), . . . , f (x)) 1 Axioms 2022, 11, 41 3 of 14 for every x 2 T, then the function f is called an algebraic combination of functions f , . . . , f . 1 m A set f f , . . . , f g is called algebraically independent if the fact that Q( f (x), . . . , f (x)) = 0 for all x 2 T implies that the polynomial Q is identically equal to zero. An inﬁnite set of functions is called algebraically independent if every ﬁnite subset is algebraically indepen- dent. Note that the algebraic independence implies the uniqueness of the representation in the form of an algebraic combination. Let A be an algebra of functions. A subset B of A is called an algebraic basis of A if each element of A can be uniquely represented as an algebraic combination of some elements of B. 2.3. Symmetric Polynomials on the Space c (C ) (m) For m 2 N, let c (C ) be the space of all sequences x = (x , . . . , x , 0, . . .), where 1 m n n n n m x , . . . , x 2 C and 0 = (0, . . . , 0) 2 C . Note that c (C ) is isomorphic to (C ) . Let m 00 (m) n ¥ n c (C ) = c (C ). m=1 00 A function f : c (C ) ! C is called symmetric if f (x s) = f (x) for every x = (x , x , . . .) 2 c (C ) and for every bijection s : N ! N, where 1 2 00 x s = x , x , . . . . s(1) s(2) n n For k 2 Z nf(0, . . ., 0)g, let us deﬁne a polynomial H : c (C ) ! C by k 00 ¥ n (s) H (x) = x , (3) k å Õ j=1 s=1 k >0 where (1) (n) (1) (n) x = x , . . . , x , x , . . . , x , . . . 2 c (C ). 1 1 2 2 n n M Let M be a nonempty ﬁnite subset of Z nf(0, . . ., 0)g. Let C be the vector space of all functions x : M ! C. Elements of the space C can be considered as j Mj-dimensional complex vectors x = (x ) , indexed by elements of M, wherej Mj is the cardinality of M. k k2 M M j Mj M Thus, C is isomorphic to C . The space C we endow with normkxk = max jx j, k2 M k where x = (x ) 2 C . k k2 M For a nonempty ﬁnite subset M of Z nf(0, . . ., 0)g, let us deﬁne a mapping p : n M c (C ) ! C by p (x) = ( H (x)) , (4) M k k2 M where x 2 c (C ). Theorem 1 ([46], Theorem 9). Let P : c (C ) ! C be a symmetric N-homogeneous polynomial. n M Let M = fk 2 Z : 1 jkj Ng. There exists a polynomial q : C ! C such that P = q p , where the mapping p is deﬁned by (4). M M N N We shall use the following lemma. m m1 Lemma 1 ([46], Lemma 11). Let K C and { : K ! C be an orthogonal projection: { (x , x , . . . , x ) = (x , . . . , x ). Let K = {(K), int K 6= Æ and for every open set U K 1 2 m 2 m 1 1 1 a set { (U) is unbounded. If polynomial Q(x , . . . , x ) is bounded on K, then Q does not depend on x . 1 Axioms 2022, 11, 41 4 of 14 3. The Main Result Let n 2 N, p , . . . , p 2 [1, +¥) and p = ( p , . . . , p ). We shall consider the Cartesian n n 1 1 power ` . . . ` of the complex spaces ` , . . . ,` as the space of all sequences p p p p n n 1 1 x = (x , x , . . .), (5) 1 2 (1) (n) (s) (s) where x = x , . . . , x 2 C for j 2 N, such that the sequence x , x , . . . belongs j j 1 to ` for every s 2 f1, . . . , ng. We endow ` . . . ` with norm p p p s 1 n 1/max p max p (s) (s) kxk = x , x , . . . , ` ...` å p pn 1 2 s=1 where k k is the norm of the space ` . Note that c (C ) is a dense subspace of p p 00 s s ` . . . ` . p p Analogically to the deﬁnition of symmetric functions on c (C ), a function f : ` . . . ` ! C is called symmetric if p p f (x s) = f (x) for every x = (x , x , . . .) 2 ` . . . ` and for every bijection s : N ! N, where p p 1 2 n x s = x , x , . . . . s(1) s(2) Let k = (k , . . . , k ) 2 Z nf(0, . . ., 0)g be such that k / p + . . . + k / p 1. Let us n n n 1 1 1 deﬁne a polynomial H : ` . . . ` ! C by p p p,k n ¥ n (s) H (x) = x . (6) p,k å Õ j=1 s=1 k >0 Note that the polynomial H is symmetric. Let us show that H is well-deﬁned p,k p,k and continuous. n (1) (n) Lemma 2. Let k 2 Z nf(0, . . ., 0)g be such that k / p + . . . + k / p 1. Let z , . . . , z 2 1 1 n n (1) (n) C be such that z 1, . . . , z 1. Then, k k p p 1 n 1 n (1) (n) (1) (n) z z z + . . . + z . Proof. Note that k / p +...+k / p 1 1 n n k / p k / p k k p 1 1 p n n p 1 n 1 n s (1) (n) (1) (n) (s) z z z z max z . 1sn (s) Since max z 1, taking into account the inequality k / p + . . . + k / p 1, n n 1sn 1 1 k / p +...+k / p 1 1 n n p p s s (s) (s) max z max z . 1sn 1sn Note that p p p s 1 n (s) (1) (n) max z z + . . . + z . 1sn Thus, k k p p 1 n 1 n (1) (n) (1) (n) z z z + . . . + z . This completes the proof. Axioms 2022, 11, 41 5 of 14 Proposition 1. The polynomial H , deﬁned by (6), is well-deﬁned and k H k n. p,k p,k Proof. Let us show that H is well-deﬁned. Let x 2 ` . . . ` be of the form (5). p p p,k 1 n p p 1 n (1) (n) ¥ ¥ Since the series x , . . . , x are convergent, it follows that there exists å å j=1 j=1 j j (1) (n) M 2 N such that x 1, . . . , x 1 for every j M. Therefore, for j M, taking j j into account the inequality k / p + . . . + k / p 1, by Lemma 2, n n 1 1 k k p p 1 n 1 n (1) (n) (1) (n) x x x + . . . + x . j j j j Consequently, ¥ ¥ k k p p 1 n 1 n (1) (n) (1) (n) x x x + . . . + x å å j j j j j= M j=1 p p 1 n (1) (1) (n) (n) = x , x , . . . + . . . + x , x , . . . < ¥. 1 2 1 2 p p 1 n (s) ¥ n Therefore, the series x is absolutely convergent. Thus, H is well- å Õ j=1 s=1 p,k deﬁned. Let us show that k H k n. Let x 2 ` . . . ` be such that kxk 1. p p p,k n ` ...` 1 p p (1) (n) Then, x 1, . . . , x 1 for every j 2 N. Therefore, for every j 2 N, by Lemma 2, j j k k p p n n 1 1 (1) (n) (1) (n) x x x + . . . + x . j j j j Consequently, ¥ ¥ k k p p 1 n 1 n (1) (n) (1) (n) j H (x)j x x x + . . . + x p,k å å j j j j j=1 j=1 p p 1 n (1) (1) (n) (n) = x , x , . . . + . . . + x , x , . . . 1 2 1 2 p p p p kxk + . . . +kxk n. ` ...` ` ...` p p p p n n 1 1 Thus, k H k n. This completes the proof. p,k For arbitrary x = (x , . . . , x , 0, . . .), y = (y , . . . , y , 0, . . .) 2 c (C ), we set 1 m 1 s 00 x y = (x , . . . , x , y , . . . , y , 0, . . .). 1 m 1 s (1) (r) n For x , . . . , x 2 c (C ), let (j) (1) (r) x = x . . . x . j=1 Note that r max p M max p (j) (j) x = x . (7) ` ...` p p ` ...` j=1 1 j=1 p p n n Note that for every k 2 Z nf(0, . . ., 0)g, (j) (j) H x = H x . (8) k å k j=1 j=1 Axioms 2022, 11, 41 6 of 14 For every m 2 N and j 2 f1, . . . , mg, we set g = exp(2pi j/m). (9) mj 1/m We set g = 0. For l = (l , . . . , l ) 2 Z nf(0, . . ., 0)g, let 01 1 n 1 n M M a = . . . (g , . . . , g ), (0, . . . , 0), . . . , (10) l l j l j 1 1 n n j =1 j =1 where l = maxf1, l g for j 2 f1, . . . , ng. j j n n Let us deﬁne a partial order on Z nf(0, . . ., 0)g in the following way. For k, l 2 n n Z nf(0, . . ., 0)g, we set k l if and only if there exists m 2 Z such that k = m l for s s s + + every s 2 f1, . . . , ng. We write k l, if k l and k 6= l. By ([46], Proposition 3), for every k, l 2 Z nf(0, . . ., 0)g, n n < l , if k l Õ Õ k /l 1 s s s=1 s=1 H (a ) = , (11) k l k >0 k =0 > s s 0, otherwise where by the deﬁnition, the product of an empty set of multipliers is equal to 1. In particular, H (a ) = 1. (12) k k (1) (n) (1) (n) n n For (l , . . . , l ) 2 C and x = x , . . . , x , x , . . . , x , . . . 2 c (C ), let 1 n 00 1 1 2 2 (1) (n) (1) (n) (l , . . . , l ) x = l x , . . . , l x , l x , . . . , l x , . . . . n n n 1 1 1 1 1 2 2 It can be easily veriﬁed that H ((l , . . . , l ) x) = H (x) l , (13) k 1 k Õ s s=1 k >0 n n where k 2 Z nf(0, . . ., 0)g. Note that max p max p (s) (s) k(l , . . . , l ) xk = l x , l x , . . . 1 n s s ` ...` 1 2 p p s=1 n n max p max p (s) (s) (s) (s) max p max p = jl j x , x , . . . jl j x , x , . . . s s å å 1 2 1 2 p 1 s=1 s=1 n n max p max p max p max p jl j kxk = kxk jl j å s n n å s ` ...` ` ...` 1 1 1 1 s=1 s=1 max p max p max p max p kxk n max jl j = nkxk max jl j . (14) s s n n ` ...` ` ...` 1 1 1 1 s2f1...,ng s2f1...,ng For k = (k , . . . , k ) 2 Z let V(k) = s 2 f1, . . . , ng : k 6= 0 and n(k) = jV(k)j. n s 1 + n n Lemma 3. Let k, l 2 Z nf(0, . . ., 0)g be such that l k. If k = 0 for some s 2 f1, . . . , ng, then l = 0. Consequently, V(l) V(k). Proof. Since l k, there exists m 2 Z such that l = m k for every s 2 f1, . . . , ng. s s s Consequently, if k = 0, then l = 0. s s Axioms 2022, 11, 41 7 of 14 If s 2 V(l), then l > 0. Therefore, k cannot be equal to zero. Consequently, s 2 V(k). s s Thus, V(l) V(k). This completes the proof. n n Lemma 4. Let k, l 2 Z nf(0, . . ., 0)g be such that l k and n(l) n(k). Then, n n l 1 k s s + . å å p max p p s s s=1 s=1 Proof. By Lemma 3, V(l) V(k). On the other hand, n(l) n(k), i.e., jV(l)j jV(k)j. Therefore, V(l) = V(k). Consequently, l = 0 if and only if k = 0. s s Since l k, there exists m 2 Z such that l = m k for every s 2 f1, . . . , ng. Since s s s l = 0 if and only if k = 0, it follows that m > 0 for every s 2 f1, . . . , ng such that k > 0. s s s s Since l 6= k, there exists s 2 f1, . . . , ng such that m 0 2. Consequently, n n n n n l k m k k (m 1)k m 0 1 1 1 s s s s s s s = = . å å å å å p p p p p p 0 p 0 max p s s s s s s s s=1 s=1 s=1 s=1 s=1 This completes the proof. For N 2 N and J 2 f1, . . . , ng, let ( J) M = l 2 Z nf(0, . . ., 0)g : l / p + . . . + l / p < 1, jlj N and n(l) J 1 1 n n [fl 2 Z nf(0, . . ., 0)g : l / p + . . . + l / p 1 and jlj Ng. (15) n n + 1 1 Note that ([46], Theorem 6) implies the following theorem. n n Theorem 2. Let M be a ﬁnite non-empty subset of Z nf(0, . . ., 0)g. Then, (m) M n (i) there exists m 2 N such that, for every x = (x ) 2 C there exists x 2 c (C ) such k k2 M x that p (x ) = x; and (ii) there exists a constant r > 0 such that if kxk < 1, then kx k n < r . M ¥ x M ` ...` 1 1 (1) By Theorem 2, for M = M , there exists r = r > 0 such that p (V ) contains the M M open unit ball of the space C with the norm kk , where 0 n V = x 2 c (C ) : kxk n < r . r ` ...` 1 1 Let V = x 2 c (C ) : kxk < r . (16) r 00 ` ...` p p n 0 Since kxk kxk n for every x 2 c (C ), it follows that V V . ` ...` 00 r p p ` ...` r n 1 1 Consequently, p (V ) also contains the open unit ball of the space C . M r ( J) Proposition 2. For J 2 f1, . . . , ng, let q (x ) be a polynomial on C . If q is bounded ( J) l2 M ( J) on p (V ), then q does not depend on x , where k 2 M is such that n(k) = J and k / p + ( J) r k 1 1 . . . + k / p < 1. n n Proof. Let k 2 Z be such that n(k) = J and k / p + . . . + k / p < 1. Let K = p (V ), ( J) 1 1 n n r K = p (V ) and { : K ! K be an orthogonal projection, deﬁned by ( J) 1 r 1 M nfkg { : (x ) 7! (x ) . ( J) ( J) l l l2 M l2 M nfkg N N Axioms 2022, 11, 41 8 of 14 Let us show that, for every ball ( J) M nfkg B(u, r) = x 2 C : kx uk < r ( J) M nfkg with center u = (u ) 2 C and radius r > 0 such that B(u, r) p (V ), ( J) ( J) r l2M nfkg M nfkg N N a set { (B(u, r)) is unbounded. Since u 2 p (V ), there exists x 2 V such that ( J) r u r M nfkg p (x ) = u. For m 2 N, we set ( J) u M nfkg x = (h(j, k ), . . . , h(j, k )) a , m n 1 k j=1 where a is deﬁned by (10) and w p h(j, s) = , for j 2 N and s 2 f1, . . . , ng, where k k w = + . . . + . p p 1 n Since 0 < w < 1, it follows that 1/w > 1. Consequently, the value z(1/w) is ﬁnite, where z() is the Riemann zeta function. Choose # such that rkx k ` ...` p p 0 < # < min 1, , 1/ max p ka k n (nz(1/w)) ` ...` 1 1 n o 1 1 1 1 N rn max ka k ,ka k z min , 1 + . k ` ...` k p p ` ...` n p p 1 w w max p Let x = (#x ) x . Let us show that x 2 V . By (7), m,# m u m,# r max p max p kx k = k(h(j, 1), . . . , h(j, n)) a k . å k ` ...` ` ...` p p p p n n 1 1 j=1 By (14), max p max p max p k(h(j, 1), . . . , h(j, n)) a k nka k max jh(j, s)j . k k n ` ...` ` ...` p p n 1 1 s2f1...,ng Note that 1 1 w p w max p 1 1 max jh(j, s)j = max = . j j s2f1...,ng s2f1...,ng Therefore, max p max p k(h(j, 1), . . . , h(j, n)) a k nka k . k k n ` ...` ` ...` p p n 1 1 1 j Consequently, Axioms 2022, 11, 41 9 of 14 max p max p kx k nka k m k n å ` ...` ` ...` p p n 1 1 1 j j=1 max p max p < nka k = nka k z(1/w). k n å k n ` ...` ` ...` 1 1 1 1 j=1 Therefore, 1/ max p kx k < ka k n (nz(1/w)) . ` ...` k ` ...` p p n 1 1 By the triangle inequality, kx k #kx k +kx k m,# m u ` ...` ` ...` ` ...` p p p p p p n n n 1 1 1 1/ max p < #ka k nz(1/w) +kx k . n ( ) k ` ...` ` ...` p p 1 1 n rkx k ` ...` p p Since # < , it follows thatkx k < r. Hence, x 2 V . m,# ` ...` m,# r 1/ max p p p ka k (nz(1/w)) ` ...` 1 1 Note that, for arbitrary l 2 Z nf(0, . . ., 0)g, by (8), H (x ) = H (h(j, 1), . . . , h(j, n)) a . l m å l k j=1 By (13), H (h(j, 1), . . . , h(j, n)) a = H (a ) h(j, s) l k l k Õ s=1 s n 1 n n å l /(w p ) å l / p s s s s w p s=1 w s=1 1 1 1 = H (a ) = H (a ) = H (a ) . l k Õ l k l k j j j s=1 Therefore, 1 n m å l / p s s w s=1 H (x ) = H (a ) . l m l k å j=1 Consequently, taking into account (8), we have 1 n å l / p m s s w s=1 jlj jlj H (x ) = # H (x ) + H (x ) = # H (a ) + H (x ). (17) l m,# l m l u l k å l u j=1 ( J) Let us show that p (x ) 2 B(u, r). For l 2 M nfkg such that l 6 k, by (11), ( J) m,# M nfkg H (a ) = 0, therefore, by (17), l k H (x ) = u . m,# l l ( J) Let l 2 M nfkg be such that l k. Consider the case l / p + . . . + l / p 1 and n n 1 1 jlj N. Since l / p + . . . + l / p 1, it follows that 1 1 n n 1 l 1 w p w s=1 Consider the case l / p + . . . + l / p < 1, jlj N and n(l) J. Since n(k) = J, it 1 1 n n follows that n(l) n(k). By Lemma 4, since l k and n(l) n(k), 1 l 1 1 + . w p w max p s=1 Axioms 2022, 11, 41 10 of 14 ( J) Thus, for l 2 M nfkg such that l k, we have l 1 1 min , 1 + . (18) p w w max p s=1 By (17), taking into account the equality H (x ) = u , we have l u l 1 n m å l / p s s w s=1 jlj j H (x ) u j # j H (a )j . l m,# l l k å j=1 jlj Since # < 1, it follows that # #. By (2), taking into account the inequality k H k n, jlj jlj we have j H (a )j nka k . Since 1 jlj N, for every b > 0, we have b l k k ` ...` p p max b, b . Therefore, n o jlj ka k max ka k ,ka k . k k ` ...` k ` ...` ` ...` p pn p p p pn n 1 1 1 Thus, n o j H (a )j n max ka k ,ka k . l k k ` ...` k p p ` ...` n p p 1 n By (18), n o 1 n 1 1 l / p min ,1+ m å s s m s=1 w w w max p 1 1 1 1 < z min , 1 + . å å j j w w max p j=1 j=1 Hence, n o 1 1 j H (x ) u j < #n max ka k ,ka k z min , 1 + . l m,# l k ` ...` k ` ...` p pn p pn 1 w w max p Since n o 1 1 # < r n max ka k ,ka k z min , 1 + , k ` ...` k p p ` ...` n p p 1 n w w max p it follows that j H (x ) u j < r, therefore, p (x ) 2 B(u, r). m,# ( J) m,# l l M nfkg By (12), H (a ) = 1, therefore, by (17), k k jkj H (x ) = # + H (x ) ! ¥ k m,# å k u j=1 as m ! +¥. Hence, { (B(u, r)) is unbounded. By Lemma 1, q does not depend on x . This completes the proof. Theorem 3. Let P : ` . . . ` ! C be an N-homogeneous symmetric continuous polyno- p p p,N mial. If N < min p, then P 0. Otherwise, there exists the polynomial q : C ! C such that (p) P = q ˆ p , where p,N n o M = k 2 Z nf(0, . . ., 0)g : k / p + . . . + k / p 1 and jkj N p,N 1 1 n n (p) ( p) p,N and p : ` . . . ` ! C is deﬁned by p (x) = ( H (x)) . p p p,k k2 M 1 n M M p,N p,N p,N Axioms 2022, 11, 41 11 of 14 ˜ ˜ Proof. Let P be the restriction of P to c (C ). Note that P is a symmetric N-homogeneous polynomial. By Theorem 1, there exists a unique polynomial q : C ! C such that P = q p . (19) Since P is continuous, P is bounded on V , deﬁned by (16). Consequently, P is bounded (1) (1) on V . Therefore, q is bounded on p (V ). Note that M = M , where M is deﬁned r r M N N N N by (15). Let us prove that q does not depend on arguments x such that k / p + . . . + k / p < 1 k 1 1 n n by induction on n(k). By Proposition 2, for J = 1, we have that q (x ) does not k k2 M depend on arguments x such that n(k) = 1 and k / p + . . . + k / p < 1. Suppose the k 1 1 n n statement holds for n(k) 2 f1, . . . , J 1g, where J 2 f2, . . . , ng, i.e., q (x ) does not k k2 M depend on arguments x such that 1 n(k) J 1 and k / p + . . . + k / p < 1. Then, 1 1 n n ( J) the restriction of q to C , by Proposition 2, does not depend on x such that n(k) = J and k / p + . . . + k / p < 1. Hence, q does not depend on x such that k / p + . . . + k / p < 1. n n n n 1 1 k 1 1 Consider the case N < min p. In this case, k / p + . . . + k / p < 1 for every k 2 M . 1 1 n n N ˜ ˜ Consequently, q is constant. Therefore, taking into account (19), P is constant. Since P is an N-homogeneous polynomial, where N > 0, it follows that P is identically equal to zero. By the continuity of P, taking into account that P is the restriction of P to the dense subspace c (C ) of the space ` . . . ` , the polynomial P is identically equal to zero. 00 p p Consider the case N min p. In this case, M 6= Æ. Since q does not depend on x p,N k such that k 2 M n M , the equality (19) implies the following equality: N p,N P = q p , (20) p,N p,N N where q ˆ is the restriction of q to C , which is the subspace of C . Let us show (p) that P = q ˆ p . Let x 2 ` . . . ` . Since c (C ) is dense in ` . . . ` , p p 00 p p n n M 1 1 p,N ¥ n there exists the sequence fx g c (C ), which is convergent to x. Since H is 00 p,k m=1 continuous and H is the restriction of H , it follows that lim H (x ) = H (x) m!¥ m k p,k k p,k (p) for every k 2 M . Therefore, lim p (x ) = p (x). Since q ˆ is the poly- p,N m!¥ M m p,N M p,N nomial on a ﬁnite dimensional space, it follows that q ˆ is continuous. Consequently, lim q ˆ p (x ) = q ˆ p (x). On the other hand, since P is continuous, (p) m!¥ M m p,N p,N taking into account (20), we have lim q ˆ p (x ) = lim P(x ) = lim P(x ) = P(x). m m m p,N m!¥ m!¥ m!¥ Therefore, P(x) = q ˆ p (x). Thus, P = q ˆ p . This completes the proof. (p) (p) M M p,N p,N Proposition 3. The set of polynomials H : k 2 Z nf(0, . . ., 0)g such that k / p + . . . + k / p 1 (21) n n p,k + 1 1 is algebraically independent. Proof. By ([46], Theorem 10), the set of polynomials n o H n : k 2 Z nf(0, . . ., 0)g (1,...,1),k + is an algebraic basis of the algebra of all symmetric continuous complex-valued polynomials on ` . . . ` . Consequently, this set of polynomials is algebraically independent. Since 1 1 Axioms 2022, 11, 41 12 of 14 every subset of an algebraically independent set is algebraically independent, the set of polynomials n o H n : k 2 Z nf(0, . . ., 0)g such that k / p + . . . + k / p 1 n n (1,...,1),k + 1 1 is algebraically independent. Since H n is the restriction of H for every k 2 (1,...,1),k p,k n n Z nf(0, . . ., 0)g such that k / p + . . . + k / p 1, it follows that the set (21) is alge- 1 1 n n braically independent. This completes the proof. Theorem 4. The set of polynomials (21) is an algebraic basis of the algebra of all symmetric continuous complex-valued polynomials on ` . . . ` . p p 1 n Proof. Let P : ` . . . ` ! C be a symmetric continuous complex-valued polynomial p p of degree at most N, where N 2 Z . Then, P = P + P + . . . + P , 0 1 N where P 2 C and P is a j-homogeneous polynomial for every j 2 f1, . . . , Ng. By the 0 j Cauchy Integral Formula for holomorphic functions on Banach spaces (see, e.g., ([48], Corollary 7.3, p. 47)), 1 P(tx) P (x) = dt j+1 2pi jtj=r t for every j 2 f1, . . . , Ng, x 2 ` . . . ` and r > 0, where t 2 C. Consequently, P is p p j 1 n symmetric and continuous for every j 2 f1, . . . , Ng. Therefore, by Theorem 3, P can be represented as an algebraic combination of elements of the set (21) for every j 2 f1, . . . , Ng. Consequently, P can be represented as an algebraic combination of elements of the set (21). 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Axioms – Multidisciplinary Digital Publishing Institute

**Published: ** Jan 21, 2022

**Keywords: **symmetric polynomial on a Banach space; continuous polynomial on a Banach space; algebraic basis; space of *p*-summable sequences

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