Zeros of real polynomials on Banach spaces

ZEROS OF REAL POLYNOMIALS ON BANACH SPACES

Jos´e G. Llavona

Abstract. This paper is an expository presentation of a large part of the results, about zeros of real polynomials on Banach spaces, that have been obtained in recent years. Also new results, for or- thogonally additive polynomials on L

spaces, are given.

1. Introduction

The study of the zeros of a complex polynomial has a long history, with results coming via complex analysis, algebraic geometry, and func- tional analysis. (See, e.g. [10], [11], [16]). Similar studies for real poly- nomials are somewhat less common. In this paper we survey a large part of the results on zeros of real polynomials that have been obtained in re- cent years. If P and P _n are k-homogeneous polynomials and {P _n } → P pointwise or uniformly on bounded sets, what about the convergence of the zero sets Z(P _n )? Section 1 is dedicated to this problem. We give results and examples to show the diference between the complex and the real case.

If P is a homogeneous polynomial, what about the dimension of sub- spaces included in Z(P ) = P ⁻¹ (0)? Section 2 deals with this question.

Specifically treated is the case where the real Banach space does not admit a positive definite 2-homogeneous polynomial.

Finally section 3 is devoted to approximation by zeros of orthogonally additive polynomials on real l _p and L _p spaces. Results about the number of zeros that are involved in the decomposition of each e _j in l _p or each characteristic function in L _p [0, 1] are given.

Throughout, X will be a Banach space over K (R or C). P( ^k X) will denote the Banach space of k-homogeneous continuous polynomials on

Received November 19, 2002.

2000 Mathematics Subject Classification: 46G25, 47H60.

Key words and phrases: polynomials, Banach spaces, zeros of polynomials.

Supported in part by DGICYT Project BFM2000-0609.

X under the norm:

 P = Sup{| P (x) |: x ∈ X,  x ≤ 1}

(see [6], [14] for a general reference on polynomials on Banach spaces).

We will denote the closure of a set A in a topological space by cl(A).

Section 1

Definition 1.1. ([7]). We say that a sequence of nonempty closed subsets of a Banach space X, {A _n }, converges in the Mosco sense to a closed subset A, [(A _n −→ A) for short ], whenever the following two ^M conditions hold:

(i) for every x ∈ A there exists a sequence {x _n }, norm convergent to x, such that x _n ∈ A n for every n,

(ii) given J ⊆ Z ⁺ cofinal, for every sequence {x n

} j∈J weakly conver- gent to x, the condition x _n

∈ A _n

for every j, implies x ∈ A.

-Kuratowski convergence is defined in the same way changing weak convergence by norm convergence in (ii).

Notation: A _n −→ A ^K

Condition (i) is equivalent to A ⊂ L i A _n and (ii), in Kuratowski convergence, is equivalent to L _s A _n ⊂ A.

Recall that x ∈ L _i A _n ⇔ ∃(x _n ) : x = lim

n x _n and x _n ∈ A _n ∀n ∈ N, and that x ∈ L _s A _n ⇔ ∃ (x _k

) , k ₁ < k ₂ < · · · such that x = lim

n x _k

and x _k

∈ A k

, ∀n ∈ N.

-We say that a sequence of nonempty closed subsets of a Banach space X, {A _n }, converges in strong (respectively, Wijsman) sense to a closed set A, provided that the sequence {λ _n } converges to λ uniformly (respectively, pointwise) on bounded sets, where λ _n and λ are defined as:

λ(x) = d(x, A), λ _n (x) = d(x, A _n ).

Notation: A _n −→ A, A ^r n −→ A. W

Remark 1.1. Mosco convergence implies Kuratowski convergence. If X is a Schur space both convergences agree. If X is a finite dimensional Banach space then M, K, W and r-convergence agree. W -convergence is weaker than r-convergence but stronger than K-convergence ([4]).

It is assumed usually that the sets in the definition of Mosco con-

vergence are convex, and consequently weakly closed. We do not, but

let us observe that without that condition a constant sequence may be non-convergent.

In [5] it is proved that if X is a Banach space then r-convergence is equivalent to the following condition: ∀r > 0, ∀ε > 0, there exists n ₀ such that: (i) A + rB ⊂ A n + εB, (ii) A _n + rB ⊂ A + εB, ∀n ≥ n 0 , where B is the unit ball of the Banach space.

Given P ∈ P( ^k X) and α ∈ K, we will denote {x ∈ X : P (x) = α} by V (P − α). When α = 0 , V (P − 0) = Z(P ).

Definition 1.2. ([7], [8]). If {P _n } and P are k-homogeneous poly- nomials on X, we say that {P n } −→ P (Mosco convergent), {P ^M n } −→ P ^K (Kuratowski convergent), {P _n } −→ P (Wijsman convergent), {P ^W _n } −→ ^r P (r-convergent) if V (P _n − α) is M-convergent, K-convergent, W -conv- ergent, r-convergent to V (P − α) for all α ∈ K, α = 0, respectively.

In [7] it is shown that K-convergence is equivalent to uniform con- vergence on compacts sets and strong convergence is equivalent to norm convergence. Also {P _n } −→ P ⇐⇒ lim ^W

n cl(P _n (B)) = cl(P (B)), for every ball B ⊆ X.

In [8] it is shown that {P n } −→ P ⇐⇒ {P ^M n } −→ P uniformly on weakly compact sets and P ∈ P( ^k X) is weakly sequentially continuous.

What about the convergence of Z(P _n )? The different behavior of the case α = 0 is related to the fact that 0 is the unique critical value of a homogeneous polynomial, and consequently it is possible to have a change of the topology of V (P − α) near α = 0.

Proposition 1.1. ([7]). Let X be a complex Banach space, P and P _n nonzero k-homogeneous polynomials on X. If {P n } → P pointwise, then Z(P _n ) −→ Z(P ). ^K

Proof. We have that L _s Z (P _n ) ⊂ Z(P ), because if P _n

(x _n

) = 0 and {x n

} converges, then P

lim k x _n



= 0.

Let’s prove that Z(P ) ⊂ L _i Z(P _n ). Let x be such that P (x) = 0. If

there does not exists a sequence {x n } converging to x such that P n (x _n ) =

0, then we may assume that there exists an ε such that Z(P _n ) ∩B(x, ε) =

φ ∀n. Now we consider a complex line L = {x + λz : λ ∈ C} such that

P is not identically 0 on L. Let h ₀ and h _n denote the restriction to L of

P and P _n respectively, and Ω = B(x, ε) ∩ L. Then h n , h ₀ : Ω → C are

1-dimensional holomorphic functions, and {h _n } converges uniformly to

h ₀ on Ω because {P _n } converges uniformly to P on the compact cl(Ω).

The fact that h _n does not have zeros in Ω give us the following al- ternative (Hurwitz’s Theorem): either h ₀ is identically 0 or it does not have zeros in Ω. Both are impossible by the choice of L and the fact that h ₀ (x) = 0.

The following easy example shows us that in the real case things are worse.

Example 1.1. ([7]). X = R ² , k = 2, P _n (x, y) = x ² + _n ¹ y ² , P (x, y) = x ² . We have  P _n = P = 1 and lim

n P _n (x, y) = P (x, y) for all (x, y) ∈ X. But {Z(P _n ) } does not converge to Z(P ) in the Kuratowski sense, because Z(P _n ) = {(0, 0)} for all n and Z(P ) = {(0, y) : y ∈ R}.

However we have:

Proposition 1.2. ([7]). Let X be a real Banach space, P and P n

nonzero k-homogeneous polynomials on X such that dP (x) = 0, ∀x = 0.

If {P n } → P pointwise, then Z(P n ) −→ Z(P ). ^K

Proof. If (P _n ) → P pointwise, it is clear that {P n (x _n ) } → P (x) for every sequence {x _n } converging to x ∈ X. Then it is clear that L _s Z (P _n ) ⊂ Z(P ). To prove that Z(P ) ⊂ L _i Z(P _n ), we consider a point x ∈ X such that P (x) = 0. If x = 0, then clearly 0 ∈ L i Z(P _n ); otherwise we choose y such that dP (x)(y) = 0 and define Q _n (r) = P _n (x + ry) and Q(r) = P (x + ry); Q satisfies that Q ^ (0) = 0. But Q n and Q are polynomials of degree k over R such that {Q _n } → Q pointwise and consequently uniformly on bounded sets.

Now the fact that 0 is a root of Q and Q ^ (0) = 0 enables us to claim that there exists a sequence {λ _n } of roots of the Q _n converging to 0. So (x + λ _n y) ∈ Z(P n ) and lim

n (x + λ _n y) = x, and the proof is finished.

Proposition 1.3. ([7]). Let X be a complex Banach space, P and P n

nonzero k-homogeneous polynomials on X. If lim

n cl(P _n (B)) = cl(P (B)) for every open ball B ⊆ X, then Z(P _n ) −→ Z(P ). ^W

Proof. We consider λ _n (x) = d (x, Z(P _n )) and λ(x) = d (x, Z(P )).

Since {P n } → P pointwise, by [Th.1.17 [7]] P n −→ P and then it is K

obvious that lim sup λ _n (x) ≤ λ(x), and consequently if the result does not hold then there exists x such that lim inf λ _n (x) < λ(x) = λ. Passing to a subsequence if necessary, we may assume that lim

n λ _n (x) = λ ^∗ < λ ^ <

λ. Let x ₀ ∈ B(0, 1) be such that P (x ₀ ) = 0. Then 0 ∈ P _n (B(x, λ ^ )) for

every n, and therefore 0 ∈ cl (P (B(x, λ ^ ))). This allows us to choose a sequence {y _n } in B(x, λ ^ ) such that lim

n P (y _n ) = 0.

Let ε = λ −λ ^ and define φ _n : D(0 : ε) → C as φ _n (ω) = P (y _n + ωx ₀ ).

A subsequence of {φ n } converges uniformly to a 1-dimensional polyno- mial φ such that φ(0) = lim

n φ _n (0) = lim

n P (y _n ) = 0. On the other hand, φ _n does not vanish, because P does not have zeros on B(x, λ) by the definition of λ. Therefore using Hurwitz’s Theorem we conclude that φ must be identically 0, contradicting the fact that P (x ₀ ) = 0.

Proposition 1.3 is false in the real case even if we have stronger hy- pothesis:

Example 1.2. ([7]). Let P, P _n : c ₀ → R defined as:

P (x) =

 ∞ k=1

1

k ² (x _k − x _k+1 ) ²

P _n (x) =

 n k=1

1

k ² (x _k − x _k+1 ) ² .

We have {P n } → P uniformly on bounded sets and P (x) = 0 ⇐⇒ x = 0; Z(P ) = {0}, Z(P _n ) = {x ∈ c ₀ : x ₁ = · · · = x _n+1 }. Therefore we have that d(e ₁ , Z(P )) = 1 and d(e ₁ , Z(P _n )) = 1/2, and consequently lim n d(e ₁ , Z(P _n )) = d(e ₁ , Z(P )). Obviously lim

n cl(P _n (B)) = cl(P (B)), since {P _n } → P uniformly on bounded sets.

Remark 1.2. ([7]). The Example 1.2 shows that the following rela- tion, which is true in the complex case, does not hold in the real case:

 B = B(x ₀ ; r), 0 ∈ cl(P (B)), η > 0 =⇒ Z(P ) ∩ B(x ₀ ; r + η) = φ ^ . It is enough to consider x ₀ = e ₁ , r = 1/2, η = 1/4. Then it is clear that 0 / ∈ P (B(e 1 ; 1/2 + 1/4)) but 0 ∈ cl(P (B)) since

z _n = (1/2 + 1/n) e ₁ +

 n j=2

(1/2 − 1/n) e j ∈ B and

0 ≤ P (z _n ) ≤ 5/n ² → 0.

Remark 1.3. ([7]) In Example 1.2, dP (x) = 0 ⇐⇒ x = 0. Then by Proposition 1.2 we have that Z(P _n ) −→ Z(P ). ^K

Stronger conditions on P give us a similar result in the real case:

Proposition 1.4. ([7]). Let X be a real Banach space, P and P n

k-homogeneous polynomials on X such that P satisfies the following condition:

( ∗) lim

n  dP (y _n ) = 0 =⇒ 0 ∈ co({y _n }).

If lim

n cl(P _n (B)) = cl(P (B)) for every open ball B ⊆ X, then Z(P n )

−→ Z(P ). W

Proof. From the condition on the polynomial P we deduce that dP (x) = 0 if and only if x = 0, and therefore, by Proposition 1.2, Z (P _n ) −→ Z(P ). As in Proposition 1.3, if lim ^K

n λ _n (x) = λ(x) we would have {y _n } ⊂ B(x, λ ^ ) such that lim

n P (y _n ) = 0. Now we define φ _n : [ −ε, ε] −→ R, φ _n (t) = P (y _n + tz _n )

where z _n are points in the unit ball such that |dP (y n ) (z _n ) | > 1/2

 dP (y _n ) . These polynomials never vanish and converge to a polyno- mial φ such that φ(0) = 0. Then necessarily φ ^ (0) = 0 and lim

n dP (y _n ) (z _n )

= lim

n φ ^ _n (0) = 0, so we conclude that lim

n  dP (y _n ) = 0, which is not possible because 0 / ∈ B(x, λ) and c _o ( {y _n }) ⊂ B(x, λ).

Remark 1.4. ([7]). Condition (∗), in Proposition 1.4, on the poly- nomial P is weaker than the property of being a separating polynomial.

Proposition 1.5. ([7]). If X is a complex Banach space and P, P n

are nonzero k-homogeneous polynomials on X such that {P _n } → P uniformly on bounded sets, then Z(P _n ) −→ Z(P ). ^r

Proof. Let B denote the unit ball of X. First we will prove that

∀r, ε > 0 there exists n ₀ such that Z (P _n ) ∩ rB ⊂ Z(P ) + εB, ∀n ≥ n ₀ . Indeed, otherwise there would exists r, ε > 0 and a sequence {x _n } contained in rB such that P _n (x _n ) = 0 but B(x _n , ε) ∩ Z(P ) = φ. Now uniform convergence on rB of {P _n } gives us lim

n P (x _n ) = 0.

Let’s choose z ₀ ∈ B such that P (z ₀ ) = 0 and define φ n : D(0, ε) → C as φ _n (ω) = P (x _n +ωz ₀ ). Then φ _n (x) = P (x _n )+  _k

1

 A (x _n , . . . , x _n , z ₀ ) ω+

· · · +  _k

k−1

 A (x _n , z ₀ , . . . , z ₀ ) ω ^k−1 + P (z ₀ )ω ^k , where A denotes the k- linear form associated to P . A subsequence of {φ _n } converges to a polynomial φ such that φ ^(k) (0) = k!P (z ₀ ) = 0. But φ(0) = lim

n φ _n (0) =

lim n P (x _n ) = 0, and this, together with the fact that φ _n does nor vanishs,

gives us (Hurwitz’s Theorem) that φ = 0, which is a contradiction.

We proceed with the other inclusion: ∀r, ε > 0 there exists n 0 such that

Z(P ) ∩ rB ⊂ Z(P n ) + εB ∀n ≥ n 0 .

If this is not the case, then there exist r > 0 and ε > 0 and {x n } be such that x _n ∈ rB, P (x _n ) = 0 and Z(P _n ) ∩ B(x _n , ε) = φ. Let z ₀ ∈ B be such that P (z 0 ) = 0, and define φ n ,  φ _n : D(0; ε) → C as φ _n (ω) = P (x _n + ωz ₀ ) ,  φ _n (ω) = P _n (x _n + ωz ₀ ).  φ _n never vanishes.

We choose now a subsequence of {φ _n } converging to a polynomial φ identically zero, such that φ(0) = 0; it is clear that the corresponding subsequence of

 φ  _n



converges to φ too. Hurwitz’s Theorem again give us the contradiction.

The Example 1.2 shows that in the real case uniformly convergence, on bounded sets, of {P n } does not imply even Z(P n ) −→ Z(P ). ^W

The following example shows us that , even assuming W -convergence, r-convergence of Z(P _n ) to Z(P ) does not follow from uniform conver- gence on bounded sets (in the real case) of the sequence {P _n }.

Example 1.3. ([7]). Let X = c ₀ , P _n =

 n k=1

1

k ² x ² _k , P (x) =

 ∞ k=1

1 k ² x ² _k . Then {P n } → P uniformly and Z(P ) = {0}, Z(P n ) = {x ∈ c 0 : x ₁ = · · · = x _n = 0 }. Z(P _n ) −→ Z(P ) since d(x, Z(P )) = x  ^W and d(x, Z(P _n )) =  x  too when n > n 0 , where n ₀ is such that

| x _n |< 1/2  x _n , ∀ _n ≥ n ₀ . However Z(P _n ) ∩ 2B is never included in Z(P ) + ¹ ₂ B, since e _n+1 ∈ Z(P _n ) ∩ 2B but e _n+1 ∈ / ¹ ₂ B, ∀n. Therefore {Z(P _n ) } does not converge to Z(P ) in the strong sense.

Remark 1.5. ([7]). In Example 1.3, dP (x) = 2  ^∞

k=1

1

k ² x _k e ^∗ _k , and therefore  dP (x)  1 = 2

 ∞ k=1

1

k ² | x _k |. Hence inf{ dP (x)  1 : x ∈ S} = 0.

This fact suggests the following result.

Proposition 1.6. ([7]). Let X be a real Banach space, P and P n

k-homogeneous polynomials such that inf { dP (x) : x = 1} > 0. If {P _n } → P uniformly on bounded sets, then Z(P _n ) −→ Z(P ). ^r

Proof. First we will prove that, if B is the unit ball of X, then

∀r, ε > 0 there exists n ₀ such that

Z(P _n ) ∩ rB ⊂ Z(P ) + εB ∀n ≥ n ₀ .

Otherwise there exists r, ε > 0 and a sequence {x n } contained in rB such that P _n (x _n ) = 0 but B(x _n , ε) ∩ Z(P ) = φ. Now uniform convergence on rB of {P n } gives us lim

n P (x _n ) = 0. Since x _n ∈ rB − εB ∀n, we have a positive constant c such that  dP (x n ) > c ∀n, and we may choose y _n in the unit sphere such that | dP (x _n )(y _n ) |> c ∀n. Let’s define ϕ _n : ( −ε, ε) → R as ϕ n (r) = P (x _n + ry _n ). A subsequence of {ϕ n } converges to a polynomial ϕ such that ϕ(0) = lim

n ϕ _n (0) = lim

n P (x _n ) = 0, and the fact that ϕ _n does nor vanish gives us that ϕ ^ (0) = 0 or equivalently that lim n dP (x _n )(y _n ) = 0 which is a contradiction.

We are now going to prove the other inclusion: ∀r, ε > 0 there exists n ₀ such that

Z(P ) ∩ rB ⊂ Z(P _n ) + εB, ∀n ≥ n ₀ .

If this is not so, then there exist r > 0 and ε > 0 and {x _n } such that x _n ∈ rB, P (x _n ) = 0 and Z(P _n ) ∩ B(x _n , ε) = φ. Considering {y _n } as in the other inclusion, we may define

φ, φ _n : ( −ε, ε) → R as : φ(r) = lim

n P (x _n + ry _n ), φ _n (r) = P _n (x _n + ry _n ) passing to a subsequence if necessary; φ _n never vanishes, and {φ _n } → φ because of the uniform convergence of the sequence {P n } to P on (r + ε)B; but φ(0) = 0 and consequently φ ^ (0) = 0 too, and so is clear that lim

n dP (x _n )(y _n ) = 0, which gives us the contradiction.

Remark 1.6. ([7]). The condition inf{ dP (x) : x = 1} > 0 is weaker than the property of being a separating polynomial but strictly stronger than property ( ∗) in Proposition 1.4, as the polynomial P in Example 1.3 proves.

Finally in the real case we cannot infer Z(P _n ) −→ Z(P ) even under ^M very strong conditions as the following example shows:

Example 1.4. ([8]). Let us suppose that k is odd (the even case is easier). Let us take ϕ ₁ , ϕ ₂ ∈ X ^∗ linearly independent, (we are only as- suming that dim X > 1). Let us define P and P _n as ϕ ^k−1 ₁ (ϕ ₁ + ϕ ₂ ) and

ϕ ^k−1 ₁ + ¹ _n ϕ ^k−1 ₂

(ϕ ₁ + ϕ ₂ ) respectively. P ∈ P( ^k X) is weakly sequen- tially continuous, P = lim

n P _n in norm but Z(P ) = ker ϕ ₁ ∪ker (ϕ 1 + ϕ ₂ ) , Z(P _n ) = ker (ϕ ₁ + ϕ ₂ ) and consequently the sequence {Z(P n ) } does not converge to Z(P ) even in the Kuratowski sense.

However, in the real case, we have the following:

Proposition 1.7. ([8]). If P ∈ P( ^k X) is weakly sequentially con- tinuous and dP (x) = 0, for all x = 0, then the uniform convergence on weakly compacts sets of the sequence {P n } to P implies Z(P n ) −→ ^M Z(P ).

Proof. Since P is weakly sequentially continuous, the proof of the second condition is trivial. To establish the first condition we have to prove that for every x ∈ Z(P ), there exists a norm convergence to x sequence (x _n ) such that P _n (x _n ) = 0. If x = 0 the constant sequence x _n = 0 works. Hence we may assume x = 0; let us consider z ∈ S E such that dP (x)(z) = 0. The following one-dimensional polynomials:

g _n (t) = P _n (x + tz), g(t) = P (x + tz)

verifies that {g _n } converges to g uniformly on the compact interval [ −1, 1], g(0) = 0 and g ^ (0) = dP (x)(z) = 0. Consequently, there exists a sequence {t _n } such that lim

n t _n = 0 and g _n (t _n ) = 0 eventually. If we define x _n = x + t _n z, the sequence {x n } fulfils the required condition.

Section 2

Plichko and Zagorodnyuk [16] have prove that for any positive in- tegers n and d, there is a positive integer m = m(n, d) such that for any complex polynomial P : C ^m → C of degree d, there is a vector subspace X _p ⊂ C ^m of dimension n such that P | _X

= P (0). For further background on related problems see [15]. In [3] the problem of finding a good bound on m = m(n, d) is treated. Estimates for degrees 2, 3 and 4 are obtained. For instance, it is not difficult to see that m(n, 2) = 2n+1, and for example the polynomial P (z ₁ , . . . , z _2n ) =

 2n j=1

z _j ² vanishes on the n-dimensional subspace generated by {e 1 + ie ₂ , . . . , e _2n−1 + ie _2n }.

Now in the real case we study four special situations. The first is a general result for real, symmetric, homogeneous polynomials of odd de- gree. We next study real 2-homogeneous polynomials P , relating the size of the subspace of zeros of P with the signature of the associated matrix.

We show a general result concerning the zeros of real 3-homogeneous

polynomials and finally we present several dichotomy results related with

existence of positive definite 2-homogeneous polynomials.

Proposition 2.1. ([2]). Let P be a symmetric homogeneous poly- nomial P : R ^m → R of odd degree. Then there is an [m/2]-dimensional subspace contained in P ⁻¹ (0).

Proof. Let P be a d-homogeneous polynomial on R ^m where d is an odd integer. It is known that the polynomials {  _m

i=1 x ^r _i | r ∈ N} form an algebraic basic for the symmetric polynomials on R ^m (see, e.g. [18], p. 79). Therefore there is a polynomial Q such that

P  _m



i=1

x _i e _i

= Q  _m



i=1

x _i , . . . ,

 m i=1

x ^d _i

. Now,

P  _m



i=1

x _i e _i

= 

i1,...,is;

i

k

+i

k

+··· ,i

k

=d

α ^k _i

^,...,k

1

,...,i

 _m



i=1

x ⁱ _i

_k

· · ·  _m



i=1

x ⁱ _i

_k

.

Note that in each expression i ₁ k ₁ + · · · + i _s k _s = d, some of the i _j ’s must be odd. Consider the [m/2]-dimensional subspace

H = 

e ₁ − e 2 , . . . , e _2[m/2]−1 − e _2[m/2] 

; clearly, if x ∈ H then P (x) = 0. So, H ⊂ P ⁻¹ (0).

The same works in the infinite dimensional case:

Proposition 2.2. ([2]). Let X be a real Banach space with a sym- metric basis and let P be a homogeneous symmetric polynomial of odd degree. Then P ⁻¹ (0) contains an infinite dimensional subspace.

Proof. Let P be a d-homogeneous polynomial on X where d is an odd integer, and let {e _n } be the symmetric basic of X. By the representation of symmetric polynomials given in [9], either P = 0 or there exists an integer N ≥ 1 such that the set of polynomials {  _∞

i=1 x ^r _i | r ≥ N} is an algebraic basis for the space of symmetric polynomials on X. Therefore, if d < N we have P = 0; otherwise, there is a polynomial Q : R ^{d−N +1} → R such that

P  _∞



i=1

x _i e _i

= Q  _∞



i=1

x ^N _i , . . . ,

 ∞ i=1

x ^d _i

. Then,

P  _∞



i=1

x _i e _i

= 

i1,...,is;

i

k

+i

k

+···+i

k

=d

α ^k _i

_i

^k

 _∞



i=1

x ⁱ _i

_k

· · ·  _∞



i=1

x ⁱ _i

_k

.

Consider now the infinite dimensional subspace:

H = [e ₁ − e ₂ , e ₃ − e ₄ , . . . , e _2n−1 − e _2n , . . . , ] .

Clearly, H is an infinite dimensional subspace contained in P ⁻¹ (0), as we required.

We turn now to searching for subspaces contained in the zero set of 2-homogeneous real polynomials. We are able to obtain a simple, general result, which depends only on the signs of the eigenvalues of the quadratic form associated to the polynomial.

We recall that a quadratic form Q on R ^k (or, equivalently, the associ- ated symmetric matrix or bilinear form A) is said to be positive definite if Q(x) > 0 (equivalently, A(x, x) > 0) for all x ∈ R ^k , x = 0. In the same way, Q will be negative definite whenever −Q is positive definite.

We denote by p(Q) (resp. n(Q), z(Q)) the number of positive (resp.

negative, zero) eigenvalues with their multiplicity.

Proposition 2.3. ([2]). Let Q ∈ P( ² R ^k ). Then, if r = min {p(Q), n(Q) } + z(Q) there is an r-dimensional subspace Y such that Y ⊂ Q ⁻¹ (0).

Proof. Consider a basis {w ₁ , . . . , w _k } with respect to which Q is diagonal. Then,

Q  _k



i=1

y _i w _i

=

 k i=1

µ _i y ² _i ,

where without loss of generality, we may assume that µ _i = ±1 or 0. We may also assume that these eigenvalues are written so that µ ₁ = 1, µ ₂ =

−1, µ ₃ = 1, . . . , µ _2s−1 = 1, µ _2s = −1, where s = min{p(Q), n(Q)}, and that µ _k−z(Q)+1 = · · · = µ _k = 0. It is easy to verify that the r-dimensional subspace

Y = 

w ₁ + w ₂ , . . . , w _2s−1 + w _2s, w _k−z(Q)+1 , . . . , w _k  is contained in Q ⁻¹ (0).

Next result shows that every 3-homogeneous polynomial P : R ^k → R vanishes on a subspace whose dimension n depends only on k, where n → ∞ as k → ∞.

Theorem 2.1. ([2]). Let P : R ^k → R be a 3-homogeneous polyno- mial and

k ≥ 3 ⁿ (6n + 5) − 1

4

for some n > 0. Then, there is an [(n/2)]-dimensional subspace con- tained in P ⁻¹ (0).

It is easy to see that every 3-homogeneous polynomial in at least k = 2 variables vanishes on an one dimensional subspace. The estimate provided by the theorem in this case is for k to be not smaller than 38. If a 2-dimensional subspace is required, so that n = 4, then the theorem gives the estimate that any 3-homogeneous polynomial in k = 587 variables has such a subspace. We remark that more, and better, information seems to be known for the analogous problem for complex polynomials. For example, every homogeneous complex polynomial(of any degree) in two or more variables vanishes on a complex line. In fact, one can show that any 3-homogeneous complex polynomial in 2 ⁿ⁻¹ (n+1) variables vanishes on a n-dimensional subspace. ([3]).

An k-homogeneous polynomial P : X → K is said to be positive definite if P (x) ≥ 0 for every x and P (x) = 0 implies that x = 0.

Proposition 2.4. ([1]). A Banach space X admits a positive defi- nite 2-homogeneous polynomial if and only if there is a 2-homogeneous polynomial P on X whose set of zeros is contained in a finite dimensional subspace of X .

Remark 2.1. ([1]). Any separable space and C(K) spaces, when K is compact and separable, admit a positive definite 2-homogeneous polynomial. On the other hand, X = c ₀ (Γ) and X = l _p (Γ), where Γ is an uncountable index set and p > 2, do not admit positive definite 2-homogeneous polynomials.

As we remarked in 2.1, the following result of interest only for non- separable spaces.

Theorem 2.2. ([1]) Let X be a real Banach space which does not admit a positive definite 2-homogeneous polynomial. Then, for every P ∈ P( ² X), there is an infinite dimensional subspace of X on which it is identically zero.

Proof. Suppose X does not admit a positive definite 2-homogeneous polynomial and that P ∈ P( ² X). Let S = {S : S is a subspace of Xand P | _S ≡ 0}. Order S by inclusion and use Zorn’s Lemma to deduce the existence of a maximal element S of S. Suppose that S is finite di- mensional. Let v ₁ , . . . , v _n be a basis for S and let T = 

x∈S ker A _x =

 _n

i=1 ker A _v

where A _x : X → R is the linear map which sends y in X

to ˘ P (x, y). We note that S ⊂ T . To see this suppose that y ∈ S. Then

for every s ∈ S, s + y is also in S. Since

0 = P (s + y) = P (s) + 2A _s (y) + P (y) = 2A _s (y) for every s ∈ S we see that y ∈ T .

Since S is finite dimensional we can write T as T = S 

Y for some subspace Y of T . It is easy to see that all the zeros of P | _T are contained in S. Therefore, either P | T or −P | T is positive definite on Y . Let us suppose, without loss of generality, that P | _T is positive definite on Y . As S is n-dimensional we can find φ ₁ , . . . , φ _n so that P +  _n

i=1 φ ² _i is positive definite on T . Note that T has finite codimension in X and hence is complemented. Let π _T be the (continuous) projection of X onto T . Then 

P +  _n

i=1 φ ² _i 

◦ π _T +  _n

i=1 A ² _v

is a positive definite polynomial on X, contradicting the fact that X does not admit such a polynomial.

Remark 2.2. In ([14]) it is shown that every C-valued polynomial P on an infinite dimensional complex Banach space X such that P (0) = 0 is identically 0 on an infinite dimensional subspace.

Theorem 2.3. ([1]) Let X be a real Banach space which does not admit a positive definite 4-homogeneous polynomial, and let (ψ _k ) ^∞ _k=1 be a sequence in X ^∗ . Then for every countable family (P _j ) ^∞ _j=1 ⊂ P( ² X), there is a non-separable subspace of

 ∞ k=1

ker ψ _k on which each P _j is iden- tically zero.

Note that if X does not admit a positive definite 4-homogeneous polynomial, then it cannot admit a positive definite 2-homogeneous one either. An example of an X satisfying the hypotheses of Theorem 2.3 is X = l _p (I), where I is an uncountable index set and p > 4.

Proof of Theorem 2.3. The argument begings in a similar way to our earlier proofs. As before, let S be a maximal element of S = {S : S is subspace of  _∞

k=1 ker ψ _k and P _j | _S ≡ 0, all j}. Suppose that S is separable, with countable dense set (v _i ) ^∞ _i=1 . Let T =  _∞

k=1 ker ψ _k ∩

 _∞

i=1

 _∞

j=1 ker (A _j ) v _i . As before, S ⊂ T . We can write T as T = S 

a Y for some subspace Y of T . Since all the common zeros of P _j | _T , j ∈ N, are contained in S,  _∞

j=1 P

j

P



is positive definite on Y . As S is separable we can find (φ _i ) ^∞ _i=1 so that  _∞

j=1 P

j

P



+  _∞

i=1 φ ⁴ _i is positive definite

on T . Then

 ∞ j=1

P _j ² j ²  P j  ² +

 ∞ i=1

φ ⁴ _i +

 ∞ i=1

 ∞ j=1

(A _j ) ⁴ _v

i

i ² j ²  (A j ) _v

i

 ⁴ +

 ∞

k=1

ψ _k ⁴ k ²  ψ _k  ⁴ is a positive definite polynomial on X, contradicting the fact that X does not admit such a polynomial.

Corollary 2.1. ([1]). Let X be a real Banach space which does not admit a positive definite 4-homogeneous polynomial. Then every P ∈ P( ³ X) is identically zero on a non-separable subspace of X.

Proof. Consider P ∈ P( ³ X) and let S be a maximal element of S = {S : S is a subspace of X and P | _S ≡ 0}. Suppose that (v _i ) ^∞ _i=1 is a countable dense subset of S. Let A _v

_,v

: X → R be the linear map which sendes x in X to ˘ P (v _i , v _j , x) and Q _v

: X → R be the continuous 2- homogeneous polynomial which sends x in X to ˘ P 

v _i , x ² 

. By Theorem

2.3,  ∞

i,j=1

ker A _v

_,v

∩  ^∞

i=1

ker Q _v

contains a non-separable subspace which we denote by T . Suppose that y ∈ T is such that P (y) = 0. Then for every x = 

i α _i v _i ∈ span S and λ ∈ R we have

P (x + λy) = P (x) + 3λA _x,x (y) + 3λ ² P (x, y, y) + λ ˘ ³ P (y)

= P (x) + 3λ 

i,j

α _i α _j A _v

_,v

(y) + 3λ ² 

i

α _i Q _v

(y) + λ ³ P (y)

= 0.

Hence, by continuity of P, P (x+λy) = 0 for every x ∈ S. By maximality of S it follows that all the zeros of P | T are contained in S. Since S is separable, we can write T as T = (S ∩ T ) 

a Y for some non-separable subspace Y of T . Since all the zeros of P | T are contained in S, P | Y

is a 3-homogeneous polynomial on an infinite dimensional space which has its only zero at the origin, an impossibility.

The final theorem of this section is a natural extension of the two preceding results.

Theorem 2.4. ([1]) Let X be a real Banach space which does not

admit a positive definite homogeneous polynomial. Then, for every poly-

nomial P on X such that P (0) = 0, there is a non-separable subspace

of X on which P is identically zero.

In [1, section 4] the special cases of C(K) and absolutely (1, 2)- summing and nuclear polynomials is considered.

Section 3

This section is concerned with the approximation by zeros of orthog- onally additive polynomials on real l _p and L _p spaces. We consider l _p and L _p [0, 1] spaces, 1 ≤ p < ∞, with the usual Banach lattice. (See [13]).

Definition 3.1. Let X be a Banach lattice. A function f : X → R is said to be orthogonally additive if f (x + y) = f (x) + f (y) for all x, y ∈ X such that x ⊥ y.

P _o ( ⁿ X) will denote the space of n-homogeneous orthogonally additive polynomials on X.

Example 3.1. ([17]). Let X = _p , 1 ≤ p < ∞. P _o ( ⁿ X) is isometri- cally isomorphic to

p−n

if n < p and to _∞ if n ≥ p. The isomorphism is given by the association P ↔ a = (a j = P (e _j )) _j≥1 .

Example 3.2. ([17]). Let X = L _p [0, 1], 1 ≤ p < ∞, and let µ be Lebesgue measure.

For 1 ≤ n < p, P ∈ P o ( ⁿ X) ⇔ There exists a unique function ξ ∈ L

such that P (x) =  ₁

0 ξx ⁿ dµ.

For n = p, P ∈ P _o ( ⁿ X) ⇔ There exists a unique function ξ ∈ L _∞ such that P (x) =  ₁

0 ξx ⁿ dµ.

For n > p, P ∈ P _o ( ⁿ X) ⇔ P ≡ 0.

For a Banach space X and an n-homogeneous polynomial P, P = 0, we consider Z = P ⁻¹ (0) and the sets D _j Z = {  _j

i=1 z _i : z _i ∈ Z, ∀i = 1, . . . , j } (j ≥ 2). Finally we consider H = span[Z], the subspace of X generated by finite linear combinations of elements of Z. Note that H = 

j≥1 D _j Z.

Theorem 3.1. ([12]). Let 1 ≤ p < ∞, n > 1 an odd integer, and P an n-homogeneous orthogonally additive polynomial on _p associated with the sequence a = (a _j ) _j≥1 .

(i) If there exist at least three j ^ s so that a _j = 0, then e _j ∈ D ₃ Z for all j ∈ N. In particular, H is dense in p .

(ii) If there are at most two j ^ s with a _j = 0, then Z is a closed hyper-

plane of _p .

Theorem 3.2. ([12]). Let 1 ≤ p < ∞, n an even integer, and P an n-homogeneous orthogonally additive polynomial on _p associated with the sequence a = (a _j ) _j≥1 .

(i) If the function sign(a _j ), a _j = 0, is constant, then Z = H = {x ∈ _p : x _j = 0 whenever a _j = 0}.

(ii) If there exists a pair i = k such that a _i , a _k = 0 and sign(a _i ) = sign(a _k ), then e _j ∈ D 2 Z for all j ∈ N. In particular H is dense in p .

From what follows we consider the usual decomposition of a function a difference of two positive maps ξ = ξ ⁺ − ξ ⁻ and we note by A ⁺ , A ⁻ the respective support of ξ ⁺ , ξ ⁻ and A ₀ the complement set of A ⁺ ∪ A ⁻ in [0, 1]

Theorem 3.3. ([12]). Let 1 ≤ p < ∞, n an odd integer, 1 < n ≤ p, and P an n-homogeneous orthogonally additive polynomial on L _p [0, 1]

associated with the function ξ ∈ L

p−n

[0, 1].

(i) If ξ ≥ 0 (or ξ ≤ 0) a.e. then, χ _E ∈ D ₃ Z, for all measurable set E ⊂ [0, 1].

(ii) If ξ is arbitrary then, χ _E ∈ D ₄ Z, for all measurable set E ⊂ [0, 1].

As a consequence of either (i) or (ii), H is dense in L _p [0, 1].

Theorem 3.4. ([12]). Let 1 ≤ p < ∞, n an even integer, 1 < n ≤ p, and P an n-homogeneous orthogonally additive polynomial on L _p [0, 1]

associated with the function ξ ∈ L

[0, 1].

(i) If ξ ≥ 0 (or ξ ≤ 0) a.e. (i.e µ(A ⁺ ).µ(A ⁻ ) = 0), then H = Z = {x ∈ L _p [0, 1] : µ(support(x) ∩ support(ξ)) = 0}

(ii) If µ(A ⁺ ).µ(A ⁻ ) > 0 and E ⊂ [0, 1] ia a measurable set, then 1. µ(A ⁺ ∩ E).µ(A ⁻ ∩ E) = 0 implies χ _E ∈ Z or χ _E ∈ D ₂ Z.

2. µ(A ⁺ ∩ E).µ(A ⁻ ∩ E) > 0 implies χ E ∈ D 3 Z.

As a consequence of either (a) or (b) in (ii), H is dense in L _p [0, 1].

We proved that given an n −homogeneous orthogonally additive poly- nomial on L _p such that H is dense, it is enough to consider D _j Z with, at most j = 4, in order to obtain the decomposition by zeros of a char- acteristic function. Now, the problem of when it is possible to consider only D ₂ Z, is treated.

Theorem 3.5. ([12]). Let p = n be an odd integer, n > 1 and P an n-homogeneous orthogonally additive polynomial on L _p [0, 1] associated with the function ξ ∈ L _∞ [0, 1].

(i) If there exists T ⊂ [0, 1] a dense set such that ξ(t ⁺ ) = 0 for all

t ∈ T, then χ _[a,b] ∈ D ₂ Z
^.

for all subinterval [a, b] ⊆ [0, 1].

(ii) If ξ = χ _F for F ⊂ [0, 1] a measurable set, then χ E ∈ D 2 Z
^.

for all measurable set E ⊂ F .

As a consequence of either (i) or (ii), H is dense in L _p [0, 1].

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Departamento de An´ alisis Matem´ atico Facultad de Matem´ aticas

Universidad Complutense de Madrid 28040 Madrid, Spain

E-mail : JL [email protected]