POSITIVITY, SUMS OF SQUARES AND THE

MULTI-DIMENSIONAL MOMENT PROBLEM

S. Kuhlmann, M. Marshall

Contents

1. Introduction

2. The case n = 1

3. Examples where (∗) fails

4. Consequences of results in [7]

5. Cylinders with compact cross-section

6. Open problems

Let K be the basic closed semi-algebraic set in ℝ^{n} defined by some finite set of polynomial inequalities g_{1} ≥ 0, …, g_{s} ≥ 0 and let T be the preordering in the polynomial ring ℝ[X] := ℝ[X_{1}, …, X_{n}] generated by g_{1}, …, g_{s}. For K compact, Schmüdgen proves in [10] that:

(∗) The K-Moment Problem has a positive solution.

(†) ∀ f ∈ ℝ[X], f ≥ 0 on K ⇒ ∀ real ϵ > 0, f + ϵ ∈ T. In the present paper, we consider the status of (∗) and (†) when K is not compact. At the same time, we consider a third property:

(‡) ∀ f ∈ ℝ[X], f ≥ 0 on K ⇒ ∃ q ∈ T such that ∀ real ϵ > 0, f + ϵq ∈ T which we prove is strictly weaker than (†) and, at the same time, which implies (∗). Many (non-compact) examples are given where (‡) holds. Many examples are given where (∗) fails. The question of whether or not (‡) and (∗) are equivalent is an open problem.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 14P10, 44A60.

This research was supported in part by NSERC of Canada.

In Section 1 we introduce the problem and the notation. In Section 2 we settle the case n = 1. In Section 3 we study property (∗) using the method developed by Berg, Christensen and Jensen in [1]. We combine this with a result of Scheiderer in [9] to prove that (∗) fails whenever K contains a cone of dimension 2. In Section 4 we mention applications of results of the second Author in [7]. In particular, we construct a large number of non-compact examples where (‡) holds. In Section 5 we prove (‡) holds for cylinders with compact cross-section. In Section 6 we provide a list of open problems.

The Authors wish to express their indebtedness to Prof. Dr. K. Schmüdgen for laying out the basic framework for a paper such as this in a series of talks given at the University of Saskatchewan in March of 2000. At the same time it is acknowledged that any shortcomings of this particular paper are due solely to the Authors.

1. Introduction

Fix an integer n ≥ 1 and denote the polynomial ring ℝ[X_{1}, …, X_{n}] by ℝ[X] for short. ∑ ℝ[X]^{2} denotes the set of finite sums ∑ f_{i}^{2}, f_{i} ∈ ℝ[X]. For S = {g_{1}, …, g_{s}}, a finite subset of ℝ[X], let K_{S} denote the basic closed semi-algebraic set in ℝ^{n} determined by S; i.e.,

K_{S} = {x ∈ ℝ^{n} | g_{i}(x) ≥ 0, i = 1, …, s}.

Let T_{S} denote the preordering in ℝ[X] generated by S, i.e., the set of all sums ∑ σ_{e}g^{e}, e = (e_{1}, …, e_{s}) running through the finite set {0, 1}^{s}, σ_{e} ∈ ∑ ℝ[X]^{2}, where g^{e} := g_{1}^{e_{1}} … g_{s}^{e_{s}}.

Denote by T_{S}^{alg} the set {f ∈ ℝ[X] | f ≥ 0 on K_{S}}. Points x in ℝ^{n} are in one-to-one correspondence with algebra homomorphisms L : ℝ[X] → ℝ via

x = (L(X_{1}), …, L(X_{n})), L(f) = f(x).

Under this correspondence, points x in K_{S} correspond to algebra homomorphisms L satisfying L(g_{i}) ≥ 0, i = 1, …, s or, equivalently, L(T_{S}) ≥ 0.

One can also consider the dual cone

K_{S}^{lin} = {L : ℝ[X] → ℝ | L is linear (≠ 0) and L(T_{S}) ≥ 0},

and the double dual cone T_{S}^{lin} = {f ∈ ℝ[X] | L(f) ≥ 0 for all L ∈ K_{S}^{lin}}; see Section 3 for a more thorough discussion of the double dual cone. Since every algebra homomorphism is, in particular, a linear map, we have that

K_{S} ↪ K_{S}^{lin} and T_{S}^{alg} ⊇ T_{S}^{lin} ⊇ T_{S}.

Note: T_{S}, T_{S}^{lin} generally depend on S. T_{S}^{alg} depends only on the basic closed set K_{S}.

Note: L ∈ K_{S}^{lin} ⇒ L(1) > 0. For 1 ∈ T_{S} implies L(1) ≥ 0. If L(1) = 0 then, for any f ∈ ℝ[X], the identity (k ± f)^{2} = k^{2} ± 2kf + f^{2} implies ±2kL(f) + L(f^{2}) ≥ 0 for all real k, so L(f) = 0.

We are interested here in the relationship between T_{S}^{alg}, T_{S}^{lin} and T_{S} for all the various choices of (finite) S in ℝ[X], n ≥ 1. In particular, we are interested in the relationship between the condition:

(∗)  T_{S}^{alg} = T_{S}^{lin}

and the condition:

(†)  ∀f ∈ ℝ[X], f ∈ T_{S}^{alg} ⇒ ∀ real ϵ > 0, f + ϵ ∈ T_{S}.

We are also interested in weak version of (†) referred to as (‡) (see below) and a strong version of (†); namely T_{S}^{alg} = T_{S}. The condition T_{S}^{alg} = T_{S} is examined in the recent paper of Scheiderer [9].

The study of the relationship between T_{S}^{alg} and T_{S} goes back at least to Hilbert and is a cornerstone of modern semi-algebraic geometry. The Positivstellensatz proved by Stengle in 1974 implies, in particular, that

∀f ∈ ℝ[X], f ∈ T_{S}^{alg} ⇔ ∃p, q ∈ T_{S} and m ≥ 0 such that pf = f^{2m} + q.

The interest in T_{S}^{lin} comes from functional analysis. For a linear functional L on ℝ[X], one is interested in when there exists a positive Borel measure μ on ℝ^{n} supported by some given closed set K in ℝ^{n} such that

∀f ∈ ℝ[X], L(f) = ∫_{ℝ^{n}} fdμ.

According to a result of Haviland [4] [5], this will be the case iff L(f) ≥ 0 holds for all f ∈ T_{S}^{alg}. The Moment Problem (or at least a version of it) is the following: When is it true that every L ∈ K_{S}^{lin} comes from a positive Borel measure μ on ℝ^{n} supported by K_{S}? In view of the result of Haviland, this is equivalent to asking: When does (∗) hold?

1.1 Remark. For each n-tuple of non-negative integers k = (k_{1}, …, k_{n}), denote the monomial X_{1}^{k_{1}} … X_{n}^{k_{n}} by X^{k} for short. The monomials form a basis for ℝ[X] so each linear functional L on ℝ[X] corresponds to a function p : (ℤ^{+})^{n} → ℝ via p(k) = L(X^{k}). We say p is positive definite if

∑_{i,j=1}^{m} p(k_{i} + k_{j})c_{i}c_{j} ≥ 0

for arbitrary (distinct) k_{1}, …, k_{m} ∈ (ℤ^{+})^{n}, c_{1}, …, c_{m} ∈ ℝ, and m ≥ 1. For g ∈ ℝ[X], g = ∑ a_{k}X^{k}, g(E)p : (ℤ^{+})^{n} → ℝ is defined by

g(E)p(ℓ) = ∑_{k} a_{k}p(k + ℓ).

The condition that L ∈ K_{S}^{lin} corresponds exactly to the condition that functions g^{e}(E)p, e ∈ {0, 1}^{s} are positive definite, see [8] or [10]. Consequently, (∗) is equivalent to the assertion that every non-zero function p : (ℤ^{+})^{n} → ℝ with g^{e}(E)p positive definite for all e ∈ {0, 1}^{s} comes from a positive Borel measure on ℝ^{n} supported by K_{S}.

Schmüdgen’s 1991 paper [10] settles the Moment Problem in the compact case. In [10], Schmüdgen proves if K_{S} is compact then (∗) and (†) both hold. In [11], Wörmann proves K_{S} is compact iff T_{S} is archimedean and uses this to obtain Schmüdgen’s result as a corollary of the Kadison-Dubois Theorem.

Denote by P_{d} the (finite dimensional) vector space consisting of all polynomials in ℝ[X] of degree ≤ 2d, and let T_{d} = T_{S} ∩ P_{d}. T_{d} is obviously a cone in P_{d}, i.e., T_{d} + T_{d} ⊆ T_{d} and ℝ^{+}T_{d} ⊆ T_{d}. Denote by \overline{T}_{d} (resp., int(T_{d})) the closure (resp., interior) of T_{d} in P_{d}. The identity pq = \frac{1}{2}((p + q)^{2} − p^{2} − q^{2}) applied to monomials p, q of degree ≤ d implies P_{d} = T_{d} − T_{d}. This means that T_{d} contains a basis of P_{d}, so int(T_{d}) ≠ ∅.

1.2 Remark. T_{S} is archimedean iff 1 ∈ int(T_{d}) for all d ≥ 0. This is just a matter of sorting out the definitions: If k ± f ∈ T_{S} for sufficiently large k ∈ ℤ, then 1 ± ϵf ∈ T_{S} for sufficiently small ϵ > 0. In fact, one even knows that T_{S} is archimedean iff k − ∑_{i=1}^{n} X_{i}^{2} ∈ T_{S} for k sufficiently large [11], so T_{S} is archimedean iff 1 ∈ int(T_{1}).

We are also interested in the condition:

(‡)  ∀f ∈ ℝ[X], f ∈ T_{S}^{alg} ⇒ ∃q ∈ T_{S} such that ∀ real ϵ > 0, f + ϵq ∈ T_{S}.

Obviously, (†) ⇒ (‡) (taking q = 1). The significance of (‡) is clear from the following:

1.3 Proposition. The following are equivalent:

(1) (‡) holds.

(2) T_{S}^{alg} = T_{S}^{lin} = ∪_{d≥0}\overline{T}_{d}.

Proof. Clearly \overline{T}_{d} ⊆ T_{S}^{lin}. For f ∈ P_{d}, q ∈ T_{d}, the condition that f + ϵq ∈ T_{S} for all real ϵ > 0 is equivalent to:

λf + (1 − λ)q ∈ T_{d} for every 1 < λ < 1,

i.e., that every point on the open line segment joining q and f belongs to T_{d}. In particular, it implies that f ∈ \overline{T}_{d}. Thus we see that (1) ⇒ (2). Now assume (2), and suppose f ∈ T_{S}^{alg}. Thus, for d sufficiently large, f ∈ \overline{T}_{d}. Pick q any point in the interior of T_{d}. (Recall: int(T_{d}) ≠ ∅.) Then obviously the open line segment joining f with q belongs to T_{d}, so (1) holds. □

Thus, condition (‡) implies not only that (∗) holds, but also that T_{S}^{lin} = ∪_{d≥0}\overline{T}_{d}. Note: The requirement in (‡) that q belongs to T_{S} is unnecessary: Using

q = (\frac{q + 1}{2})^{2} − (\frac{q − 1}{2})^{2},

we see that if f + ϵq ∈ T_{S}, q ∈ ℝ[X], then f + ϵ(\frac{q+1}{2})^{2} ∈ T_{S}.

See Section 4 for additional discussion of (‡) and for examples where (‡) holds. See Section 5 for examples where (‡) holds but (†) does not hold. No examples are known where (∗) holds but (‡) does not hold. No examples are known where T_{S}^{lin} ≠ ∪_{d≥0}\overline{T}_{d}. In fact, there is a general shortage of examples.

Instead of working with T_{S}, one can work with the ∑ ℝ[X]^{2}-module M_{S} generated by S, i.e., the set of all sums σ_{0} + ∑_{i=1}^{s} σ_{i}g_{i}, σ_{i} ∈ ∑ ℝ[X]^{2}, i = 0, …, s, and M_{S}^{lin}, the double dual cone of M_{S}. One is interested in the analog of (∗):

(∗_{M})  T_{S}^{alg} = M_{S}^{lin},

the analog to (†):

(†_{M})  ∀f ∈ ℝ[X], f ≥ 0 on K_{S} ⇒ ∀ real ϵ > 0, f + ϵ ∈ M_{S},

and the analog to (‡):

(‡_{M}) ∀f ∈ ℝ[X], f ∈ T_{S}^{alg} ⇒ ∃q ∈ M_{S} such that ∀ real ϵ > 0, f + ϵq ∈ M_{S}.

Again, T_{S}^{alg} = M_{S} implies (†_{M}) which, in turn, implies (‡_{M}) and (‡_{M}) is equivalent to T_{S}^{alg} = M_{S}^{lin} = ∪_{d≥0}\overline{M}_{d} where \overline{M}_{d} denotes the closure in P_{d} of M_{d} := M_{S} ∩ P_{d}. (†_{M}) is known to hold in a few cases, for example, if K_{S} is compact and either n = 1 [6] or the polynomials g_{1}, …, g_{s} are linear [6] [8] or s ≤ 2 [6]. In [6], Jacobi and Prestel develop a powerful valuation-theoretic method for testing the validity of (†_{M}) when K_{S} is compact.

2. The case n = 1

In this section we consider the case n = 1, extending the work of Berg and Maserick in [3].

2.1 Theorem. Suppose n = 1. If K_{S} is compact, then (†) and (†_{M}) hold. If K_{S} is not compact, then the conditions (†), (‡), (∗) and T_{S}^{alg} = T_{S} are equivalent and the conditions (†_{M}), (‡_{M}), (∗_{M}) and T_{S}^{alg} = M_{S} are equivalent.

Proof. If K_{S} is compact then (†) holds by Schmüdgen’s result and (†_{M}) holds by [6, Remark 4.7]. We defer the proof of the rest of Theorem 2.1 to Section 3 where it is an immediate consequence of Theorem 3.5. □

Note: Already in the case n = 1, we see the dichotomy between the compact case and the non-compact case. If K_{S} is compact then (†) (resp., (†_{M})) holds but, unlike what happens in the non-compact case, this does not imply that T_{S}^{alg} = T_{S} (resp., T_{S}^{alg} = M_{S}) holds. For example, if S = {(1 − X^{2})^{3}} then T_{S}^{alg} ≠ T_{S}; see [8, Example 4.3.3].

The second part of Theorem 2.1 is only useful if we know when T_{S}^{alg} = T_{S} (resp., T_{S}^{alg} = M_{S}) holds. For the rest of the section we concentrate our attention on answering this question.

2.2 Theorem. Suppose n = 1. If K_{S} is not compact and T_{S}^{alg} = T_{S}, then:

(i) If K_{S} has a smallest element, call it a, then r(X − a) ∈ S for some real r > 0.

(ii) If K_{S} has a largest element, call it a, then r(a−X) ∈ S for some real r > 0. (iii) For every a, b ∈ K_{S} with a < b and (a, b) ∩ K_{S} = ∅, r(X − a)(X − b) ∈ S

for some real r > 0.

Conversely, for any K_{S}, if conditions (i), (ii) and (iii) hold, then T_{S}^{alg} = T_{S}.

2.3 Notes.

(1) Since K_{S} is a closed semi-algebraic set in ℝ, it is the union of finitely many closed intervals and points (including possibly closed intervals of the form (−∞, a] or [a, ∞)).

(2) If K ⊆ ℝ is any closed semi-algebraic set then one checks easily that K = K_{S} for S the set of polynomials defined as follows:

— If a ∈ K and (−∞, a) ∩ K = ∅, then X − a ∈ S.

— If a ∈ K and (a, ∞) ∩ K = ∅, then a − X ∈ S.

— If a, b ∈ K, a < b, (a, b) ∩ K = ∅, then (X − a)(X − b) ∈ S.

— S has no other elements except these. We call S the natural choice of generators for K.

(3) Theorem 2.2 proves that, for K = K_{S} not compact, T_{S}^{alg} = T_{S} holds iff S contains the natural choice of generators (up to scalings by positive reals).

(4) If K_{S} is compact then T_{S}^{alg} = T_{S} can hold without conditions (i), (ii) and (iii) holding. For example, if S = \{1 − X^{2}\}, then 1 ± X = \frac{1}{2}(1 ± X)^{2} + \frac{1}{2}(1 − X^{2}) ∈ T_{S} so T_{S}^{alg} = T_{S}.

Proof. Suppose T_{S}^{alg} = T_{S}, K_{S} not compact. We can assume that the elements of S have degree ≥ 1. Since K_{S} is not compact, it either contains an interval of the form [a, ∞) or it contains an interval of the form (−∞, a]. Replacing X by −X if necessary, we can assume we are in the first case. Thus each g_{i} ∈ S is non-negative on some interval [a, ∞), so has positive leading coefficient. Thus, for any element p = ∑ σ_{e}g_{1}^{e_{1}} … g_{s}^{e_{s}} of T_{S}, the degree of p is equal to the maximum of the degrees of the terms σ_{e}g_{1}^{e_{1}} … g_{s}^{e_{s}}. Also, since each σ_{e} is a sum of squares, it has even degree.

Suppose that K_{S} has a smallest element, a say. Let p = X − a. Then p ≥ 0 on K_{S} so p ∈ T_{S}. Since p has degree 1, p is a sum of terms of the form σ, σ ∈ ℝ^{+}, and σg_{i} with σ ∈ ℝ^{+} and g_{i} ∈ S linear. Each such g_{i} is ≥ 0 at a (since a ∈ K_{S}). Consequently, at least one linear g_{i} in S is equal to zero at a, so g_{i} = r(X − a) as required.

Suppose now that a, b ∈ K_{S} are such that a < b and (a, b) ∩ K_{S} = ∅. Let p = (X − a)(X − b). Then p ≥ 0 on K_{S} so p ∈ T_{S}. Since p has degree 2, it is a sum of terms of the form σ ∈ ∑ ℝ[X]^{2} of degree 0 or 2, σg_{i} with σ ∈ ℝ^{+} and g_{i} ∈ S linear or quadratic, and σg_{i}g_{j} with σ ∈ ℝ^{+} and g_{i}, g_{j} ∈ S linear. Since any linear g_{i} ∈ S is increasing and g_{i}(a) ≥ 0, g_{i} is positive on the interval (a, b). Thus p ≥ σ_{1}g_{1} + ⋯ + σ_{t}g_{t} on (a, b) where g_{1}, …, g_{t} are quadratics in S which assume at least one negative value on (a, b), and σ_{1}, …, σ_{t} are positive reals. Define the width of a quadratic g to be r_{2} − r_{1} where r_{1} ≤ r_{2} are the roots of g (or 0 if g has no roots). Each g_{i} opens upward and is non-negative at a, b (since a, b ∈ K_{S}), has its roots between a and b and consequently has width at most b − a, i = 1, …, t. It suffices to show that g_{i} has width exactly b − a for some i ∈ {1, …, t}, for then g_{i} necessarily has the form r(X − a)(X − b) for some real r > 0. Since the width of p is equal to b − a and σ_{i}g_{i} has the same width as g_{i}, this is a consequence of the following elementary result:

2.4 Lemma. If f_{1}, f_{2} are quadratics with positive leading coefficients, then

width(f_{1} + f_{2}) ≤ max{width(f_{1}), width(f_{2})}.

Proof. Without loss of generality we can assume width(f_{1}) ≥ width(f_{2}) and that f_{1} has positive width. Shifting and scaling, we can assume f_{1} = X^{2} − X, f_{2} = c(X − a)(X−(a+b)), c > 0, 0 ≤ b ≤ 1. Thus f_{1}+f_{2} = (c+1)X^{2}−(2ac+bc+1)X+ca(a+b) and

width(f_{1} + f_{2}) = \frac{\sqrt{(2ac + bc + 1)^{2} − 4(c + 1)ca(a + b)}}{c + 1}.

Thus we are reduced to showing that

(1)  \frac{(2ac + bc + 1)^{2} − 4(c + 1)ca(a + b)}{(c + 1)^{2}} ≤ 1,

i.e., that

(2)  (2ac + bc + 1)^{2} ≤ (c + 1)^{2} + 4(c + 1)ca(a + b).

Expanding and canceling, this reduces to showing that

(3)  b^{2}c + 4a + 2b ≤ c + 2 + 4a^{2} + 4ab

or, equivalently, that

(4)  (1 − b^{2})(c + 1) + (2a + b − 1)^{2} ≥ 0.

Since 0 ≤ b ≤ 1 and c > 0, this is clear. □

Note: Equality holds iff b = 1 and a = 0, i.e., iff f_{2} = cf_{1}.

Suppose now that (i) (ii) (iii) hold. If K_{S} has a smallest element a then X − a ∈ T_{S} so X − d = (X − a) + (a − d) ∈ T_{S} for any d ≤ a. Similarly, if K_{S} has a largest element a then a − X ∈ T_{S}, so d − X = (a − X) + (d − a) ∈ T_{S} for any d ≥ a. Also, if a, b ∈ K_{S} are such that a < b and (a, b) ∩ K_{S} = ∅, then (X − a)(X − b) ∈ T_{S}. Moreover, by the proof of [3, Lemma 4], if a ≤ c ≤ d ≤ b then (X − c)(X − d) is in the preordering generated by (X − a)(X − b), so (X − c)(X − d) ∈ T_{S}.

Suppose f ∈ ℝ[X], f ≥ 0 on K_{S}. We prove f ∈ T_{S} by induction on the degree. If f has degree zero it is clear. If f ≥ 0 on ℝ then f ∈ ∑ ℝ[X]^{2} so, in particular, f ∈ T_{S}. Thus we can assume that f(c) < 0 for some c. There are three possibilities: Either K_{S} has a least element a and c < a or K_{S} has a largest element a and c > a or there exist a, b ∈ K_{S}, a < b with (a, b) ∩ K_{S} = ∅, and a < c < b. In the first case f has a least root d in the interval (c, a], X − d ∈ T_{S}, f = (X − d)g for some g ∈ ℝ[X] and one checks that g ≥ 0 on K_{S}. In the second case f has a greatest root d in the interval [a, c), d − X ∈ T_{S}, f = (d − X)g and again g ≥ 0 on K_{S}. Similarly, in the third case, f has greatest root d in the interval [a, c) and a least root e in the interval (c, b], (X − d)(X − e) ∈ T_{S}, f = (X − d)(X − e)g and g ≥ 0 on K_{S}. Thus, in any case, the result follows by induction on the degree. □

The question of when T_{S}^{alg} = M_{S} holds for K = K_{S} not compact is more complicated. Theorem 2.2 provides an obvious necessary condition: S must contain the natural choice of generators (up to scalings by positive reals). But this necessary condition is sufficient only in very special cases:

2.5 Theorem. Suppose K is not compact and S is the natural choice of generators for K. Then T_{S}^{alg} = M_{S} holds iff either |S| ≤ 1 or |S| = 2 and K has an isolated point.

Proof. If |S| ≤ 1, then M_{S} = T_{S} so T_{S}^{alg} = M_{S} by Theorem 2.2; also see [3, Theorem 1]. Suppose |S| = 2 and K has an isolated point. If this isolated point is the least element of K, then S = {g_{1}, g_{2}}, g_{1} = X − b, g_{2} = (X − b)(X − c), b < c. Then one checks that

g_{1}g_{2} = (X − c)^{2}g_{1} + (c − b)g_{2} ∈ M_{S},

so T_{S} = M_{S}. The case where the isolated point is the greatest element of K is similar. The remaining case is where S = {g_{1}, g_{2}}, g_{1} = (X − a)(X − b), g_{2} = (X − b)(X − c), a < b < c. In this last case one checks that

g_{1}g_{2} = \frac{b − a}{c − a}(X − c)^{2}g_{1} + \frac{c − b}{c − a}(X − a)^{2}g_{2} ∈ M_{S}.

Thus, in all cases, M_{S} = T_{S} so T_{S}^{alg} = M_{S} by Theorem 2.2.

Suppose now that either |S| = 2 and K has no isolated points, or |S| ≥ 3. Say S = {g_{1}, …, g_{s}}. Replacing X by −X if necessary, we can assume that K contains an interval of the form [a, ∞). Reindexing, we can suppose g_{1} is either linear of the form g_{1} = X − b (corresponding to the case where K has a least element b) or g_{1} = (X − a)(X − b), a < b where (a, b) is the left-most gap of K (corresponding to the case where K has no least element) and that g_{2} = (X − c)(X − d) where (c, d) is the right-most gap of K. By our hypothesis, b < c. We claim that g_{1}g_{2} ∉ M_{S}. Since g_{1}g_{2} ≥ 0 on K, this will complete the proof. Suppose, to the contrary, that

g_{1}g_{2} = σ_{0} + σ_{1}g_{1} + ⋯ + σ_{s}g_{s}, σ_{i} ∈ ∑ ℝ[X]^{2}.

Comparing degrees, we see that σ_{i} has degree 2 or 0 if i ≥ 1 and σ_{0} has degree 4, 2 or 0. Let g = g_{2} − σ_{1}. Thus g_{1}g = σ_{0} + σ_{2}g_{2} + ⋯ + σ_{s}g_{s}. Thus g_{1}(c)g(c) ≥ 0, g_{1}(d)g(d) ≥ 0, so g(c) ≥ 0, g(d) ≥ 0. On, the other hand, from g = g_{2} − σ_{1}, it follows that g(c) ≤ 0, g(d) ≤ 0. Since g has degree ≤ 2, this implies g = 0, i.e., g_{2} = σ_{1}, contradicting the fact that g_{2} is strictly negative on (c, d). □

3. Examples where (∗) fails

In this section we study condition (∗) using an extension of the method introduced by Berg, Christensen and Jensen in [1]. A major tool here is the recent work of Scheiderer in [9].

As we have seen in Section 1:

T_{S}^{alg} = T_{S} ⇒ (†) ⇒ (‡) ⇒ (∗).

If T_{S}^{lin} = T_{S} we can obviously say more:

3.1 Proposition. If T_{S}^{lin} = T_{S} then the conditions (†), (‡), (∗) and T_{S}^{alg} = T_{S} are equivalent.

And similarly, for M_{S}:

3.2 Proposition. If M_{S}^{lin} = M_{S} then the conditions (†_{M}), (‡_{M}), (∗_{M}) and T_{S}^{alg} = M_{S} are equivalent.

Proof. This is clear. □

We recall results from the theory of locally convex vector spaces. Any vector space V over ℝ comes equipped with a unique finest locally convex topology [2, Sect. 1.1.9]. For a cone C in V, C^{lin} denotes the double dual cone of C, i.e., the set of all x ∈ V such that L(x) ≥ 0 holds for all linear functionals L on V with L(C) ≥ 0. We need the following two results:

3.3 Lemma. If C is a cone in a vector space V over ℝ then (1) C^{lin} is the closure of C.

(2) C^{lin} = C ⇔ C is closed.

Proof. According to the Bipolar Theorem [2, Theorem 1.3.6], C^{lin} is the smallest closed convex set in V containing C. Since the closure of a cone is a cone (so, in particular, is convex), (1) is clear. (2) is immediate from (1). □

3.4 Lemma. Suppose C is a cone in a vector space V over ℝ and C_{i} = C∩V_{i} where V_{i}, i ≥ 0 are finite dimensional subspaces of V with V_{i} ⊆ V_{i+1} and V = ∪_{i≥0}V_{i}. Then the following are equivalent:

(1) C is closed in V.

(2) C_{i} is closed in V_{i} for each i ≥ 0.

Proof. See [2, Lemma 6.3.3]. □

Our next result provides examples where T_{S} and M_{S} are closed.

3.5 Theorem. If K_{S} contains a cone of dimension n then T_{S} and M_{S} are closed.

Note: Theorem 3.5 extends [1, Theorem 3], [2, Lemma 6.3.9] and [3, Lemma 3]. Note: Theorem 3.5 applies, in particular, if n = 1 and K_{S} is not compact.

Combining this with Lemma 3.3(2) and Propositions 3.1 and 3.2 completes the proof of Theorem 2.1.

Proof. The first assertion follows from the second, replacing S by the set of all products g^{e}, e ∈ {0, 1}^{s}, e ≠ (0, …, 0). By Lemma 3.4, to prove the second assertion, it suffices to show that M_{i} is closed in P_{i}, for each i ≥ 0. Let S = {g_{1}, …, g_{s}}. We can assume each g_{i} is not zero. By our hypothesis, K_{S} contains a cone C with non-empty interior. Making a linear change in coordinates if necessary, we can assume the vertex of C is at the origin.

For f any non-zero element of M_{S}, let f = f_{0} + ⋯ + f_{d} be its homogeneous decomposition. Observe that, for any a ∈ C and any λ > 0,

f(λa) = f_{0} + λf_{1}(a) + ⋯ + λ^{d}f_{d}(a).

Since λa ∈ C ⊆ K_{S} for all positive λ it follows that, if f_{d}(a) ≠ 0, then f_{d}(a) > 0.

Now suppose

p = σ_{0} + σ_{1}g_{1} + ⋯ + σ_{s}g_{s},

σ_{i} ∈ ∑ ℝ[X]^{2}. Let d denote the maximum of the degrees of σ_{0}, σ_{1}g_{1}, …, σ_{s}g_{s}. Using the fact that C has non-empty interior, we can pick a point a ∈ C such that the homogeneous parts of highest degree of the various polynomials in question do not vanish, so the homogeneous part of p having degree d is positive when evaluated at a (so, in particular, it is not zero). To summarize: if p has degree d, then σ_{0} has degree ≤ d and each σ_{i}, i ≥ 1 has degree ≤ d−deg(g_{i}). Thus any p ∈ M_{m} is expressible as above with σ_{i} = ∑ f_{ij}^{2}, deg(f_{0j}) ≤ m, deg(f_{ij}) ≤ m − \frac{1}{2}deg(g_{i}), i ≥ 1. Also, as in [1], we can assume σ_{i} = ∑_{j=1}^{N} f_{ij}^{2}, where N is the dimension of P_{m}.

Making an additional linear change of coordinates, we can also assume that the point (1, …, 1) lies in the interior of C and none of the g_{i} vanish at this point. Thus, for real distinct a_{1}, …, a_{2m+1} sufficiently close to 1, the set

H = {(a_{j_{1}}, …, a_{j_{2m+1}}) | 1 ≤ j_{k} ≤ 2m + 1, k = 1, …, 2m + 1}

is contained in C and g_{i}(a) > 0 for each a ∈ H. As explained in [1], two elements of P_{m} are equal iff they are equal on H, and the topology on P_{m} coincides with the topology of pointwise convergence on H. Also, for any a ∈ H, f_{0j}(a)^{2} ≤ σ_{0}(a) ≤ p(a), and f_{ij}(a)^{2} ≤ σ_{i}(a) ≤ p(a)/D_{i} where D_{i} is the minimum of the g_{i}(a), a ∈ H, if i ≥ 1. Thus, if p_{ℓ} is a sequence in M_{m} converging to some p ∈ P_{m} then, as in [1], there exists a subsequence p_{ℓ_{k}} with the associated f_{ℓ,i,j} converging pointwise on H to some f_{ij} ∈ P_{m} of the appropriate degree, so p ∈ M_{m}. □

Theorem 3.5 does not cover the case where K_{S} is a curve. At the same time T_{S} is closed for many curves, at least, for the “right” choice of S. We give two examples to illustrate.

3.6 Example. Consider the curve

C = {(x, y) ∈ ℝ^{2} | y = q(x)}

in ℝ^{2}, q ∈ ℝ[X]. Then C = K_{S} where S = {−(Y − q)^{2}}. We claim that (†) holds in this choice of S. Suppose f ≥ 0 on C. Then f(X, Y) = f(X, q(X)) + h(X, Y) with h vanishing on C. Also f(X, q(X)) ≥ 0 on ℝ so f(X, q(X)) ∈ ∑ ℝ[X]^{2}. Thus we are reduced to showing that ϵ + h ∈ T_{S} for each real ϵ > 0. Since Y − q is irreducible, h = (Y − q)h_{1} for some h_{1} ∈ ℝ[X, Y], so

ϵ + h = ϵ(1 + \frac{h_{1}}{2ϵ} (Y − q))^{2} − \frac{h_{1}^{2}}{4ϵ} (Y − q)^{2} ∈ T_{S}.

Note: Y − q ∉ T_{S}. If Y − q = σ_{0} − σ_{1}(Y − q)^{2}, σ_{0}, σ_{1} ∈ ∑ ℝ[X, Y]^{2}, then σ_{0} vanishes on C so (Y − q)^{2} divides σ_{0}. Dividing through by Y − q this implies Y − q divides 1, a contradiction. Thus T_{S}^{alg} ≠ T_{S}, so T_{S} is not closed for this choice of S. On the other hand, the curve C is described more naturally as C = K_{S} where S = {Y − q, −(Y − q)}, and, for this choice of S, T_{S}^{alg} = T_{S} (so, in particular, T_{S} is closed). To prove this, just write h_{1} = r^{2} − s^{2}, r = (h_{1} + 1)/2, s = (h_{1} − 1)/2, so

f = f(X, g(X)) + r^{2}(Y − q) − s^{2}(Y − q) ∈ T_{S}.

3.7 Example. Consider the curve

C = {(x, y) ∈ ℝ^{2} | y^{2} = q(x)}

where q ∈ ℝ[X] is not a square. Thus C = K_{S} where S = {Y^{2} − q, −(Y^{2} − q)}. Assume that C is not compact. Replacing X by −X if necessary, we can assume that q(x) > 0 for x sufficiently large (so the leading coefficient of q is positive). We claim that, under this assumption, T_{S} is closed. Since T_{S} = ∑ ℝ[X, Y]^{2} + (Y^{2} − q), where (Y^{2} −q) denotes the ideal generated by Y^{2} −q, it suffices to prove that ∑ A^{2} is closed in A, where A denotes the coordinate ring of C, i.e.,

A = \frac{ℝ[X, Y]}{(Y^{2} − q)}.

Every f ∈ A is expressible uniquely as f = g + h\sqrt{q}, g, h ∈ ℝ[X]. Let Q_{d} denote the subspace of A consisting of all f = g + h\sqrt{q} ∈ A with deg(g), deg(h) ≤ 2d. According to Lemma 3.4, it suffices to show that (∑ A^{2}) ∩ Q_{d} is closed in Q_{d} for each d ≥ 0. We need certain degree estimates. If f = g + h\sqrt{q}, g, h ∈ ℝ[X], then f^{2} = (g^{2} + h^{2}q) + (2gh)\sqrt{q}. Thus every element of ∑ A^{2} has the form

p = ∑_{i=1}^{k}(g_{i} + h_{i}\sqrt{q})^{2} = ∑_{i=1}^{k}(g_{i}^{2} + h_{i}^{2}q) + 2 ∑_{i=1}^{k} g_{i}h_{i}\sqrt{q}.

Thus, if p = p_{1} + p_{2}\sqrt{q}, p_{1}, p_{2} ∈ ℝ[X], then

deg(p_{1}) = max{2deg(g_{i}), 2deg(h_{i}) + deg(q) | 1 ≤ i ≤ k}.

In particular, if p ∈ Q_{d}, then deg(g_{i}) ≤ d and deg(h_{i}) ≤ d− deg(q)/2, i = 1, …, k. With these estimates in hand, the proof that (∑ A^{2}) ∩ Q^{d} is closed in Q_{d} follows along the same lines as in the proof of Theorem 3.5 and can safely be left to the reader.

Since T_{S} is closed, the various conditions (†), (‡), (∗) and T_{S}^{alg} = T_{S} are equivalent. We claim that these conditions hold if deg(q) = 1 or 2 and that they fail if deg(q) ≥ 3. The first assertion is elementary; e.g., see [9, Proposition 2.17]. Suppose now that deg(q) ≥ 3. If q is not in ∑ ℝ[X]^{2} then q takes on negative values so there exists a quadratic p ∈ ℝ[X] which takes on negative values with p ≥ 0 on C. If p ∈ T_{S} then p = ∑_{i=1}^{k}(g_{i}^{2} + h_{i}^{2}q) for some g_{i}, h_{i} ∈ ℝ[X]. Then ∑_{i=1}^{k} h_{i}^{2} ≠ 0, so deg(p) ≥ deg(q), a contradiction. If q ∈ ∑ ℝ[X]^{2}, then q = r^{2} + s^{2}, r, s ∈ ℝ[X]. Take p = p_{1} +Y where p_{1} ∈ ℝ[X] has degree ≤ max{deg(r), deg(s)}+1 and is such that p_{1} ≥ |r| + |s| on ℝ. Then

p(x, ±\sqrt{q(x)}) = p_{1}(x) ± \sqrt{q(x)} ≥ |r(x)| + |s(x)| ± \sqrt{r(x)^{2} + s(x)^{2}} ≥ 0

for each x ∈ ℝ so p ≥ 0 on C. If p ∈ T_{S}, then p_{1} = ∑_{i=1}^{k}(g_{i}^{2} +h_{i}^{2}q), 1 = 2 ∑_{i=1}^{k} g_{i}h_{i} for some g_{i}, h_{i} ∈ ℝ[X]. Then ∑ h_{i}^{2} ≠ 0, so deg(p_{1}) ≥ deg(q). Since deg(q) = max{2deg(r), 2deg(s)}, this contradicts the assumption that deg(q) ≥ 3.

Note: According to [9, Theorem 3.4], if C is non-singular and deg(q) ≥ 3 then the preordering T_{S}^{alg} is not even finitely generated.

We turn now to the case where dim(K_{S}) ≥ 2. In [1, Theorem 4], Berg, Christensen and Jensen use their weak version of Theorem 3.5 to show that (∗) fails for S = ∅ (so K_{S} = ℝ^{n}, T_{S} = ∑ ℝ[X]^{2}) if n ≥ 2. In [2, Theorem 6.3.9] this same method is used to show that (∗) fails for S = {X_{1}, …, X_{n}} if n ≥ 2. We proceed to generalize these results. In general, for T_{S} closed, showing (∗) fails is equivalent to showing ∃ p ∈ T_{S}^{alg}, p ∉ T_{S}. Such a polynomial always exists when dim(K_{S}) is large enough, e.g., the Motzkin polynomial if S = ∅, n ≥ 2. The following general result is proved in [9]:

3.8 Theorem. If dim(K_{S}) ≥ 3 then there exists a polynomial p ≥ 0 on ℝ^{n} such that p ∉ T_{S}.^{1}

Proof. See [9, Prop. 6.1]. □

Theorem 3.8 implies, in particular, that (∗) fails whenever n ≥ 3 and K_{S} contains a cone of dimension n. Another result in [9] allows us to greatly improve on this:

1Apparently this is not true for dim(K_{S}) = 2. In the “Added to proof” at the end of [9], Scheiderer reports the discovery of a smooth compact surface K_{S} with T_{S}^{alg} = T_{S}.

3.9 Theorem. If n = 2 and K_{S} contains a 2-dimensional cone, then there exists a polynomial p ≥ 0 on ℝ^{2} such that p ∉ T_{S}.

Proof. See [9, Remark 6.7]. □

Note: The proof of Theorem 3.9 given in [9] is highly non-trivial. It uses deep results on existence of “enough” psd regular functions on non-compact non-rational curves [9, Section 3] in conjunction with an extension theorem for extending such functions to the ambient space [9, Section 5].

3.10 Corollary. (∗) fails whenever n ≥ 2 and K_{S} contains a cone of dimension 2.

Note: We are not claiming now that T_{S} is closed.

Proof. Changing coordinates, we may assume the cone is given by

X_{1} ≥ 0, X_{2} ≥ 0, X_{i} = 0, i ≥ 3.

Define S′ ⊆ ℝ[X_{1}, X_{2}] by S′ = {g′_{1}, …, g′_{s}} where g′_{i} = g_{i}(X_{1}, X_{2}, 0, …, 0). Thus K_{S′} contains the cone defined by X_{1} ≥ 0, X_{2} ≥ 0 so, by Theorem 3.5, T_{S′} is closed. Use Theorem 3.9 to pick p = p(X_{1}, X_{2}) such that p ≥ 0 on ℝ^{2} and p ∉ T_{S′}. Then p ∈ T_{S}^{alg}. We claim that p ∉ T_{S}^{lin}. For suppose p ∈ T_{S}^{lin}. Any linear functional L on ℝ[X_{1}, X_{2}] with L(T_{S′}) ≥ 0 extends to a linear functional L′ on ℝ[X] with L′(T_{S}) ≥ 0 via L′(f) = L(f(X_{1}, X_{2}, 0…. , 0)). This is clear. Thus, for any such L, L(p) = L′(p) ≥ 0, so p ∈ T_{S′}^{lin}. Since T_{S′}^{lin} = T_{S′}, this contradicts the choice of p. □

The method of Corollary 3.10 applies in other cases as well:

3.11 Example. Take n = 2, S = {X, 1 − X, Y^{3} − X^{2}}. K_{S} consists of the points (x, y) ∈ ℝ^{2} on the vertical strip 0 ≤ x ≤ 1 with y ≥ x^{2/3}. We claim that (∗) fails for S. Let S′ ⊆ ℝ[Y] be defined by S′ = {Y^{3}}. Then K_{S′} = [0, ∞) so, by Theorem 3.5, T_{S′} is closed. One checks that Y ∉ T_{S′}. (Degree considerations show that Y = σ_{0} + σ_{1}Y^{3}, σ_{0}, σ_{1} ∈ ∑ ℝ[Y]^{2} is not possible.) Clearly Y ≥ 0 on K_{S}. We claim Y ∉ T_{S}^{lin}. We argue as in the proof of Corollary 3.10: Every linear functional L on ℝ[Y] with L(T_{S′}) ≥ 0 extends to a linear functional L′ on ℝ[X, Y] with L′(T_{S}) ≥ 0 via L′(f) = L(f(0, Y)). Thus, if Y ∈ T_{S}^{lin} then L(Y) = L′(Y) ≥ 0 for every such L, so Y ∈ T_{S′}^{lin} = T_{S′}.

4. Consequences of the results in [7]

In this section we recall results from [7], and mention applications of these results to the problem at hand. The results in [7] extend Schmüdgen’s result in the compact case, replacing the assumption that K_{S} is compact with the assumption that p ∈ 1 + T_{S} is chosen so that the coordinate functions X_{1}, …, X_{n} are bounded on K_{S} by kp^{ℓ} for k, ℓ sufficiently large. The method of proof in [7] is a generalization of Wörmann’s method in [11].

Note: Such a polynomial p always exists, e.g., p = 1 + ∑_{i=1}^{n} X_{i}^{2}. The ‘best’ choice of p may depend on K_{S}. For example:

— If K_{S} is compact, we can take p = 1.

— If n = 2 and S = {X, 1 − X}, we can take p = 1 + Y^{2}.

Fixing such a polynomial p, we have:

4.1 Theorem. For any f ∈ ℝ[X] there exist integers k, ℓ such that kp^{ℓ} ± f ∈ T_{S}.

Proof. See [7, Corollary 1.4]. □

This gives better control over the element q appearing in condition (‡). We can choose q to be a power of p if we want. In more detail, we have:

4.2 Corollary. The following are equivalent:

(1) (‡) holds.

(2) ∀f ∈ ℝ[X], f ∈ T_{S}^{alg} ⇒ ∃ℓ ≥ 0 such that ∀ real ϵ > 0, f + ϵp^{ℓ} ∈ T_{S}.

Proof. (2) ⇒ (1) is clear. For (1) ⇒ (2), use Theorem 4.1 to pick k, ℓ so that kp^{ℓ} − q ∈ T_{S} and use the identity f + ϵp^{ℓ} = (f + \frac{ϵ}{k} q) + \frac{ϵ}{k} (kp^{ℓ} − q). □

The second main result in [7] is:

4.3 Theorem. For any f ∈ ℝ[X], the following are equivalent:

(1) f ∈ T_{S}^{alg}.

(2) For any sufficiently large integer m and all rational ϵ > 0, there exists an integer ℓ ≥ 0 such that p^{ℓ}(f + ϵp^{m}) ∈ T_{S}.

Proof. See [7, Corollary 3.1]. □

Let

T′_{S} = {f ∈ ℝ[X] | p^{ℓ}f ∈ T_{S} for some ℓ ≥ 0}.

For d ≥ 0, let T′_{d} = T′_{S} ∩ P_{d}.

4.4 Corollary. T_{S}^{alg} = (T′_{S})^{lin} = ∪_{d≥0}\overline{T′_{d}}.

Proof. Suppose f ∈ T_{S}^{alg}. By Theorem 4.3 there exists m ≥ 0 such that, for all ϵ > 0, f + ϵp^{m} ∈ T′_{S}. This implies f ∈ \overline{T′_{d}} for d sufficiently large. □

Note: In view of the result of Haviland in [4] [5], Corollary 4.4 implies, in particular, that a linear functional L on ℝ[X] comes from a positive Borel measure on ℝ^{n} supported by K_{S} iff L(T′_{S}) ≥ 0; also see [8].

Note: The preordering T′_{S} is not finitely generated in general. This is the case, for example, if K_{S} contains a cone of dimension 2; see Corollary 3.10.

Theorem 4.3 also yields a large number of non-compact examples where (‡) holds. Denote by ℝ[X, Y] the polynomial ring in n + 1 variables X_{1}, …, X_{n}, Y and consider the finite set S′ = S ∪ {1 − pY, −(1 − pY)} in ℝ[X, Y]. K_{S′} consists of those points on the hypersurface

H = {(x, y) ∈ ℝ^{n+1} | p(x)y = 1}

in ℝ^{n+1} which map to K_{S} under the projection (x, y) ↦ x.

4.5 Corollary. (‡) holds for S′.

Proof. Suppose f ∈ T_{S′}^{alg}. Thus f(X, \frac{1}{p}) ≥ 0 on K_{S}. Writing f(X, \frac{1}{p}) = \frac{g}{p^{k}}, g ∈ ℝ[X], we see that g ≥ 0 on K_{S} so there exists m ≥ 0 such that for each ϵ > 0, p^{ℓ}(g + ϵp^{m}) ∈ T_{S} for some ℓ ≥ 0. Since p − 1 ∈ T_{S}, there is no harm in assuming m ≥ k. Thus

(1)  \frac{g}{p^{k}} + ϵp^{m−k} = \frac{h}{p^{k+ℓ}}

for some h ∈ T_{S}. This implies that the polynomial f + ϵp^{m−k} − hp^{k+ℓ}Y^{2(k+ℓ)} vanishes on H so

(2)  f + ϵp^{m−k} = hp^{k+ℓ}Y^{2(k+ℓ)} + j(1 − pY)

for some j ∈ ℝ[X, Y]. Using the identity j = (\frac{j+1}{2})^{2} − (\frac{j−1}{2})^{2} this implies f + ϵp^{m−k} ∈ T_{S′} for all real ϵ > 0. □

Thus, for example, although (∗) fails for the plane (Corollary 3.10), (‡) holds for the surface Z = \frac{1}{1+X^{2}+Y^{2}} in ℝ^{3}.

4.6 Corollary. For a non-zero linear functional L on ℝ[X], the following are equivalent:

(1) L comes from a positive Borel measure on ℝ^{n} supported by K_{S}.

(2) L extends to a linear functional on L′ on ℝ[X, Y] satisfying L′(T_{S′}) ≥ 0.

Proof. (1) ⇒ (2). Define

L′(f) = ∫_{ℝ^{n}} f(x, \frac{1}{p(x)})dμ(x)

where μ is the measure. Since p ≥ 1 on K_{S}, L′ is well-defined. (2) ⇒ (1). If f ∈ T_{S}^{alg} then f ∈ T_{S′}^{alg} = T_{S′}^{lin}, so L(f) = L′(f) ≥ 0. (1) follows now, by Haviland’s result. □

Remark: T′_{S} and T_{S′} are related by T′_{S} = T_{S′} ∩ ℝ[X].

4.7 Example. We use Corollary 4.5 to construct examples where dim(K_{S}) = 1, K_{S} is not compact, (‡) holds and T_{S} is not closed.

(1) Take n = 1, S = {X^{3}}, p = 1 + X^{2}. Thus K_{S′} is defined by Y = 1/(1 + X^{2}), X ≥ 0. We claim T_{S′} is not closed. Otherwise, by Corollary 4.5 and Proposition 3.1, T_{S′}^{alg} = T_{S′} so, intersecting with ℝ[X], T_{S}^{alg} = T′_{S}. Since X ≥ 0 on K_{S}, this implies (1 + X^{2})^{ℓ}X ∈ T_{S} for some ℓ ≥ 0, i.e, (1 + X^{2})^{ℓ}X = σ_{0} + σ_{1}X^{3}, σ_{i} ∈ ∑ ℝ[X]^{2}, i = 0, 1. Evaluating at X = 0, we see that X^{2} divides σ_{0}. Consequently, X divides (1 + X^{2})^{ℓ}, a contradiction.

(2) Take n = 2, S = {Y^{2} − X^{3}, −(Y^{2} − X^{3})}. Since Y^{2} = X^{3} on K_{S}, we can take p = 1 + X^{2}. Thus K_{S′} is defined by Z = 1/(1 + X^{2}), Y^{2} = X^{3}. We claim that T_{S′} is not closed. Otherwise, as in (1), (1 + X^{2})^{ℓ}X ∈ T_{S} for some ℓ ≥ 0 so, by the method of Example 3.7, (1 + X^{2})^{ℓ}X = σ_{0} + σ_{1}X^{3} for some σ_{i} ∈ ∑ ℝ[X]^{2}, i = 0, 1.

5. Cylinders with compact cross-section

We denote by ℝ[X, Y] the polynomial ring in n + 1 variables X_{1}, …, X_{n}, Y. In this section we consider a subset S = {g_{1}, …, g_{s}} of ℝ[X, Y] where the polynomials g_{1}, …, g_{s} involve only the variables X_{1}, …, X_{n}, so K_{S} has the form K×ℝ, K ⊆ ℝ^{n}. We further assume that K is compact. We describe this situation by saying that K_{S} is a cylinder with compact cross-section. We aim to prove:

5.1 Theorem. If K_{S} is a cylinder with compact cross-section, then (‡) holds.

Proof. Suppose f ∈ ℝ[X, Y] is such that f ≥ 0 on K_{S}. Fix d ≥ 1 so that the degree of f as a polynomial in Y is ≤ 2d. Let ϵ > 0 be given. Let f_{1} = f + ϵ + ϵY^{2d}, say f_{1} = a_{0} + a_{1}Y + ⋯ + a_{2d}Y^{2d}, a_{i} ∈ ℝ[X]. f_{1} is strictly positive on K × ℝ and its leading coefficient is strictly positive on K. Denote by A the ring of all continuous functions from K to ℝ and consider ϕ = ã_{0} + ã_{1}Y + ⋯ + ã_{2d}Y^{2d} ∈ A[Y] where ã_{i} is the continuous function corresponding to a_{i}. Since the roots of ϕ depend continuously on the coefficients, ϕ factors as ϕ = ã_{2d}ψ\overline{ψ} where the coefficients of ψ are continuous complex-valued functions on K and “bar” denotes complex conjugation. (At each point of K, take ψ = ∏_{i=1}^{d}(Y − z_{i}) where the z_{1}, …, z_{d} are the roots with positive imaginary part, counting multiplicities.) Decomposing ψ as ψ_{1} + iψ_{2}, ψ_{1}, ψ_{2} ∈ A[Y] this yields

(1)  ϕ = ã_{2d}(ψ_{1}^{2} + ψ_{2}^{2}).

Now use the compactness of K to approximate the coefficients of ψ_{1}, ψ_{2} closely by polynomials in ℝ[X]. In this way we obtain

(2)  f + ϵ + ϵY^{2d} = a_{2d}(h_{1}^{2} + h_{2}^{2}) + ∑_{i=0}^{2d} b_{i}Y^{i},

where h_{1}, h_{2} are polynomials in X, Y and b_{0}, …, b_{2d} are polynomials in X with |b_{i}| < ϵ on K, i = 0, …, 2d. By Schmüdgen’s Theorem, a_{2d} ∈ T and ϵ ± b_{i} ∈ T where T is the preordering in ℝ[X] generated by g_{1}, …, g_{s}. This yields

(3)  ϵY^{i} + b_{i}Y^{i} ∈ T_{S}, for i even.

For i odd, say i = 2k + 1, use the identity Y^{2k+1} = \frac{1}{2}Y^{2k}((Y + 1)^{2} − Y^{2} − 1) plus the fact that ϵY^{2k}(Y + 1)^{2} + b_{i}Y^{2k}(Y + 1)^{2}, ϵY^{2k}Y^{2} − b_{i}Y^{2k}Y^{2} and ϵY^{2k} − b_{i}Y^{2k} all belong to T_{S} to obtain

(4)  ϵ(Y^{i+1} + Y^{i} + Y^{i−1}) + b_{i}Y^{i} ∈ T_{S}, for i odd.

Thus, adding ϵ(2 + Y + 3Y^{2} + Y^{3} + 3Y^{4} + ⋯ + 2Y^{2d}) to each side of (2), we see finally that

(5)  f + ϵ(3 + Y + 3Y^{2} + Y^{3} + ⋯ + 3Y^{2d}) ∈ T_{S}.

Since the polynomial q = 3 + Y + 3Y^{2} + Y^{3} + ⋯ + 3Y^{2d} depends only on d (not on ϵ), the proof is complete. □

Cylinders with compact cross-section also provide examples where (‡) holds but (†) does not hold:

5.2 Example. Take n = 2. Take S ⊆ ℝ[X, Y] defined by S = {X^{3}(1 − X)^{3}}. Then K_{S} is the strip [0, 1] × ℝ and (‡) holds by Theorem 5.1. Consider f = XY^{2}. Clearly f ≥ 0 on K_{S}. We claim that ∀ϵ > 0, f + ϵ ∉ T_{S}. This will show, in particular, that (†) does not hold. For suppose

f + ϵ = σ_{0} + σ_{1}X^{3}(1 − X)^{3}, σ_{0}, σ_{1} ∈ ∑ ℝ[X, Y]^{2}.

Comparing degrees in Y, we see that σ_{0}, σ_{1} have the form

σ_{0} = ∑(a_{i} + b_{i}Y)^{2}, σ_{1} = ∑(c_{j} + d_{j}Y)^{2},

with a_{i}, b_{i}, c_{j}, d_{j} ∈ ℝ[X]. Equating coefficients of Y^{2}, this gives

X = ∑ b_{i}^{2} + ∑ d_{j}^{2}X^{3}(1 − X)^{3}.

Evaluating at X = 0, we see that X divides each b_{i}. Dividing through by X, we see that X divides 1, a contradiction.

5.3 Notes.

(1) Example 5.2 generalizes to arbitrary dimension, taking f = pY^{2} with p ∈ ℝ[X], p ≥ 0 on K and p ∉ T. Such a polynomial p will exist (regardless of the description of K) if dim(K) is sufficiently large; see Theorem 3.8.

(2) In Example 5.2 we are not using the “natural” description of the strip 0 ≤ X ≤ 1. The question of whether or not (†) holds for the natural description (i.e. S = {X, 1 − X}) is open.

Theorem 5.1 also extends to sets of the form K × [0, ∞) with K compact:

5.4 Corollary. Suppose S′ = S∪{Y }, S as in Theorem 5.1 (so K_{S′} = K×[0, ∞)). Then (‡) holds for S′.

Note: The result is false if we take the “wrong” description of K × [0, ∞), e.g., for S′ = S ∪ {Y^{3}} (∗) fails. (Just modify the proof of Example 3.11.)

Proof. Suppose f = f(X, Y) is non-negative on K_{S′}. Then f(X, Y^{2}) is nonnegative on K_{S}. By the proof of Theorem 5.1, for q = 3 + Y + 3Y^{2} + ⋯ + 3Y^{2d}, d the degree of f in Y, we have, for all real ϵ > 0,

(1)  f(X, Y^{2}) + ϵq = ∑_{e} σ_{e}g^{e},

with σ_{e} ∈ ∑ ℝ[X, Y]^{2}. Replacing Y by −Y, adding, and dividing by 2, using the standard identity

(2)  \frac{1}{2}(g(Y)^{2} + g(−Y)^{2}) = h(Y^{2})^{2} + k(Y^{2})^{2}Y^{2}

(where, if g(Y) = ∑_{i} a_{i}Y^{i}, then h(Y^{2}) = ∑_{j} a_{2j}Y^{2j}, k(Y^{2}) = ∑_{j} a_{2j+1}Y^{2j}), this yields

(3)  f(X, Y^{2}) + ϵq′ = ∑_{e}(γ_{e} + δ_{e}Y^{2})g^{e}

where q′ = 3 + 3Y^{2} + ⋯ + 3Y^{2d} and γ_{e}, δ_{e} ∈ ∑ ℝ[X, Y^{2}]^{2}. Replacing Y^{2} by Y gives the desired conclusion. □

6. Open problems

1. Does Corollary 5.4 generalize to sets of the form K × L, K compact, L closed semi-algebraic in ℝ (with the natural description)? We know this is true for L of the form (−∞, ∞), [a, ∞) or (−∞, a]. Obviously, it is also true for L compact, regardless of the description.

2. Consider the generalized polyhedron K_{S} in ℝ^{n} defined by S = {ℓ_{1}, …, ℓ_{s}} where ℓ_{1}, …, ℓ_{s} are linear. If K_{S} is compact, we know that (†) holds. (In fact, even (†_{M}) holds.) What if K_{S} is not compact? If K_{S} contains a cone of dimension 2 then, by Corollary 3.10, (∗) fails. This will always be the case if s < n. If s ≥ n then it may happen that K_{S} does not contain a cone of dimension 2. Is it true in this case that (‡) holds? We know this is true if s = n (resp., s = n + 1) by Theorem 5.1 (resp., Corollary 5.4).

3. Find examples where (∗) holds but (‡) does not hold. Find examples where T_{S}^{lin} ≠ ∪_{d≥0}\overline{T}_{d}. Could it be true that (∗) ⇒ (‡)? This seems highly unlikely.

4. Find a smooth irreducible curve C in ℝ^{n} with ∑ ℝ[C]^{2} not closed in ℝ[C]. Here, ℝ[C] denotes the coordinate ring of C. According to the “Added in proof” at the end of [9], any such C is necessarily not compact. Note: Our Example 4.7(2) is not smooth.

5. It should be possible to use the method of Corollary 4.5 to construct a curve where (‡) holds but (†) fails. We were not able to do this. In general, it would be nice to know more about the curve case.

6. Find examples where K_{S} is not compact, dim(K_{S}) ≥ 2, and (†) holds. So far, we only know examples where (‡) holds. Does (†) hold for the strip 0 ≤ X ≤ 1 in ℝ^{2} with the presentation S = {X, 1 − X}?

7. For n = 1 and K_{S} compact, whether T_{S} is closed or not depends on the presentation. For example, for S = {1 + X, 1 − X}, T_{S} is closed, but for S = {(1 − X^{2})^{3}}, T_{S} is not closed. In both cases, K_{S} = [−1, 1]. If K_{S} = [−1, 1]^{n}, n ≥ 3, then T_{S} is never closed, regardless of the presentation (Theorem 3.8). What if n = 2? We know T_{S} is not closed for the presentation S = {(1−X^{2})^{3}, (1−Y^{2})^{3}}. What about the presentation S = {1 + X, 1 − X, 1 + Y, 1 − Y }?

8. Certain results are independent of the presentation, i.e., depend only on K_{S}: If K_{S} is compact then (†) holds, regardless of the presentation; if n ≥ 2 and K_{S} contains a cone of dimension 2 then (∗) fails, regardless of the presentation; if K_{S} contains a cone of dimension n then T_{S} is closed, regardless of the presentation. Other results depend on the particular presentation. In general, one would like to know to what extent results are independent of the presentation.

References

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Department of Mathematics & Statistics, University of Saskatchewan, Saskatoon, SK Canada, S7N 5E6

E-mail address: skuhlman@math.usask.ca, marshall@math.usask.ca