Positivity, sums of squares and the multi-dimensional moment problem II

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1 Positivity, sums of squares and the multi-dimensional moment problem II S. Kuhlmann, M. Marshall and N. Schwartz Abstract The paper is a continuation of work initiated by the first two authors in [K M]. Section 1 is introductory. In Section 2 we give new proofs of results of Scheiderer in [S1] [S2] in the compact case; see Corollaries 2.3, 2.4 and 2.5. The main tool in Section 2, Lemma 2.1, is also used in Section 3 where we continue the examination of the case n = 1 initiated in [K M], concentrating on the compact case. In Section 4 we prove certain uniform degree bounds for representations in the case n = 1 which we then use in Section 5 to prove that ( ) holds for basic closed semi-algebraic subsets of cylinders with compact cross-section provided the generators satisfy certain conditions; see Theorem 5.3 and Corollary 5.5. Theorem 5.3 provides a partial answer to a question raised by Schmüdgen in [Sc2]. We also show that for basic closed semi-algebraic subsets of cylinders with compact cross-section the necessary conditions for (SMP) given in [Sc2] are also sufficient; see Corollary 5.2(b). In Section 6 we prove a module variant of the result in [Sc2] in the same spirit as Putinar s variant [Pu] of the result in [Sc1] in the compact case; see Theorem 6.1. We then apply this to basic closed semi-algebraic subsets of cylinders with compact cross-section; see Corollary 6.4. In Section 7 we apply the results from Section 5 to solve two of the open problems listed in [K M]; see Corollary 7.1 and Corollary 7.5. In Section 8 we consider a number of examples in the plane. In Section 9 we list some open problems. 1 Introduction Let R[X] denote the polynomial ring in n variables X = (X 1,..., X n ) with real coefficients, and consider a finite set S = {g 1,, g s } of polynomials in R[X]. We 2000 Mathematics Subject Classification: Primary 14P10, Secondary 44A60. Partially supported by an NSERC research grant. Partially supported by an NSERC research grant. Partially supported by the European RTN Network RAAG, Contract No. HPRN-CT

2 denote by K S the basic closed semi-algebraic set in R n defined by: K S = {x R n : g 1 (x) 0,..., g s (x) 0}. We denote by T S the preordering in R[X] generated by S: T S = { σ e g e : σ e R[X] 2 }, e {0,1} s where g e := g e 1 1 g e s s if e = (e 1,, e s ), and R[X] 2 denotes the preordering of R[X] consisting of sums of squares. Note: K = R n, T = R[X] 2. By the Positivstellensatz [St], K S = 1 T S T S = R[X]. Set and Note: T S T lin S T alg T alg S = {f R[X] : f 0 on K S }, T S = {L : R[X] R : L is linear ( 0) and L(T S ) 0}, T lin S = T S = {f R[X] : L(f) 0 for all L T S }. T alg S. TS, TS lin depend on T S. T alg S depends only on K S. S is a preordering, called the saturation of T S. We say T S is saturated if T alg S = T S. TS lin is the closure of T S in R[X], giving R[X] the unique finest locally convex topology [Po S, p. 76]. TS lin is a preordering [Po S, Lem. 1.2]. We say T S is closed if TS lin = T S. The Moment Problem is the following: For a linear functional L on R[X], when is there a positive Borel measure µ or R n such that f R[X] L(f) = R n fdµ? The following result is due to Haviland [H1] [H2]. For a proof based on the Riesz representation theorem, see [M2, Th. 3.1] for example. Theorem 1.1 For a linear functional L on R[X] and a closed set K of R n, the following are equivalent: (i) a positive Borel measure µ on K such that f R[X], L(f) = K fdµ. (ii) f R[X], f 0 on K L(f) 0. Since T alg S is in general not finitely generated, one is interested in approximating it by T S. Therefore one studies the following concrete version of the Moment Problem: When is it true that every L TS comes from a positive Borel measure on K S? By Theorem 1.1, this is equivalent to asking when the following condition holds: (SMP) T alg S = T lin S. Given a pair of sets S, S with K S K S one can also ask: When is it true that every L TS comes from a positive Borel measure supported by K S? By Theorem 2

3 1.1 this is equivalent to asking if T alg S T lin S. This property is most often considered in the special case S = : (MP) T alg T lin S. In [Sc1], the author proves (SMP) when the basic closed semi-algebraic set K S is compact. Moreover, for compact K S, he gets the following substantial improvement of the Positivstellensatz: ( ) f R[X], f 0 on K S real ɛ > 0, f + ɛ T S. See [M1] or [P D] for a proof of ( ) based on the Kadison-Dubois representation theorem. Note: (i) T S is saturated ( ) holds (SMP) holds (MP) holds. (ii) If T S is closed then T S is saturated ( ) holds (SMP) holds. In [K M], the authors consider the condition: ( ) f R[x], f 0 on K S q R[X] such that real ɛ > 0, f + ɛq T S. It is shown that ( ) is strictly weaker than ( ), but implies (SMP). The authors show that ( ) holds in various non-compact cases, e.g., for basic closed semi-algebraic sets obtained by a certain natural dimension extension process [K M, Cor. 4.5] and for cylinders with compact cross-section [K M, Th. 5.1]. Motivated by condition ( ) we define T S = {f R[X] : q R[X] such that real ɛ > 0, f + ɛq T S }. Note: (i) T S T S TS lin. (ii) ( ) holds iff T alg S to also consider the set T S = {f R[X] : real ɛ > 0, f + ɛ T S }, = T S. (iii) It would seem natural but it is not clear that T S is closed under multiplication in general. It follows from the proof of [K M, Cor. 4.2] that the element q appearing in the description of T S can be chosen from the set {p l : l 0} where p 1 + T S is any polynomial which grows sufficiently rapidly on K S, e.g., p = 1 + X 2 always works. Here, X 2 := n i=1 X 2 i. If K S is compact then p = 1 works. Denote by P d the (finite dimensional) vector space consisting of all polynomials in R[X] of degree 2d, and by 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 T d the closure of T d in P d. Proposition 1.2 (i) T S = d 0 T d. (ii) T S is a preordering. 3

4 Proof: Assertion (i) is clear from the proof of [K M, Prop. 1.3]. For polynomials f, g, the coefficients of f + g and fg are polynomial functions (in particular, continuous functions) of the coefficients of f and of g. Assertion (ii) is clear from this, using (i). The question of whether (SMP) is strictly weaker than ( ), posed as a problem in [K M], is still open. It seems likely that this is the case, but no examples are known. In fact, no examples are known where T S TS lin. Observe that ( ), ( ), (SMP) and (MP) all depend on T S. Consequently, if ( ), resp., ( ), resp., (SMP), resp., (MP) holds then we say that ( ), resp., ( ), resp., (SMP), resp., (MP) holds for T S or that T S satisfies ( ), resp., ( ), resp., (SMP), resp., (MP). The case n = 1 is already of considerable interest. In fact, early work on the Moment Problem dealt almost exclusively with this case. If K R is a non-empty closed semi-algebraic set then one checks easily that K = K N for N the set of polynomials defined as follows: If a K and (, a) K =, then X a N. If a K and (a, ) K =, then a X N. If a, b K, a < b, (a, b) K =, then (X a)(x b) N. N has no other elements except these. We call N the natural set of generators for K. We adopt the convention that the natural set of generators of the empty set is { 1}. The following can be easily deduced from [K M] and [Sc1]: Theorem 1.3 Suppose n = 1. (a) If K S is not compact, then T S is closed, for any finite set of generators S. Moreover the following are equivalent: (i) T S is saturated (ii) T S contains the natural set of generators (iii) S contains the natural set of generators (up to scalings by positive reals). (b) If K S is compact, then ( ) holds for T S, for any finite set of generators S. Moreover the following are equivalent: (i) T S is saturated (ii) T S is closed (iii) T S contains the natural set of generators. Note: Unlike the non-compact case (a), T S may be saturated in the compact case (b) even when S does not contain a scaling of the natural set of generators (cf. [K-M, Notes 2.3(4)]). See Section 3 for a useful criterion for determining when T S is saturated (i.e., closed) in the compact case when n = 1. For n 2 there is no general criterion for determining when T S is closed. We do have the following: 4

5 Proposition 1.4 (i) If K S is compact and dim(k S ) 3 then T S is not closed. (ii) If K S contains a cone of dimension n then T S is closed. Assertion (i) is immediate from [S1, Prop. 6.1] using [Sc1, Cor. 2]. (ii) is proved in [K M, Th. 3.5] using a method introduced earlier in [B C J]. See [Po S, Prop. 3.7] for another proof. Proposition 1.5 If K S contains a cone of dimension 2 then (MP) fails for T S. The statement of Proposition 1.5 is stronger than that of [K M, Cor. 3.10] but the proof is the same. Namely, one uses [K M, Th. 3.5] along with [S1, Remark 6.7] to produce p T alg, p / TS lin. See [Po S, Cor. 3.10] for a more general criterion. Note: For n 3 the condition that K S contains a cone of dimension 2 does not imply that T S is closed. Example 1.6 Let n 1 and pick S 1 in R[X] such that K S1 is compact and T S1 is not closed. Define S in R[X, Y, Z] by S = S 1. K S = K S1 R 2 contains a 2- dimensional cone. Also TS lin 1 TS lin and T S1 = T S R[X] so if p TS lin 1, p / T S1, then p TS lin, p / T S. The possible choices for S 1 are necessarily a bit artificial if n = 1, e.g., S 1 = {X 3 (1 X)} will do. For n = 2 it seems that less contrived choices should exist. For n 3 any S 1 with dim(k S1 ) 3 will do [S1, Prop. 6.1]. In [Sc2], the author considers the following situation: Suppose h 1,..., h d R[X] are bounded on K S (so N d i=1 h 2 i 0 on K S for N sufficiently large). For λ = (λ 1,..., λ d ) R d define S λ = S {h 1 λ 1, (h 1 λ 1 ),..., h d λ d, (h d λ d )}. Thus K Sλ = K S C λ where C λ denotes the algebraic set in R n defined by the equations h i (X) = λ i, i = 1,..., d. By the assumption K Sλ = (i.e., T Sλ = R[X]) for λ N. The author proves: Theorem 1.7 Suppose h 1,..., h d R[X] are bounded on K S. If (SMP) (resp., (MP)) holds for T Sλ for each λ then (SMP) (resp. (MP)) holds for T S. Theorem 1.7 can be applied in a variety of cases, e.g., to basic closed semi-algebraic subsets of cylinders with compact cross-section; see Corollary 5.2 below. As explained in [K M], one can also consider the (generally smaller) quadratic module s M S = { σ i g i : σ i R[X] 2 }, i=0 5

6 generated by S (here g 0 := 1) and the corresponding objects M S = {L : R[X] R L is linear ( 0) and L(M S ) 0}, M lin S = M S = {f R[X] : L(f) 0 for all L M S } and M S = d 0 M d where M d := M S P d. M lin S is the closure of M S in R[X]. Theorems 1.4 and and 1.5 and Proposition 1.2 continue to hold in this situation. One can also consider the corresponding conditions (SMP ) T alg S = M lin S, (MP ) T alg M lin S, ( ) f R[X], f 0 on K S real ɛ > 0, f + ɛ M S and ( ) f R[X], f 0 on K S q R[X] such that real ɛ > 0, f + ɛq M S. Again, (MP ) is strictly weaker than (SMP ) and ( ) is strictly weaker than ( ), but implies (SMP ). If ( ), resp., ( ), resp., (SMP ), resp., (MP ), holds then we say that ( ), resp., ( ), resp., (SMP), resp., (MP) holds for M S or that M S satisfies ( ), resp., ( ), resp., (SMP), resp., (MP). According to [Pu], also see [J], if K S is compact and there exists N such that N X 2 M S, then ( ) holds for M S. In certain cases the condition that there exists N such that N X 2 M S is automatically fulfilled; see [J P]. In [M2] various results in [K M], in particular, the results for dimension extension and for cylinders with compact cross-section are extended under suitable assumptions to quadratic modules; see [M2, Cor. 4.3, Cor 5.3]. 2 Basic Lemma The following general result is useful: Lemma 2.1 Let C be a compact Hausdorff space, A a commutative ring with 1 with 1 A for some integer n 2 and φ : A Cont(C, R) a (unitary) ring n homomorphism. Suppose f, g A are such that φ(f) 0, φ(g) 0 and sf +tg = 1 for some s, t A. Then there exist σ, τ A such that σf + τg = 1 and φ(σ), φ(τ) are strictly positive. 6

7 Proof: We suppress φ from the notation. Let s, t A be such that 1 = sf + tg. On the compact set L 1 := {p C s(p) 0}, tg = 1 sf 1. Thus g > 0 on L 1 so, for N sufficiently large, s + Ng > 0 on L 1. On C\L 1 this is obviously also true. Define s 1 = s + Ng, t 1 = t Nf. Thus 1 = s 1 f + t 1 g in A and s 1 > 0 on C. Choose a positive rational δ A so small that δfg < 1 on C. Choose a positive rational ɛ A so small that, on the compact set L 2 = {p C : g(p) ɛ}, f > 0, 1 > ɛδf and ɛt 1 + s 1 f > 0. Choose k so large that, on the set L 2, ɛt 1 + s 1 f > s 1 f(1 ɛδf) k and, on the set L 3 = {p C : g(p) ɛ}, s 1 f(1 ɛδf) k < 1. Choose r = s 1 δf k 1 i=0 (1 δfg) i. Choose σ = s 1 rg, τ = t 1 +rf. Thus 1 = σf + τg in A. It remains to verify that σ, τ > 0 on C. Using the identity (1 z) k 1 i=0 z i = 1 z k, we see that, on C, On L 2, On L 3, σ = s 1 rg = s 1 s 1 δfg k 1 (1 δfg) i i=0 = s 1 s 1 (1 (1 δfg) k ) = s 1 (1 δfg) k > 0. τ = t 1 + rf = t 1 + s 1 δf 2 k 1 (1 δfg) i i=0 t 1 + s 1 δf 2 k 1 (1 ɛδf) i = t 1 + (s 1 f/ɛ)(1 (1 ɛδf) k ) > 0. i=0 τ = t 1 + rf = t 1 + (f/g)rg = t 1 + (f/g)s 1 (1 (1 δfg) k ) = t 1 + s 1 f/g (s 1 f/g)(1 δfg) k = 1/g (s 1 f/g)(1 δfg) k > 0. This completes the proof. We apply Lemma 2.1 to C = K S and A = R[X]/I where I is some ideal of polynomials vanishing on K S. In the next section we will be interested in the case where n = 1 and I = {0}. Note: For K S compact, σ, τ > 0 on K S implies, by [Sc1, Cor. 3], that σ, τ T S. Corollary 2.2 Let K S be compact, f, g 0 on K S. Assume that f and g are relatively prime modulo T S T S and that fg T S. Then f T S and g T S. 7

8 Proof: Apply Lemma 2.1 with C = K S, A = R[X]/(T S T S ) together with [Sc1, Cor. 3] to obtain σ, τ T S, υ T S T S such that 1 = σf + τg + υ. Then f = σf 2 + τfg + fυ T S and g = σfg + τg 2 + gυ T S. Corollary 2.3 Assume f, g R[X] are relatively prime. K {fg} is compact and f, g 0 on K {fg}. Then (i) There exist σ, τ R[X] 2 such that 1 = σf + τg. (ii) M {fg} = M {f,g} = T {fg} = T {f,g}. Assume further that See [S1, Prop. 4.8] for another proof of Corollary 2.3 (i) in the case n = 1. Proof: Apply Lemma 2.1 with C = K {fg} and A = R[X] and [Sc1, Cor. 3] to obtain 1 = σf + τg, σ, τ T {fg}. Thus there exist σ i, τ i R[X] 2, i = 0, 1 such that σ = σ 0 + σ 1 fg, τ = τ 0 + τ 1 fg. Then 1 = σf + τg = (σ 0 + σ 1 fg)f + (τ 0 + τ 1 fg)g = (σ 0 + τ 1 g 2 )f + (τ 0 + σ 1 f 2 )g. This proves (i): 1 = σf + τg, σ, τ R[X] 2. Thus fg = σf 2 g + τg 2 f M {f,g}. Also, f = σf 2 + τfg T {fg} = M {fg} and, similarly, g T {fg} = M {fg}. This proves (ii). Lemma 2.1 is also closely related to results in [S2]. We illustrate by giving a proof of two key results in [S2]: Corollary 2.4 Let K S be compact, f 0 on K S, and suppose f T S + f (f) + (T S T S ). Then f T S. Proof: f = σ + fg, σ T S, g k = af + τ, k 1, a R[X], τ T S T S. af + (1 g k ) = 1 τ so f, 1 g are relatively prime modulo T S T S. Also f(1 g) = σ T S. Using this and the definition of g one checks that 1 g 0 on K S. Corollary 2.2 yields f T S. Corollary 2.5 Let K S be compact, f 0 on K S and suppose f = σ + τb, where σ, τ T S and b is such that f = 0 b > 0 on K S. Then f T S. Proof: By [Sc1, Cor. 3] applied to the set S { f 2 }, b = α βf 2, α, β T S. Then f = σ + τb = σ + τ(α βf 2 ) = (σ + τα) τβf 2. The conclusion follows by Corollary

9 3 Compact subsets of Lines A proper understanding of the case n = 1 is essential in dealing with the fiber sets that arise in the study of subsets of cylinders with compact cross-section. In this section we continue the analysis of this case begun in [K M, Sect. 2], focusing on the compact case. For related results see [Po R] and [S2]. Theorem 1.3 (b) is unsatisfactory unless we have a practical criterion on S for determining when T S is saturated. The following is such a criterion. Here, g denotes the derivative of g. Theorem 3.1 Let K S = k j=0[a j, b j ], b j 1 < a j, j = 1,..., k, S = {g 1,, g s }. Then T S is saturated if and only if the following two conditions hold: (a) for each endpoint a j i {1,, s} such that g i (a j ) = 0 and g i(a j ) > 0, (b) for each endpoint b j i {1,, s} such that g i (b j ) = 0 and g i(b j ) < 0. In fact, Theorem 3.1 is just a special case of a general criterion for curves proved in [S2, Th. 5.17]. For completeness we include our own proof. Proof: We prove the necessity of conditions (a) and (b). Assume T S is saturated. Suppose there is no i such that condition (a) holds for a j. Let a = a j. Fix c such that (c, a) K S =. Then (X a)(x c) T S, say (X a)(x c) = e σ e g e, σ e R[X] 2. Let E 1 = {e {0, 1} s g e (a) > 0}, E 2 = {e {0, 1} s g e (a) = 0}. Evaluating at X = a we see that (X a) 2 σ e for each e E 1. Consider the exponents e E 2 such that (X a) 2 does not divide g e. For each such e, exactly one of the g i appearing is divisible by X a (and not by (X a) 2 ) say g i = (X a)h i, and h i (a) = g i(a) 0. Consequently, dividing the equation (X a)(x c) = e σ e g e by X a and then evaluating at X = a we obtain a c 0, a contradiction. This proves the necessity of condition (a). The proof of the necessity of condition (b) is similar. We prove the sufficiency of conditions (a) and (b). Assume (a) and (b) hold. According to Theorem 1.3 (b) it suffices to show that T S contains the natural generators. We begin with the linear generators. Let a = a 0. By (a) there exists i such that g i = (X a)h i, h i (a) = g i(a) > 0. Then h i and X a are each 0 on K S and gcd(h i, X a) = 1 so, by Corollary 2.2, X a T S. A similar argument shows that b X T S where b = b k. Now consider a generator of the form (X a)(x b), a = a l, b = b l 1, l {1,..., k}. By (a) and (b) there exists i, j such that g i (a) = 0, g i(a) > 0, g j (b) = 0, g j(b) < 0. Factor g i, g j as g i = (X a)h i, g j = (X b)h j. Then (X a)(x b)((x b)h i + (X a)h j ) = (X b) 2 g i + (X a) 2 g j T S, and (X a)(x b) and (X b)h i +(X a)h j are each 0 on K S and are relatively prime, so by Corollary 2.2, (X a)(x b) T S. 9

10 Corollary 3.2 Let K S be compact, S = {g 1,, g s }. Assume that K S has no isolated points. Then T S is saturated if and only if for each endpoint a K S there exists i {1,, s} such that x a divides g i but (x a) 2 does not. Corollary 3.3 Let K be a compact semi-algebraic set in R without isolated points, N its set of natural generators, and π the product of the elements of N. Then M {π} = M N = T {π} = T N. Proof: Applying Corollary 3.2 to S = {π} we see that T {π} is saturated. Since K {π} = K = K N, and T N is saturated, this implies T {π} = T N. It remains to show that π M N. Decompose π as π = fq 1... q t where f = (X a)(b X) is the product of the linear natural generators and q 1,..., q t are the quadratic natural generators, t 0. Suppose t = 0. Then π = (X a)(b X) M {X a,b X} by Corollary 2.3. Suppose t > 0. Then π = π q t where π = fq 1... q t 1. By Corollary 2.3, π M {π,q t}. By induction on t, π M N where N = {X a, b X, q 1,..., q t 1 }. Thus π M N {q t } = M N. We need no isolated points in Corollaries 3.2 and 3.3: Take K = {0}, so N = {X, X} and π = (X)( X) = X 2. M { X 2 } M {X, X}. On the other hand, writing X = a 2 b 2, a, b R[X], we see that X 2 = (a 2 b 2 )X = a 2 X + b 2 ( X) M {X, X}, so M {X, X} = T {X, X}. This suggests that M N = T N may hold even when K has isolated points. In fact this is the case. Theorem 3.4 Let K be a compact semi-algebraic set in R, N its set of natural generators. Then M N = T N. Note: If K is not compact then M N = T N holds only in a few cases: if N = 1 or if N = 2 and K has an isolated point; see [K M, Th. 2.5]. Proof: Let N = {g 1,..., g s }. It suffices to show that M N is closed under multiplication, equivalently, that g i g j M N for all i j. Case 1: g i, g j are both linear. g i = X a, g j = b X, a b. If a = b then g j = g i. Write g j = f 2 g 2, f, g R[X]. Then g i g j = f 2 g i g 2 g i = f 2 g i + g 2 g j M {gi,g j }. If a < b then {g i, g j } are the natural generators of the compact set [a, b] so g i g j M {gi,g j } by Corollary

11 Case 2: g i is linear, g j is quadratic. Replacing X by X if necessary, we can assume g i = X a. g j = (X c)(x d), a c < d. If a = c then g i g j M {gi,g j } by [K M, Th. 2.5]. Suppose a < c. Fix β R so large that β X > 0 on K. β X M N by [J P, Remark 4.7]. Applying Corollary 3.3 to [a, c] [d, β] we see that g i g j = (X a)(x c)(x d) M {X a,(x c)(x d),β X} M N. Case 3: g i = (X c)(x d), g j = (X c )(X d ), c < d c < d. If c = d then g i g j M {gi,g j } by [K M, Th. 2.5]. Suppose c > d. Use [J P, Remark 4.7] to choose α < c and β > d so that X α, β X M N. Applying Corollary 3.3 to [α, c] [d, c ] [d, β] yields g i g j M {X α,(x c)(x d),(x c )(X d ),β X} M N. Corollary 3.5 For n = 1, K S compact, the following are equivalent: (i) M S is saturated, i.e., M S = T alg S (ii) M S contains the natural set of generators of K S (iii) M S is closed. Proof: By [J P, Remark 4.7] ( ) holds for M S so, in particular, (SMP) holds for M S. (i) (iii) follows from this fact. (i) (ii) is clear. (ii) (i) follows from Theorem 3.4 and Theorem 1.3(b). Corollary 3.5 can be viewed as the module analog of Theorem 1.3 (b). One would also like to have a module analog of Theorem Uniform Degree Bounds We continue to assume n = 1. Theorems 4.1 and 4.5 and Corollary 4.3 in this section are used in the next section, in the proof of Theorem 5.3 and Corollary 5.5. These results, suitably formulated, hold over an arbitrary real closed field. Fix a non-empty basic closed semi-algebraic set K in R with N = {s 1,..., s t } its set of natural generators. T N is saturated by Theorem 1.3 so each P R[X] non-negative on K is expressible as (1) P = σ e s e s e t t e {0,1} t with σ e R[X] 2. 11

12 Theorem 4.1 Each P nonnegative on K has a presentation (1) with the degree of each term σ e s e s et t bounded by the degree of P. Proof: This is just a matter of keeping track of degrees in the proof of [K M, Th. 2.2]. By the proof of [K M, Th. 2.2] there are four cases to consider: Case 1. P 0 on R. In this case P is a sum of squares so there is nothing to prove. Case 2. K has a least element a and P factors as P = (X d)q with d a, Q 0 on K. In this case, Q has a presentation of the required form, by induction on the degree. Replacing Q by this presentation in the identity P = (X a)q + (a d)q and rearranging terms then yields a presentation of P of the required form. Note: We are using the fact that X a is one of the natural generators of K. Case 3. K has a greatest element a and P factors as P = (d X)Q with d a, Q 0 on K. In this case proceed as in Case 2, writing P = (a X)Q + (d a)q. Case 4. There exist a, b K, a < b, with (a, b) K = and P factors as P = (X d)(x e)q with a d < e b and Q 0 on K. In this case choose λ > 0 so that F := (X d)(x e) λ(x a)(x b) 0 on R. This is possible, e.g., choose λ so that the equation F = 0 has a root of multiplicity two. Then F is a sum of squares, and we can proceed as in Case 2, using the identity P = F Q + λ(x a)(x b)q. Note: We are using the fact that (X a)(x b) is one of the natural generators of K. Corollary 4.2 Suppose K = K {g1,...,g k } and T {g1,...,g k } is saturated. Then each P nonnegative on K has a presentation (2) P = τ i g i g i k i {0,1} k τ i R[X] 2, where the degrees of the terms are bounded by some number of the form deg(p ) + N, with N depending only on g 1,..., g k. k, Proof: There exist presentations s j = ρ jp g p g p k k, p {0,1} k ρ jp R[X] 2, for each j. Substituting these into (1) and rearranging terms one obtains (2) with the degree of each term at most deg(p ) + t max{deg(ρ jp g p g p k) : j, p}. k 12

13 It does not seem possible to choose N in Corollary 4.2 so as to depend only on the degrees of g 1,..., g k. Unfortunately, this limits what we are able to prove later in Section 5, in Theorem 5.3. We now consider the corresponding question for modules. Is it always possible to find presentations P = σ 0 + σ 1 s σ t s t, σ i R[X] 2, with the terms of bounded degree? For t 1 it is trivially true by Theorem 4.1 (with deg(p ) serving as the bound). If K is not compact and t 2 then such a presentation may not exist. (M N is saturated iff either t 1 or if t = 2 and K has an isolated point; see [K M, Th. 2.5].) If such a presentation does exist, then deg(p ) serves as a degree bound. If K is compact and t = 2 then one checks that deg(p ) + 1 serves as a degree bound. To summarize: Corollary 4.3 If either t 1 or t = 2 and K either has an isolated point or is compact then every polynomial P non-negative on K has a presentation P = σ 0 + σ 1 s σ t s t with the degree of each term bounded by deg(p ) + 1. What if K is compact and t 3? In view Theorem 4.1, this reduces to the question of whether a degree bound exists for the products s i s j, i < j. Such a bound exists if s i, s j are both linear or if at least one of s i, s j is quadratic and K {si,s j } has an isolated point. In remaining cases (i.e, the cases where at least one of s i, s j is quadratic and K {si,s j } has no isolated point) there is no degree bound in general. Example 4.4 (a) Suppose s 1 = X a, s 2 = (X b)(x c), where a < b < c are fixed reals. Then for any real β > c, s 1 s 2 has a presentation (3) s 1 s 2 = σ 0 + σ 1 s 1 + σ 2 s 2 + σ 3 (β X) with σ i R[X] 2 depending on β. We claim that for any such presentation, as β the maximum of the degrees of the σ i necessarily tends to infinity. Otherwise, by the Transfer Principle, for any real closed field R R and any β R, β > c, we would have a presentation (3) with σ i R[X] 2. Choose any such (nonarchimedean) R, e.g., take R to be the field of formal Puiseux series with real coefficients. Denote by v the unique finest valuation on R compatible with the ordering, and choose β to be positive and infinitely large relative to R, i.e., v(β) < 0. Write σ i = f 2 ij, f ij R[X]. Choose d R, d > 0, with v(d) equal to the minimum of the values of the coefficients of the f ij, i {0, 1, 2} and the βf 3j. Case 1: v(d) 0. Then (3) pushes down to the residue field giving an equation of the form s 1 s 2 = α 0 + α 1 s 1 + α 2 s 2, α i R[X] 2. This contradicts [K-M, Th. 2.5]. Case 2: v(d) < 0. Then dividing (3) by d 2 and pushing down to the residue field yields an equation of the form 0 = α 0 + α 1 s 1 + α 2 s 2, α i R[X] 2, α i 0 for 13

14 some i. Since s 1 and s 2 are both strictly positive on the infinite set (a, b), this is not possible. (b) A similar remark applies to presentations of the form s 1 s 2 = σ 0 + σ 1 s 1 + σ 2 s 2 + σ 3 (β + X) + σ 4 (β X), σ i R[X] 2, where s 1 = (X a)(x b), s 2 = (X c)(x d), a < b < c < d are fixed reals, β > max{ a, d}. We turn now to the case where K is the empty set. We use the next result in the proof of both Theorem 5.3 and Corollary 5.5. Actually, for the proof of Theorem 5.3 we need only a weaker preordering version of the result. The weaker preordering version holds even in the multivariable situation and is a consequence of the Positivstellensatz. Theorem 4.5 Given a positive integer k and non-negative integers d 1,..., d k, there exists a positive integer N such that for any real closed field R and any set of polynomials S = {g 1,..., g k } in the polynomial ring R[X] with deg(g j ) d j, j = 1,..., k, if K S =, then there exist σ i R[X] 2 of degree N such that 1 = k i=0 σ i g i. (Convention: g 0 = 1.) Proof: If 1 / M S then, by [Br, Satz 1.8], there exists a real prime ideal p in R[X] such that the quadratic form φ which is obtained from φ = 1, g 1,..., g k by deleting the entries belonging to p is strongly anisotropic over the residue field, call it k(p), of R[X] at p. (This works even in the multivariable situation, in fact it works for any commutative ring.) The point is, since we are in the 1-variable situation, k(p) has transcendence degree 1 over R, so there exists an ordering of k(p) making φ positive definite [P, Th. 9.4]. By the Transfer Principle this implies K S. The degree bounds follow from a standard ultrapower argument: If the result is false then for each integer i 1 there exists a real closed field R i and S i = {g i1,..., g ik } R i [X] with deg(g ij ) d j, j = 1,..., k, K Si = but with 1 not expressible as k j=0 σ j g ij with all σ j R i [X] 2 of degree i. Let R = ( i N R i )/U, U a non-principal ultrafilter of N. R is a real closed field. In the polynomial ring R[X] we have the set S = { g 1,..., g k } defined in the obvious way from the sets S i. If g j (ỹ) 0, j = 1,..., k holds for some ỹ R, ỹ = (y i ) i N mod U, then the set of i N satisfying g ij (y i ) 0, j = 1,..., k belongs to U. Since this set is empty, this is a contradiction. This proves K S =. Thus 1 M S, i.e., we have a presentation 1 = k j=0 σ j g j, with each σ j R[Y ] 2. This yields polynomials σ ij R i [X] 2 with bounded degrees (bounded by the degrees of the σ j ) with the 14

15 property that the set of i N such that 1 = k j=0 σ ij g ij holds belongs to U. In particular, 1 = k j=0 σ ij g ij holds for infinitely many i, a contradiction. Note: Using the identity P = ( P +1) 2 ) 2 ( P 1 2 )2, we also get degree bounds for presentations P = k i=0 σ i g i depending only on k and the degree of P and the degrees of g 1,..., g k, for each P R[X]. 5 Subsets of Cylinders We fix the terminology. We consider the polynomial ring R[X, Y ] in n+1 variables X 1,..., X n, Y with coefficients in R. We suppose S is a finite subset of R[X, Y ]. When is it true that (SMP) (or ( ) or ( )) holds for T S? Our first result gives a general necessary condition. Let L R n+1 be any line, λ : R = L R n+1 a linear isomorphism onto this line, and λ : R[X, Y ] R[Z] the corresponding map between the polynomial rings. Thus λ 1 (K S ) is the basic closed semialgebraic subset of R defined by the set λ (S). Theorem 5.1 Suppose that (SMP) holds for T S and λ 1 (K S ) is not compact. Then λ (S) contains the natural generators of λ 1 (K S ) (up to scaling by positive reals). Note: If λ 1 (K S ) is compact then T λ (S) satisfies ( ). Thus, one can conclude that T λ (S) satisfies ( ) in any case. Proof: By Theorem 1.3, it suffices to prove that the (closed) preordering T λ (S) contains the natural generators of λ 1 (K S ). To simplify the notation needed in the proof, the coordinates of R n+1 are arranged so that L is the Y -axis and λ : R L R n+1 is the isomorphism defined by y (0, y). Thus, λ : R[X, Y ] R[Y ] is given by X i 0, i = 1,..., n, Y Y. If λ 1 (K S ) = R there is nothing to prove. So assume that λ 1 (K S ) R. Case 1: Suppose that λ 1 (K S ) contains a smallest element, say a. (The case of a largest element is handled in exactly the same way.) The natural generator belonging to a is the linear polynomial Y a. It is claimed that Y a belongs to T λ 1 (S). For each 1 n N define b n = a 1, c n n = b n n. Then: (b n ) n increases monotonically with limit a; (c n ) n decreases monotonically with limit ; for all n, b n > c n. The interval [c n, b n ] L is compact and does not meet the closed set K S. Thus, the distance δ n = dist([c n, b n ], K S ) is positive. For each n, let E n be the ellipsoid with center 1 2 (b n + c n ), one axis on the line L, the other perpendicular to L. The half axis on L has length 1 2 (b n c n ), the 15

16 perpendicular half axis has length δ n. The ellipsoid is given by the polynomial ( Y 1 2 P n = (b ) 2 n + c n ) 1 (b + 2 n c n ) The interior of the ellipsoid is the set n ( ) 2 Xi 1. i=1 δ n N(P n ) = {(x, y) R n+1 : P n (x, y) < 0}. Every point of N(P n ) has a distance less than δ n from the line segment [c n, b n ], hence N(P n ) K S =. This means that P n is positive semidefinite on K S. Each linear functional L 0 on R[Y ] non-negative on T λ (S) lifts to a linear functional L 1 = L 0 λ on R[X, Y ] non-negative on T S. Since T S satisfies (SMP), L 0 (λ (P n )) = L 1 (P n ) 0 for all such L 0. It follows that λ (P n ) Tλ lin (S) = T λ (S). One computes ( ( λ 4 1 (P n ) = Y b ) ( )) n Y b n b2 n b n c n b n c n b n c n b n c n Since 1 b n c n converges to 0 as n one concludes that the sequence ( ) bn c n λ (P n ) 4 of polynomials converges to Y a. Once again, closedness of T λ (S) implies Y a T λ (S). Case 2: Suppose that a, b λ 1 (K S ), a < b, (a, b) λ 1 (K S ) =. It is claimed that the natural generator (Y a)(y b) belongs to T λ (S). The method of proof is much the same as in Case 1; let n 0 N be such that b a > 2 n 0. For n n 0 one defines a n = a + 1 n, b n = b 1 n. Then: a n < b n ; a n a; b n b; the compact line segment [a n, b n ] does not meet K S. Set δ n = dist([a n, b n ], K S ), and let ( Y 1 2 P n = (b ) 2 n + a n ) 1 (b + 2 n a n ) n n ( ) 2 Xi 1. Again, N(P n ) is the interior of the associated ellipsoid E n, and N(P n ) K S =. Thus, P n 0 on K S. Again, since T S satisfies (SMP) and T λ (S) is closed one gets λ (P n ) T λ (S). The sequence converges to i=1 ( (1 ) 2 2 (b n a n ) λ (P n )) δ n ( Y 1 2 (b + a)) 2 ( 1 2 (b a)) 2 = Y 2 (b + a)y + ab = (Y a)(y b). 16 n

17 Once again, since T λ (S) is closed the limit also belongs to T λ (S). We assume now that N X 2 0 on K S for some N (so K S is a subset of a cylinder of the form B R where B := {x R n : x N}). For each a B let L a = λ a (R) where λ a : R R n+1 is defined by y (a, y). The line L a lies in the cylinder B R. The corresponding map λ a : R[X, Y ] R[Y ] is given by X i a i, Y Y. λ 1 a (K S ) is the basic closed set in R defined by λ a(s). Theorem 5.1 and [Sc2, Th. 1] (see Theorem 1.7) combine to yield the following: Corollary 5.2 Suppose K S is a subset of a cylinder of the form B R, where B := {x R n : x N}, N 0. For each a B define λ a : R R n+1 as above. Then: (a) T S satisfies (MP). (b) The following are equivalent: (i) T S satisfies (SMP). (ii) For each non-compact λ 1 a (K S ) (up to scalings by positive reals). λ 1 a (K S ), λ a(s) contains the natural generators for Proof: (a) Since every polynomial non-negative on R is a sum of squares, (MP) holds for each T λ a (S), so (a) is immediate from [Sc2, Th. 1]. (b) The implication (i) (ii) follows from Theorem 5.1. (ii) (i): To be able to apply [Sc2, Th. 1] we must show that each T λ a (S) satisfies (SMP). For λ 1 a (K S ) compact this is always the case. For λ 1 a (K S ) non-compact it is a consequence of assumption (ii). See Corollary 6.4 below for a module version of Corollary 5.2. We are unable to prove that Corollary 5.2(b) holds with (SMP) replaced by ( ). However, we are able to prove the following weaker result: Theorem 5.3 Suppose the following conditions hold: (a) There exists N such that N X 2 T S and (b) For each non-empty λ 1 a (K S ), λ a(s) contains the natural generators for λ 1 a (K S ) (up to scaling by positive reals). Then T S satisfies ( ), i.e., for each P R[X, Y ] such that P 0 on K S, there exists Q R[X, Y ] such that for all real ɛ > 0, P + ɛq T S. See Corollary 5.5 below for a module version of Theorem 5.3. It follows from [Sc1, Cor. 3] that condition (a) of Theorem 5.3 holds iff there exist elements f 1,..., f t T S R[X] such that the subset K {f1,...,f t } of R n is compact. 17

18 Proof: Suppose that P R[X, Y ] belongs to T alg S one has P a := λ a(p ) = k i=0 p i (a)y i T alg λ a(s) (b), T λ a (S) = T alg λ a(s) (4) P a =. Writing P = k i=0 p i (X)Y i for each a B. Because of condition for each a. Thus, for each a there is a presentation i {0,1} t σ a,i λ a(s 1 ) i1... λ a(s t ) it (where S = {s 1,..., s t } and the σ a,i R[Y ] are sums of squares). By assumption (b) we may assume there is a uniform bound for the degrees of the sums of squares occuring in these representations that is independent from a. If λ 1 a (K S ) then the degrees can be bounded by k by Theorem 4.1. For λ 1 a (K S ) = a uniform bound exists by the note following Theorem 4.5. Pick any ε > 0. For each a, consider σ a,i s i s it t, P σ a,i s i s it t i {0,1} t i {0,1} t as polynomials in the variable Y with coefficients from R[X]. There is some large even number D that bounds the Y degrees of these polynomials and is independent from a. Write D σ a,i s i s it t = p a,d Y d, P D σ a,i s i s it t = q a,d Y d. i {0,1} t d=0 i {0,1} t d=0 Note that q a,0 (a) =... = q a,d (a) = 0 for each a B. Hence there is an open semi-algebraic neighborhood U a of a in R n such that D q a,d (x) < ε d=0 2 for every x U a. By compactness of B, there is a finite subset E of B such that B a E U a. On B there is a continuous semi-algebraic partition of unity subordinate to this cover, say e a, a E, with 0 e a 1 and e a (x) = 1 for each x B. Each e a has a nonnegative square root f a, which is also a continuous semi-algebraic function. Using Stone Weierstrass approximation on the compact set B one finds polynomials F a, a E, so close to the continuous semi-algebraic functions f a that p a,d (fa 2 Fa 2 )(x) < ε 2 a E for every d = 0,..., D and every x B. The polynomial P can be written in the a E 18

19 form P = P 1 + P 2 where ( ) P 1 = σ a,i Fa 2 i {0,1} t a E ( P 2 = D r d Y d = D d=0 + D d=0 ( d=0 s i s i t t T S, q a,d Fa 2 a E ) q a,d (fa 2 Fa 2 ) a E ) Y d ( Y d + D d=0 p a,d (fa 2 Fa 2 ) a E The choice of the approximations yields r d (x) = q a,d fa 2 (x) + p a,d (fa 2 F 2 a )(x) a E a E q a,d fa 2 (x) + p a,d (fa 2 F 2 a )(x) a E a E a E q a,d (x) + a E p a,d (fa 2 Fa 2 )(x) < ε for each x B. But then [Sc1, Cor. 3] implies that ε ± r d T {N X 2 } for every d = 0,..., D. The argument in [K M, Th. 5.1] shows that εy 2 + (ε + r d )Y + ε (Y + 1) 2 T {N X 2 } + (Y 2 + 1)T {N X 2 }. As in loc.cit., one adds εq where Q = 2 + Y + 3Y 2 + Y 3 + 3Y Y D 2 + Y D 1 + 2Y D to both sides of P = P 1 + P 2 and obtains ) Y d. P + εq = i={0,1} t + D d=0 d even ( ) σ a,i Fa 2 a E (ε + r d )Y d + D s i s i t t d=0 d odd εy d+1 + (ε + r d )Y d + εy d 1 T S. Since the polynomial Q is independent from ε, the proof is finished. There is one case for which the results proved so far give even a necessary and sufficient condition for ( ) to hold: Corollary 5.4 Suppose the following conditions hold: (a) There exists N such that N X 2 T S and (b) Each nonempty λ 1 a (K S ) is unbounded. Then the following conditions are equivalent: (i) T S satisfies condition ( ). (ii) For each non-empty λ 1 a (K S ), λ a(s) contains the natural generators for λ 1 a (K S ) (up to scaling by positive reals). 19

20 If we look carefully at the proof of Theorem 5.3 we see that in fact we have proved more than is stated: We have shown that for each ɛ > 0, P = P 1 + P 2, P 1 T S and P 2 + ɛq k T {N X 2 }Y 2k + k T {N X 2 }Y 2k (1 + Y ) 2. We may not need all the various sorts of terms in T S to represent P 1. If S = {s 1,..., s k } and e i {0, 1}, e i > 1, we need the term involving s e s e k k only if there exists a such that the λ a(s i ) with e i = 1 actually occur as (distinct) natural generators of λ 1 a (K S ). For example, if the hypothesis of Corollary 4.3 holds for each non-empty fiber, then we can arrange things so P 1 M S. Corollary 5.5 Suppose the following conditions hold: (a) There exists N such that N X 2 M S. (b) For each non-empty λ 1 a (K S ), λ a(s) contains the natural generators for λ 1 a (K S ) (up to scaling by positive reals). (c) Each non-empty λ 1 a (K S ) has the form (, ) or (, p], [q, ), (, p] [q, ) (perhaps with a discrete point added) or [p, q]. Then M S satisfies ( ), i.e., for each P R[X, Y ] such that P 0 on K S, there exists Q R[X, Y ] such that for all real ɛ > 0, P + ɛq M S. 6 Module Version of Schmüdgen s Theorem We prove the following module version of the Theorem in [Sc2] quoted in the introduction: Theorem 6.1 Let h 1,..., h d R[X]. Suppose there exists N such that N di=1 h 2 i M S. Denote by V h R d the algebraic set associated to the R-algebra R[h] := R[h 1,..., h d ]. Suppose that the image of K S under the map R n V h, x (h 1 (x),..., h d (x)), is dense in the set {z V h : g(z) 0 for all g(h) R[h] M S }. If (SMP) (resp., (MP)) holds for M Sλ for each λ in the closed ball of radius N about the origin in R d, then (SMP) (resp., (MP)) holds for M S. The density assumption in Theorem 6.1 is somewhat restrictive. At the same time, this assumption holds in a wide variety of interesting cases. The proof of Theorem 6.1 follows along the same lines as the proof given in [Sc2]. We need two lemmas. Lemma 6.2 Suppose p R[h], p 0 on K S. Then p + ɛ M S holds for all real ɛ > 0. 20

21 Proof: The hypothesis N d i=1 h 2 i M S implies the quadratic module R[h] M S is Archimedean in R[h]. Let z V h be such that g(z) 0 for all g(h) R[h] M S. Write p = f(h), f R[Z 1,..., Z d ]. By hypothesis there is a sequence x (j) in K S, (h 1 (x (j) ),..., h d (x (j) ) z, so p(x (j) ) = f(h 1 (x (j) ),..., h d (x (j) )) f(z) as j. It follows that f(z) 0 for any such z. By Jacobi s generalization of the Kadison- Dubois Theorem [J, Th. 4], this implies p + ɛ = f(h) + ɛ M S for any real ɛ > 0. We use the notation of [Sc2]. L is a linear functional on C[X], L(M S ) 0. N is the ideal in C[X] defined by N = {p C[X] : L(pp) = 0}. D L = C[X]/N comes equipped with an involution p + N p + N and a scalar product p 1 + N, p 2 + N := L(p 1 p 2 ). Each p C[X] defines a linear operator π L (p) on D L by π L (p)(q + N ) = pq + N. We use the following easy version of [Sc2, Prop. 2]: Lemma 6.3 If p C[h] then the operator π L (p) on D L is bounded and π L (p) p K where p K := sup{ p(x) : x K S }. Proof: Using π L (p) 2 = π L (pp) and p 2 = pp, we are reduced to the case p R[h]. Fix ɛ > 0 and let ρ 2 = p 2 K + ɛ. By Lemma 6.2, ρ 2 p 2 M S so, for q C[X], ρ 2 qq p 2 qq = (ρ 2 p 2 )qq M S. It follows that ρ 2 L(qq) L(p 2 qq), i.e., ρ 2 q + N 2 π L (p)(q + N ) 2. Since q C[X] is arbitrary, this implies ρ π L (p). Since ɛ > 0 is arbitrary, the result follows. It follows from Lemma 6.3 that the operators π L (h 1 ),..., π L (h d ) are bounded. One finishes the proof of Theorem 6.1 now, exactly as in [Sc2]. In terms of basic closed semi-algebraic subsets of cylinders with compact crosssection, set-up as in Section 5, Theorem 6.1 yields the following: Corollary 6.4 Let S R[X, Y ] where R[X, Y ] denotes the polynomial ring in n + 1 variables X 1,..., X n, Y. Suppose N X 2 M S and, for each x R n, x / π(k S ), there exists g M S R[X] such that g(x) < 0, where π : R n+1 R n denotes the projection (x, y) x. Then: (a) (MP) holds for M S. (b) If (SMP) holds for M λ a (S) holds for M S. 7 Applications for each non-compact fiber λ 1 a (K S ) then (SMP) In [K S] the authors asked whether [K M, Cor. 5.4] generalizes to sets of the form K L, K compact, L non-compact closed semialgebraic in R (with the natural description). We are now able to prove this. See [Po, Cor. 3] for another proof. 21

22 Denote by R[X, Y ] the polynomial ring in n + 1 variables X 1,..., X n, Y. 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. Given L closed semialgebraic in R let N R[Y ] be the natural set of generators for L. In this situation we have the following: Corollary 7.1 Let S L := S N (so K SL = K L). Then ( ) holds for T SL. Proof: Immediate from Theorem 5.3. Our next application is to generalized polyhedra. Assume that K S is the basic closed semi-algebraic set in R n is defined by S = {l 1,..., l s } where l 1,..., l s are linear, so K S is a closed polyhedron. If K S is compact then, by [J P, Th. 4.2], ( ) holds for M S. What if K S is not compact? If K S contains a cone of dimension 2 then, by Theorem 1.5 (MP) fails for T S. In [K M] the authors ask whether ( ) holds in the remaining case, i.e., when K S is not compact and does not contain a cone of dimension 2. We are now in a position to settle this question completely. We need two preliminary lemmas: Lemma 7.2 Let n 1. Let P be a closed polyhedron in R n and let a P. Then P is compact if and only if there is no infinite half line starting at a and contained in P. Proof: We may assume that a = (0,..., 0). Suppose P is not compact. Then, for each positive integer i there exists p i P with p i i. Using the compactness of the unit sphere S n 1 there exists a subsequence {p ij } of {p i } with lim j p ij p ij = q Sn 1. For t R, t 0, t i j for j sufficiently large, so t p i j p ij is on the line segment joining p ij and the origin so is in P. Thus tq is a limit of elements of P so is in P. This proves that the half line {tq t R, t 0} lies in P. The other assertion is clear. Lemma 7.3 Let n 1. Let P be a non-compact closed polyhedron in R n+1 and let a P. Then the following are equivalent: (i) There is exactly one infinite half line starting at a and contained in P. (ii) There is a subset S 1 of the set of linear polynomials defining P such that K S1 is a cylinder with compact cross-section. 22

23 Proof: (ii) (i) is clear. (i) (ii): Let S = {g 1,..., g s } be the set of linear polynomials defining P. We may assume a = (0,..., 0). By Lemma 7.2, {tr t R, t 0} P for some r R n+1, r = 1. Making an affine change of coordinates, we can assume r = (0,..., 0, 1). Let q be any vector on the hyperplane through the origin orthogonal to r with q = 1. For each positive integer k, the half-line {t(q + kr) t R, t 0} does not lie in P, i.e., it meets the set {b R n+1 g i (b) < 0} for some i {1,..., s}. Fixing q and letting k vary, using the fact that S is finite, this means there exists i {1,..., s} such that the hyperplane {b R n+1 g i (b) = 0} intersects the plane generated by q and r in a line parallel to r (i.e., Y does not appear in g i ) of the form {αq + tr t R}, α 0 and {b R n+1 g i (b) 0} intersects the plane generated by q and r in the region {t 1 q + t 2 r t 1, t 2 R, t 1 α}. This holds for any q S n orthogonal to r. It follows, using Lemma 7.2, that if {i 1,..., i k } = {i 1 i s, Y does not appear in g i } then K {gi1,...,g ik } is a cylinder in R n+1 with compact base. Corollary 7.4 Let n 1. Let P be a closed polyhedron in R n+1 defined by a finite set S of linear polynomials. If P is not compact and does not contain a 2-dimensional cone then ( ) holds for M S. Proof: According to Lemmas 7.2 and 7.3, after a suitable affine change in coordinates, there exists a subset S 1 S R[X 1,..., X n ] such that K S1 is a cylinder with compact cross-section. The result follows, applying Corollary 5.5. The fact that condition (a) of Corollary 5.5 holds follows by applying [J P, Theorem 4.2] to the compact polyhedron K S1 R n. In summary then, we have the following: Theorem 7.5 Let n 1. Let P be a closed polyhedron in R n+1 defined by a finite set S of linear polynomials. (i) If P is compact then ( ) holds for M S. (ii) If P is not compact but does not contain a 2-dimensional cone then ( ) holds for M S. (iii) If P contains a 2-dimensional cone then (MP) fails for T S. 8 Examples in the Plane Example 8.1 Consider {Y X 2 } R[X, Y ]. K {Y X 2 } is the region above the parabola Y = X 2 in R 2. K {Y X 2 } does not contain a 2 dimensional cone. On the other hand, the isomorphism φ : R 2 R 2, (x, y) (x, y x 2 ), 23

24 carries K {Y X 2 } onto K {Y } which does contain a 2 dimensional cone. By Propositions 1.4 and 1.5, T {Y } = T{Y lin } T alg. Applying the induced algebra isomorphism φ : R[X, Y ] R[X, Y ], X X, Y Y X 2, yields T {Y X 2 } = T lin {Y X 2 } T alg, so T {Y X 2 } is closed and (MP) fails for T {Y X 2 }. Example 8.2 Let S 1 = {X, 1 X, Y 1, 1 XY }, S 2 = {X, 1 X, XY 1}. By Corollary 5.5 M S1 and M S2 satisfy condition ( ). Example 8.3 Let S 1 = {X, 1 X, Y 2 X 3 }, S 2 = {X, 1 X, Y 3 X 2 }. By Corollary 5.5, M S1 satisfies ( ). By Theorem 5.1, T S2 does not satisfy (SMP). By Corollary 6.4, M S2 does satisfy (MP). In [K S] the authors ask for examples where (SMP) holds but ( ) does not hold, and for examples where TS lin T S. Although we are unable to settle this issue yet, we note that TS lin = T S does hold when n = 1. Also, in the special case where S is a set of linear polynomials TS lin = T S holds when n = 2 (using Theorem 7.5 and Proposition 1.4 (ii)) and also when n = 3 except possibly in the case when K S contains a 2-dimensional cone but does not contain a 3-dimensional cone. What happens in this exceptional case is not clear. Also, in this regard, the following two examples are interesting. Example 8.4 Let S = {X, Y, 1 XY }. The polynomial XY is bounded on K S and one checks that (SMP) holds for M S {XY λ, (XY λ)} for each 0 λ 1. Thus by Theorem 6.1, (SMP) holds for M S. It is not known if ( ) holds for T S. This example is due to K. Schmüdgen. We note the following: Proposition 8.5 Let S and S R[X] (any number of variables) and assume that K S = K S. Assume that (SMP) holds for T S and that S TS lin. Then (SMP) holds for T S. Proof: But T alg S If S T lin S = T alg S then T S TS lin lin (since T S. It follows that T alg = T lin S S lin is a preordering), therefore TS T lin = T S lin. S. Example 8.6 In Example 8.2, the set K S2 was defined using the set S 2 = {X, 1 X, XY 1}. But it can also be defined by the set S 3 = {1 X, Y, XY 1}. According to Theorem 1.7, (SMP) holds for T S3. Is it true that ( ) holds for T S3? This would be true by Theorem 5.3 if we could show that X + N T S3 for N sufficiently large. Unfortunately, we have the following: 24

25 Claim 1: If N is a positive integer, then X + N / T S3. contradiction that X + N T S3, there is a representation Assuming by way of X + N = 1 i,j,k=0 σ ijk (1 X) i (XY 1) j Y k with σ ijk sums of squares in R[X, Y ]. Since the variable Y does not occur on the left hand side one concludes that σ ijk = 0 if j + k 1, hence X + N = σ 0 + σ 1 (1 X), where σ 0, σ 1 are sums of squares in R[X]. But then (1+σ 1 )(X+N) = σ 0 +(N +1)σ 1, hence X + N is a sum of squares in R(X) which is false. On the other hand, we have: Claim 2: X + εy T S3 for every 0 < ε R. Note that X + εy = (X + 1 Y ) + (ε n 1 )Y T n S 3 if 1 n N such that 1 ε and X + 1 Y T n n S 3. Thus it suffices to prove that there are arbitrarily large numbers n N with nx + Y 1 T S3. The proof is by induction on n. n = 1: (XY 1)(1 X) = XY 1 X 2 Y + X T S3. Adding X 2 Y T S3 and Y (1 X) T S3 one gets X + Y 1 T S3. n n + 1: (n + 1)X + Y 1 nx 2 XY = (nx + Y 1)(1 X) T S3. Adding nx 2 T S3, XY 1 T S3 and 1 T S3 one obtains (n + 1)X + Y 1 T S3. By Claim 2, X T S 3, so T S2 T S 3 TS lin 3. Using Proposition 8.5 this provides a second proof that (SMP) holds for T S3. The question of whether or not ( ) holds for T S3 remains open. 9 Some Open Problems 1. In Theorem 5.3 is it possible to replace assumption (a) with the (apparently weaker) assumption that there exists N such that N X 2 0 on K S? Note: If the answer is no then we have an example where (SMP) holds for T S but ( ) fails for T S. This would answer Open Problem 3 in [K M, Sect. 6]. 2. In Theorem 5.3 are the strong assumptions (b) on the fibers actually necessary? If they are necessary then again we have the answer to Open Problem 3 in [K M]. 3. Does the converse of Theorem 1.7 hold? This seems unlikely. By Theorem 5.1 it is true when the fibers are linear. To the extent that the converse of Theorem 1.7 is true, it provides a way of studying the question of when (SMP) and (MP) hold by induction on the dimension. 4. Suppose n = 2, S = {X, Y, 1 (X 1)(Y 1)}. Does (SMP) hold for T S? Note: The only polynomials which are bounded on K S are the constants, so Theorem 1.7 does not apply in this case. This problem is due to K. Schmüdgen. 25