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 Rn dened by some nite set of polynomial inequalities g1 0:::gs 0 and let T be the preordering in the polynomial ring R X] := R X1:::Xn] generated by g1:::gs. For K compact, Schmudgen proves in 10] that:
( ) TheK-Moment Problem has a positive solution.
(y)8 f 2R X], f 0 on K ) 8 real >0, f +2T.
In the present paper, we consider the status of ( ) and (y) when K is not compact.
At the same time, we consider a third property:
(z)8 f 2R X], f 0 on K ) 9 q 2T such that 8 real >0, f+q 2T which we prove is strictly weaker than (y) and, at the same time, which implies ( ). Many (non-compact) examples are given where (z) holds. Many examples are given where ( ) fails. The question of whether or not (z) and ( ) are equivalent is an open problem.
1991 Mathematics Subject Classication. 14P10, 44A60.
This research was supported in part by NSERC of Canada.
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In Section 1 we introduce the problem and the notation. In Section 2 we settle the casen= 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 wheneverK 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 (z) holds. In Section 5 we prove (z) 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. Schmudgen 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 X1:::Xn] by R X] for short. PR X]2 denotes the set of nite sums Pfi2, fi 2 R X]. For S =
fg1:::gsg, a nite subset of R X], let KS denote the basic closed semi-algebraic set in Rn determined byS i.e.,
KS =fx2Rn jgi(x)0i= 1:::sg:
Let TS denote the preordering in R X] generated by S, i.e., the set of all sums
Pege e = (e1:::es) running through the nite set f01gs, e 2 PR X]2, where ge :=g1e1:::gses.
Denote by TSalg the set ff 2 R X] j f 0 on KSg. Points x in Rn are in one-to-one correspondence with algebra homomorphismsL:R X] !R via
x= (L(X1):::L(Xn)) L(f) =f(x):
Under this correspondence, points x in KS correspond to algebra homomorphisms L satisfying L(gi)0, i= 1:::s or, equivalently, L(TS)0.
One can also consider the dual cone
KSlin =fL:R X] !R jL is linear (6= 0) and L(TS)0g
and the double dual coneTSlin =ff 2R X] jL(f)0 for allL2KSling 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
KS ,!KSlin and TSalg TSlin TS:
Note: TS,TSlin generally depend onS. TSalg depends only on the basic closed set KS.
Note: L 2KSlin ) L(1)>0. For 12TS implies L(1)0. If L(1) = 0 then, for any f 2R X], the identity (kf)2 =k22kf +f2 implies 2kL(f) +L(f2)0 for all real k, so L(f) = 0.
We are interested here in the relationship betweenTSalg, TSlin and TS for all the various choices of (nite) S in R X], n1. In particular, we are interested in the relationship between the condition:
( ) TSalg =TSlin
and the condition:
(y) 8f 2R X]f 2TSalg )8 real >0f+ 2TS:
We are also interested in weak version of (y) referred to as (z) (see below) and a strong version of (y) namely TSalg = TS. The condition TSalg = TS is examined in the recent paper of Scheiderer 9].
The study of the relationship betweenTSalg andTS 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
8f 2R X]f 2TSalg ,9pq 2TS andm0 such that pf =f2m+q:
The interest in TSlin comes from functional analysis. For a linear functional L on
R X], one is interested in when there exists a positive Borel measure on Rn supported by some given closed set K in Rn such that
8f 2R X]L(f) =Z
Rnfd:
According to a result of Haviland 4] 5], this will be the case i L(f) 0 holds for all f 2TSalg. The Moment Problem (or at least a version of it) is the following:
When is it true that every L2KSlin comes from a positive Borel measure on Rn supported by KS? 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 = (k1:::kn), denote the monomial X1k1:::Xnkn by Xk for short. The monomials form a basis forR X] so each linear functional L on R X] corresponds to a function p : (Z+)n ! R via p(k) =L(Xk). We say p is positive denite ifm
X
ij=1p(ki+kj)cicj 0
for arbitrary (distinct) k1:::km 2 (Z+)n, c1:::cm 2 R, and m 1. For g2R X], g=PakXk, g(E)p: (Z+)n !R is dened by
g(E)p(`) =X
k akp(k+`):
The condition that L 2 KSlin corresponds exactly to the condition that functions ge(E)p, e 2f01gs are positive denite, see 8] or 10]. Consequently, ( ) is equiv- alent to the assertion that every non-zero function p : (Z+)n ! R with ge(E)p positive denite for all e 2 f01gs comes from a positive Borel measure on Rn supported by KS.
Schmudgen's 1991 paper 10] settles the Moment Problem in the compact case.
In 10], Schmudgen proves if KS is compact then ( ) and (y) both hold. In 11], Wormann proves KS is compact i TS is archimedean and uses this to obtain Schmudgen's result as a corollary of the Kadison-Dubois Theorem.
Denote by Pd the (nite dimensional) vector space consisting of all polynomials in R X] of degree 2d, and let Td = TS \Pd. Td is obviously a cone in Pd, i.e., Td +Td Td and R+Td Td. Denote by Td (resp., int(Td)) the closure (resp., interior) ofTd inPd. The identitypq = 12((p+q)2;p2;q2) applied to monomials pq of degree d implies Pd =Td;Td. This means thatTd contains a basis of Pd, so int(Td)6=.
1.2 Remark.
TS is archimedean i 1 2 int(Td) for all d 0. This is just a matter of sorting out the denitions: If k f 2 TS for suciently large k 2 Z, then 1f 2 TS for suciently small > 0. In fact, one even knows that TS is archimedean ik;Pni=1Xi2 2TS fork suciently large 11], soTS is archimedean i 12 int(T1).We are also interested in the condition:
(z) 8f 2R X]f 2TSalg )9q 2TS such that 8 real >0f+q 2TS:
Obviously, (y) ) (z) (taking q = 1). The signicance of (z) is clear from the following:
1.3 Proposition.
The following are equivalent:(1) (z) holds.
(2) TSalg =TSlin =d 0Td.
Proof. Clearly Td TSlin. For f 2 Pd, q 2 Td, the condition that f +q 2 TS for all real >0 is equivalent to:
f + (1;)q 2Td for every 1< < 1
i.e., that every point on the open line segment joining q and f belongs to Td. In particular, it implies that f 2 Td. Thus we see that (1) ) (2). Now assume (2), and suppose f 2 TSalg. Thus, for d suciently large, f 2 Td. Pick q any point in the interior of Td. (Recall: int(Td) 6= .) Then obviously the open line segment joiningf with q belongs to Td, so (1) holds.
Thus, condition (z) implies not only that ( ) holds, but also thatTSlin =d 0Td. Note: The requirement in (z) that q belongs to TS is unnecessary: Using
q = (q+ 1
2 )2;(q;1 2 )2
we see that if f+q 2TS, q 2R X], then f+(q+12 )2 2TS.
See Section 4 for additional discussion of (z) and for examples where (z) holds.
See Section 5 for examples where (z) holds but (y) does not hold. No examples are known where ( ) holds but (z) does not hold. No examples are known where TSlin 6=d 0Td. In fact, there is a general shortage of examples.
Instead of working withTS, one can work with the PR X]2-moduleMS gener- ated by S, i.e., the set of all sums0+Psi=1igi, i 2PR X]2, i= 0:::s, and MSlin, the double dual cone of MS. One is interested in the analog of ( ):
( M) TSalg =MSlin
the analog to (y):
(yM) 8f 2R X]f 0 on KS )8 real >0f + 2MS and the analog to (z):
(zM) 8f 2R X]f 2TSalg )9q 2MS such that 8 real > 0f+q 2MS:
Again, TSalg =MS implies (yM) which, in turn, implies (zM) and (zM) is equivalent to TSalg =MSlin =d 0Md whereMd denotes the closure inPd of Md :=MS\Pd. (yM) is known to hold in a few cases, for example, if KS is compact and either n= 1 6] or the polynomials g1:::gs are linear 6] 8] or s 2 6]. In 6], Jacobi and Prestel develop a powerful valuation-theoretic method for testing the validity of (yM) when KS 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 KS is compact, then (y) and (yM) hold. If KSis not compact, then the conditions (y), (z), ( ) and TSalg =TS are equivalent and the conditions (yM), (zM), ( M) and TSalg =MS are equivalent.
Proof. If KS is compact then (y) holds by Schmudgen's result and (yM) 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 KS is compact then (y) (resp., (yM)) holds but, unlike what happens in the non-compact case, this does not imply that TSalg =TS
(resp., TSalg = MS) holds. For example, if S =f(1;X2)3g then TSalg 6= TS see 8, Example 4.3.3].
The second part of Theorem 2.1 is only useful if we know whenTSalg =TS (resp., TSalg = MS) holds. For the rest of the section we concentrate our attention on answering this question.
2.2 Theorem.
Suppose n= 1. If KS is not compact and TSalg =TS, then:(i) If KS has a smallest element, call it a, then r(X ;a) 2 S for some real r >0.
(ii) If KS has a largest element, call ita, thenr(a;X)2S for some real r >0. (iii) For every ab 2 KS with a < b and (ab)\KS =, r(X ;a)(X ;b) 2 S
for some real r > 0.
Conversely, for any KS, if conditions (i), (ii) and (iii) hold, then TSalg =TS.
2.3 Notes.
(1) SinceKS is a closed semi-algebraic set in R, it is the union of nitely many closed intervals and points (including possibly closed intervals of the form (;1a] or a1)).
(2) IfK R is any closed semi-algebraic set then one checks easily that K =KS
forS the set of polynomials dened as follows:
| Ifa 2K and (;1a)\K =, then X;a2S.
| Ifa 2K and (a1)\K =, then a;X 2S.
| Ifab 2K, a < b, (ab)\K =, then (X;a)(X;b)2S.
|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 = KS not compact, TSalg = TS holds i S contains the natural choice of generators (up to scalings by positive reals).
(4) IfKS is compact thenTSalg =TS can hold without conditions (i), (ii) and (iii) holding. For example, if S =f1;X2g, then 1X = 12(1X)2+12(1;X2)2TS
so TSalg =TS.
Proof. Suppose TSalg =TS, KS not compact. We can assume that the elements of S have degree 1. Since KS is not compact, it either contains an interval of the form a1) or it contains an interval of the form (;1a]. Replacing X by ;X if necessary, we can assume we are in the rst case. Thus eachgi 2S is non-negative on some interval a1), so has positive leading coecient. Thus, for any element p=Peg1e1:::gses ofTS, the degree ofpis equal to the maximum of the degrees of the terms eg1e1:::gses. Also, since each e is a sum of squares, it has even degree.
Suppose that KS has a smallest element, a say. Let p=X ;a. Then p 0 on KS so p 2 TS. Since p has degree 1, p is a sum of terms of the form , 2 R+, and gi with 2 R+ and gi 2 S linear. Each such gi is 0 at a (since a 2 KS).
Consequently, at least one linear gi in S is equal to zero at a, so gi =r(X ;a) as required.
Suppose now that ab 2 KS are such that a < b and (ab) \KS = . Let p = (X ;a)(X ;b). Then p 0 on KS so p 2 TS. Since p has degree 2, it is a sum of terms of the form 2 PR X]2 of degree 0 or 2, gi with 2 R+ and gi 2S linear or quadratic, and gigj with 2R+ and gigj 2 S linear. Since any linear gi 2S is increasing and gi(a) 0, gi is positive on the interval (ab). Thus p 1g1 ++tgt on (ab) where g1:::gt are quadratics in S which assume at least one negative value on (ab), and 1:::t are positive reals. Dene the width of a quadraticg to ber2;r1 where r1 r2 are the roots of g (or 0 if g has
no roots). Each gi opens upward and is non-negative at ab (since ab 2KS), has its roots betweena andband consequently has width at most b;a,i= 1:::t. It suces to show that gi has width exactly b;a for some i2f1:::tg, for thengi
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 igi has the same width as gi, this is a consequence of the following elementary result:
2.4 Lemma.
If f1f2 are quadratics with positive leading coecients, then width(f1+f2)maxfwidth(f1)width(f2)g:Proof. Without loss of generality we can assume width(f1)width(f2) and thatf1 has positive width. Shifting and scaling, we can assumef1 =X2;X, f2 =c(X; a)(X;(a+b)),c >0, 0b1. Thusf1+f2 = (c+1)X2;(2ac+bc+1)X+ca(a+b) and width(f1+f2) =
p(2ac+bc+ 1)2;4(c+ 1)ca(a+b)
c+ 1 :
Thus we are reduced to showing that
(1) (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) b2c+ 4a+ 2bc+ 2 + 4a2+ 4ab or, equivalently, that
(4) (1;b2)(c+ 1) + (2a+b;1)2 0: Since 0b1 and c >0, this is clear.
Note: Equality holds i b= 1 and a = 0, i.e., i f2 =cf1.
Suppose now that (i) (ii) (iii) hold. IfKS has a smallest elementa thenX;a 2 TS soX;d= (X;a) + (a;d)2TS for any d a. Similarly, if KS has a largest
element a thena;X 2TS, sod;X = (a;X)+(d;a)2TS for any da. Also, if ab 2 KS are such that a < b and (ab)\KS = , then (X;a)(X ;b) 2 TS. Moreover, by the proof of 3, Lemma 4], if a cd bthen (X;c)(X;d) is in the preordering generated by (X ;a)(X;b), so (X;c)(X;d)2TS.
Supposef 2R X], f 0 on KS. We prove f 2 TS by induction on the degree.
If f has degree zero it is clear. If f 0 on R then f 2PR X]2 so, in particular, f 2TS. Thus we can assume that f(c)<0 for somec. There are three possibilities:
Either KS has a least elementa and c < aorKS has a largest elementa andc > a or there exist ab 2 KS, a < b with (ab)\KS = , and a < c < b. In the rst case f has a least root d in the interval (ca], X ;d2 TS, f = (X;d)g for some g 2 R X] and one checks that g 0 on KS. In the second case f has a greatest root d in the interval ac), d;X 2 TS, f = (d;X)g and again g 0 on KS. Similarly, in the third case, f has greatest root d in the interval ac) and a least root e in the interval (cb], (X;d)(X;e)2TS, f = (X;d)(X ;e)g and g 0 on KS. Thus, in any case, the result follows by induction on the degree.
The question of when TSalg = MS holds for K = KS 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 sucient only in very special cases:
2.5 Theorem.
SupposeK is not compact and S is the natural choice of generators for K. Then TSalg = MS holds i either jSj 1 or jSj = 2 and K has an isolated point.Proof. If jSj 1, then MS = TS so TSalg = MS by Theorem 2.2 also see 3, Theorem 1]. Suppose jSj= 2 and K has an isolated point. If this isolated point is the least element of K, then S =fg1g2g,g1 =X;b, g2 = (X;b)(X;c), b < c. Then one checks that
g1g2 = (X;c)2g1+ (c;b)g2 2MS
so TS = MS. The case where the isolated point is the greatest element of K is similar. The remaining case is where S = fg1g2g, g1 = (X ; a)(X ;b), g2 = (X;b)(X;c), a < b < c. In this last case one checks that
g1g2 = b;a
c;a(X;c)2g1+ c;b
c;a(X ;a)2g2 2MS: Thus, in all cases, MS =TS so TSalg =MS by Theorem 2.2.
Suppose now that either jSj= 2 and K has no isolated points, or jSj 3. Say S =fg1:::gsg. ReplacingX by ;X if necessary, we can assume thatK contains an interval of the form a1). Reindexing, we can suppose g1 is either linear of the formg1 =X;b (corresponding to the case where K has a least element b) or g1 = (X;a)(X;b),a < bwhere (ab) is the left-most gap ofK (corresponding to the case whereK has no least element) and that g2 = (X;c)(X;d) where (cd) is the right-most gap of K. By our hypothesis, b < c. We claim that g1g2 2= MS. Since g1g2 0 on K, this will complete the proof. Suppose, to the contrary, that
g1g2 =0+1g1++sgs i 2XR X]2:
Comparing degrees, we see that i has degree 2 or 0 if i 1 and 0 has degree 4, 2 or 0. Let g = g2;1. Thus g1g = 0+2g2 ++sgs. Thus g1(c)g(c) 0, g1(d)g(d) 0, so g(c) 0, g(d) 0. On, the other hand, from g = g2 ;1, it follows that g(c) 0, g(d) 0. Since g has degree 2, this implies g = 0, i.e., g2 =1, contradicting the fact that g2 is strictly negative on (cd).
3. Examples where ( ) fails
In this section we study condition ( ) using an extension of the method intro- duced 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:
TSalg =TS )(y) ) (z) ) ( ): If TSlin =TS we can obviously say more:
3.1 Proposition.
If TSlin =TS then the conditions (y), (z), ( ) and TSalg =TS are equivalent.And similarly, forMS:
3.2 Proposition.
IfMSlin =MS then the conditions (yM), (zM), ( M) andTSalg = MS 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 nest locally convex topology 2, Sect. 1.1.9]. For a cone C in V, Clin denotes the double dual cone of C, i.e., the set of all x 2 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) Clin is the closure of C.(2) Clin =C , C is closed.
Proof. According to the Bipolar Theorem 2, Theorem 1.3.6], Clin 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.
SupposeC is a cone in a vector spaceV overR andCi =C\Vi where Vi, i 0 are nite dimensional subspaces of V with Vi Vi+1 and V = i 0Vi. Then the following are equivalent:(1) C is closed in V.
(2) Ci is closed in Vi for each i0. Proof. See 2, Lemma 6.3.3].
Our next result provides examples where TS andMS are closed.
3.5 Theorem.
If KS contains a cone of dimension n then TS and MS 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 KS 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 rst assertion follows from the second, replacing S by the set of all products ge, e 2 f01gs, e 6= (0:::0). By Lemma 3.4, to prove the second assertion, it suces to show that Mi is closed in Pi, for each i 0. Let S =
fg1:::gsg. We can assume eachgi is not zero. By our hypothesis,KS contains a coneCwith 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 MS, let f = f0 ++fd be its homogeneous decomposition. Observe that, for any a 2C and any > 0,
f(a) =f0+f1(a) ++dfd(a):
Since a2C KS for all positive it follows that, if fd(a)6= 0, then fd(a)>0.
Now suppose
p=0+1g1++sgs
i 2 R X]2. Let d denote the maximum of the degrees of 01g1:::sgs. Using the fact that C has non-empty interior, we can pick a point a 2C such that the homogeneous parts of highest degree of the various polynomials in question do not vanish, so the homogeneous part ofphaving degreedis positive when evaluated at a (so, in particular, it is not zero). To summarize: if p has degree d, then 0 has degree d and each i, i 1 has degree d;deg(gi). Thus any p 2 Mm
is expressible as above with i = Pfij2, deg(f0j) m, deg(fij) m; 12deg(gi), i 1. Also, as in 1], we can assume i = PNj=1fij2, where N is the dimension of Pm.
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 gi vanish at this point.
Thus, for real distinct a1:::a2m+1 suciently close to 1, the set H =f(aj1:::aj2m+1)j1jk2m+ 1k = 1:::2m+ 1g
is contained in C and gi(a) >0 for each a 2H. As explained in 1], two elements of Pm are equal i they are equal on H, and the topology onPm coincides with the topology of pointwise convergence on H. Also, for any a 2 H, f0j(a)2 0(a) p(a), and fij(a)2 i(a)p(a)=Di where Di is the minimum of thegi(a), a2H, ifi1. Thus, if p` is a sequence in Mm converging to some p2Pm then, as in 1], there exists a subsequencep`k with the associatedf`ij converging pointwise onH to some fij 2Pm of the appropriate degree, so p2Mm.
Theorem 3.5 does not cover the case where KS is a curve. At the same time TS is closed for many curves, at least, for the \right" choice of S. We give two examples to illustrate.
3.6 Example.
Consider the curveC =f(xy)2R2 jy =q(x)g
in R2, q 2R X]. Then C =KS where S =f;(Y ;q)2g. We claim that (y) holds in this choice of S. Suppose f 0 on C. Then f(XY) = f(Xq(X)) +h(XY) withh vanishing onC. Also f(Xq(X))0 on R so f(Xq(X))2PR X]2. Thus we are reduced to showing that +h 2 TS for each real > 0. Since Y ;q is irreducible, h= (Y ;q)h1 for some h1 2R XY], so
+h=(1 + h1
2(Y ;q))2; h21
4(Y ;q)2 2TS:
Note: Y ; q =2 TS. If Y ;q = 0 ;1(Y ;q)2, 01 2 R XY]2, then 0 vanishes onC so (Y ;q)2 divides0. Dividing through by Y ;q this impliesY ;q divides 1, a contradiction. Thus TSalg 6= TS, so TS is not closed for this choice of S. On the other hand, the curve C is described more naturally as C = KS where S =fY ;q;(Y ;q)g, and, for this choice ofS, TSalg =TS (so, in particular,TS is closed). To prove this, just write h1 =r2;s2, r = (h1+ 1)=2, s= (h1;1)=2, so
f =f(Xg(X)) +r2(Y ;q);s2(Y ;q)2TS:
3.7 Example.
Consider the curveC =f(xy)2 R2 jy2 =q(x)g
where q 2 R X] is not a square. Thus C = KS where S = fY2;q;(Y2;q)g. Assume that C is not compact. Replacing X by ;X if necessary, we can assume thatq(x)>0 for xsuciently large (so the leading coecient of q is positive). We claim that, under this assumption,TS is closed. SinceTS =PR XY]2+(Y2;q), where (Y2;q) denotes the ideal generated by Y2;q, it suces to prove thatPA2 is closed in A, where A denotes the coordinate ring of C, i.e.,
A= R XY] (Y2;q):
Every f 2 A is expressible uniquely as f = g+hpq, gh 2 R X]. Let Qd denote the subspace of A consisting of all f = g+hpq 2 A with deg(g), deg(h) 2d. According to Lemma 3.4, it suces to show that (PA2)\Qd is closed in Qd for each d 0. We need certain degree estimates. If f = g+hpq, gh 2 R X], then f2 = (g2+h2q) + (2gh)pq. Thus every element of PA2 has the form
p=Xk
i=1(gi+hipq)2 =Xk
i=1(g2i +h2iq) + 2Xk
i=1gihipq:
Thus, if p=p1+p2pq, p1p2 2R X], then
deg(p1) = maxf2deg(gi)2deg(hi) + deg(q)j1ikg:
In particular, if p2Qd, then deg(gi)d and deg(hi)d; deg(q)=2, i= 1:::k. With these estimates in hand, the proof that (PA2)\Qd is closed in Qd follows
along the same lines as in the proof of Theorem 3.5 and can safely be left to the reader.
Since TS is closed, the various conditions (y), (z), ( ) and TSalg =TS are equiv- alent. We claim that these conditions hold if deg(q) = 1 or 2 and that they fail if deg(q) 3. The rst assertion is elementary e.g., see 9, Proposition 2.17]. Sup- pose now that deg(q) 3. If q is not in PR X]2 then q takes on negative values so there exists a quadratic p 2R X] which takes on negative values with p 0 on C. If p2TS then p =Pki=1(g2i +h2iq) for some gihi 2R X]. Then Pki=1h2i 6= 0, so deg(p) deg(q), a contradiction. If q 2PR X]2, thenq =r2+s2, rs2R X].
Takep=p1+Y wherep1 2R X] has degreemaxfdeg(r)deg(s)g+1 and is such that p1 jrj+jsj on R. Then
p(xpq(x)) =p1(x)pq(x)jr(x)j+js(x)jpr(x)2+s(x)2 0 for eachx2R sop0 onC. Ifp2TS, thenp1 =Pki=1(gi2+h2iq), 1 = 2Pki=1gihi
for some gihi 2 R X]. Then Ph2i 6= 0, so deg(p1) deg(q). Since deg(q) = maxf2deg(r)2deg(s)g, 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 TSalg is not even nitely generated.
We turn now to the case where dim(KS) 2. In 1, Theorem 4], Berg, Chris- tensen and Jensen use their weak version of Theorem 3.5 to show that ( ) fails for S = (so KS = Rn, TS = PR X]2) if n 2. In 2, Theorem 6.3.9] this same method is used to show that ( ) fails for S =fX1:::Xng if n 2. We proceed to generalize these results. In general, for TS closed, showing ( ) fails is equivalent to showing 9 p 2 TSalg, p =2 TS. Such a polynomial always exists when dim(KS) 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(KS) 3 then there exists a polynomial p 0 on Rn such that p =2TS.1Proof. See 9, Prop. 6.1].
Theorem 3.8 implies, in particular, that ( ) fails whenevern3 andKScontains a cone of dimension n. Another result in 9] allows us to greatly improve on this:
1Apparently this is not true for dim(KS) = 2. In the \Added to proof" at the end of 9], Scheiderer reports the discovery of a smooth compact surfaceKS withTSalg=TS.
3.9 Theorem.
If n = 2 and KS contains a 2-dimensional cone, then there exists a polynomial p0 on R2 such that p =2TS.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 KS contains a cone of dimension 2.Note: We are not claiming now thatTS is closed.
Proof. Changing coordinates, we may assume the cone is given by X1 0 X2 0 Xi = 0 i3:
Dene S0 R X1X2] by S0 = fg01:::gs0g where g0i = gi(X1X20:::0). Thus KS0 contains the cone dened byX1 0,X2 0 so, by Theorem 3.5, TS0 is closed.
Use Theorem 3.9 to pick p =p(X1X2) such that p 0 on R2 and p =2TS0. Then p 2 TSalg. We claim that p =2 TSlin. For suppose p 2 TSlin. Any linear functional L on R X1X2] with L(TS0) 0 extends to a linear functional L0 on R X] with L0(TS) 0 via L0(f) = L(f(X1X20::::0)). This is clear. Thus, for any such L, L(p) = L0(p) 0, so p 2 TSlin0 . Since TSlin0 = TS0, 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 =fX1;XY3;X2g. KS consists of the points (xy)2 R2 on the vertical strip 0 x 1 with y x2=3. We claim that ( ) fails forS. Let S0 R Y] be dened byS0 =fY3g. ThenKS0 = 01) so, by Theorem 3.5, TS0 is closed. One checks that Y =2 TS0. (Degree considerations show that Y =0+1Y3,01 2PR Y]2 is not possible.) Clearly Y 0 on KS. We claim Y =2TSlin. We argue as in the proof of Corollary 3.10: Every linear functional L onR Y] with L(TS0)0 extends to a linear functionalL0 onR XY] with L0(TS)0 via L0(f) = L(f(0Y)). Thus, if Y 2 TSlin then L(Y) = L0(Y) 0 for every such L, so Y 2TSlin0 =TS0.
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 Schmudgen's result in the compact case, replacing the assumption that KS is compact with the assumption that p 2 1 +TS is chosen so that the coordinate functions X1:::Xn are bounded on KS
by kp` for k` suciently large. The method of proof in 7] is a generalization of Wormann's method in 11].
Note: Such a polynomial p always exists, e.g., p = 1 +Pni=1Xi2. The `best' choice of p may depend on KS. For example:
| IfKS is compact, we can take p= 1.
| Ifn= 2 and S =fX1;Xg, we can take p= 1 +Y2. Fixing such a polynomialp, we have:
4.1 Theorem.
For any f 2R X] there exist integers k` such that kp`f 2TS. Proof. See 7, Corollary 1.4].This gives better control over the element q appearing in condition (z). 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) (z) holds.
(2) 8f 2R X], f 2TSalg ) 9`0 such that 8 real > 0, f +p` 2TS.
Proof. (2) ) (1) is clear. For (1) ) (2), use Theorem 4.1 to pick k` so that kp`;q2TS and use the identity f +p` = (f + kq) + k(kp`;q).
The second main result in 7] is:
4.3 Theorem.
For any f 2R X], the following are equivalent:(1) f 2TSalg.
(2) For any suciently large integer m and all rational > 0, there exists an integer `0 such that p`(f +pm)2TS.
Proof. See 7, Corollary 3.1].
Let TS0 =ff 2R X] jp`f 2TS for some `0g: For d 0, let Td0 =TS0 \Pd.
