An Exact Jacobian SDP Relaxation for Polynomial Optimization

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1 An Exact Jacobian SDP Relaxation for Polynomial Optimization Jiawang Nie May 26, 2011 Abstract Given polynomials f(x), g i (x), h j (x), we study how to imize f(x) on the set S = {x R n : h 1 (x) = = h m1 (x) = 0, g 1 (x) 0,..., g m2 (x) 0}. Let f be the imum of f on S. Suppose S is nonsingular and f is achievable on S, which are true generically. This paper proposes a new type semidefinite programg (SDP) relaxation which is the first one for solving this problem exactly. First, we construct new polynomials ϕ 1,..., ϕ r, by using the Jacobian of f, h i, g j, such that the above problem is equivalent to x R n f(x) s.t. h i (x) = 0, ϕ j (x) = 0, 1 i m 1, 1 j r, g 1 (x) ν1 g m2 (x) νm 2 0, ν {0, 1} m2. Second, we prove that for all N big enough, the standard N-th order Lasserre s SDP relaxation is exact for solving this equivalent problem, that is, its optimal value is equal to f. Some variations and examples are also shown. Key words deterantal varieties, ideals, ors, polynomials, nonsingularity, semidefinite programg, sum of squares AMS subject classification 14P99, 65K05, 90C22 1 Introduction Consider the optimization problem x R n f(x) s.t. h 1 (x) = = h m1 (x) = 0 g 1 (x) 0,..., g m2 (x) 0 (1.1) where f(x), g i (x), h j (x) are polynomial functions. When m 1 = 0 (resp. m 2 = 0), there are no equality (resp. inequality) constraints. Let S be its feasible set and f be its global imum. We are interested in computing f. The problem is NP-hard [16]. Department of Mathematics, University of California, 9500 Gilman Drive, La Jolla, CA njw@math.ucsd.edu. The research was partially supported by NSF grants DMS , DMS

2 A standard approach for solving (1.1) is the hierarchy of semidefinite programg (SDP) relaxations proposed by Lasserre [16]. It is based on a sequence of sum of squares (SOS) type representations of polynomials that are nonnegative on S. The basic idea is, for a given integer N > 0 (called relaxation order), solve the SOS program max γ s.t. f(x) γ = m 1 φ i h i + m 2 σ j g j, i=1 j=1 deg(φ i h i ), deg(σ j g j ) 2N i, j, σ 1,..., σ m2 are SOS. (1.2) In the above, g 0 (x) 1, the decision variables are the coefficients of polynomials φ i and σ j. Here a polynomial is SOS if it is a sum of squares of other polynomials. The SOS program (1.2) is equivalent to an SDP problem (see [16]). Let p N be the optimal value of (1.2). Clearly, p N f for every N. Using Putinar s Positivstellensatz [21], Lasserre proved p N f as N, under the archimedean condition. A stronger relaxation than (1.1) would be obtained by using cross products of g j, which is max γ s.t. f(x) γ = φ i h i + σ ν g ν, i=1,...,m 1 ν {0,1} m 2 deg(φ i h i ) 2N, deg(σ ν g ν ) 2N i, ν, σ ν are all SOS. (1.3) In the above, g ν = g ν 1 1 gνm 2 m 2. Let q N be the optimal value of (1.3). When S is compact, Lasserre showed q N f as N goes to infinity, using Schmügen s Positivstellensatz [24]. An analysis for the convergence speed of p N, q N to f is given in [19, 25]. Typically, (1.2) and (1.3) are not exact for (1.1) with a finite N. Scheiderer [23] proved a very surprising result: whenever S has dimension three or higher, there always exists f such that f(x) f does not have a representation required in (1.3). Thus, we usually need to solve a big number of SDPs until convergence is met. This would be very inefficient in many applications. Furthermore, when S is not compact, typically we do not have the convergence p N f or q N f. This is another difficulty. Thus, people are interested in more efficient methods for solving (1.1). Recently, the author, Demmel and Sturmfels [18] proposed a gradient type SOS relaxation. Consider the case of (1.1) without constraints, i.e., m 1 = m 2 = 0. If the imum f is achieved at a point u, then f(u) = 0, and the problem is equivalent to f(x) s.t. x R n f = = f = 0. (1.4) x 1 x n In [18], Lasserre s relaxation is applied to solve (1.4), and it was shown that a sequence of lower bounds converging to f would be obtained. It has finite convergence if the gradient ideal, generated by the partial derivatives of f(x), is radical. More recently, Demmel, the author and Powers [7] generalized the gradient SOS relaxation to solve (1.1) by using the Karush-Kuhn-Tucker (KKT) conditions of (1.1) m 1 m 2 f(x) = λ i h i (x) + µ j g j (x), µ j g j (x) = 0, j = 1,..., m 2. i=1 j=1 2

3 If a global imizer of (1.1) is a KKT point, then (1.1) is equivalent to x,λ,µ f(x) s.t. h 1 (x) = = h m1 (x) = 0, f(x) = m 1 λ i h i (x) + m 2 µ j g j (x), i=1 j=1 µ j g j (x) = 0, g j (x) 0, j = 1,..., m 2. (1.5) Let {v N } be the sequence of lower bounds for (1.5) obtained by applying Lasserre s relaxation of type (1.3). It was shown in [7] that v N f, no matter S is compact or not. Furthermore, it holds that v N = f for a finite N when the KKT ideal is radical, but it was unknown in [7] whether this property still holds without the KKT ideal being radical. A drawback for this approach is that the involved polynomials are in (x, λ, µ). There are totally n + m 1 + m 2 variables, which makes the resulting SDP very difficult to solve in practice. Contributions This paper proposes a new type SDP relaxation for solving (1.1) via using KKT conditions but the involved polynomials are only in x. Suppose S satisfies a nonsingularity assumption (see Assumption 2.2 for its meaning) and f is achievable on S, which are true generically. We construct new polynomials ϕ 1 (x),..., ϕ r (x), by using the ors of the Jacobian of f, h i, g j, such that (1.1) is equivalent to x R n s.t. f(x) h i (x) = 0 (1 i m 1 ), ϕ j (x) = 0 (1 j r), g 1 (x) ν1 g m2 (x) ν m 2 0, ν {0, 1} m 2. Then we prove that for all N big enough, the standard N-th order Lasserre s relaxation for the above returns the imum f. That is, an exact SDP relaxation for (1.1) is obtained by using the Jacobian. This paper is organized as follows. Section 2 gives the construction of this exact SDP relaxation by using Jacobian. Its exactness and genericity are proved in Section 3. Some efficient variations are proposed in Section 4. Some examples of how to apply it are shown in Section 5. Some conclusions and discussions are made in Section 6. Finally, we attach an appendix introducing some basics of algebraic geometry and real algebra that are used in the paper. Notations The symbol N (resp., R, C) denotes the set of nonnegative integers (resp., real numbers, complex numbers). For any t R, t denotes the smallest integer not smaller than t. For integer n > 0, [n] denotes the set {1,..., n}, and [n] k denotes the set of subsets of [n] whose cardinality is k. For a subset J of [n], J denotes its cardinality. For x R n, x i denotes the i-th component of x, that is, x = (x 1,..., x n ). For α N n, denote α = α α n. For x R n and α N n, x α denotes x α 1 1 xα n n. The symbol R[x] = R[x 1,..., x n ] (resp. C[x] = C[x 1,..., x n ]) denotes the ring of polynomials in (x 1,..., x n ) with real (resp. complex) coefficients. A polynomial is called a form if it is homogeneous. The R[x] d denotes the subspace of polynomials in R[x] of degrees at most d. For a general set T R n, int(t ) denotes its interior, and T denotes its boundary in standard Euclidean topology. For a symmetric matrix X, X 0 (resp., X 0) means X is positive semidefinite (resp. positive definite). For u R N, u 2 denotes the standard Euclidean norm. 3

4 2 Construction of the exact Jacobian SDP relaxation Let S be the feasible set of (1.1) and m = {m 1 + m 2, n 1}. (2.1) For convenience, we denote h(x) = (h 1 (x),..., h m1 (x)) and g(x) = (g 1 (x),..., g m2 (x)). For a subset J = {j 1,..., j k } [m 2 ], denote g J (x) = (g j1 (x),..., g jk (x)). Let x be a imizer of (1.1). If J is the active index set at x such that g J (x ) = 0 and the KKT conditions hold at x, then there exist λ i and µ j (j J) such that h(x ) = 0, g J (x ) = 0, f(x ) = i [m 1 ] λ i h i (x ) + j J µ j g j (x ). The above implies the Jacobian matrix of (f, h, g J ) is singular at x. For a subset J [m 2 ], denote the deterantal variety of (f, h, g J ) s Jacobian being singular by G J = { x C n : rank B J (x) m 1 + J }, B J (x) = [ f(x) h(x) g J (x) ]. (2.2) Then, x V (h, g J ) G J where V (h, g J ) := { x C n : h(x) = 0, g J (x) = 0 }. This motivates us to use g J (x) = 0 and G J to get tighter SDP relaxations for (1.1). To do so, a practical issue is how to get a nice description for G J? An obvious one is that all its maximal ors vanish. But there are totally ( n m 1 +k+1) such ors (if m1 + k + 1 n), which is huge for big n, m 1, k. Can we define G J by a set of the smallest number of equations? Furthermore, the active index set J is usually unknown in advance. Can we get an SDP relaxation that is independent of J? 2.1 Minimum defining equations for deterantal varieties Let k n and X = (X ij ) be a n k matrix of indeterants X ij. Define the deterantal variety } D n,k t 1 {X = C n k : rank X < t. For any index set I = {i 1,..., i k } [n], denote by det I (X) the (i 1,..., i k ) (1,..., k)-or of matrix X, i.e., the deterant of the submatrix of X whose row indices are i 1,..., i k and column indices are 1,..., k. Clearly, it holds that { } D n,k k 1 = X C n k : det I (X) = 0 I [n] k. The above has ( n k) defining equations of degree k. An interesting fact is that we do not need ( n k) equations to define D n,k k 1. Actually, this number would be significantly smaller. There is very nice work on this issue. Bruns and Vetter [3] showed that nk t equations are enough to define D n,k t 1. Later, Bruns and Schwänzl [2] showed that nk t2 + 1 is the smallest number of equations for defining D n,k t 1. Typically, nk t2 +1 ( n k) for big n and k. A general 4

5 method for constructing nk t defining polynomial equations for D n,k t 1 was described in Chapt. 5 of [3]. Here we briefly show how it works for D n,k k 1. Let Γ(X) denote the set of all k-ors of X (assume their row indices are strictly increasing). For convenience, for i 1 < < i k, denote by [i 1,..., i k ] the (i 1,..., i k ) (1,..., k)-or of X. Define a partial ordering on Γ(X) as follows: [i 1,..., i k ] < [j 1,..., j k ] i 1 j 1,..., i k j k, i i k < j j k. If I = {i 1,..., i k }, we also write I = [i 1,..., i k ] as a or in Γ(X) for convenience. For any I Γ(X), define its rank as { } rk(i) = max l : I = I (l) > > I (1), every I (i) Γ(X). The maximum or in Γ(X) is [n k + 1,..., n] and has rank nk k For every 1 l nk k 2 + 1, define η l (X) = det I (X). (2.3) I [n] k,rk(i)=l Lemma 2.1 (Lemma (5.9), Bruns and Vetter [3]). It holds that } D n,k k 1 {X = C n k : η l (X) = 0, l = 1,..., nk k When k = 2, D n,2 1 would be defined by 2n 3 polynomials. The biggest or is [n 1, n] and has rank 2n 3. For each l = 1, 2,..., 2n 3, we clearly have η l (X) = [i 1, i 2 ]. 1 i 1 <i 2 n: i 1 +i 2 =l+2 Every 2-or of X is a summand of some η l (X). When k = 3, D n,3 2 can be defined by 3n 8 polynomials of the form η l (X). For instance, when n = 6, the partial ordering on Γ(X) is shown in the following diagram: In the above, an arrow points to a bigger or. Clearly, we have the expressions η 1 (X) = [1, 2, 3], η 2 (X) = [1, 2, 4], η 3 (X) = [1, 2, 5] + [1, 3, 4], η 4 (X) = [1, 2, 6] + [1, 3, 5] + [2, 3, 4], η 5 (X) = [1, 3, 6] + [1, 4, 5] + [2, 3, 5], η 6 (X) = [1, 4, 6] + [2, 3, 6] + [2, 4, 5], η 7 (X) = [1, 5, 6] + [2, 4, 6] + [3, 4, 5], η 8 (X) = [2, 5, 6] + [3, 4, 6], η 9 (X) = [3, 5, 6], η 10 (X) = [4, 5, 6]. 5

6 Every above η i (X) has degree 3. Note that the summands [i 1, i 2, i 3 ] from the same η i (X) have a constant summation i 1 + i 2 + i 3. Thus, for each l = 1,..., 3n 8, we have η l (X) = [i 1, i 2, i 3 ]. 1 i 1 <i 2 <i 3 n: i 1 +i 2 +i 3 =l+5 When k > 3 is general, D n,k k 1 can be defined by nk k2 +1 polynomials of the form η l (X). For each l = 1, 2,..., nk k 2 + 1, we similarly have the expression η l (X) = [i 1,..., i k ]. 1 i 1 < <i k n: i 1 + +i k =l+( k+1 2 ) The exact Jacobian SDP relaxation For every J = {j 1,..., j k } [m 2 ] with k m m 1, by applying formula (2.3), let ( ) η1 J,..., ηlen(j) J len(j) := n(m 1 + k + 1) (m 1 + k + 1) be the set of defining polynomials for the deterantal variety G J defined in (2.2) of the Jacobian of (f, h, g J ) being singular. For each i = 1,..., len(j), define ( ) ϕ J i (x) = ηi J B J (x) g j (x), where J c = [m 2 ]\J. (2.4) j J c Using the product j J c g j(x) in the above is motivated by a characterization of critical points in [15]. For simplicity, list all possible ϕ J i in (2.4) sequentially as ϕ 1, ϕ 2,..., ϕ r, where r = len(j). (2.5) J [m 2 ], J m m 1 Now define the variety W = {x C n : h 1 (x) = = h m1 (x) = ϕ 1 (x) = = ϕ r (x) = 0}. (2.6) If the imum f of (1.1) is achieved at a KKT point, then (1.1) is equivalent to x R n s.t. f(x) h i (x) = 0 (1 i m 1 ), ϕ j (x) = 0 (1 j r), g ν (x) 0, ν {0, 1} m 2. (2.7) In the above, each g ν = g ν 1 1 gνm 2 m 2. Let f be the imum of (2.7). If (1.1) has a imizer that is a KKT point, then f = f. If (1.1) does not have a imizer (i.e., f is not achievable in (1.1)), or if (1.1) has one or several imizers but none of them is a KKT point, then we might have f < f (e.g., for f(x) = x (1 x 1x 2 ) 2 and S = R 2, f = 0 < f = 1 [18]). But in any case the imum f is always finite as will be shown by Theorem 2.3, while f might not. 6

7 To construct an SDP relaxation for (2.7), we need to define localizing moment matrices. Let q(x) be a polynomial with deg(q) 2N. Define symmetric matrices A (N) α such that q(x)[x] d [x] T d = A (N) α x α, where d = N deg(q)/2. α N n : α 2N Then the N-th order localizing moment matrix of q is defined as L (N) q (y) = A (N) α y α. α N n : α 2N Here y is a moment vector indexed by α N n with α 2N. Moreover, denote L f (y) = f α y α for f(x) = f α x α. α N n : α deg(f) The N-th order Lasserre s relaxation for (2.7) is the SDP α N n : α deg(f) f (1) N := L f (y) s.t. (y) = 0 (1 i m 1 ), L (N) (y) = 0 (1 j r), L (N) h i L (N) ϕ j g ν (y) 0, ν {0, 1} m 2, y 0 = 1. (2.8) Compared to Schmüdgen ( type Lasserre s ) relaxation (1.3), the number of new constraints in (2.8) is r = O 2 m2 n (m 1 + m 2 ) by (2.5). That is, r is of linear order in nm 1 for fixed m 2, but is exponential in m 2. So, when m 2 is small or moderately large, (2.8) is practical; but for big m 2, (2.8) becomes more difficult to solve numerically. Now we present the dual of (2.8). Define the truncated preordering P (N) generated by g j as P (N) = ν {0,1} m 2 σ ν g ν deg(σ νg ν ) 2N, (2.9) and the truncated ideal I (N) generated by h i and ϕ j as m 1 r I (N) = p i h i + q j ϕ j deg(p i h i ) 2N deg(q j ϕ j ) 2N i=1 j=1 i j. (2.10) Then, as shown in Lasserre [16], the dual of (2.8) is the following SOS relaxation for (2.7): f (2) N := max γ s.t. f(x) γ I (N) + P (N). (2.11) Note the relaxation (2.11) is stronger than (1.3). Then, by weak duality, we have f (2) N f (1) N f. (2.12) We are going to show that when N is big enough, (2.8) is an exact SDP relaxation for (2.7), i.e., f (2) N = f (1) N = f. For this purpose, we need the following assumption. 7

8 Assumption 2.2. (i) m 1 n. (ii) For any u S, at most n m 1 of g 1 (u),..., g m2 (u) vanish. (iii) For every J = {j 1,..., j k } [m 2 ] with k n m 1, the variety V (h, g J ) = {x C n : h(x) = 0, g J (x) = 0} is nonsingular (its Jacobian has full rank on V (h, g J )). Theorem 2.3. Suppose Assumption 2.2 holds. Let f be the imum of (2.7). Then f > and there exists N N such that f (1) N = f (2) N = f for all N N. Furthermore, if the imum f of (1.1) is achievable, then f (1) N = f (2) N = f for all N N. The proof of Theorem 2.3 is based on a new kind of Positivstellensatz using Jacobians of the objective and constraining polynomials: there exists an integer N > 0, which only depends on polynomials f, h i, g j, such that for all ɛ > 0 f(x) f + ɛ I (N ) + P (N ). Note that the order N in the above is independent of ɛ. This new Positivstellensatz is given by Theorem 3.4 in the next section. Theorem 2.3 as well as this Positivstellensatz will be proved in Section 3. We would like to remark that in Theorem 2.3, for N > N, the optimal value of (2.11) might not be achievable (e.g., see Example 5.1), while the imum of (2.8) is always achievable if (1.1) has a imizer (e.g., [x ] 2N is one for any optimizer x of (1.1)). When the feasible set S of (1.1) is compact, the imum f is always achievable. Thus, Theorem 2.3 implies the following. Corollary 2.4. Suppose Assumption 2.2 holds. If S is compact, then f (1) N = f (2) N = f for all N big enough. A practical issue in applications is how to identify whether (2.8) is exact for a given N. This would be possible by applying the flat-extension condition (FEC) [6]. Let y be a imizer of (2.8). We say y satisfies FEC if rank L (N) g 0 (y ) = rank L (N d S) g 0 (y ), where d S = max i [m 1 ],j [r],ν {0,1} m 2 { deg(h i )/2, deg(ϕ j )/2, deg(g ν )/2 Note that g 0 1 and L (N) g 0 (y ) reduces to an N-th order moment matrix. When FEC holds, (2.8) is exact for (1.1), and a finite set of global imizers would be extracted from y. We refer to [13] for a numerical method on how to do this. A very nice software for solving SDP relaxations from polynomial optimization is GloptiPoly 3 [14] which also provides routines for finding imizers if FEC holds. Now we discuss how general the conditions of Theorem 2.3 are. Define { } B d (S) = f R[x] d : inf f(u) >. u S Clearly, B d (S) is convex and has nonempty interior. Define the projectivization of S as { S prj = x R n+1 : h 1 ( x) = = h } m1 ( x) = 0, g 1 ( x) 0,..., g m2 ( x) 0. (2.13) }. 8

9 Here p denotes the homogenization of p, and x = (x 0, x 1,..., x n ), i.e., p( x) = x deg(p) 0 p(x/x 0 ). We say S is closed at if S prj {x 0 0} = closure ( S prj {x 0 > 0} ). Under some generic conditions, Assumption 2.2 holds and the imum f of (1.1) is achievable. These conditions are expressed as non-vanishing of the so-called resultants Res or discriants, which are polynomial in the coefficients of f, h i, g j. We refer to Appendix for a short introduction about Res and. Theorem 2.5. Let f, h i, g j be the polynomials in (1.1), and S be the feasible set. (a) If m 1 > n and Res(h i1,..., h in+1 ) 0 for some {i 1,..., i n+1 }, then S =. (b) If m 1 n and for every {j 1,..., j n m1 +1} [m 2 ] then item (ii) of Assumption 2.2 holds. Res(h 1,..., h m1, g j1,..., g jn m1 +1 ) 0, (c) If m 1 n and for every {j 1,..., j k } [m 2 ] with k n m 1 then item (iii) of Assumption 2.2 holds. (h 1,..., h m1, g j1,..., g jk ) 0, (d) Suppose S is closed at and f B d (S). If the resultant of any n of h hom i, gj hom nonzero (only when m 1 + m 2 n), and for every {j 1,..., j k } with k n m 1 1 is (f hom, h hom 1,..., h hom, g hom,..., gj hom k ) 0, m 1 then there exists v S such that f = f(v). Here p hom denotes p s homogeneous part of the highest degree. (e) If f B d (R n ) and (f hom ) 0, then the imum of f(x) in R n is achievable. Theorem 2.5 will be proved in Section 3. Now we consider the special case of (1.1) having no constraints. If f > is achievable, then (1.1) is equivalent to (1.4). The item (e) of Theorem 2.5 tells us that this is generically true. The gradient SOS relaxation for solving (1.4) proposed in [18] is a special case of (2.11). The following is an immediate consequence of Theorem 2.3 and item (e) of Theorem 2.5. Corollary 2.6. If S = R n, f(x) has imum f >, and (f hom ) 0, then the optimal values of (2.8) and (2.11) are equal to f if N is big enough. Corollary 2.6 is stronger than Theorem 10 of [18], where the exactness of gradient SOS relaxation for a finite order N is only shown when the gradient ideal is radical. j 1 9

10 3 Proof of exactness and genericity This section proves Theorems 2.3 and 2.5. First, we give some lemmas that are crucially used in the proofs. Lemma 3.1. Let K be the variety defined by the KKT conditions K = (x, λ, µ) Cn+m 1+m 2 f(x) = m 1 λ i h i (x) + m 2 µ j g j (x) i=1 j=1 h i (x) = µ j g j (x) = 0, (i, j) [m 1 ] [m 2 ]. (3.1) If Assumption 2.2 holds, then W = K x where K x = {x C n : (x, λ, µ) K for some λ, µ}. Proof. First, we prove W K x. Choose an arbitrary u W. Let J = {j [m 2 ] : g j (u) = 0} and k = J. By Assumption 2.2, m 1 + k n. Recall from (2.2) that B J (x) = [ f(x) h(x) g J (x) ]. Case m 1 + k = n By Assumption 2.2, the matrix H(u) = [ h(u) g J (u) ] is nonsingular. Note that H(u) is now a square matrix. So, H(u) is invertible, and there exist λ i and µ j (j J) such that f(u) = m 1 i=1 λ i h i (u) + j J Define µ j = 0 for j J, then we have u K x. µ j g j (u). (3.2) Case m 1 + k < n By the construction of polynomials ϕ i (x) in (2.5), some of them are ϕ J i (x) := η i (B J (x)) g j (x), i = 1,..., n(m 1 + k + 1) (m 1 + k + 1) j J c So the equations ϕ i (u) = 0 imply every above ϕ J i (u) = 0 (see its definition in (2.4)). Hence B J (u) is singular. By Assumption 2.2, the matrix H(u) is nonsingular. So there exist λ i and µ j (j J) satisfying (3.2). Define µ j = 0 for j J, then we also have u K x. Second, we prove K x W. Choose an arbitrary u K x with (u, λ, µ) K. Let I = {j [m 2 ] : g j (u) = 0}. If I =, then µ = 0, and [ f(u) h(u) ] and B J (u) are both singular, which implies all ϕ i (u) = 0 and u W. If I, write I = {i 1,..., i t }. Let J = {j 1,..., j k } [m 2 ] be an arbitrary index set with m 1 + k m. Case I J At least one j J c belongs to I. By choice of I, we know from (2.4) ϕ J i (u) = η i (B J (u)) g j (u) = 0. j J c Case I J Then µ j = 0 for all j J c. By definition of K, the matrix B J (u) must be singular. All polynomials ϕ J i (x) vanish at u by their construction. Combining the above two cases, we know all ϕ J i (x) vanish at u, that is, ϕ 1(u) = = ϕ r (u) = 0. So u W. 10

11 Lemma 3.2. Suppose Assumption 2.2 holds. Let W be defined in (2.6), and T = {x R n : g j (x) 0, j = 1,..., m 2 }. Then there exist disjoint subvarieties W 0, W 1,..., W r of W and distinct v 1,..., v r R such that W = W 0 W 1 W r, W 0 T =, W i T, i = 1,..., r, and f(x) is constantly equal to v i on W i for i = 1,..., r. Proof. Let K = K 1 K r be a decomposition of irreducible varieties. Then f(x) is equaling a constant v i on each K i, as shown by Lemma 3.3 in [7]. By grouping all K i on which v i are same into a single variety, we can assume all v i are distinct. Let Ŵi be the projection of K i into x-space, then by Lemma 3.1 we get W = Ŵ1 Ŵr. Let W i = Zar(Ŵi). Applying Zariski closure in the above gives W = Zar(W ) = W 1 W r. Note that f(x) still achieves a constant value on each W i. Group all W j for which W j T = into a single variety W 0 (if every W j T we set W 0 = ). For convenience, we still write the resulting decomposition as W = W 0 W 1 W r. Clearly, W 0 T =, and for i > 0 the values v i are real and distinct (because = W i T R n and f(x) has real coefficients). Since f(x) achieves distinct values on different W i, we know W i must be disjoint from each other. Therefore, we get a desired decomposition for W. Lemma 3.3. Let I 0, I 1,..., I k be ideals of R[x] such that V (I i ) V (I j ) = for distinct i, j, and I = I 0 I 1 I k. Then there exist a 0, a 1,..., a k R[x] satisfying a a 2 k 1 I, a i I j. i j {0,...,k} Proof. We prove by induction. When k = 1, by Theorem A.2, there exist p I 0, q I 1 such that p + q = 1. Then a 0 = p, a 1 = q satisfy the lemma. Suppose the lemma is true for k = t. We prove it is also true for k = t + 1. Let J = I 0 I t. By induction, there exist b 0,..., b t R[x] such that b b 2 t 1 J, b i I j, i = 0,..., t. i j {0,...,t} Since V (I t+1 ) is disjoint from V (J) = V (I 0 ) V (I t ), by Theorem A.2, there exist p I t+1 and q J such that p + q = 1. Let a i = b i p for i = 0,..., t and a t+1 = q. Then a i I j, i = 0,..., t + 1. i j {0,...,t+1} Since (p + q) 2 = 1, I = I t+1 J, we have pq I, (b b2 t 1)p 2 I, and which completes the proof. a a a 2 t+1 1 = (b b 2 t 1)p 2 2pq I, 11

12 Theorem 3.4. Suppose Assumption 2.2 holds and let f be the imum of (2.7). Then f > and there exists N N such that for all ɛ > 0 f(x) f + ɛ I (N ) + P (N ). (3.3) Proof. Note that the feasible set of (2.7) is contained in the variety W defined by (2.6). Decompose W as in Lemma 3.2. Thus, f(x) achieves finitely many values on W and f = {v 1,..., v r } >. So, we can generally assume f = 0. Reorder W i such that v 1 > v 2 > > v r = 0. The ideal I W = h 1,..., h m1, ϕ 1,..., ϕ r (3.4) has a primary decomposition (see Sturmfels [28, Chapter 5]) I W = E 0 E 1 E r such that each ideal E i R[x] has variety W i = V (E i ). When i = 0, we have V R (E 0 ) T = (T is defined in Lemma 3.2). By Theorem A.3, there exist SOS polynomials τ ν satisfying 1 τ ν g ν (x) mod E 0. ν {0,1} m 2 Thus, from f = 1 4 (f + 1)2 1 4 (f 1)2, we have f 1 4 (f + 1)2 + (f 1) 2 τ ν g ν mod E 0 ν {0,1} m 2 ν {0,1} m 2 τ ν g ν mod E 0 for certain SOS polynomials τ ν. Let σ 0 = ɛ + ν {0,1} m 2 τ ν g ν. Clearly, if N 0 > 0 is big enough, then σ 0 P (N0) for all ɛ > 0. Let q 0 = f + ɛ σ 0 E 0, which is independent of ɛ. For each i = 1,..., r 1, v i > 0 and vi 1 f(x) 1 vanishes on W i. By Theorem A.1, there exists k i > 0 such that (vi 1 f(x) 1) k i E i. Thus, it holds that s i (x) := ( v i 1 + ( v 1 f(x) 1 )) 1/2 vi i k i 1 j=0 ( 1/2 j ) (v 1 i f(x) 1) j mod E i. Let σ i = s i (x) 2 + ɛ, and q i = f + ɛ σ i E i, which is also independent of ɛ > 0. 12

13 When i = r, v r = 0 and f(x) vanishes on W r. By Theorem A.1, there exists k r > 0 such that f(x) k r E r. Thus we obtain that s r (x) := ɛ ( 1 + ɛ 1 f(x) ) k 1/2 r 1 ɛ j=0 ( 1/2 Let σ r = s r (x) 2, and q r = f + ɛ σ r E r. Clearly, we have j ) ɛ j f(x) j mod E r. q r (x) = k r 2 j=0 c j (ɛ)f(x) k r+j for some real scalars c j (ɛ). Note each f(x) k r+j E r. Applying Lemma 3.3 to E 0, E 1,..., E r, we can find a 0,..., a r R[x] satisfying a a 2 r 1 I W, a i E j. Let σ = σ 0 a σ 1a σ ra 2 r, then f(x) + ɛ σ = i j {0,1,...,r} r (f + ɛ σ i )a 2 i + (f + ɛ)(1 a 2 0 a 2 r). i=0 Since q i = f + ɛ σ i E i, it holds that (f + ɛ σ i )a 2 i r E j = I W. For each 0 i < r, q i is independent of ɛ. There exists N 1 > 0 such that for all ɛ > 0 j=0 (f + ɛ σ i )a 2 i I (N 1), i = 0, 1,..., r 1. For i = r, q r = f + ɛ σ r depends on ɛ. By the choice of q r, it holds that (f + ɛ σ r )a 2 r = k r 2 j=0 c j (ɛ)f kr+j a 2 r. Note each f kr+j a 2 r I W, since f kr+j E r. So, there exists N 2 > 0 such that for all ɛ > 0 (f + ɛ σ r )a 2 r I (N 2). Since 1 a 2 1 a2 r I W, there also exists N 3 > 0 such that for all ɛ > 0 (f + ɛ)(1 a 2 1 a 2 r) I (N 3). Combining the above, we know if N is big enough, then f(x) + ɛ σ I (N ) for all ɛ > 0. From the constructions of σ i and a i, we know their degrees are independent of ɛ. So, σ P (N ) for all ɛ > 0 if N is big enough, which completes the proof. 13

14 Theorem 3.4 is a kind of Positivstellensatz of representing f(x) f + ɛ, which is positive on S for all ɛ > 0, by the preordering generated by g j modulo the ideal I W in (3.4) of variety W. Usually, we can not conclude f(x) f I (N ) + P (N ) by setting ɛ = 0, because the coefficients of the representing polynomials of f(x) f + ɛ in I (N ) + P (N ) go to infinity as ɛ 0 (see s r (x) in the proof). It is possible that f(x) f I (N) + P (N) for every N > 0. Such a counterexample is Example 5.1. However, Theorem 3.4 shows that the degree bound N required for representing f(x) f + ɛ is independent of ɛ. This is a crucial property justifying the exactness of the SDP relaxation (2.8). Now we present its proof below. Proof of Theorem 2.3 that for all ɛ > 0 By Theorem 3.4, we know f > and there exists N N such f(x) (f ɛ) I (N ) + P (N ). Since f (1) N, f (2) N are the optimal values of (2.8) and (2.11) respectively, we know f ɛ f (2) N f (1) N f. Because ɛ > 0 is arbitrary, the above implies f (1) N = f (2) N = f. Since the sequence {f (2) N } is monotonically increasing and every f (2) N f (1) N f by (2.12), we get f (1) N = f (2) N = f for all N N. If the imum f of (1.1) is achievable, then there exists x S such that f = f(x ). By Assumption 2.2, we must have x W. So x is feasible for (2.7), and f = f. Thus, we also have f (1) N = f (2) N = f for all N N. Last we prove Theorem 2.5 by using the properties of resultants and discriants described in Appendix. Proof of Theorem 2.5 (a) If Res(h i1,..., h in+1 ) 0, then the polynomial system h i1 (x) = = h in+1 (x) = 0 does not have a complex solution. Hence, V (h) = and consequently S =. (b) For a contradiction, suppose n m of g j vanish at u S, say, g j1,..., g jn m1 +1. Then the polynomial system h 1 (x) = = h m1 (x) = g j1 (x) = = g jn m1 +1 (x) = 0 has a solution, which contradicts Res(h 1,..., h m1, g j1,..., g jn m1 +1 ) 0. (c) For every J = {j 1,..., j k } [m 2 ] with k n m 1, if then the polynomial system (h 1,..., h m1, g j1,..., g jk ) 0, h 1 (x) = = h m1 (x) = g j1 (x) = = g jk (x) = 0 has no singular solution, i.e., the variety V (h, g J ) is smooth. (d) Let f 0 (x) = f(x) f. Then f 0 lies on the boundary of the set P d (S) = { p B d (S) : p(x) 0 x S }. 14

15 Since S is closed at, by Prop. 6.1 of [20], f 0 P d (S) implies 0 = f 0 ( x). x S prj, x 2 =1,x 0 0 Let ũ = (u 0, u 1,..., u n ) 0 be a imizer of the above, which must exist because the feasible set is compact. We claim that u 0 0. Otherwise, suppose u 0 = 0. Then u = (u 1,..., u n ) 0 is a imizer of 0 = f hom (x) s.t. h hom 1 (x) = = h hom m 1 (x) = 0, g1 hom (x) 0,..., gm hom 2 (x) 0. Let j 1,..., j k [m 2 ] be the indices of active constraints. By Fritz-John optimality condition (see Sec in [1]), there exists (λ 0, λ 1,..., λ m1, µ 1,..., µ k ) 0 satisfying λ 0 f hom (u) + m 1 λ i h hom (u) + + k µ l gj hom l (u) = 0, i=1 i f hom (u) = h hom 1 (u) = = h hom (u) = g hom (u) = = gj hom k (u) = 0. Thus, the homogeneous polynomial system m 1 f hom (x) = h hom 1 (x) = = h hom (x) = g hom (x) = = gj hom k (x) = 0 m 1 j 1 j 1 l=1 has a nonzero singular solution. Since the resultant of any n of h hom i, gj hom must have m 1 + k n 1. So the discriant is nonzero, we (f hom, h hom 1,..., h hom, g j1,..., g hom is defined and must vanish, which is a contradiction. So u 0 0. Let v = u/u 0, then ũ S prj implies v S and f(v) f = u d 0 f 0 (ũ) = 0. Clearly, (e) is true since it is a special case of (d). 4 Some variations This section presents some variations of the exact SDP relaxation (2.8) and its dual (2.11). m 1 j k ) 4.1 A refined version based on all maximal ors An SDP relaxation tighter than (2.8) would be obtained by using all the maximal ors to define the deterantal variety G J in (2.2), while the number of equations would be significantly larger. For every J = {j 1,..., j k } [m 2 ] with m 1 + k m, let τ1 J,..., τ l J be all the maximal ors of B J (x) defined in (2.2). Then define new polynomials ψi J := τi J g j, i = 1,..., l. (4.1) j J c List all such possible ψ J i as ψ 1, ψ 2,..., ψ t, where t = J [m 2 ], J m m 1 ( ) n. J + m

16 Like (2.7), we formulate (1.1) equivalently as x R n s.t. f(x) h i (x) = ψ j (x) = 0, i [m 1 ], j [t], g ν (x) 0, ν {0, 1} m 2. The standard N-th order Lasserre s relaxation for the above is s.t. L f (y) (y) = 0, L (N) (y) = 0, i [m 1 ], j [t], L (N) h i ψ j L (N) g ν (y) 0, ν {0, 1} m 2, y 0 = 1. (4.2) (4.3) Note that every ϕ J i in (2.4) is a sum of polynomials like ψi J in (4.1). So the equations L (N) ψ j (y) = 0 in (4.3) implies L ϕ (N) j (y) = 0 in (2.8). Hence, (4.3) is stronger than (2.8). Its dual is an SOS program like (2.11). Theorem 2.3 then implies the following. Corollary 4.1. Suppose Assumption 2.2 is true, and the imum f of (1.1) is achievable. If N is big enough, then the optimal value of (4.3) is equal to f. 4.2 A Putinar type variation without using cross products of g j If the imum f of (1.1) is achieved at a KKT point, then (1.1) is equivalent to x R n s.t. f(x) h i (x) = ϕ j (x) = 0, i [m 1 ], j [r], g 1 (x) 0,..., g m2 (x) 0. The standard N-th order Lasserre s relaxation for (4.4) is s.t. L f (y) (y) = 0, L (N) (y) = 0, i [m 1 ], j [r], L (N) h i L (N) ϕ j g i (y) 0, i = 0, 1,..., m 2, y 0 = 1. (4.4) (4.5) The difference between (4.5) and (2.8) is that the cross products of g j (x) are not used in (4.5), which makes the number of resulting LMIs much smaller. Similar to P (N), define the truncated quadratic module M (N) generated by g i as { m2 } M (N) = σ i g i deg(σ i g i ) 2N. (4.6) i=0 The dual of (4.5) would be shown to be the following SOS relaxation for (4.4): max γ s.t. f(x) γ I (N) + M (N). (4.7) Clearly, for the same N, (4.7) is stronger than the standard Lasserre s relaxation (1.2). To prove (4.5) and (4.7) are exact for some N, we need the archimedean condition (AC) for S, i.e., there exist R > 0, φ 1,..., φ m1 R[x] and SOS s 0,..., s m2 R[x] such that m 1 m 2 R x 2 2 = φ i h i + s j g j. i=1 j=0 16

17 Theorem 4.2. Suppose Assumption 2.2 and the archimedean condition hold. If N is big enough, then the optimal values of (4.5) and (4.7) are equal to f. To prove Theorem 4.2, we need the following. Theorem 4.3. Suppose Assumption 2.2 and the archimedean condition hold. Let f be the optimal value of (4.4). Then there exists an integer N > 0 such that for every ɛ > 0 f(x) f + ɛ I (N ) + M (N ). (4.8) Proof. The proof is almost same as for Theorem 3.4. We follow the same approach used there. The only difference occurs for the case i = 0 and V R (E 0 ) T =. By Theorem A.3, there exist SOS polynomials η ν satisfying 2 η ν g ν mod E 0. ν {0,1} m 2 Clearly, each η m 2 ν g ν 1 1 gν m 2 m 2 is positive on S. Since AC holds, by Putinar s Positivtellensatz (Theorem A.4), there exist SOS polynomials θ ν,i such that Hence, it holds that m 2 + η ν g ν = ν {0,1} m 2 m 2 i=0 m 2 i=0 ( ) 1 2 m + η ν g 2 ν ν {0,1} m 2 θ ν,i θ ν,i g i mod h 1,..., h m1. g i mod E 0. The second equivalence above is due to the relation h 1,..., h m1 I W E 0. Letting τ i = ν {0,1} m 2 θ ν,i, which is clearly SOS, we get mod h 1,..., h m1 + E 0 1 τ 0 + τ 1 g τ m2 g m2 mod E 0. The rest of the proof is almost same as for Theorem 3.4. Proof of Theorem 4.2 For convenience, still let f (1) N, f (2) N be the optimal values of (4.5) and (4.7) respectively. From Theorem 4.3, there exists an integer N such that for all ɛ > 0 f(x) (f ɛ) I (N ) + M (N ). Like in the proof of Theorem 2.3, we can similarly prove f (1) N = f (2) N = f for all N N. Since AC holds, the set S must be compact. So the imum f of (1.1) must be achievable. By Assumption 2.2, we know f = f, and the proof is complete. 17

18 4.3 A simplified version for inactive constraints Suppose in (1.1) we are only interested in a imizer making all the inequality constraints inactive. Consider the problem x R n f(x) s.t. h 1 (x) = = h m1 (x) = 0, g 1 (x) > 0,..., g m2 (x) > 0. (4.9) Let u be a imizer of (4.9). If V (h) is smooth at u, there exist λ i such that Thus, u belongs to the deterantal variety f(u) = λ 1 h 1 (u) + + λ m1 h m1 (u). G h = { x : rank [ f(x) h(x) ] m 1 }. If m 1 < n, let φ 1,..., φ s be a imum set of defining polynomials for G h by using formula (2.3). If m 1 = n, then G h = R n and we do not need these polynomials; set s = 0, and [s] is empty. Then, (4.9) is equivalent to x R n s.t. f(x) h i (x) = φ j (x) = 0, i [m 1 ], j [s], g 1 (x) > 0,..., g m2 (x) > 0. (4.10) The difference ) between (4.10) and (2.7) is that the number of new equations in (4.10) is s = O (nm 1, which is much smaller than r in (2.7). So, (4.10) is preferable to (2.7) when the inequality constraints are all inactive. The N-th order Lasserre s relaxation for (4.10) is s.t. L f (y) (y) = 0, L (N) (y) = 0, i [m 1 ], j [s], L (N) h i φ j L (N) g j (y) 0, j = 1,..., m 2, y 0 = 1. (4.11) A tighter version than the above using cross products of g j is s.t. L f (y) (y) = 0, L (N) (y) = 0, i [m 1 ], j [s], L (N) h i φ j L (N) g ν (y) 0, ν {0, 1} m 2, y 0 = 1. (4.12) Define the truncated ideal J (N) generated by h i (x) and φ j as m 1 s J (N) deg(p = p i h i + q j φ i h i ) 2N j deg(q j φ j ) 2N i=1 The dual of (4.11) is the SOS relaxation j=1 i j. max γ s.t. f(x) γ J (N) + M (N). (4.13) 18

19 The dual of (4.12) is the SOS relaxation max γ s.t. f(x) γ J (N) + P (N). (4.14) The exactness of the above relaxations is summarized as follows. Theorem 4.4. Suppose the variety V (h) is nonsingular and the imum f of (4.9) is achieved at some feasible u with every g j (u) > 0. If N is big enough, then the optimal values of (4.12) and (4.14) are equal to f. If, in addition, the archimedean condition holds for S, the optimal values of (4.11) and (4.13) are also equal to f for N big enough. Proof. The proof is almost same as for Theorems 2.3 and 4.2. We can first prove a decomposition result like Lemma 3.2, and then prove there exists N > 0 such that for all ɛ > 0 (like in Theorem 3.4) f(x) f + ɛ J (N ) + P (N ). Furthermore, if AC holds, we can similarly prove there exists N > 0 such that for all ɛ > 0 (like in Theorem. 4.3) f(x) f + ɛ J (N ) + M (N ). The rest of the proof is almost same as for Theorems 2.3 and 4.2. Due to its repeating, we omit the details here for cleanness of the paper. 5 Examples This section presents some examples on how to apply the SDP relaxation (2.8) and its dual (2.11) to solve polynomial optimization problems. The software GloptiPoly 3 [14] is used to solve (2.8) and (2.11). First, consider unconstrained polynomial optimization. Then the resulting SOS relaxation (2.11) is reduced to the gradient SOS relaxation in [18], which is a special case of (2.11). Example 5.1. Consider the optimization problem x 8 x R x x x 4 1x x 2 1x x 6 3 3x 2 1x 2 2x 2 3. This example was studied in [18]. Its global imum is zero. The SDP relaxation (2.8) and its dual (2.11) for this problem are equivalent to gradient SOS relaxations in [18] (there are no constraints). We apply (2.8) of order N = 4, and get a lower bound The resulting SDP (2.8) has a single block of size The imizer (0, 0) is extracted. In [18], it was shown that f(x) is not SOS modulo its gradient ideal I grad. But for every ɛ > 0, f(x)+ɛ s ɛ (x) modulo I grad for some SOS polynomial s ɛ (x), whose degree is independent of ɛ (see equation (10) of [18]). But its coefficients go to infinity as ɛ 0. This shows that the optimal value of (2.11) might not be achievable. On the other hand, if (1.1) has a imizer (say, x ) that is a KKT point (then x W ), then its dual problem (2.8) always achieves its optimal value f for N big enough, e.g., [x ] 2N is a imizer. Thus, for this example, the imum of the dual (2.8) is achievable. 19

20 Second, consider polynomial optimization having only equality constraints: f(x) s.t. h 1 (x) = = h m (x) = 0. (5.1) x R n When V (h) is nonsingular, its equivalent version (2.7) reduces to x R n f(x) s.t. h 1 (x) = = h m (x) = 0, I [n] m+1 sum(i)=l det I F (x) = 0, l = ( m+2 2 ),..., (n m 2 )(m + 1). (5.2) In the above sum(i) denotes the summation of the indices in I, F (x) = [ f(x) h(x) ], and det I F (x) denotes the maximal or of F (x) whose row indices are in I. When m n, there are no or equations in (5.2). Example 5.2. Consider the optimization problem x R 3 x x x x 2 1x 2 2x 2 3 (x 2 1(x x 4 3) + x 2 2(x x 4 1) + x 2 3(x x 4 2)) s.t. x 1 + x 2 + x 3 1 = 0. The objective is the Robinson polynomial, which is nonnegative everywhere but not SOS [22]. So the imum f = 0. We apply SDP relaxation (2.8) of order N = 4, and get a lower bound The resulting SDP (2.8) has a single block of size The imizer (1/3, 1/3, 1/3) is also extracted. Applying Lasserre s relaxation (1.2) of orders N = 3, 4, 5, 6, 7, we get lower bounds respectively , , , , We can see that (1.2) is weaker than (2.8). It is not clear whether the sequence of lower bounds returned by (1.2) converges to zero or not, because it is not guaranteed when the feasible set is noncompact, which is the case in this example. The objective f(x) here is not SOS modulo the constraint. Otherwise, suppose there exist polynomials σ(x) being SOS and φ(x) such that f(x) = σ(x) + φ(x)(x 1 + x 2 + x 3 1). In the above, replacing every x i by x i /(x 1 + x 2 + x 3 ) gives f(x) = (x 1 + x 2 + x 3 ) 6 σ(x/(x 1 + x 2 + x 3 )). So, there exist polynomials p 1,..., p k, q 1,..., q l such that f(x) = p p 2 k + q 2 1 (x 1 + x 2 + x 3 ) q 2 l (x 1 + x 2 + x 3 ) 2l. Since the objective f(x) does not have any pole, every q i must vanish on the plane x 1 + x 2 + x 3 = 0. Thus q i = (x 1 + x 2 + x 3 ) i w i for some polynomials w i. Hence, we get is SOS, which is a contradiction. f(x) = p p 2 r + w w 2 l 20

21 Third, consider polynomial optimization having only a single inequality constraint. f(x) s.t. g(x) 0. (5.3) x R n Its equivalent problem (2.7) becomes x R n s.t. f(x) g(x) f(x) ( f(x) x i i+j=l x i = 0, i = 1,..., n, g(x) 0, g(x) x j f(x) x j There are totally 3(n 1) new equality constraints. Example 5.3. Consider the optimization problem g(x) x i ) = 0, l = 3,..., 2n 1. x R 3 x 4 1x x 2 1x x 6 3 3x 2 1x 2 2x 2 3 s.t. x x x (5.4) The objective is the Motzkin polynomial which is nonnegative everywhere but not SOS [22]. So its imum f = 0. We apply SDP relaxation (2.8) of order N = 4, and get a lower bound The resulting SDP (2.8) has two blocks of sizes and The imizer (0, 0, 0) is also extracted. Now we apply Lasserre s relaxation (1.2). For orders N = 4, 5, 6, 7, 8, (1.2) returns the lower bounds respectively , , , , Clearly, (1.2) is weaker than (2.8). The sequence of lower bounds given by (1.2) converges to zero for this example, because the archimedean condition holds. The feasible set has nonempty interior. Hence, for every N, there is no duality gap between (1.2) and its dual, and (1.2) has an optimizer. The objective does not belong to the preordering generated by the ball condition. This fact was kindly pointed out to the author by Claus Scheiderer (implied by his proof of Prop. 6.1 in [23], since the objective is a nonnegative but non-sos form vanishing at origin). Therefore, for every N, the optimal value of (1.2) as well as its dual is strictly smaller than the imum f. Example 5.4. Consider Example 5.3 but the constraint is the outside of the unit ball: x R 3 x 4 1x x 2 1x x 6 3 3x 2 1x 2 2x 2 3 s.t. x x x Its imum is still 0. We apply SDP relaxation (2.8) of order N = 4, and get a lower bound (its sign is not correct due to numerical issues). The resulting SDP (2.8) has two blocks of sizes and Now we compare it with Lasserre s relaxation (1.2). When N = 4, (1.2) is not feasible. When N = 5, 6, 7, 8, (1.2) returns the following lower bounds respectively , , , So we can see (1.2) is much weaker than (2.8). Again, it is not clear whether the sequence of lower bounds given by (1.2) converges to zero or not, because it is only guaranteed when the feasible set is compact, which is not the case in this example. 21

22 Last, we show some general examples. Example 5.5. Consider the following polynomial optimization x R 2 x x2 2 s.t. x , x 2 1 Mx 1x 2 1 0, x Mx 1x This problem was studied in [7, 12]. Its global imum is M(M + M 2 + 4). Let M = 5 here. Applying (2.8) of order N = 4, we get a lower bound which equals the global imum, and four global imizers (±5.1926, ±1.0000). The resulting SDP (2.8) has eight blocks whose matrix lengths are 15, 10, 10, 10, 6, 6, 6, 3 respectively. However, if we apply the Lasserre s relaxation either (1.2) or (1.3), the best lower bound we would obtain is 2, no matter how big the relaxation order N is (see Example 4.5 of [7]). The reason for this is that its feasible set is noncompact, while Lasserre s relaxations (1.2) or (1.3) are only guaranteed to converge for compact sets. Example 5.6. Consider the polynomial optimization x R 3 x 4 1 x2 2 + x4 2 x2 3 + x4 3 x2 1 3x2 1 x2 2 x2 3 s.t. 1 x 2 1 0, 1 x2 2 0, 1 x The objective is a nonnegative form being non-sos [22, Sec. 4c]. Its imum f = 0. We apply SDP relaxation (2.8) of order N = 6, and get a lower bound A imizer (0, 0, 0) is also extracted. The resulting SDP (2.8) has eight blocks whose matrix lengths are 84, 56, 56, 56, 35, 35, 35, 20 respectively. Now we apply Lasserre s relaxation of type (1.3). Let be the optimal value of (1.3) for an order N, and fn mom be the optimal value of its dual optimization problem (an analogue of (2.8) by using moments and localizing matrices, see Lasserre [16]). Since the feasible set here has nonempty interior, the dual optimization problem of (1.3) has an interior point, so (1.3) always has an optimizer and there is no duality f smg N gap, i.e., f smg N = fn mom for every order N (cf. [16]). For N = 6, 7, 8, (1.3) returns the lower bounds f smg N respectively , , The sequence of lower bounds given by (1.3) converges to zero, since the feasible set is compact. However, the objective does not belong to the preordering generated by the constraints, which is implied by the proof of Prop. 6.1 of [23] (the objective is a nonnegative but non-sos form vanishing at origin). Because (1.3) has an optimizer for every order N, this implies that f smg N = fn mom < f = 0 for every N. Thus, the relaxation (1.3) and its dual could not be exact for any order N. 6 Some conclusions and discussions This paper proposes an exact type SDP relaxation (2.8) and its dual (2.11) for polynomial optimization (1.1) by using the Jacobian of its defining polynomials. Under some generic 22

23 conditions, we showed that the imum of (1.1) would be found by solving the SDP (2.8) for a finite relaxation order. A drawback of the proposed relaxation (2.8) and its dual (2.11) is that there are totally O(2 m2 n (m 1 + m 2 )) new constraints. This would make the computation very difficult to implement if either m 1 or m 2 is big. Thus, this method is more interesting theoretically. However, this paper discovers an important fact: it is possible to solve the polynomial optimization (1.1) exactly by a single SDP relaxation, which was not known in the prior existing literature. Currently, it is not clear for the author whether the number of newly introduced constraints would be significantly dropped while the exactness is still wanted. This is an interesting future research topic. On the other hand, the relaxation (2.8) is not too bad in applications. For problems that have only a few constraints, the method would also be efficiently implemented. For instance, in all the examples of Section 5, they are all solved successfully, and the advantages of the method over the prior existing ones are very clear. Thus, this method would also be computationally attractive in such applications. The results of this paper improve the earlier work [7, 18], where the exactness of gradient or KKT type SOS relaxations for a finite relaxation order is only proved when the gradient or KKT ideal is radical. There are other conditions like boundary hessian condition (BHC) guaranteeing this property, like in [15, 17]. In [17], Marshall showed that the gradient SOS relaxation is also exact for a finite relaxation order by assug BHC, in unconstrained optimization. In this paper, the exactness of (2.8) and (2.11) for a finite N is proved without the conditions like radicalness or BHC. The only assumptions required are nonsingularity of S and the imum f being achievable (the earlier related work also requires this), but they are generically true as shown by Theorem 2.5. We would like to point out that the KKT type SOS relaxation proposed in [7] using Lagrange multipliers is also exact for a finite order, no matter the KKT ideal is radical or not. This would be proved in a similar way as we did in Section 3. First, we can get a similar decomposition for the KKT variety like Lemma 3.2. Second, we can prove a similar representation for f(x) f + ɛ like in Theorem 3.4, with degree bounds independent of ɛ. Based on these two steps, we can similarly prove its exactness for a finite relaxation order. Since the proof is almost a repeating of Section 3, we omit it for cleanness of the paper. The proof of the exactness of (2.8) provides a representation of polynomials that are positive on S through using the preordering of S and the Jacobian of all the involved polynomials. A nice property of this representation is that the degrees of the representing polynomials are independent of the imum value. This is presented by Theorem 3.4. A similar representation result using the quadratic module of S is given by Theorem 4.3. An issue that is not addressed by the paper is that the feasible set S has singularities. If a global imizer x of (1.1) is singular on S, then the KKT condition might no longer hold, and x W. In this case, the original optimization (1.1) is not equivalent to (2.7), and the SDP relaxation (2.8) might not give a correct lower bound for f. It is not clear how to handle singularities generally in an efficient way. Another issue that is not addressed by the paper is the imum f of (1.1) is not achievable, which happens only if S is noncompact. For instance, when S = R 2, the polynomial x (x 1x 2 1) 2 has imum 0 but it is not achievable. If applying the relaxation (2.8) for this instance, we would not get a correct lower bound. Generally, this case will not happen, as shown by items (d), (e) of Theorem 2.5. In unconstrained optimization, when 23

24 f is not achievable, excellent approaches are proposed in [9, 10, 26]. It is an interesting future work to generalize them to constrained optimization. An important question is for what concrete relaxation order N the SDP relaxation (2.8) is exact for solving (1.1). No good estimates for N in Theorem 2.3 are available currently. Since the original problem (1.1) is NP-hard, any such estimates would be very bad if they exist. This is another interesting future work. Acknowledgement The author is grateful to Bernd Sturmfels for pointing out the references on imum defining equations for deterantal varieties. The author thanks Bill Helton for fruitful discussions. A Some basics in algebraic geometry and real algebra In this appendix, we give a short review on basic algebraic geometry and real algebra. More details would be found in the books [4, 11]. An ideal I of R[x] is a subset such that I R[x] I and I + I I. Given polynomials p 1,..., p m R[x], p 1,, p m denotes the smallest ideal containing every p i, which is the set p 1 R[x] + + p m R[x]. The ideals in C[x] are defined similarly. An algebraic variety is a subset of C n that are common complex zeros of polynomials in an ideal. Let I be an ideal of R[x]. Define V (I) = {x C n : p(x) = 0 V R (I) = {x R n : p(x) = 0 p I}, p I}. The V (I) is called an algebraic variety or just a variety, and V R (I) is called a real algebraic variety or just a real variety. Every subset T C n is contained in a variety in C n. The smallest one containing T is called the Zariski closure of T, and is denoted by Zar(T ). In the Zariski topology on C n, the varieties are called closed sets, and the complements of varieties are called open sets. A variety V is irreducible if there exist no proper subvarieties V 1, V 2 of V such that V = V 1 V 2. Every variety is a finite union of irreducible varieties. Theorem A.1 (Hilbert s Strong Nullstellensatz). Let I R[x] be an ideal. vanishes on V (I), then p k I for some integer k > 0. If p R[x] If an ideal I has empty variety V (I), then 1 I. This is precisely the Hilbert s weak Nullstellensatz. Theorem A.2 (Hilbert s Weak Nullstellensatz). Let I R[x] be an ideal. If V (I) =, then 1 I. Now we consider I to be an ideal generated by polynomials having real coefficients. Let T be a basic closed semialgebraic set. There is a certificate for V R (I) T =. This is the so-called Positivstellensatz. Theorem A.3 (Positivstellensatz, [27]). Let I R[x] be an ideal, and T = {x R n : g 1 (x) 0,..., g r (x) 0} be defined by real polynomials g i. If V R (I) T =, then there exist SOS polynomials σ ν such that 1 σ ν g ν1 1 gνr r mod I. ν {0,1} r 24