Polynomials non negative on a strip and some applications

HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS NGUYEN THI BICH LOAN POLYNOMIALS NON-NEGATIVE ON A STRIP AND SOME APPLICATIONS GRADUATION THESIS Hanoi, 2019 HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS NGUYEN THI BICH LOAN POLYNOMIALS NON-NEGATIVE ON A STRIP AND SOME APPLICATIONS GRADUATION THESIS Speciality: Analysis Supervisor: Ho Minh Toan Hanoi, 2019 Confirmation This dissertation has been written on the basis of my research works carried at Hanoi Pedagogical University 2, under the supervision of Ph.D Ho Minh Toan The presented results have never been published by others Acknowledgment Before presenting the main content of the thesis, I would like to express my gratitude to the mathematics teachers, Hanoi Pedagogical University 2, teachers in the calculus group as well as the teachers involved Teaching has dedicatedly conveyed valuable knowledge and created favorable conditions for her to successfully complete the course and thesis In particular, I would like to express my deep respect and gratitude to Ph.D Ho Minh Toan, who directly instructed, just told to help me so that I could complete this thesis Due to limited time, capacity and conditions, the discourse cannot avoid errors Therefore, I look forward to receiving valuable comments from teachers and friends Contents Table of Notations ii Introduction 1 Polynomials Non-negative on a Strip 1.1 Preliminary reductions 1.2 The additional ingredients 1.3 The main theorem 17 1.3.1 The idea of the proof 18 1.3.2 The end of the proof 19 Some Applications 24 Conclusions 27 References 28 i Table of Notations R the field of real numbers, R[x] the ring of polynomials of one variables with real coefficient, R[x1 , , xn ] the ring of polynomials of n variables with real coefficient, A2 the set of all (finite) sums of squares of elements of A, C the field of complex numbers, C((x)) the formal power series field over C ii Introduction Let f is a polynomials of two variables in R[x, y] Obviously, if f has a presentation of the form f (x, y) = σ(x, y) + τ (x, y)x(1 − x), (1) where σ(x, y) and τ (x, y) are sums of squares in R[x, y], then f ≥ for all (x, y) ∈ [0, 1]×R, because sums of squares is always ≥ and x(1−x) ≥ 0, ∀x ∈ [0, 1] It is natural to ask the following: QUESTION: Is the converse true, i.e., is it true that if f (x, y) ≥ on [0, 1] × R then f is expressible as Equation above? In 1988, Hilbert proved that if f is a polynomial non-negative of two variables has degree then f is always expressible as sums of squares Easy to see that if f is expressible as sums of squares then f has the presentation (1) above, where τ (x, y) = for all (x, y) In general case, Hilbert showed that there are polynomials f (x, y) ∈ R[x, y] (necessarily of degree ≥ 6) which are non-negative on all of R2 , but are not expressible as a sums of squares in R[x, y] The best-known example is the Motzkin’s polynomial: f (x, y) = − 3x2 y + x4 y + x2 y Definition 0.1 Let A be a commutative ring with A preordering of A is a subset T of A satisfying T + T ⊆ T, T T ⊆ T, and f ∈ T for all f ∈ A Denote Then A2 is the set consisting of all finite sums a2i , ∈ A A2 is the unique smallest preordering of A Let T is the preordering of A If T generated by finitely many elements g1 , , gs ∈ A then σ i g i | σi ∈ T = T (g1 , , gs ) = A2 i where g i := g1i1 gsis , i := (i1 , , is ) running through the set {0, 1}s For example, The preordering of A generated by g ∈ A is T = T (g) = σ0 + σ1 g | σ0 , σ1 ∈ A2 The preordering of A generated by g1 , g2 ∈ A is T = T (g1 , g2 ) = σ0 + σ1 g1 + σ2 g2 + σ3 g1 g2 | σi ∈ A2 , i = 0, In a ring R[x1 , , xn ], T is a finitely generated preordering We define KT := {a ∈ Rn | ∀g ∈ T, g(a) ≥ 0} If g1 , , gs are generators of T , then KT is the subset of Rn defined by the polynomial inequalities gi ≥ 0, i = 1, , s, i.e., KT = a ∈ Rn | gi (a) ≥ 0, i = 1, s Definition 0.2 T is said to be saturated if, for all f ∈ R[x1 , , xn ], f ≥ on KT we have f ∈ T In R[x, y], T is the preordering generated by x(1 − x), we have T = T (x(1 − x)) = {σ0 + σ1 x(1 − x) | σi are sums of squares in R[x, y]} The identities x = x2 + x(1 − x) and − x = (1 − x)2 + x(1 − x) imply that T = T (x(1 − x)) = T (x, − x) In this case, KT = (x, y) ∈ R2 | x(1 − x) ≥ = (x, y) ∈ R2 | ≤ x ≤ = [0, 1] × R By the result in previous, the preordering of R[x, y] generated by x(1 − x) is saturated We list here some well-known problems in Real Algebraic Geometry: In R[x1 , , xn ] Characterize these sets S = KT to have one of the following properties: (P1) Every polynomial f ≥ on S belongs to T (g1 , , gm ) (Nichtnegativstellensatz); or a weaker property: (P2) Every polynomial which is strictly positive on S is an element of T (g1 , , gm ) (Positivstellensatz) A closely related problem is the moment problem Given a linear functional L on R[X], assume that there exists a positive Borel measure µ with support in S such that L(f ) = f dµ, ∀f ∈ R[x1 , , xn ] S (2) Then, that f (x1 , , xn ) is ≥ on S implies L(f ) ≥ Conversely, given a linear functional L on R[x1 , , xn ], if L(f ) is non-negative for every polynomial f with f ≥ on S then L has the representation (2) above In case K is compact, Schmă udgen [3] has proved that any polynomial, which is positive on K, is in the preordering generated by the gi ’s If K is not compact, the above characterizations not hold in general and can depend on the choice of generators In fact, Scheiderer has shown that Schmă udgens Positivstellensăatz does not hold if K is not compact and dim K ≥ 3, or dim K = and K contains a 2-dimensional cone In this thesis, we give the main results from [1] and some applications The thesis consist two chapters Chapter ”Polynomials non-negative on a strip” presents the main results of the paper [1], in particular, the theorem of the representation of a polynomials of two variables which is non-negative with real coefficients Chapter ”Some applications”, the main content is some applications of the main theorem of [1] in polynomial inequalities and optimization for each ω ∈ µn , where the are complex numbers The zω are complex analytic functions of x n in a neighborhood of zero The coefficient of p are elementary symmetric function of the roots, so are complex analytic functions of x in some neighborhood of zero Denote by p¯ the polynomial in C((x))[y] obtained from p conjugating coefficients in the obvious way p¯ is an irreducible factor of f If p¯ = p, then z1 coincides with one of the ∞ i ω i x n zω := i=0 This implies, in turn, that there are (two) real half-branches of f coming from p Note: If n is odd the real half-branches (one with x > 0, one with x < 0) are given by y = i τ i x n , where τ is the unique element of µn satisfying τ = ω If n is even there are two elements τ, −τ ∈ µ2n satisfying τ = (−τ )2 = ω If τ, −τ ∈ µn the real half-branches (both with x > 0) i i τ i x n and y = are given by y = (−τ )i x n If τ, −τ ∈ / µn the real half-branches (both with x < 0) are given by y = y= i τ i (−x) n and i (−τ )i (−x) n Since p changes sign at any such half-branches , p must appear in f with even multiplicity in this case Thus f has a factorization of the form k f = a(x) pi pi i=1 where each pi is irreducible and a(x) is the leading coefficient 14 Then f = g¯ g where g = a(x)p1 pk Decomposing g as g = √ g1 + g2 −1, g1 , g2 ∈ R((x))[y], this yields f = g12 + g22 Lemma 1.6 [1, Lemma 4.4] Suppose f ∈ R[x, y] is non-negative on the strip [0, 1] × R, and the leading coefficient of f is positive on the interval [0, 1] Then (1) For each r ∈ (0, 1), there exists g1 , g2 polynomials in y with coefficients analytic functions in x in some neighborhood of r, such that f = g12 + g22 holds for x sufficiently close to r (2) There exists gij , i, j = 1, 2, polynomial in y with coefficients analytic functions in x in some neighborhood of 0, such that 2 gi1 f= gi2 x + i=1 i=1 holds for x sufficiently close to (3) There exists gij , i, j = 1, 2, polynomial in y with coefficients analytic functions in x in some neighborhood of 1,such that 2 gi1 f= gi2 (1 − x) + i=1 i=1 holds for x sufficiently close to Proof For (1), apply Lemma 1.5, viewing f as a polynomial in x − r and y 15 For (2), apply Lemma 1.5, viewing f as a polynomial in √ x and y, to obtain f = g12 + g22 with gi a polynomial in y with coefficients analytic √ in x, i = 1, Decomposing each of the coefficient , using √ k a2l+1 xl x ak x = a2l xl + k l l √ yields gi = gi1 + gi2 x, where the gij are polynomials in y with coefficient analytic functions in x near x = Expanding gi2 , i = 1, then yields 2 gi1 f= + i=1 2 gi2 x i=1 so 2 gi1 f= gi2 x + i=1 √ gi1 gi2 x, i=1 and +2 i=1 √ gi1 gi2 x = i=1 The proof of (3) is similar to the proof of (2) Proposition 1.2 Suppose φ, ψ : [0, 1] → R are continuous functions, φ(x) ≤ ψ(x) for all x ∈ [0, 1], and φ(x) < ψ(x) for all but finitely many x ∈ [0, 1] If φ and ψ are analytic at each point a ∈ [0, 1] where φ(a) = ψ(a) then there exists a polynomial p(x) ∈ R[x] such that φ(x) ≤ p(x) ≤ ψ(x) holds for all x ∈ [0, 1] Proof Induct on the number of points a ∈ [0, 1] satisfying φ(a) = ψ(a) If there are no such points, existence of p(x) follows from the Weierstrass Approximation Theorem 16 Suppose a ∈ [0, 1] satisfying φ(a) = ψ(a) Let k be the vanishing order of ψ − φ at a If a ∈ (0, 1) then k is even In this case, φ(x) = f (x) + (x − a)k φ1 (x) ψ(x) = f (x) + (x − a)k ψ1 (x) where f (x) ∈ R[x], φ1 (x), ψ1 (x) are analytic at a, and φ1 (x) < ψ1 (x) Extend φ1 , ψ1 to continuous functions φ1 , ψ1 : [0, 1] → R by defining φ1 (x) = ψ(x) − f (x) φ(x) − f (x) , ψ (x) = for x = a (x − a)k (x − a)k Then φ1 (x) ≤ ψ1 (x) for all x ∈ [0, 1], and ∀b ∈ [0, 1], φ1 (b) = ψ1 (b) if and only if φ(b) = ψ(b) and b = a By induction, we have p1 (x) ∈ R[x] such that φ1 (x) ≤ p1 (x) ≤ ψ1 (x) on [0, 1] Take p(x) = f (x) + (x − a)k p1 (x), so φ(x) ≤ p(x) ≤ ψ(x) The case where a = and the case where a = are dealt with in a similar fashion (Note: xk > and (1−x)k > for all x ∈ [0, 1], k > 0) 1.3 The main theorem Theorem 1.1 Suppose f (x, y) ∈ R[x, y] is non-negative on the strip [0, 1] × R Then f (x, y) is expressible as f (x, y) = σ(x, y) + τ (x, y)x(1 − x), where σ(x, y) and τ (x, y) are sums of squares in R[x, y] 17 1.3.1 The idea of the proof Consider the case where the polynomial f (x, y) = 2d i i=0 (x)y is positive on [0, 1] × R and a2d (x) > on [0, 1] Let 2d (x)y i z 2d−i , F (x, y, z) := i=0 then y F (x, y, z) = z 2d f (x, ) if z = z Therefore F is positive for ≤ x ≤ 1, (y, z) = (0, 0), so it achieves a positive minimum on the compact set [0, 1] × S1 := (x, y, z)|0 ≤ x ≤ 1, y + z Then on the strip [0, 1] × R, we have F x, y + y2 , (1 + y )d + y2 2d d = (1 + y ) (x) i=0 2d d = (1 + y ) i=0 yi ( + y )i ( + y )2d−i yi (x) (1 + y )d 2d (x)y i = F (x, y, 1), = i=0 so f (x, y) = F (x, y, 1) = F x, y + y2 , 1 + y2 Using this, f (x, y) has the required presentation 18 (1 + y )d ≥ (1 + y )d In the general case, one cannot possibly have such an The idea is to replace by a polynomial (x) Specifically, we look for a polynomial (x) ∈ R[x] such that f (x, y) ≥ (x)(1 + y )d holds on the strip, (x) ≥ on [0, 1] and ∀x ∈ [0, 1], (x) = if and only if f (x, y) = for some y ∈ R It always possible to find such a polynomial (x), assuming that a2d (x) > on [0, 1] and f (x, y) has only finitely many zeros in the strip So, we can show that f (x, y) has the required presentation, with 1.3.2 replaced by (x) The end of the proof Proof Let 2d (x)y i , (x) ∈ R[x], d ≥ f (x, y) = i=0 By Lemma 1.1 and Lemma 1.2, we can assume a2d (x) > on the interval [0, 1] and f (x, y) has only finitely many zeros in [0, 1] × R By Lemma 1.4, ∃ a polynomial (x) ∈ R[x] such that f (x, y) ≥ (x)(1 + y )d on the strip, (x) ≥ on [0, 1] and (x) = if and only if ∃y ∈ R with f (x, y) = Let f1 (x, y) := f (x, y) − (x)(1 + y )d Then f1 is ≥ on the strip Replacing (x) by (x) N , if necessary, we can assume f1 has degree 2d (as 19 the polynomials in y) and the leading coefficient of f1 is positive on [0, 1] By Lemma 1.6, for each r ∈ [0, 1], there exists an open neighborhood U (r) of r in R such that f1 decomposes as 2 f1 = g0j (r) + j=1 2 g2j (r)2 (1 − x) g1j (r) x + j=1 j=1 on U (r) × R, where the gij (r) are polynomials in y (of degree ≤ d) whose coefficients are analytic functions of x, for x ∈ U (r) (Applying Lemma 1.6, we can choose the gij (r) so that g2j (r) = 0, j = 1, if r = 0; g1j (r) = 0, j = 1, if r = and g1j (r) = g2j (r) = 0, j = 1, if < r < 1.) By compactness of [0, 1], finitely many of the U (x) cover [0, 1], say as U (r1 ), , U (rl ) cover [0, 1] Choose a continuous partition of unity = v1 + + vl on [0, 1], with ≤ vk ≤ on [0, 1] and supp(vk ) ⊆ U (rk ) for k = 1, , l, having the additional property that, for each root r of (x) in [0, 1], there is just one k such that vk (x) = close to r (so vk (x) = for x close to r) One way to ensure the last property is to shrink the covering sets U (rk ) ahead of time so that each root r of (x) in [0, 1] lies in some unique U (rk ) Then f1 decomposes as l f1 = l vk f1 = k=1 φ20jk k=1 + j=1 φ21jk x j=1 φ22jk (1 − x) + j=1 on [0, 1] × R, where φijk denotes the polynomial of degree ≤ d in y whose coefficients are the functions from [0, 1] to R obtained by extending the √ corresponding coefficient of vk gij (rk ) by zero off U (rk ) 20 The coefficients of the φijk are continuous on [0, 1] and analytic at each of the roots of (x) in [0, 1] (since vk is constantly os in a neighborhood of each of these roots) By Proposition 1.2, for each real N > 0, and each triple i, j, k, ∃ a polynomial hijk of degree ≥ d in y with coefficients in R[x], such that, for each coefficient u of φijk , the corresponding coefficient w of hijk satisfies u(x) − (x) (x) ≤ w(x) ≤ u(x) + , N N for each x ∈ [0, 1] At this point, approximating the coefficients of the φijk closely by polynomials (by taking N sufficiently large), to obtain polynomials hijk of degree ≤ d in y with coefficients in R[x] such that f1 (x, y) l 2 2 h0jk (x, y) + = j=1 k=1 h2jk (x, y)2 (1 − x) h1jk (x, y) x + j=1 j=1 2d bi (x)y i , + i=0 where bi (x) ∈ R[x], |bi (x)| ≤ (x) on [0, 1], i = 0, , 2d Combining this with f (x, y) = f1 (x, y) + (x)(1 + y )d yields f (x, y) = s1 (x, y) + s2 (x, y) + s3 (x, y), where l s1 (x, y) := ( k=1 j=1 2 2 h0jk (x, y) + h2jk (x, y)2 (1 − x)), h1jk (x, y) x + j=1 j=1 21 s2 (x, y) := (x)(2 + y + 3y + y + 3y + + y 2d−1 + 2y 2d ) + 2d bi (x)y i , i=0 s3 (x, y) := (x)[(1 + y )d − (2 + y + 3y + y + 3y + + y 2d−1 + 2y 2d )] Let T denote the preordering of R[x, y] generated by x(1 − x) As pointed out earlier x, − x ∈ T Clearly, s1 (x, y) ∈ T We need to show s2 (x, y) ∈ T and s3 (x, y) ∈ T Since |bi (x)| ≤ (x) on [0, 1], then (x)±bi (x) ∈ T (by [2, pages 37,38]), for = 0, , 2d This yields (x)y i + bi (x)y i ∈ T, for i even (1.1) For i is odd, say i = 2m + 1, we have y i = y 2m+1 = y 2m [(y + 1)2 − y − 1], 2 (x)y 2m (y + 1)2 + bi (x)y 2m (y + 1)2 ∈ T, (x)y 2m y − bi (x)y 2m y ∈ T, (x)y 2m − bi (x)y 2m ∈ T So, we obtain (x)(y i+1 + y i + y i−1 ) + bi (x)y i ∈ T, for i odd (1.2) Adding together the various terms of type (1.1) and (1.2), i = 0, , 2d, we can see s2 (x, y) ∈ T The fact that s3 (x, y) ∈ T follows from the 22 identity (1 + y )d − (2 + y + 3y + y + 3y + + y 2d−1 + 2y 2d ) = (1 + y )d + (1 + y + + y 2d−2 )(1 − y)2 − (y + y + + y 2d−2 ) − (1 + y 2d ) d−1 d 2i 2d−2 = (1 + y + + y )(1 − y) + − y i i=1 Therefore, f (x, y) = s1 (x, y) + s2 (x, y) + s3 (x, y) ∈ T 23 Chapter Some Applications We will presents some applications of the main theorem in Chapter in polynomial inequalities and optimization Remark 2.1 By Theorem 1.1, we can write: f ≥ on [0, 1] × R ⇔ f (x, y) = σ(x, y) + τ (x, y)x(1 − x), for σ(x, y) and τ (x, y) are sums of squares in R[x, y] Corollary 2.1 Suppose f is the polynomial in R[x, y] The following are equivalent: (1) f ≥ on [a, b] × R (2) f has a presentation of the form f (x, y) = σ(x, y) + τ (x, y)(x − a)(b − x), where σ(x, y) and τ (x, y) are sums of squares in R[x, y] Proof Making a change of variables u = (b − a)−1 (x − a), v = y, we get back the usual strip [0, 1] × R 24 Example 2.1 Prove that f (x, y) = + x + x3 + xy − x4 − x2 y is non-negative on [0, 1] × R Find the expression f as f (x, y) = σ(x, y) + τ (x, y)x(1 − x), where σ(x, y) and τ (x, y) are sums of squares in R[x, y] For x ∈ [0, 1], we have x3 − x4 = x3 (1 − x) ≥ and xy − x2 y = xy (1 − x) ≥ So f ≥ on [0, 1] × R Then f is expressible as f (x, y) = + x + x3 + xy − x4 − x2 y = + x(1 + x2 − x3 + y − xy ) = + x[(1 − x) + x2 (1 − x) + y (1 − x) + x] = + x2 + (1 + x2 + y )x(1 − x) See + x2 and + x2 + y are sums of squares in R[x, y] Example 2.2 Find the minimum of f (x, y) = 2+3y +2xy −x2 y +x2 y on [−1, 3] × R if it exists We have f (x, y) = + 3y + 2xy − x2 y + x2 y = + x2 y + y (3 + 2x − x2 ) = + x2 y + y (x + 1)(3 − x) 25 + x2 y ≥ and y (x + 1)(3 − x) ≥ on [−1, 3] × R Therefore, f ≥ on [−1, 3] × R and f (x, 0) = Thus −1≤x≤3,y∈R f (x, y) = Example 2.3 Let f (u, v) = v −v −2uv +v +2uv +u2 v −v −u2 +u Prove that f (u, v) ≥ 0, ∀(u, v) ∈ D := (u, v) ∈ R2 | v ≤ u ≤ v + Making a change of variables x = u − v , y = v, we get back the usual strip [0, 1] × R, and f (x, y) is expression as f (x, y) = x2 y + y + x(1 − x) Thus f (u, v) ≥ 0, ∀(u, v) ∈ D 26 Conclusions The thesis present the main results: Theorem: f (x, y) ≥ on [0, 1]×R then f (x, y) = σ(x, y)+τ (x, y)x(1−x), for σ(x, y) and τ (x, y) are sums of squares in R[x, y] The main content is some applications of the main theorem in polynomial inequalities and optimization 27 Bibliography [1] M.Marshall, ”Polynomials non-negative on a strip”, Proceedings of the American Mathematical Society, 138 (2010), no 5, 1559-1567 [2] M.Marshall, Positive polynomials and sum of squares, Mathematical Surveys and Monographs, 146, American Mathematical Society, 2008 [3] K Schmă udgen, The K-moment problem for compact semi-algebraic sets, Math Ann 289 (1991), 203–206 28 ...HANOI PEDAGOGICAL UNIVERSITY DEPARTMENT OF MATHEMATICS NGUYEN THI BICH LOAN POLYNOMIALS NON-NEGATIVE ON A STRIP AND SOME APPLICATIONS GRADUATION THESIS Speciality: Analysis Supervisor:... representation of a polynomials of two variables which is non-negative with real coefficients Chapter Some applications , the main content is some applications of the main theorem of [1] in polynomial... give the main results from [1] and some applications The thesis consist two chapters Chapter Polynomials non-negative on a strip presents the main results of the paper [1], in particular, the