Polynomials non negative on a strip and some applications

HANOI PEDAGOGICAL UNIVERSITY 2 DEPARTMENT OF MATHEMATICSNGUYEN THI BICH LOAN POLYNOMIALS NON-NEGATIVE ON A STRIP AND SOME APPLICATIONS GRADUATION THESIS Hanoi, 2019... HANOI PEDAGOGICAL

HANOI PEDAGOGICAL UNIVERSITY 2 DEPARTMENT OF MATHEMATICS

NGUYEN THI BICH LOAN

POLYNOMIALS NON-NEGATIVE ON

A STRIP AND SOME APPLICATIONS

GRADUATION THESIS

Hanoi, 2019

HANOI PEDAGOGICAL UNIVERSITY 2 DEPARTMENT OF MATHEMATICS

NGUYEN THI BICH LOAN

POLYNOMIALS NON-NEGATIVE ON

A STRIP AND SOME APPLICATIONS

GRADUATION THESISSpeciality: Analysis

Supervisor: Ho Minh Toan

Hanoi, 2019

Before presenting the main content of the thesis, I would like toexpress my gratitude to the mathematics teachers, Hanoi PedagogicalUniversity 2, teachers in the calculus group as well as the teachers in-volved Teaching has dedicatedly conveyed valuable knowledge and cre-ated favorable conditions for her to successfully complete the course andthesis

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 sothat I could complete this thesis

Due to limited time, capacity and conditions, the discourse cannotavoid errors Therefore, I look forward to receiving valuable commentsfrom teachers and friends

1.1 Preliminary reductions 5

1.2 The additional ingredients 9

1.3 The main theorem 17

1.3.1 The idea of the proof 18

1.3.2 The end of the proof 19

Table of Notations

R[x] the ring of polynomials of one variables with real

co-efficient,R[x1, , xn] the ring of polynomials of n variables with real coeffi-

cient,

P A2 the set of all (finite) sums of squares of elements of A,

C((x)) the formal power series field over C

∀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) ≥ 0 on[0, 1] × R then f is expressible as Equation 1 above?

In 1988, Hilbert proved that if f is a polynomial non-negative of twovariables has degree 4 then f is always expressible as sums of squares.Easy to see that if f is expressible as sums of squares then f has thepresentation (1) above, where τ (x, y) = 0 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) = 1 − 3x2y2 + x4y2 + x2y4

Definition 0.1 Let A be a commutative ring with 1 A preordering of A

is a subset T of A satisfying

T + T ⊆ T, T T ⊆ T, and f2 ∈ T for all f ∈ A

Denote P A2 is the set consisting of all finite sums P a2

i, ai ∈ A.Then P 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

T = T (g1, , gs) =

(X

T = T (g1, g2) =

n

σ0 + σ1g1 + σ2g2 + σ3g1g2 | σi ∈ XA2, i = 0, 3

o

In a ring R[x1, , xn], T is a finitely generated preordering Wedefine

KT := {a ∈ Rn | ∀g ∈ T, g(a) ≥ 0}

If g1, , gs are generators of T , then KT is the subset of Rn defined bythe 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 ≥ 0 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 + σ1x(1 − x) | σi are sums of squares in R[x, y]} The identities x = x2 + x(1 − x) and 1 − x = (1 − x)2 + x(1 − x) implythat T = T (x(1 − x)) = T (x, 1 − x) In this case,

We list here some well-known problems in Real Algebraic Geometry:

In R[x1, , xn] Characterize these sets S = KT to have one of thefollowing properties:

(P1) Every polynomial f ≥ 0 on S belongs to T (g1, , gm) tivstellensatz); or a weaker property:

(Nichtnega-(P2) Every polynomial which is strictly positive on S is an element of

Then, that f (x1, , xn) is ≥ 0 on S implies L(f ) ≥ 0 Conversely, given

a linear functional L on R[x1, , xn], if L(f ) is non-negative for everypolynomial f with f ≥ 0 on S then L has the representation (2) above

In case K is compact, Schm¨udgen [3] has proved that any nomial, which is positive on K, is in the preordering generated by the

poly-gi’s

If K is not compact, the above characterizations do not hold ingeneral and can depend on the choice of generators In fact, Scheidererhas shown that Schm¨udgen’s Positivstellens¨atz does not hold if K is notcompact and dim K ≥ 3, or dim K = 2 and K contains a 2-dimensionalcone

In this thesis, we give the main results from [1] and some tions The thesis consist two chapters

applica-Chapter 1 ”Polynomials non-negative on a strip” presents the main sults of the paper [1], in particular, the theorem of the representation of apolynomials of two variables which is non-negative with real coefficients.Chapter 2 ”Some applications”, the main content is some applications ofthe main theorem of [1] in polynomial inequalities and optimization

Suppose that f ∈ R[x, y], f ≥ 0 on the strip [0, 1] × R We need toshow f has a presentation f = σ + τ x(1 − x), σ, τ are sums of squares inR[x, y].

Proposition 1.1 [2, Proposition 1.2.1] Suppose f is a non-zeros

Lemma 1.1 [1, Lemma 2.1] We may assume a2d > 0 on [0, 1].

Proof Factor a2d as a2d = ¯a˜a, where ¯a, ˜a ∈ R[x], ¯a is > 0 on [0, 1]

and ˜a is ±1 times a product of linear factors x − r, r ∈ [0, 1] Then

˜2d−1f = ˜a2d−1(a2dy2d + a2d−1y2d−1 + + a1 + a0)

= ¯a(˜ay)2d + a2d−1(˜ay)2d−1 + + a1˜2d−1y + a0˜2d−1

Let g := ¯ay2d + a2d−1y2d−1 + + a1˜2d−2y + a0˜2d−1 Using the fact that

˜

a ≥ 0 on [0, 1] and the set of points (r, s) in the strip satisfying ˜a(r) 6= 0

is dense in the strip, one have g ≥ 0 on the strip [0, 1] × R

If we are able to show that g = σ + τ x(1 − x), σ, τ sums of squares

in R[x, y], then

˜

a(x)2d−1f (x, y) = σ(x, ˜a(x)y) + τ (x, ˜a(x)y)x(1 − x)

Thus we are reduced to showing that if b(x)f (x, y) has a presentation

b(x)f (x, y) = σ(x, y) + τ (x, y)x(1 − x),for some sums of squares σ(x, y),τ (x, y), where b(x) ≥ 0 on the interval[0, 1] and b(x) is ±1 times a product of linear factors x − r, r ∈ [0, 1],then f (x, y) also has such a presentation

The proof is by induction on the degree of b(x) Suppose x − r is afactor of b(x), 0 ≤ r ≤ 1

First, suppose 0 < r < 1 Then b(x) = ¯b(x)(x − r)2, also σ(x, y) and

τ (x, y) vanish at x = r, so

σ(x, y) = ¯σ(x, y)(x − r)2 and τ (x, y) = ¯τ (x, y)(x − r)2,

where ¯σ(x, y),¯τ (x, y) sums of squares in R[x, y],

and

¯b(x)f (x, y) = ¯σ(x, y) + ¯τ (x, y)x(1 − x)

If r = 0, then b(x) = ¯b(x)x and σ(x, y) = ¯σ(x, y)x, ¯σ(x, y) a sum of

squares in R[x, y], and

¯b(x)f (x, y) = ¯σ(x, y)x + τ (x, y)(1 − x).

Using x = x2 + x(1 − x) and 1 − x = (1 − x)2 + x(1 − x), we have

¯b(x)f (x, y) = σ0(x, y) + τ0(x, y)x(1 − x),where σ0(x, y) and τ0(x, y) are sums of squares in R[x, y]

Since x and 1 − x do not divide f (because they do not divide a2d),

f has only finitely many zeros on the boundary of the strip If someirreducible factor p of f has infinitely many zeros in the interior of thestrip then p has nonsingular zero (a, b) in the interior of the strip which

is not a zero of any other irreducible factor of f [2, pages 130,131]

Then p changes sign at (a, b) but each other irreducible factor of

f has constant sign in a neighborhood of (a, b), this is contradicts with

f ≥ 0 on the strip

It follows that f has only finitely many zeros in the strip

1.2 The additional ingredients

We establish the additional results

Lemma 1.3 [1, Proposition 4.1] Suppose f ∈ R[x, y] is non-negative on

a strip of the form [0, ) × R,  > 0, f has only finitely many zeros inthis strip, and the leading coefficient of f is positive on the interval [0, ).Then there exists a real constant C > 0, an even interval m ≥ 0, and

a real number δ, 0 < δ ≤ , such that f (x, y) ≥ Cxm holds on the strip[0, ) × R If f has no real zeros on the y-axix, we may take m = 0

We denote by k((x)) the formal power series field over a field k, i.e.,the field of fractions of the formal power series ring k[[x]]

Proof The leading coefficient of f is positive at zero; in particular, it is

is another root of f, and ¯z 6= z

(Note: If all the ai were real then the equation y := P∞

i=0aixni woulddefine a real half-branch in the zero set of f , for x close to zero, x > 0,

contradicting our assumption that f has only finitely many zeros in thestrip [0, ) × R.)

n be the order of z2 at zero, i.e., the least ni such that

ci 6= 0 For any real x > 0 close to zero, and any y ∈ R,

x k is a real number close to the non-zero real constant ci0

This implies there exists a real constant C > 0 such that

(y − z)(y − ¯z)

x2k > Cfor all real x > 0 sufficiently close to zero and all real y

Note: If f has no real zeros on the y-axis then a0 is not real, i.e, c0 6= 0,

where a(x) is the leading coefficient, this yields rationals ki ≥ 0 and realconstants Ci > 0 and

f (x, y)

Qd i−1x2ki = a(x)

for all real y and all real x > 0 sufficiently close to zero

Finally, f (x, y) ≥ Cxm for any real x ≥ 0 sufficiently close to zeroand any y ∈ R, where C is Qd

i=1Ci times the minimum value of a(x) on[0, 2] and m is the least even integer ≥ 2Pd

i=1ki

Lemma 1.4 [1, Lemma 4.2] Suppose f (x, y) = P2d

i=0ai(x)yi is negative on the strip [0, 1] × R, f (x, y) has only finitely many zeros inthe strip, and a2d(x) is positive on the interval [0,1] Then there exists apolynomial (x) ∈ R[x], (x) ≥ 0 on [0, 1], such that f (x, y) ≥ (x)(1+y2)dholds on the strip and, for each x ∈ [0, 1], (x) = 0 if and only if thereexists y ∈ R such that f (x, y) = 0

non-Proof For each r ∈ [0, 1], by Lemma 1.3, applied to the new variable

t = x − r and t = r − x, or just to t = x − r if r = 0, just to t = r − x if

r = 1, we have a real constant C > 0 and an even integer m ≥ 0 (with

m = 0 if f has no real zeros on the line x = r), such that

f (x, y) ≥ C(x − r)mholds for all (x, y) in the strip, x sufficient close to r

Since [0, 1] is compact, by compactness of the interval, there arefinitely many 0 ≤ r1 < < rk ≤ 1 and finitely many positive constant

Ci and even integer mi ≥ 0 (with mi = 0 if f has no real zeros on theline x = ri) such that, for each (x, y) in the strip,

f (x, y) ≥ Ci(x − ri)mi,for some i

We may assume each Ci is ≤ 1 Then

for each x ∈ [0, 1] and each i

Thus f (x, y) ≥ 1(x) holds on the strip, where

a real constant M > 0 such that f (x, y) > 0 if |y| ≥ M , 0 ≤ x ≤ 1.Arguing with the form

(1 + M2)d ≥ 1(x)

(1 + M2)d

If |y| ≥ M ,

f (x, y)(1 + y2)d ≥ C ≥ C

D1(x)where D := max {1(x) | x ∈ [0, 1]}

So, in any case, we have

f (x, y)(1 + y2)d ≥ (x)holds on the strip, where

(x) := min



1(1 + M2)d, C

Proof Let p be an irreducible factor of f in C((x))[y] which is monic

By Puiseux’s Theorem, p factors in C((x1n))[y], where n is degree of p, as

ω∈µ n

(y − zω)where µn denotes the group of complex n-th roots of 1, and

for each ω ∈ µn, where the ai are complex numbers.

The zω are complex analytic functions of x1n in a neighborhood ofzero 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 conjugatingcoefficients in the obvious way ¯p is an irreducible factor of f

If ¯p = p, then z1 coincides with one of the

y = P ai(−τ )i(−x)ni

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 theform

Then f = g¯g where g = pa(x)p1 pk Decomposing g as g =

g1 + g2

√

−1, g1, g2 ∈ R((x))[y], this yields f = g2

1 + g22.Lemma 1.6 [1, Lemma 4.4] Suppose f ∈ R[x, y] is non-negative on thestrip [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 cients analytic functions in x in some neighborhood of r, such that

coeffi-f = g21 + g22holds for x sufficiently close to r

(2) There exists gij, i, j = 1, 2, polynomial in y with coefficients analyticfunctions in x in some neighborhood of 0, such that

(3) There exists gij, i, j = 1, 2, polynomial in y with coefficients analyticfunctions in x in some neighborhood of 1,such that

Proof For (1), apply Lemma 1.5, viewing f as a polynomial in x − r andy

For (2), apply Lemma 1.5, viewing f as a polynomial in √

x, where the gij are polynomials in y with coefficientanalytic functions in x near x = 0

Expanding g2i, i = 1, 2 then yields

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) thenthere exists a polynomial p(x) ∈ R[x] such that φ(x) ≤ p(x) ≤ ψ(x) holdsfor 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 WeierstrassApproximation Theorem

Suppose a ∈ [0, 1] satisfying φ(a) = ψ(a) Let k be the vanishingorder of ψ − φ at a.

If a ∈ (0, 1) then k is even In this case,

φ(x) = f (x) + (x − a)kφ1(x)ψ(x) = f (x) + (x − a)kψ1(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) = φ(x) − f (x)

(x − a)k , ψ1(x) = ψ(x) − f (x)

(x − a)k for x 6= aThen φ1(x) ≤ ψ1(x) for all x ∈ [0, 1], and ∀b ∈ [0, 1], φ1(b) = ψ1(b) if andonly if φ(b) = ψ(b) and b 6= 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)kp1(x), so φ(x) ≤ p(x) ≤ ψ(x)

The case where a = 0 and the case where a = 1 are dealt with in asimilar fashion (Note: xk > 0 and (1−x)k > 0 for all x ∈ [0, 1], k > 0)

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]

1.3.1 The idea of the proof

Consider the case where the polynomial f (x, y) = P2d

i=0ai(x)yi ispositive on [0, 1] × R and a2d(x) > 0 on [0, 1]

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 + y2)dholds on the strip, (x) ≥ 0 on [0, 1] and ∀x ∈ [0, 1], (x) = 0 if andonly if f (x, y) = 0 for some y ∈ R It always possible to find such apolynomial (x), assuming that a2d(x) > 0 on [0, 1] and f (x, y) has onlyfinitely many zeros in the strip So, we can show that f (x, y) has therequired presentation, with  replaced by (x)

1.3.2 The end of the proof

By Lemma 1.4, ∃ a polynomial (x) ∈ R[x] such that

f (x, y) ≥ (x)(1 + y2)d on the strip, (x) ≥ 0 on [0, 1] and (x) = 0 if andonly if ∃y ∈ R with f (x, y) = 0

Let

f1(x, y) := f (x, y) − (x)(1 + y2)d.Then f1 is ≥ 0 on the strip

Replacing (x) by (x)N , if necessary, we can assume f1 has degree 2d (as

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 hood U (r) of r in R such that f1 decomposes as

0, j = 1, 2 if r = 1 and g1j(r) = g2j(r) = 0, j = 1, 2 if 0 < 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

1 = v1 + + vl on [0, 1], with 0 ≤ vk ≤ 1 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) 6= 0 close to r (so vk(x) = 1for x close to r) One way to ensure the last property is to shrink thecovering sets U (rk) ahead of time so that each root r of (x) in [0, 1] lies

vkgij(rk) by zero off U (rk)