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PRESERVING POSITIVE POLYNOMIALS AND BEYOND JULIUSBORCEA,ALEXANDERGUTERMAN,ANDBORISSHAPIRO Abstract. Following the classical approach of Po´lya-Schur theory [14] we initiateinthispaperthestudyoflinearoperatorsactingonR[x]andpreserving eitherthesetofpositiveunivariatepolynomialsorsimilarsetsofnon-negative 8 andellipticpolynomials. 0 0 2 n a Contents J 2 1. Introduction and main results 1 2 2. Some preliminaries on the considered classes of preservers 3 3. The case of diagonal transformations 6 ] 3.1. Known correct results 6 A 3.2. Known wrong results 8 C 4. Linear ordinary differential operators of finite order 9 . h 5. Linear ordinary differential operators with constant coefficients 12 t References 14 a m [ 2 1. Introduction and main results v 9 Let R[x] denote the ring of univariate polynomials with real coefficients and 4 denotebyR [x]itslinearsubspaceconsistingofallpolynomialsofdegreelessthan n 7 or equal to n. 1 In what follows we will discuss the following five important types of univariate . 1 polynomials: 0 8 Definition 1.1. A polynomial p(x)∈R[x] is called 0 - hyperbolic, if all its roots are real; : v - elliptic, if it does not have reals roots; Xi - positive, if p(x)>0 for all x∈R; - non-negative, if p(x)≥0 for all x∈R; r a - a sum of squares, if there is a positive integer k and there are polynomials p (x),...,p (x)∈R[x] such that p(x)=p2(x)+...+p2(x). 1 k 1 k Note that the term “elliptic” is sometimes used to define other types of polynomials,see,e.g.,[10,12]. Thesetofnon-negativepolynomialsisclassicallycompared with the set of sums of squares which is a subset of the latter. Moreover, a well- knownresultclaims that in the univariate case these twoclasses coincide,see, e.g., [17, p. 132]. Proposition 1.2. A polynomial p(x) ∈ R[x] is non-negative if and only if there exist p (x),p (x)∈R[x] such that p(x)=p2(x)+p2(x). 1 2 1 2 2000 Mathematics Subject Classification. Primary: 12D15,15A04;Secondary: 12D10. Key words and phrases. Positive, non-negative and elliptic polynomials, linear preservers, Po´lya-Schurtheory. 1 2 J.BORCEA,A.GUTERMAN,ANDB.SHAPIRO Remark 1.3. Note that the situation is quite different for polynomials in several variables. In particular, even in 2 variables not all non-negative polynomials can be represented as sums of squares. One of the simplest examples of this kind is the polynomial p(x,y) = x2y2(x2+y2−3)+1 which is non-negative but can not be represented as the sum of squares, see [11] for details. In general, this topic is related to the Hilbert 17-th problem, see [13]. Definition 1.4. LetV denoteeitherR [x]orR[x]. We saythatamapΦ:V →V n preserves acertainsetM ⊂V ifforanyp(x)∈M itsimageΦ(p(x))belongstoM. InthispaperwestudylinearoperatorsonR[x]orR [x]whichpreserveoneofthe n classes of polynomials introduced above. Namely, we call a linear operator acting on R[x] or R [x] a hyperbolicity-, ellipticity-, positivity-, non-negativity-preserver n if it preserves the sets of hyperbolic, elliptic, positive, non-negative polynomials respectively. The classical case of (linear) hyperbolicity-preserverswhich are diagonal in the monomial basis of R[x] was thoroughly studied about a century ago by Po´lyaandSchur[14]. Its substantialgeneralizationsbothintheunivariateandthe multivariate cases can be found in [1, 2, 3]. Followingtheset-upof[14]weconcentratebelowontheremainingthreeclassesof preservers(restrictingourattentionmainlytolinearordinarydifferentialoperators of finite order, see Remark 1.6). In short, it turns out that there are much fewer such linear operators than those preserving hyperbolicity. More precisely, our two main results are as follows. Theorem A. Let U : R[x] → R[x] be a linear ordinary differential operator of Q order k ≥1 with polynomial coefficients Q=(q (x),q (x),...,q (x)), q (x)∈R[x], 0 1 k i i=0,...,k, q (x)6≡0, i.e., k d d2 dk U =q (x)+q (x) +q (x) +...+q (x) . (1) Q 0 1 dx 2 dx2 k dxk Then for any coefficient sequence Q the operator U does not preserve the set of Q non-negative (resp., positive or elliptic) polynomials of degree 2k. Corollary. There are no linear ordinary differential operators of positive finite order which preserve the set of non-negative (resp., positive or elliptic) polynomials in R[x]. Remark 1.5. Notice that by contrast with the above situation there are many hyperbolicity-preservers which are finite order linear differential operators with polynomial coefficients. In fact, such examples exist even among operators with constant coefficients, see Remark 1.6 and [2]. Remark 1.6. Any linear operatoron R[x] andC[x] canbe representedas a linear ordinary differential operator of, in general, infinite order, i.e., as a formal power series in d with polynomial coefficients, see, e.g., [2]. Thus the subclass of finite dx order linear differential operators, i.e., those belonging to the Weyl algebra A is 1 a natural object of study. Note that unlike the case of finite order operators there existplentyoflineardifferentialoperatorsofinfiniteorderwhichpreservepositivity. Apparently, the simplest example of this kind is d −1 d d2 1− =1+ + +... (2) (cid:18) dx(cid:19) dx dx2 More generally, the inverse of any finite order differential operator with constant coefficients and positive constant term whose symbol is a hyperbolic polynomial yields an example of such an operator. PRESERVING POSITIVE POLYNOMIALS AND BEYOND 3 Luckily the case of infinite order linear differential operators with constant coefficients can be handled completely. Namely, slightly generalizing a one hundred years old result of Remak [16] and Hurwitz [7] (see also Problem 38 in [15, Ch. 7]) one obtains the following statement. Theorem B. Let α=(α ,α ,...,α ,...,) be an infinite sequence of real numbers. 0 1 k Consider the infinite order linear ordinary differential operator d d2 dk U =α +α +α +...+α +... (3) α 0 1dx 2dx2 kdxk with constant coefficients. Then the operator U preserves positivity (resp., non- α negativity) if and only if one of the following two equivalent conditions holds: (1) foranypositive(resp.,non-negative)polynomialp(x)=a xk+...+a x+a k 1 0 one has that U (p(0))=a α +a α +...+k!a α >0 (resp., ≥0); α 0 0 1 1 k k (2) the following infinite Hankel matrix α 1!α 2!α ... l!α ... 0 1 2 l 1!α1 2!α2 3!α3 ... (l+1)!αl+1 ... 2!α 3!α 4!α ... (l+2)!α ... 2 3 4 l+2    ... ... ... ... ... ...    l!α (l+1)!α (l+2)!α ... (2l)!α ...  l l+1 l+2 2l   ... ... ... ... ... ...   is positive definite (resp., positive semi-definite), i.e. all its principal mi- nors are positive (resp. non-negative). To illustrate the latter result notice that for the operator (2) above one has 1 = α = α = α = α = ... and ∆ = Πl (j!)2, l = 0,1,... where ∆ is the 0 1 2 3 l j=1 l corresponding (l+1)×(l+1) principal minor, see [6]. Remark1.7. Themajorremainingchallengeinthisareaistoclassifyallpositivity- preservers. We finish our introduction with this question. Problem 1. Find a complete classification of positivity-preservers. We also state a more concrete and (hopefully) simpler question. Problem 2. Isittruethatanypositivity-preserverwhichisaninfiniteorderlinear differential operator with constant coefficients has a hyperbolicity-preserver as its inverse? Acknowledgments. The authors are grateful to Petter Bra¨nden for important references. The second author is sincerely grateful to the Swedish Royal Academy of Sciences and the Mittag-Leffler Institute for supporting his visit to Stockholm in Spring 2007 when a substantial part of this project was carried out. The first and third authors would like to thank the American Institute of Mathematics for its hospitality in May 2007. 2. Some preliminaries on the considered classes of preservers Belowwe discuss the relationships betweenthe classes ofellipticity-, positivity-, and non-negativity-preservers. As we mentioned in the introduction the set of all univariate non-negative polynomials coincides with the set of sums of squares and therefore linear preservers of the latter set do not require separate consideration. Onthe otherhand, itis obviousthatthe setsofelliptic, positive, andnon-negative polynomialsaredistinct. Inthissectionweanswerthequestionabouthowdifferent 4 J.BORCEA,A.GUTERMAN,ANDB.SHAPIRO are the corresponding sets of ellipticity-, positivity- and non-negativity-preservers, respectively, see Theorems 2.4 and 2.5 below. We start with the following lemma showing that the assumption that a linear operator Φ is a non-negativity-preserveris quite strong. Lemma 2.1. Let Φ : R[x] → R[x] be a linear operator preserving the set of non- negative polynomials. If Φ(1)≡0 then Φ≡0. Proof. Assume that Φ(1) ≡ 0. First we show that for any polynomial p(x) = a xn +... of even degree n = 2m one has that if a > 0 then Φ(p(x)) is non- n n negative and if a < 0 then Φ(p(x)) is non-positive. Indeed, if n is even and n a > 0 then p(x) has a global minimum, say M. Thus p(x)+|M| ≥ 0 for all n x ∈ R. Therefore, Φ(p(x)+|M|) ≥ 0 for all x ∈ R. However, by linearity and the assumption Φ(1)≡0 we get that Φ(p(x))=Φ(p(x))+|M|Φ(1)=Φ(p(x)+|M|)≥0 for all x∈R. For a <0 the result follows by linearity. n Now let us show that Φ(1) ≡ 0 implies that Φ ≡ 0. Let q(x) be a polynomial such that Φ(q(x)) = a xn + ... + a xi with the smallest possible non-negative n i value of i such that a 6= 0. Let p(x) be a monic real polynomial of even degree i satisfying the condition deg(q(x)) < deg(p(x)). Thus p(x) +µq(x) is monic for any µ ∈ R. The above argument shows that Φ(p(x)+µq(x)) is non-negative for all µ ∈ R. Notice that our choice of q(x) implies that the polynomial Φ(p(x)) has vanishing coefficients at the degrees 0,...,i−1. Hence Φ(p(x))=b xl+...+b xi l i for some positive integer l and some coefficients b ,...,b ∈R. Then for any given l i µ there exists g (x) ∈ R[x] such that Φ(p(x))+µΦ(q(x)) = xig (x). Obviously, µ µ the constant term of g (x) equals b + µa . Since a 6= 0 there exists µ ∈ R µ i i i 0 such that b +µ a = g (0) < 0 and by continuity it follows that there exists a i 0 i µ0 neighborhood N(0) of the origin such that g (x)<0 for all x∈N(0). Therefore, µ0 thereexists06=x ∈N(0)suchthatxi >0,hencexig (x )<0. Thiscontradicts 0 0 0 µ0 0 the assumptionthat Φ is a non-negativity-preserver. Thus Φ(q(x)) hasa vanishing termofdegreei,whichcontradictsthe choiceofq(x). We deduce thatΦ(q(x))≡0 for all q(x)∈R[x]. (cid:3) Theorem 2.2. Let Φ : R[x] → R[x] be a linear operator. Then the following conditions are equivalent: (1) Φ preserves the set of elliptic polynomials; (2) either Φ or −Φ preserves the set of positive polynomials. Also each of these conditions implies that (3) either Φ or −Φ preserves the set of non-negative polynomials. Proof. Note that the identically zero operator satisfies neither condition (1) nor condition (2). Therefore, we will assume that Φ6≡0. (1) ⇒ (2). Assume that Φ preserves the set of elliptic polynomials and that neither Φ nor −Φ preserves positivity. In other words, since Φ is an ellipticity- preserver this means that there exist positive polynomials p(x),q(x) ∈ R[x] such that Φ(p(x))> 0 and Φ(q(x)) <0 for all x∈R. Note that no elliptic polynomials can be annihilated by Φ since 0 is not an elliptic polynomial. We consider the following two subcases: A.Thereexisttwopositivepolynomialsp(x),q(x)asabovesuchthatdegΦ(p(x))6= degΦ(q(x)). Wlog we can assume that degΦ(p(x)) > degΦ(q(x)). Since Φ(p(x)) is a positive polynomial it has even degree and positive leading coefficient. Thus for any µ ∈ R the polynomial Φ(p(x))+µΦ(q(x)) has the same properties, i.e., is of even degree and has positive leading coefficient. Hence there exists x (µ) ∈ R 0 such that Φ(p(x))+µΦ(q(x))>0 for all x with |x|>x (µ). 0 PRESERVING POSITIVE POLYNOMIALS AND BEYOND 5 Now set y := Φ(p)(0) and z := Φ(q)(0). Obviously, y > 0 and z < 0 since 0 0 0 0 Φ(p) is positive and Φ(q) is negative. Let µ = 2y0 > 0. Then p(x)+µ q(x) is 0 −z0 0 positive since it is the sum of two positive polynomials. At the same time for its image we have that Φ(p+µ q)(x) >0 for x> x (µ ). However, at the origin one 0 0 0 has 2y 0 Φ(p+µ q)(0)=y + z =y −2y =−y <0, 0 0 0 0 0 0 −z 0 so by continuity Φ(p+µ q)(x) must have at least one real zero, which is a contra- 0 diction. B.Itremainstoconsiderthe casewhenthe imagesofallpositivepolynomialshave the same degree, say m. Let Φ(p(x)) =a xm+..., Φ(q(x)) =b xm+.... Since m m Φ(p(x)) > 0 it follows that a > 0, and since Φ(q(x)) < 0 one has b < 0. Thus m m the polynomial −b p(x)+a q(x) is positive. However, its image is of the degree m m less than m, which is a contradiction. (2)⇒(1). If Φ is a positivity-preserver then by linearity Φ is also a negativity- preserver,and thus Φ preserves the set of elliptic polynomials as well. (2) ⇒ (3). Assume that Φ preserves positivity. Take p(x) ∈ R[x], p(x) ≥ 0. Thenfor anyε>0,p(x)+ε>0. Thus Φ(p(x))+εΦ(1)=Φ(p(x)+ε)>0. Taking the limit when ε→0 we get that Φ(p(x))≥0. (cid:3) The following example shows that, in general, (3) does not imply (1) and (2). Example 2.3. Let Φ : R[x] → R[x] be defined as follows: Φ(1) = x2, Φ(xi) = 0 for all i > 0. Obviously, Φ preserves the set of non-negative polynomials but does not preserve the set of positive polynomials since 1 is mapped to x2 which is only non-negative. We are now going to show that in fact this example is in some sense the only possibility,i.e.,itessentiallydescribesthewholedistinctionbetweenpositivity-and non-negativity-preservers. Theorem 2.4. Let Φ : R[x] → R[x] be a non-negativity-preserver. Then either Φ is a positivity-preserver (and therefore an ellipticity-preserver as well) or Φ(1) is a polynomial which is only non-negative but not positive. Moreover, in the latter case for any positive polynomial p(x)∈R[x] the zero locus of Φ(p(x)) is a subset of the zero locus of Φ(1). Proof. Assume that Φ is a non-negativity-preserver. Then Φ sends positive polynomials to non-negative ones. Let us assume that p(x) ∈ R[x] is positive but its image Φ(p(x)) has real zeros. Since p(x) is positive there exists ε > 0 such that p(x)−ε>0 for all x. Thus for its image we have Φ(p(x)−ε)=Φ(p(x))−εΦ(1)≥0. Set g(x) := Φ(p(x)), f(x) := Φ(p(x) − ε), and h(x) := Φ(1). Since all three polynomials are non-negative and g(x) = εh(x)+f(x), it follows that for any x 0 such that g(x ) = 0 one has that f(x ) = h(x ) = 0. Since h(x) is a polynomial 0 0 0 then either h(x) ≡ 0, or h(x) has a finite number of zeros. However, the first possibility is ruled out by Lemma 2.1, since h(x) = Φ(1). The second possibility impliesthatallpositivepolynomialswhoseimagesarenon-negativebutnotpositive havealtogetheronlyafinitenumberofzerosbelongingtothezerolocusofΦ(1). (cid:3) Corollary 2.5. Let Φ:R[x]→R[x] be a linear operator such that Φ(1)>0. Then the conditions (1), (2) and (3) of Theorem 2.2 are equivalent. In exactly the same way we can show the following. 6 J.BORCEA,A.GUTERMAN,ANDB.SHAPIRO Theorem 2.6. Let Φ : R [x] → R [x] be a linear operator with Φ(1) > 0. Then n n the following conditions are equivalent: (1) Φ preserves the set of elliptic polynomials of degree ≤n; (2) either Φ or −Φ preserves the set of positive polynomials of degree ≤n; (3) either Φor −Φpreserves theset of non-negativepolynomials of degree ≤n. Remark 2.7. Corollary 2.5 and Theorem 2.6 will allow us to reduce the investi- gation of non-negativity-, positivity-, and ellipticity-preservers (both in the finite- dimensionaland the infinite-dimensional cases)to just one of these three classes of preservers. Forthe sakeofcompleteness noticethatfor non-linearoperatorsthe situationis different from the one above as the following simple examples show. Example 2.8. 1. The bijective map Φ :R[x]→R[x] defined by 1 Φ (p(x))=p(x)+c 1 where c is a positive constant, preserves both positivity and non-negativity but does not preserve ellipticity. 2. The bijective map Φ :R[x]→R[x] defined by 2 p(x) ∀p(x)∈R[x]\{x2+1,−x2−1} Φ (p(x))=−x2−1 if p(x)=x2+1 2  x2+1 if p(x)=−x2−1 preserves ellipticity, butdoes not preserve positivity and non-negativity. 3. The bijective map Φ :R[x]→R[x] defined by 3 p(x) ∀p(x)∈R[x]\{±x2} Φ (p(x))=−x2 if p(x)=x2 3  x2 if p(x)=−x2 preserves ellipticity and positivity, but does not preserve non-negativity. 3. The case of diagonal transformations Theaimofthis sectionis twofold. Firstlywe wanttorecallwhatwaspreviously knownaboutpositivity-andnon-negativity-preserversintheclassicalcaseoflinear operators acting diagonally in the standard monomial basis of R[x] and secondly we want to point out some (known to the specialists in the field, [4], [5]) mistakes in the important treatise [8]. 3.1. Known correct results. Let T :R[x]→ R[x] be a linear operator defined ∞ by T(xi)=λ xi for i=0,1,... (4) i and, analogously, let T :R [x]→R [x] be a linear operator defined by n n n T (xi)=λ xi for i=0,1,...,n. (5) n i Denote them by {λ }∞ and {λ }n , respectively. We will refer to such opera- i i=0 i i=0 tors as diagonal transformations or diagonal sequences. Diagonal transformations preserving the set of positive polynomials are referred to as Λ-sequences in the lit- erature,see[4,5,9]. ReservingthesymbolΦforgenerallinearoperatorsweusein this section the notation T ∈ Λ to emphasize that T is a diagonal transformation preserving positivity. Multiplying if necessary all elements of our sequence with −1, we can assume that λ ≥0. 0 PRESERVING POSITIVE POLYNOMIALS AND BEYOND 7 Remark 3.1. Notice that in the finite-dimensional case we only need to consider transformations acting on R [x] for n even since there are no positive polynomials n of odd degree and a sequence {λ }2k+1 preserves the set of positive polynomials in i i=0 R if and only if {λ }2k preserves the set of positive polynomials in R . 2k+1 i i=0 2k Let us establishsome immediate consequences of the fact that a diagonaltrans- formation T is a positivity-preserver. Lemma 3.2. Assumethat a transformation T ={λ }α , α∈N∪{∞}, preserves α i i=0 positivity. Then (1) λ ≥0 for any even i; i (2) λ2 ≤λ λ for any i. i 0 2i Proof. To settle (1) consider the polynomial p(x) = xi+1 which is positive if i is even. Thus T(x) = λ xi+λ should be positive as well. Since λ > 0, the result i 0 0 follows. To settle (2) consider the polynomial p(x) = x2i+axi+b with a2 < 4b. Then p(x)ispositiveaswellasitsimageq(x):=Φ(p(x))=λ x2i+aλ xi+bλ . Ifiisodd 2i i 0 then the positivity of q(x) is equivalent to the negativity of its discriminant, i.e., D := a2λ2−4bλ λ < 0, which implies λ2 ≤ λ λ since a2 < 4b. Finally, if i is q i 0 2i i 0 2i eventhenq(x) ispositive iffeitherD <0orD >0andifadditionallybothroots q q of the quadratic polynomial λ z2+aλ z +bλ are negative. In the first subcase 2i i 0 one has λ2 ≤λ λ as for i odd. In the second subcase we obtain that the positive i 0 2i polynomialx2i−axi+bistransformedtothepolynomialλ x2i−aλ xi+bλ which 2i i 0 has some real roots. To check this notice that the roots of λ z2−aλ z+bλ are 2i i 0 opposite to that of λ z2+aλ z+bλ andare,therefore, positive. Thus extracting 2i i 0 their i-th root one will get some positive roots as well. This contradiction shows that the inequality λ2 ≤λ λ is necessary for the positivity-preservation. (cid:3) i 0 2i Diagonal transformations which are positivity-preservers are known to be very closely related to the following class of sequences of real numbers. Definition 3.3. A sequence {λ }α is called positive definite if for any positive i i=0 polynomial p(x) = xn +a xn−1 +...+a x+a ∈ R [x] one has that λ + n−1 1 0 α n a λ +...+a λ +a λ >0, i.e., T (p)(1)>0. n−1 n−1 1 1 0 0 α In the infinite-dimensional case the following characterizations of the set of diagonal positivity-preserversis known. Theorem 3.4. ([4], [5, Theorem 1.7]) Let {λ }∞ be a sequence of real numbers. k k=0 Then the following conditions are equivalent: (1) {λ }∞ ∈Λ; k k=0 (2) {λ }∞ is a positive definite sequence; k k=0 λ λ ... λ 0 1 k (cid:12) λ1 λ2 ... λk+1 (cid:12) (3) |(λi+j)|=(cid:12)(cid:12)(cid:12) ... ... ... ... (cid:12)(cid:12)(cid:12)>0, for k =0,1,2... (cid:12) (cid:12) (cid:12) λ λ ... λ (cid:12) (cid:12) k k+1 2k (cid:12) (4) There exist(cid:12)s a non-decreasing funct(cid:12)ion µ(t) with infinitely many points of (cid:12) (cid:12) increase such that for all k =0,1,2,... one has ∞ λ = tkdµ(t). k Z −∞ Proof. Theequivalences(2)⇔(3)⇔(4)aresettledin[17,p.132]independentlyof condition (1). The implication (1) ⇒ (2) is evident so we only have to concentrate 8 J.BORCEA,A.GUTERMAN,ANDB.SHAPIRO on the remaining implication (2) ⇒ (1). Take a positive polynomial g(x)=a xl+ l l ...+a x+a andsetq(x):=Λ(g(x))= λ a xi. We wanttoshowthatq(x)>0 1 0 i i iP=0 for all real x. ∞ Condition (4) implies that λ = tidα(t), where α(t) is a monotone non- i −R∞ decreasing function with infinitely many points of increase. Hence ∞ ∞ ∞ l l q(x)= tidα(t) xi = a tixidα(t)= g(xt)dα(t)>0 i Xi=0(cid:16)−Z∞ (cid:17) −Z∞ Xi=0 −Z∞ since g(xt) > 0 for all t. Notice that the above integrals are convergent for any fixed value of x. Thus q(x)>0 and the lemma follows. (cid:3) In the finite-dimensional case one has a similar statement. Theorem 3.5. [9] Let {λ }n be a sequence of real numbers. Then the following k k=0 conditions are equivalent: (1) {λ }n ∈Λ; k k=0 (2) {λ }n is a positive definite sequence; k k=0 λ λ ... λ 0 1 k (cid:12) λ1 λ2 ... λk+1 (cid:12) (3) |(λi+j)|=(cid:12)(cid:12)(cid:12) ... ... ... ... (cid:12)(cid:12)(cid:12)>0, for k =0,1,2...,n2. (cid:12) (cid:12) (cid:12) λ λ ... λ (cid:12) (cid:12) k k+1 2k (cid:12) (4) Thereexist(cid:12)sanon-decreasingfuncti(cid:12)onµ(t)withatleastnpointsofincrease (cid:12) (cid:12) such that ∞ λ = tkdµ(t), k =0,1,2,...,n; k Z −∞ Proof. Repeats that of Theorem 3.4. (cid:3) 3.2. Known wrong results. To present some erroneous results from [8] and the corresponding counterexamples we need to introduce the following classes of diagonal transformations. Definition3.6. WesaythatT ,α∈N∪∞,or,equivalently,thesequence{λ }α , α i i=0 is a hyperbolicity-preserver, if forany hyperbolicp(x)∈R [x] its imageT (p(x)) is α α hyperbolic. We denote this class of transformations by H or H. α Clearly, this class is the restriction of the earlier defined class of hyperbolicity- preservers to diagonal transformations. Theorem 4.6.14 of [8] states that T ∈ Λ if and only if T−1 ∈ H. We will now show that this statement is wrong in both directions. Proposition 3.7. There exist (i) T ∈Λ such that T−1 ∈/ H; (ii) T ∈H such that T−1 ∈/ Λ. Proof. We present below 3 concrete examples verifying the above claims. To illustrate (i) consider the diagonal transformation T : R [x] → R [x] defined by 4 4 4 the sequence (λ ,λ ,λ ,λ ,λ ) = 1 , 1 , 1 , 1 , 1 . By the determinant cri- 0 1 2 3 4 29 68 123 200 305 teria (3) of Theorem 3.5 the opera(cid:0)tor T4 preserves po(cid:1)sitivity. However, one can checkthatitsinversesendsthenon-negativepolynomial(x+1)4 tothepolynomial (x+1)(305x3+495x2+243x+29) possessing two real and two complex roots. PRESERVING POSITIVE POLYNOMIALS AND BEYOND 9 Thisexampleshowsthatinthe finite-dimensionalcasethereisadiagonaltrans- formation which preserves positivity, but whose inverse does not preserve hyperbolicity. We can extend this example to the infinite-dimensional case as follows. By [4, Proposition 3.5] that there exists an infinite sequence {λ }∞ ∈ Λ such i i=0 ∞ 1 that the sequence of inverses ∈/ H. As an explicit example one can take (cid:26)λ (cid:27) i i=1 1 λ = . i i3+5i2+33i+29 An example illustrating (ii) is given in [4, p. 520], see also [5, Example 1.8]. Namely, the sequence {1 + i + i2}∞ corresponds to a diagonal transformation i=0 ∞ 1 preserving hyperbolicity. However, the sequence of inverses leads (cid:26)1+i+i2(cid:27) i=0 to a diagonal transformation which is not a positivity-preserver. (cid:3) Definition 3.8. We say that a diagonaltransformationT , α∈N∪∞, generated α by the sequence {λ }α is a complex zero decreasing sequence (CZDS for short), if i i=0 for any polynomial p(x) ∈ R [x] the polynomial T(p) has no more non-real roots α (counted with multiplicities) than p. We denote the set of all CZDS by R. Remark 3.9. Obviously, any CZDS preserves hyperbolicity, i.e., R ⊂ H. For a whileitwasbelievedthatR=HuntilCravenandCsordasfoundacounterexample [4]. Additionally, one can see directly from the definition that the inverse of any positive CZDS is a Λ-sequence, that is, a diagonal positivity-preserver. Finally, Theorem 4.6.13 of [8] states that T ∈ Λ if and only if T−1 ∈ R, which we disprove below. Proposition 3.10. There exist T ∈Λ such that T−1 ∈/ R. Proof. Use the first two counterexamples from the proof of Proposition3.7. (cid:3) 4. Linear ordinary differential operators of finite order Our aim in this section is to prove Theorem A, i.e., to show that there are no positivity-, non-negativity-, and ellipticity-preservers which are linear differential operators of finite positive order. In fact we are going to show that for any linear differentialoperatorU oforderk ≥1thereexistsanintegernsuchthatU :R [x]→ n R [x] is not a non-negativity preserver. Moreover, we show that one can always n choose n = 2k. Since any positivity-preserver is automatically a non-negativity- preserver and any ellipticity-preserver is a positivity-preserver up to a sign change we will get Theorem A in its complete generality from the above statement. Denote by S[s ,...,s ] the ring of symmetric polynomials with real coefficients 1 k in the variables s ,...,s . Let σ be the l-th elementary symmetric function, i.e., 1 k l σ = s ···s ∈S[s ,...,s ], l=1,...,k. l j1 jl 1 k j1<X...<jl We will need the following technical fact. Proposition 4.1. Let p(x) = (x−x )2···(x−x )2 ∈ R[x,x ,...,x ]. Consider 1 k 1 k the following two families of rational functions: p(l)(x) w =w (x,x ,...,x )= , l=1,...,k; l l 1 k p(x) 1 1 u =σ ,..., , l =1,...,k. (6) l l (cid:18)x−x x−x (cid:19) 1 k Then 10 J.BORCEA,A.GUTERMAN,ANDB.SHAPIRO 1 1 1. w ∈S ,..., , l=1,...,k. l (cid:20)x−x x−x (cid:21) 1 k 2. For any l=1,...,k one has that w =2l!u +g (u ,...,u ), l l l 1 l−1 whereg ∈R[y ,...,y ],l=1,2,...,k,arecertainpolynomials(thatcanbefound l 1 l−1 explicitly but we will not need their explicit form; in particular, g ≡0). 1 Proof. Set p (x)=(x−x )···(x−x ). Then one can immediately check that 1 1 k (i) p (x) 1 =i!u i p (x) 1 for all i=1,...,k. Using the Leibniz rule we get p(l)(x)=(p2(x))(l) =2p(l)(x)p (x)+ c p(i)(x)p(j)(x), 1 1 1 i,j 1 1 i,j≥X1,i+j=l where c ≥0 are certain binomial coefficients. Thus i,j p(l)(x) p(l)(x) p(i)(x) p(j)(x) w = =2 1 + c 1 · 1 =2l!u + c u u . l i,j l i,j i j p(x) p (x) p (x) p (x) 1 i,j≥X1,i+j=l 1 1 i,j≥X1,i+j=l The result follows. (cid:3) We are now ready to prove the first main result of this paper. Theorem4.2. LetU :R[x]→R[x]bealinearordinarydifferentialoperatorofor- Q der k ≥1 of the form (1) with polynomial coefficients Q=(q (x),q (x),...,q (x)), 0 1 k q ∈ R[x], q (x) 6≡ 0. Then for any such coefficient sequence Q the operator U i k Q does not preserve the set of non-negative polynomials of degree 2k. Proof. We assume that U 6≡ 0. Since U (1) = q (x) an obvious necessary con- Q Q 0 dition for the operator U to preserve non-negativity is that q (x) itself is a non- Q 0 negative polynomial. Moreover,q (x) does not vanish identically by Lemma 2.1. 0 We will now construct a non-negative polynomial p(x):=p (x)=(x−x )2(x−x )2···(x−x )2 ∈R[x,x ,...,x ]. x1,...,xk 1 2 k 1 k such that its image under the action of U attains negative values. For this we Q define U (p(x)) p′(x) p′′(x) p(k)(x) Q R(x,x ,...,x )= =q (x)+q (x) +q (x) +...+q (x) . 1 k 0 1 2 k p(x) p(x) p(x) p(x) Then in the notationof Proposition4.1 we have that R(x,x ,...,x )=q (x)+ 1 k 0 q (x)w +...+q (x)w . Let us fix x ∈ R such that x 6= 0 and for any i = 1 1 k k 0 0 0,1,...,k either q (x) ≡ 0 or q (x ) 6= 0. Set α = q (x ), i = 1,...,k, and i i 0 i i 0 r(x ,...,x )=R(x ,x ,...,x ). Thenr(x ,...,x )isalinearforminw′,...,w′, 1 k 0 1 k 1 k 1 k k wherew′ =w (x ). Thusthereexista ,...,a ∈Rsuchthatq (x )+ α a <0. i i 0 1 k 0 0 i i iP=1 (Notice that by our choice of x one has q (x )>0.) 0 0 0 Let now b ,...,b ∈R be defined by 1 k 1 b =a , b = (a −g (b ,...,b )), i=2,...,k, (7) 1 1 i i i 1 i−1 2i! where g are defined in Proposition 4.1. Consider the system of equations i 1 1 b =σ ,..., , i=1,...,k, (8) i i (cid:18)x −t x −t (cid:19) 0 1 0 k