A characterization of polynomials whose high powers have non-negative coefficients

Let $f \in \mathbb{R}[x]$ be a polynomial with real coefficients. We say that $f$ is eventually non-negative if $f^m$ has non-negative coefficients for all sufficiently large $m \in \mathbb{N}$. In this short note, we give a classification of all eventually non-negative polynomials. This generalizes a theorem of De Angelis, and proves a conjecture of Bergweiler, Eremenko and Sokal


Introduction
In this short note we study the following basic problem about iterated convolutions of sequences of real numbers.
For what sequences S = (c 0 , . . . , c d ) are all "high" convolutions S * S * · · · * S non-negative? Of course, this is the same as asking "for what polynomials f ∈ R[x] does f m have non-negative coefficients for all large m". In this note we give a classification of these polynomials by showing that two natural necessary conditions are indeed sufficient.
Such questions reach back to the work of Poincaré [7], who studied polynomials f for which there exists a polynomial p so that pf has all positive coefficients. He gave a full characterization of such polynomials by showing f satisfies this condition if and only if f (r) > 0 for all r > 0. Pólya [8] later proved a multivariate analogue which in the univariate case shows p can be taken to be a sufficiently high power of 1 + z. This line of results culminated in a pair papers by Handelman [4,5] who gave necessary and sufficient conditions on the pair (p, f ) so that p(z) m f (z) has non-negative coefficients for some large m, provided p has non-negative coefficients. Interestingly, Handelman [6] also proved that if a monic polynomial with f (1) > 0 has f m non-negative coefficients for some m, then it in fact has non-negative coefficients for all sufficiently large m.
More similar to our problem, De Angelis [3] studied polynomials f for which f m has strictly positive coefficients for all large m. This was also studied by Bergweiler, Eremenko and Sokal [2] and a key step in their work on empirical distributions of polynomials with non-negative coefficients [1]. For their results, these authors used an interesting notion of positivity: We say that a non-zero polynomial is strongly for all z ∈ C \ R ≥0 . It is easy to see that if a polynomial f m has non-negative coefficients then f is strongly positive, as long as f = g(z ℓ ), for ℓ ≥ 2. Indeed, if f m has all non-negative coefficients, for some m, then for all z ∈ C \ R ≥0 by the triangle inequality, and therefore |f (z)| < f (|z|). De Angelis gave the following classification of eventually positive polynomials, that is polynomials for which all of the coefficients a i of f m are non-negative for all i ∈ {0, 1, . . . , m deg(f )}.
). Let f (z) = a 0 + a 1 z + · · · + a d z d be a polynomial with real coefficients. Then f is eventually positive if and only if f satisfies |f (z)| < f (|z|) for all z ∈ C \ R ≥0 and a 0 , a d > 0, a 1 , a d−1 > 0.
This note concerns an extension of this result to the situation when coefficients of f m are just non-negative, rather than positive. Interestingly, this problem appears to be non-trivial; a special case of this problem was conjectured to hold by Bergweiler, Eremenko and Sokal [2] (See p. 11). Tan and To [9,10], who have recently extended Theorem 1 to the multivariate setting, also state that new ideas will be necessary to tackle the case of non-negative coefficients. For this, let us say that a polynomial f is eventually non-negative if f m has all positive coefficients for all large m.
For our classification, we will use the notion of "strong positivity" but will additionally need to supply a more sophisticated "boundary condition" on the coefficients: for example, we cannot assume that a 1 > 0, as De Angelis does. To get a feel for what we need, let us consider the simplest non-example, when a 0 = −1.
As we iterate the polynomial the constant term simply alternates between −1, 1, as there is no interaction with any of the other terms. For a slightly more complicated example, consider the polynomial that starts f = 1 + z 2 − z 3 + · · · . Here, it is not possible for the coefficient of z 3 to become non-negative, as there is no way for the positive terms to 1, z 2 "add up" and reach z 3 . Now, one might hope that such examples could simply be excluded by the strong positivity hypothesis. However, it is not hard to construct examples that are additionally strongly positive. Indeed, Bergweiler, Eremenko and Sokal observed that for ε > 0 sufficiently small, the family of polynomials 1 + z 3 + z 4 − εz 5 + z 6 + z 7 + z 10 is strongly positive, but all powers have a negative coefficient of z 5 .
Our "boundary condition" simply excludes these "obvious" ways of ensuring that f m has negative coefficients.
Definition 2. Let f (z) = a 0 + a 1 z + · · · + a d z d be a degree d polynomial with a 0 = 0 and real coefficients. Let S + (f ) := {i : a i > 0} denote the indices of the positive coefficients and and let S − (f ) := {i : a i < 0} denote the indices of the negative. We say that f has the one-sided positive covering property if where the right-hand-side denotes the d-times iterated sumset. We then define f to have the positive covering property if both f (z), z d f (1/z) have the one-sided covering property.
We now state the main theorem in this paper, which says that the necessary conditions of strong positivity and the positive covering property are sufficient.
Theorem 3. Let f ∈ R[x] be a polynomial of degree d, then f is eventually nonnegative if and only if f = z k g(z ℓ ), where k, ℓ ∈ N and g is strongly positive and satisfies the positive covering property.
Our short proof has two steps. In the first step we use the positive covering property alone to show that the first and last m 1−ε coefficients of f m must be non-negative. In the second step we show that we can lift this proof to show that all [0, δm] coefficients in f m are non-negative, by using a saddle-point method. We can thus finish the proof of Theorem 3 by appealing to a result of De Angelis.
Our proof of non-negativity of coefficients in the range [0, m 1−ε ] for ε ∈ (0, 1) is purely combinatorial, and relies solely on the positive covering hypothesis. Our goal will be to prove the following lemma. For a polynomial f we write [z k ]f to denote the coefficient of z k in f .
To understand the combinatorics at play here, it is useful to look at a basic example first. Consider the polynomial f (z) = 1 + z 2 + z 3 − Az 5 for some A > 0, and examine the coefficient of z 5 in f m : expanding the polynomial directly shows The idea is that the freedom of choosing more terms of smaller degree-i.e. z 2 and z 3 rather than z 5 in this example-gives a larger contribution than swapping out some number of small degree terms for a larger degree term, even if it has a large negative coefficient.
Our method for proving Lemma 4 will be to push this idea much further. Before jumping in, we need an elementary bound on ratios of binomial coefficients; this will quantify how much is won when choosing multiple small degree terms over a single larger degree term.
Proof. Put c := (4d) −d and let (a) b = a(a − 1) · · · (a − b + 1) denote the falling factorial. We have for sufficiently large a. Using the choice of c completes the proof.
Each term in the expansion of f m contributing to [z n ]f m can be associated to a partition λ = (1 λ1 , 2 λj , . . . , d λ d ) of n in which we choose the monomial a j z j , λ j times. Then the contribution associated to given partition λ is equal to We may thus write where the sum is over all integer partitions of n.
In the next lemma, we shall show that for each n (in an appropriate range), we can define a map M = M n,m,f , so that |Cont(M (λ))| > d|Cont(λ)| and the preimage of each element has size at most d. It shall then follow that [z n ]f is positive.
To define this map, we need to define a simple combinatorial quantity. For a polynomial f = k a k z k we set S + (f ) := {k : a k < 0} and S − (f ) := {k : a k > 0}, as before. If f has the positive covering property then for every k ∈ S − (f ) we may express k = b 1 + · · · + b t , for some t ∈ N and b 1 , . . . , b t ∈ S + (f ) \ {0}. Let us define the weight of k to be the maximum t for which this holds. That is, We then define w(f ) = min To define the map M , let f be a polynomial with the positive covering property and let m, n ∈ N. For each λ ∈ {λ : Cont n,m (λ) < 0}, there must be some minimal j for which a j < 0 and λ j > 0. By the covering property, we know that there exists a partition µ = (1 µ1 , 2 µ2 , . . . , d µ d ) of j for which µ i > 0 implies a j > 0 for each j ∈ {0, 1, . . . , d} and i µ i ≥ w(f ). Then define a partitionλ of n byλ j = λ j − 1 andλ i = λ j + µ j , for i = j. Finally set M (λ) :=λ.
We pause to make a simple observation about this map.
Lemma 7. Let f ∈ R[z] be a polynomial of degree d with the positive covering property and let w := w(f ) be as defined above. For all sufficiently large m ∈ N and n ∈ [0, m 1−3/(2w) ] let λ ⊢ n be a partition of n for which where M is the map defined above.
Proof. First observe that Cont(λ) < 0 implies Cont(M (λ)) > 0, by construction. Hence we only need to show that |Cont(M (λ))| > d|Cont(λ)|. For this, we let j andλ j be as in the definition of M (λ) and we write and similarly for Cont(λ). So using thatλ i = λ i + µ i , for i = j andλ j = λ j − 1 we have . Now note that since µ is a partition of j each µ i ∈ [0, d] and therefore there is a constant C = C(f ) for which We now use that n ≤ m 1−3/(2w) and therefore m−λ 1 −· · ·−λ i−1 ≥ m/2, for m large. Moreover, we have that λ 1 + · · · + λ i − λ 1 + · · · +λ i ≤ j by Observation 6 and thus we may apply Lemma 5 to obtain So from (2) we obtain if m is sufficiently large compared to C(f ). For the penultimate inequality, we have used that i µ i ≥ w = w(f ), which holds by the definition of w(f ). Proof. Let M = M f,n,m denote the map defined above. We have that where the penultimate inequality holds due to the fact that each element of {λ : Cont(λ) ≥ 0} is mapped to by at most d distinct partitions. The last inequality holds by applying Lemma 7. Lemma 4 follows from applying Corollary 8 twice: first we apply it to see that w(f m ) gets arbitrarily large as m gets large. We then apply Corollary 8 again to f m , f m+1 , f m−1 to prove Lemma 4. For this we need a simple observation. Proof. We first see that for x = km, k ∈ N, we may find a solution (a 0 , b 0 , c 0 ) where a 0 , b 0 , c 0 ∈ {⌊k/3⌋, ⌈k/3⌉}. Then, for x = km + ℓ we find a solution (a, b, c) by setting a = a 0 − ℓ and b = b 0 + ℓ, c = c 0 .
We now apply Corollary 8 to find a m 0 so that for all m ≥ m 0 we have [z k ]f m ≥ 0 for all k ∈ {0, 1, . . . , dw 0 }. We claim that w(f m ) ≥ w 0 for all m ≥ m 0 . To see this, let n be such that [z n ]f m < 0 and note that we must have n > dw 0 . Now let S(f ) = {k : [z k ]f = 0} be the support of f and since n is in the support S(f m ), we may write n = b 1 +· · ·+b t , where b 1 , . . . , b t ∈ S(f )\{0} and t ≥ w 0 . Since f has the positive covering property we can write Now, choose m 1 = m 0 + 1 and given x ≥ 8(m 1 ) 2 we may apply Observation 9 to write x = am 1 + b(m 1 + 1) + c(m 1 − 1), where a, b, c ≥ x/(4m 1 ) and thus we may write Now crucially note that each of f m1 , f m1−1 , f m1+1 has the positive covering property and that w(f m1 ), w(f m1−1 ), w(f m1+1 ) ≥ w 0 . Therefore we may apply ] where x 0 = min{a, b, c} ≥ x/(4m 1 ) and x is sufficiently large. This completes the proof. Lemma 10. Let f (z) = 1 + a 1 z + · · · + a d z d have the positive covering property and assume that f = g(z ℓ ) for g ∈ R[z], ℓ ≥ 2. Then there exists a δ > 0 so that for all sufficiently large m we have [z n ]f m ≥ 0 for all n ∈ [0, δm].
One we have proved Lemma 10 we essentially be finished, by appealing to the following result of De Angelis [3].
Lemma 11 ( [3]). Let f (z) = a 0 + · · · + a d z d be strongly positive with a 0 , a d > 0. Then for every δ > 0, there exists an M = M (δ) so that for all m ≥ M , For the proof of Lemma 10 we will again only require the positive covering property, and not strong-positivity. The proof is entirely analytic, and makes use of the saddle-point method : the coefficient [z n ]f (z) m is written as a contour integral using Cauchy's integral formula; by a careful choice of the contour, we may minimize the oscillation in the integrand and show that the integral is dominated by the piece of the contour nearest to the positive real-axis. This method was used by Bergweiler-Eremenko-Sokal [2] in their proof of Theorem 1 and is similar to De Angelis' proof [3]. Our new ingredient starts with the observation that the positive covering property implies a quantitative version of strong positivity for z in a neighborhood of the origin. This observation is captured in the following lemma.
Proof. By compactness of the set [θ 0 , π] and continuity of the function g(θ) := f (re iθ ), it is sufficient to show that (4)  Note that if T = ∅ then f (z) = g(z ℓ ) for ℓ = 2π/θ, contradicting the condition on f . So we can assume T = ∅ and thus we may set t 0 := min T . We now show that with a t0 > 0, as r → 0. Note that this is sufficient to imply Lemma 12. Now expanding the left-hand-side of (5), we have In this case, we can see that the smallest power of r that appears in this sum is r t0 : suppose j is such that r j has a non-zero coefficient in (6). Then there exist p, q so that p + q = j and a p a q (1 − cos((p − q)θ)) = 0. If both p, q ∈ T * then (1 − cos((p − q)θ)) = 0. So one of p, q ∈ T and thus j ≥ t 0 . Now note that t 0 = t 0 + 0 is the unique such expression for t 0 = p + q, for which one of p, q ∈ T . Thus |f (r)| 2 − |f (re iθ )| 2 = 2a t0 r t0 + o(r t0 ).
We now see that a t0 > 0 by using the positive covering hypothesis. If a t0 < 0, then we can write t 0 and a sum of indices with positive coefficients t 0 = i 1 + i 2 + · · · + i ℓ . But minimality of t 0 implies that i 1 , . . . , i ℓ ∈ S * . This implies that e it0φ = e i( ip)φ = 1 and therefore t 0 ∈ T , which is a contradiction.
We may now prove Lemma 10.
Proof of Lemma 10. Define k to be the minimum ℓ ≥ 1 for which a ℓ = 0. Thus we may write f (z) = 1 + a k z k + · · · + a d z d . We may also assume that a k = 1 k , by possibly rescaling z; i.e. z → λz. We now write z = ρe iθ and use Cauchy's integral formula to express (7) [ We will find asymptotics as m → ∞ of the right hand of this equation for an appropriate choice of ρ. For this, set α := n/m, and note that by Lemma 4 we know that [z n ]f (z) m ≥ 0 for all α ≤ m −1/(d+1) and sufficiently large m. Thus, it is sufficient to assume α ≥ m −1/(d+1) . Define so that Our first claim shows that I 1 is large and positive.
where C f > 0 is a constant depending only on f .
To obtain a lower bound on the real part of the integrand, we obtain an upper bound Im(mR(θ)) for θ ∈ (0, η) by using (13), to get Now, using the expansion at (12) and the choice of η = (k 2 mρ k ) −1/3 , we obtain a lower bound where we have used mρ k → ∞ and set which is a constant, depending only on f . This completes the proof of the claim. The next two claims show that I 2 and I 3 are small.
Where c 2 > 0 is a constant depending only on f .
Proof of Claim 14. Note that for |θ| ≤ 1 k , we have Thus, using (12) and the triangle inequality, we have Proof of Claim 15. To bound I 3 , we can apply Lemma 12 with θ 0 = 1 k to get a constant c > 0 so that for |θ| > 1 k we have Using that ρ k ≥ cα and therefore, ρ d ≥ cα d/k ≥ cα d , we have , where the last inequality follows from our assumption that α ≥ m −1/(d+1) .
This gives an upper bound on . This completes the proof of Claim 15.
We now simply apply Claims 13, 14 and 15 to estimate (7) and finish. Indeed, we have Proof of Theorem 3. We write f (z) = z k g(z ℓ ) and assume that g cannot be expressed in this form. Now f is eventually non-negative if and only if g is. By the discussion in the introduction we see that if g is eventually non-negative, then g is strongly positive and satisfies the positive covering property.
To see the converse, assume that d = deg(g) and apply Lemma 10 to g(z) and z d g(1/z) to obtain a δ = δ(f ) > 0 to learn that for all n ∈ [0, δm] ∪ [(1 − δ)dm, dm] we have [z n ]f ≥ 0, for m sufficiently large. We then finish by applying Lemma 11.