It was asked at the Bulletin of the American Mathematical Society Volume 64, Number 2, 1958, as a Research Problem, if a Hurwitz polynomial with real coefficients (i.e. all of its zeros have negative real parts) can be divided into the arithmetic sum of two or three polynomials, each of which has positive coefficients and only nonpositive real roots.

I would like to know if the following problem is known and how it can be solved:

Can any polynomial with complex coefficients and degree $n$ be divided into the arithmetic sum of three complex polynomials, each of which has degree at most $n$ and only real roots?

Any help would be appreciated.

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    $\begingroup$ is it clear that two polynomials with real roots are not enough? $\endgroup$ Mar 10, 2018 at 11:29
  • $\begingroup$ @FedorPetrov it's not clear at all. Maybe two polynomials are enough. $\endgroup$
    – jack
    Mar 10, 2018 at 20:16
  • $\begingroup$ @jack: I think what Fedor Petrov probably had in mind was just expectations based on naive counting of the degrees of freedom. A 5th-order polynomial's roots are described by the 10 real and imaginary parts, so it has 10 d.f. (ignoring scaling). A 5th-order polynomial with real roots has 5 d.f. Therefore it seems reasonable to hope that two 5th-order polynomials with real roots have enough d.f. to allow us to form a linear combination of them and get an arbitrary 5th-order polynomial. Alexandre Eremenko's answer verifies this conjecture. $\endgroup$
    – user21349
    Mar 10, 2018 at 20:35
  • $\begingroup$ ah, yes, it is clear:) posted as an answer $\endgroup$ Mar 11, 2018 at 6:52

3 Answers 3


To address the original question about the polynomials with complex coefficients.

Given a polynomial $P\in\mathbb C[x]$ of degree $n$, write $P=Q+iR$ with $Q,R\in\mathbb R[x]$, and fix arbitrarily a polynomial $S\in\mathbb R[x]$ of degree $n$ with all roots real and pairwise distinct. As observed in Eremenko's answer, for any $\varepsilon\ne 0$ sufficiently small in absolute value, the polynomials $Q_\varepsilon:=S+\varepsilon Q$ and $R_\varepsilon:=S+\varepsilon R$ will also have all their roots real, and then \begin{align*} P &= Q+iR \\ &= \varepsilon^{-1}(Q_\varepsilon-S)+i\varepsilon^{-1}(R_\varepsilon-S) \\ &= \varepsilon^{-1}Q_\varepsilon+i\varepsilon^{-1}R_\varepsilon-(1+i)\varepsilon^{-1}S \end{align*} is the required representation of $P$ as a sum of three polynomials of degree at most $n$ with all their roots real.


Every real polynomial $f$ is the sum of TWO polynomials $g,h$ with all zeros real (of the same degree). Indeed, take any $g_1$ of the same degree as $f$, whose roots are real and simple. Small perturbation does not destroy this property. Therefore $h_1=g_1+\epsilon f$ also has real zeros when $\epsilon$ is small and real. Now take $g=-g_1/\epsilon, \; h=h_1/\epsilon$. We obtain $f=g+h$, and both $g,h$ have real zeros.

A stronger result is available: Every real entire function $f$ of exponential type and satisfying $$\int_{-\infty}^\infty\frac{\log|f(x)|}{1+x^2}dx<\infty$$ (evidently this class contains all real polynomials) is a sum of two entire functions of the same exponential type with all zeros real (and moreover, with all $\pm1$ points real).

Your conjecture is not a formal corollary from this result, because it does not say that the summands are polynomials if $f$ is a polynomial, but if you look at the proof you see that when applied to a polynomial it gives polynomials of the same degree (the general idea of the proof is the same as in my proof in the first paragraph).

The reference is MR0751391 Katsnelʹson, V. È. On the theory of entire functions of the Cartwright class. (Russian) Teor. Funktsiĭ Funktsional. Anal. i Prilozhen. No. 42 (1984), 57–62. English transl: MR0862002 American Mathematical Society Translations, Series 2, Vol. 131. Ten papers in analysis. AMS Translations, Ser. 2, 131.American Mathematical Society, Providence, RI, 1986. viii+120 pp. ISBN: 0-8218-3106-2

Remark 1. There is a lot of arbitrary choice in the construction in the first paragraph. One reasonable choice of $g_1$ would be the Chebyshev polynomial of degree $d=\mathrm{deg}\, f$. It has $d+1$ points of alternance, where the critical values are $\pm1$. So if we take $\epsilon<1/\| f\|,$ where $\| f\|$ is the $\sup$-norm on $[-1,1]$, then the construction works, because $g_1+\epsilon f$ has at least $d$ sign changes on $[0,1]$ thus all roots are real. So the coefficients of $g$ and $h$ can be estimated in terms of $\| f\|$ or in terms of the sup-norm on any interval. One can probably state and solve some extremal problem about this.

Remark 2. In the original problem (see the reference on BAMS in the question), $f$ is a Hurwitz polynomial (zeros in the left-half-plane), and it is required that $g,h$ be also Hurwitz. This can be easily achieved even without the condition that $f$ is Hurwitz. Indeed, the argument in Remark 1 yields $g,h$ whose zeros lie on an arbitrary prescribed interval.

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    $\begingroup$ "satisfying" what? You probably mean the integral should be $<\infty$? $\endgroup$
    – Wolfgang
    Mar 10, 2018 at 14:02
  • $\begingroup$ @Wolfgang: Yes, thanks for the correction. $\endgroup$ Mar 10, 2018 at 14:08
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    $\begingroup$ As a result of the observation in the first paragraph, every polynomial with complex coefficients of degree $n$ is a sum of at most FOUR polynomials with all their roots real and the degree of each one at most $n$. This follows by writing the original polynomial as $P+iQ$ where $P$ and $Q$ are polynomials of degree at most $n$ with real coefficients, and then splitting each of them into a sum of two polynomials with real coefficients and real roots. $\endgroup$
    – Seva
    Mar 10, 2018 at 18:39
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    $\begingroup$ I am afraid that two polynomials are not enough, posted this as an answer. $\endgroup$ Mar 11, 2018 at 6:51
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    $\begingroup$ @FedorPetrov It looks like Eremenko's answer considers a general real polynomial $f$ and writes it as a sum $f=g+h$ of real polynomials. Your answer, and the question, is about a complex polynomial. $\endgroup$ Mar 11, 2018 at 9:19

This is the answer only to my own question in the comments.

Two polynomials with real roots are not enough in general.

For seeing this, assume that $f(z)=\lambda g(z)+\mu h(z)$ for complex numbers $\lambda$, $\mu$ and real polynomials $g,h$. Then $\bar{f}=\bar{\lambda}g+\bar{\mu}h$, and provided that $f/\bar{f}\ne \rm{const}$, we solve this $2\times 2$ system for $g, h$ and see that $g$ is a linear combination of $f,\bar{f}$. In other words, $g$ is a real linear combination of $\operatorname{Re}f, \operatorname{Im}f$. But it often happens that no real linear combination of two given real polynomials has only real roots. Indeed, take the polynomials like $x^{100}+1$ and $x^{17}+x^5$, by sign change (Descartes?) estimate their linear combination can not have only real roots.


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