This is a lemma I wanted in order to solve Patrizio Neff's conjecture. It turned out to be the wrong way to think about it, but I am still curious if it is true.

Let $z^n+a_{n-1} z^{n-1} + \cdots + a_1 z + a_0$ and $z^n + b_{n-1} z^{n-1} + \cdots + b_1 z + b_0$ be two polynomials all of whose roots are real, satisfying $a_k \geq b_k > 0$ for $1 \leq k \leq n-1$ and $a_0=b_0 > 0$. Is there a family of polynomials $c(t)(z) = z^n + c_{n-1}(t) z^{n-1} + \cdots + c_1(t) z + c_0(t)$ such that $c(t)$ has real roots for all $t$, and the function $c_k$ decreases monotonically from $c_k(0)=a_k$ to $c_k(1)=b_k$?

  • $\begingroup$ Without thinking much about it, so maybe I'm missing something: Write the two polynomials as $f=\prod_i (X-\alpha_i),\,\, g = \prod_i(X-\beta_i)$. Let $\gamma_i(t)$ be a path from $\alpha_i$ to $\beta_i$ and set $c(t) = \prod_i (X-\gamma_i(t))$. Since coefficients in degree $i$ are the values of a polynomial $p_i$ in the roots, we find $a_i = p_i(\alpha_1,...,\alpha_n),\,b_i = p_i(\beta_1,...,\beta_n), c_i(t)=p_i(\gamma_1(t),...,\gamma_n(t))$. In particular, $c_i(0)=a_i$ and $c_i(1)=b_i$. $\endgroup$
    – tj_
    May 29, 2015 at 21:06
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    $\begingroup$ @tj_ I want the $c_j$ to be decreasing functions. If you choose linear paths $\gamma$, I can give examples where the $c_j$ aren't monotone, and the same is true if I take the $\gamma_j$ to be linear in $\log$ coordinates. $\endgroup$ May 29, 2015 at 21:09
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    $\begingroup$ @Wolfgang: Simple linear interpolation between $x^3+3x^2+3x+1$ and $x^3 + 5x^2 + (17/4)x+1$ doesn't work since $x^3+4x^2+(29/8)x+1$ has non-real roots. $\endgroup$
    – Ken Fan
    Jun 4, 2015 at 22:25
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    $\begingroup$ You want the family to be a continuous path, of course. An easy observation: Call the first polynomial f, the second g. The answer is yes if f and g differ only in the coefficient of z^k: Then z f' - k f interlaces f in the sense of Definition 4.1 in [arxiv.org/pdf/1304.4132.pdf] (this uses that f has no positive roots). So does z g' - k g interlace g. But z f' - k f = z g' - k g. Hence f and g have a common interlacing. Now apply Lemma 4.5 in [arxiv.org/pdf/1304.4132.pdf]. $\endgroup$ Jun 6, 2015 at 20:48
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    $\begingroup$ One possibly helpful thing is to see on papers by Dedieu sciencedirect.com/science/article/pii/002240499290060S sciencedirect.com/science/article/pii/0022404994000034 Main result from there answers the question for the linear interpolation case. $\endgroup$
    – probably
    Feb 25, 2016 at 18:40

1 Answer 1


Without the condition that all coefficients are positive there is the following ounterexample: $$a(x)=(x+1)(x+2)(x+50).$$ For $a(x)-tx$ all zeros are real when $t\leq 0$ and when $t>300$. But for $t\in (100,200)$ only one zero is real. Two zeros disappeared from the negative ray, and after some time re-appear on the positive ray. Take $b(x)=a(x)-400x.$

Here is the graph of $a(x)/x$:


  • $\begingroup$ I'm sorry I don't see the equivalence to the special case you are considering. Why couldn't I have $x^2+a_1 x + a_2$ and $x^2 + b_1 x + b_2$ with positive real roots, but neither $x^2+a_1 x + b_2$ nor $x^2+b_1 x + a_2$ does? $\endgroup$ May 27, 2018 at 20:29
  • $\begingroup$ @David E Speyer: I edited my answer. I actually misunderstood what was asked. Now I hope it is correct. Sorry for the confusion. $\endgroup$ May 27, 2018 at 21:37

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