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In the paper "The Kadison-Singer Problem" by Marcin Bownik (https://arxiv.org/pdf/1702.04578.pdf), the following Lemma (3.8) is proven:

Lemma: Let $p, q\in \mathbb{R}[x]$ be stable monic polynomials of the same degree. Suppose that every convex combination $(1 − t)p + tq, 0 \leq t \leq 1$, is also stable. Then for any $0 \leq t_0 \leq 1$, $\operatorname{maxroot}(((1 − t_0)p + t_0q)$ lies between $\operatorname{maxroot}(p)$ and $\operatorname{maxroot}(q)$.

Proof: WLOG, we can assume $0<t_0<1$ and $\operatorname{maxroot}(p)\leq\operatorname{maxroot}(q).$ The goal is then to show that

$m_p := \operatorname{maxroot}(p) \leq \operatorname{maxroot}(((1 − t_0)p + t_0q) \leq maxroot(q) =: m_q.$ The second inequality is easy. The first is proven by contradiction:

Suppose that $(1 − t_0)p + t_0q$ has no roots in $[m_p, m_q]$. This implies that $(1−t_0)p+t_0q > 0$ for all $x \geq m_p$. Then $q(m_p)>0$ and $q$ must have (counting multiplicity) at least 2 roots to the right of $m_p$. Letting now $D$ be the open disk in $\mathbb{C}$ centered at $\frac{m_p+m_q}{2}$ and radius $\frac{m_q-m_p}{2}$. It is claimed then that

$$ ((1 − t)p + tq)(z) \neq 0 \text{ for all } z \in\partial D \text{ and } t_0 \leq t \leq 1. $$

As by assumption, the convex combination of the polynomials is also real-stable, it is sufficient to check this for $z=m_p$ and $z=m_q$. As $p(m_p)=0<q(m_p)$, this is clear for $m_p$ as $t_0>0$. However, for $t=1$, we have that

$$ ((1 − t)p + tq)(z)\mid_{t=1,z=m_q} = q(m_q)=0. $$

Clearly contradicting the statement that is made in the paper that

$$ \inf_{(z,t)\in \partial D\times [t_0,1]} \vert ((1 − t)p + tq)(z)\vert>0,$$ which then itself is used to obtain a contradiction to the assumption that there are no zeros in $[m_p, m_q]$ via Rouche's Theorem.

So I neither see how they can claim that the infimum is positive nor how the proof should be adapted to correct for this perceived mistake. Thanks for any help.

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Indeed, the proof in the linked paper does not seem valid.

Here is a "real" proof of the lemma in question, without using complex analysis:

Let $m:=$maxroot and $q_t:=(1-t)p+tq$, so that $q_0=p$ and $q_1=q$. As in your post, without loss of generality (wlog) $m(p)\le m(q)$. If $m(p)=m(q)$, then $m(p)$ is a common root of both $p$ and $q$, and hence $m(p)$ is a root of $q_t$ for all $t$, so that $m(q_t)\ge m(p)$, as desired. So, wlog $m(p)<m(q)$, and then, by shifting and rescalling, wlog \begin{equation*} m(p)=0,\quad m(q)=1. \end{equation*} We want to show that \begin{equation*} m(q_t)\ge0 \end{equation*} for all $t\in[0,1]$. Let \begin{equation*} t_*:=\min\{t\in[0,1]\colon m(q_s)\ge0\ \forall s\in[t,1]\}. \end{equation*} This $\min$ exists in view of the continuity of $q_s$ in $s$ and the compactness of $[0,1]$.

Moreover, for each $t\in[0,t_*]$, $q_t$ is a convex combination of $p=q_0$ and $q_{t_*}$. So, wlog $t_*=1$, so that \begin{equation*} m(q_{t_n})<0 \tag{1} \end{equation*} for some sequence $(t_n)$ such that $t_n\uparrow1$.

One of the following two cases must occur:

Case 1: there is a root $r\in(0,1)$ of $q$ of an odd multiplicity. Then $q_1=q$ changes its sign at $r$. So, for all $t\in(0,1)$ close enough to $1$, $q_t$ will change its sign in some small enough neighborhood of $r$ and thus will have a root close enough to $r$, which will contradict (1).

Case 2: there is no root $r\in(0,1)$ of $q$ of an odd multiplicity. Then $q$ does not change its sign on $(0,1)$ from $+$ to $-$ or vice versa. Therefore and because $q(0)>0$ (as was shown in your post), we see that $q\ge0$ on $[0,1]$ (and $q>0$ on $(1,\infty)$). Moreover, $1$ is a root of $q$. Also, $q_t>0$ on $[0,\infty)$ for all $t>0$, because $p>0$ on $(0,\infty)$. So, switching from $q$ to $q_t$, we will lose the (multiple) root $1$ of $q$, as well as all the other roots of $q$ in $(0,\infty)$ (all of them of an even multiplicity). Moreover, if $t$ is close enough to $1$, this loss cannot be regained elsewhere. Thus, we get a contradiction with the assumption that $q_t$ is stable (that is, with the assumption that all roots of $q_t$ are real).


Details on the loss and no gains in Case 2: We have $q_t/t=q+sp$, where $s:=(1-t)/t$ and $t:=t_n$. So, $t\uparrow1$ and hence $s\downarrow0$. For a real polynomial $P$ and an interval $I\subseteq\mathbb R$, let $N_I(P)$ denote the number of roots of $P$ in $I$ (counting the multiplicities), so that $N_{\mathbb R}(P)$ is the number of real roots of $P$; moreover, $N_{\mathbb R}(P)$ equals the degree of $P$ if $P$ is stable.

In view of (1), $N_{(0,\infty)}(q+sp)=0$, whereas $N_{(0,\infty)}(q)\ge2$ (this is what was referred to as the loss). On the other hand, the condition that the $q_t$'s are stable monic polynomials of the same degree implies that $N_{\mathbb R}(q+sp)=N_{\mathbb R}(q)$. So, to get a contradiction, it suffices to show that \begin{equation*} N_{(-\infty,0]}(q+sp)\le N_{(-\infty,0]}(q) \tag{2} \end{equation*} for all small enough $s>0$ (this is what was referred to as the no-gain).

By standard and easy bounds on the roots of a polynomials and by compactness, eventually (that is, for all small enough $s>0$) every root of $q+sp$ will be in arbitrarily small neighborhood of a root of $q$. Let $v$ be any (necessarily real) root of $q$, so that \begin{equation*} q(x)=(x-v)^l q_1(x) \end{equation*} for all real $x$, where $l$ is the natural number equal the multiplicity of the root $v$ of $q$, so that $q_1$ is a polynomial with $b:=q_1(v)\ne0$. Let then \begin{equation*} p(x)=(x-v)^k p_1(x) \end{equation*} for all real $x$, where $k$ is the unique nonnegative integer such that $p_1$ is a polynomial with $a:=p_1(v)\ne0$.

One of the following two cases must occur:

Case 2.1: $l\le k$. Then, with $u:=x-v$, \begin{equation*} q(x)+sp(x)=u^l(q_1(x)+su^{k-l}p_1(x))\sim bu^l \end{equation*} as $s\to0$ and $x\to v$, so that $N_V(q+sp)=l=N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$.

Case 2.2: $l>k$. Then, again with $u:=x-v$, \begin{equation*} q(x)+sp(x)=u^kp_1(x)\big(R(u)u^{l-k}+s\big), \end{equation*} where \begin{equation*} R(u):=\frac{q_1(v+u)}{p_1(v+u)}. \end{equation*} Note that $(R(u)u^{l-k})'=R(u)(l-k)u^{l-k-1}+R'(u)u^{l-k}\sim\dfrac ba\,(l-k)u^{l-k-1}$ as $u\to0$. So, $R(u)u^{l-k}$ is monotonic with nonzero derivative in $u$ in a right neighborhood of $0$ and in a left neighborhood of $0$. It follows that the equation $R(u)u^{l-k}+s=0$ has no more than two roots $u$ in a neighborhood of $0$ if $l-k$ is even (and hence $\ge2$, by the Case 2.2 condition) and no more than one root $u$ in a neighborhood of $0$ if $l-k$ is odd. So, $N_V(q+sp)\le k+2\le l=N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$ if $l-k$ is even, and $N_V(q+sp)\le k+1\le l=N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$ if $l-k$ is odd.

So, in all cases $N_V(q+sp)\le N_V(q)$ for some neighborhood $V$ of $v$ and all small enough $s>0$. Thus, (2) is proved. $\Box$

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  • $\begingroup$ Thanks a lot. If I understand correctly, the argument for case 1 seems a continuity argument of sort $\lim_{t_n\to t_*} m(q_{t_n})=m(q_{t_*})>m_p=0$, which leads to the contradiction. Did I understand this correctly? If so, is it clear that $m()$ is continuous and secondly, why wouldn't this hold for case 2 as well? $\endgroup$
    – Strickland
    Mar 23 at 6:41
  • $\begingroup$ @Strickland : Thank you for your comment. Somehow, I thought only of the roots of $q$ in $(0,1)$ of odd multiplicities. This is now fixed. $\endgroup$ Mar 23 at 14:11
  • $\begingroup$ Thanks for the clarification. If still have some questions though, if you do not mind. 1. When you argue that $R(u)u^{l-k}+s=0$ has no more than two roots $u$ resp. one, I assume you base this on monotonicity. However, could it not have a root of multiplicity 4 for example? 2. It is immediately assumed that $m(p)<m(q)$. What if they are equal? $\endgroup$
    – Strickland
    Mar 24 at 6:41
  • $\begingroup$ @Strickland : (i) Such multiple roots cannot exist, because $R(u)u^{l-k}$ has a nonzero derivative in those right and left neighborhoods. I have now added this additional explanation. (ii) I have also added details on $m(p)<m(q)$. $\endgroup$ Mar 24 at 13:59

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