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Suppose that we have two complex polynomials $p(z)=\sum_{k=0}^n p_kz^k$ and $q(z)=\sum_{k=0}^n q_kz^k$ and also that we have $|p_k|<|q_k|$ for $k=0,1,...,n$.

We say that a polynomial $r$ is between $p$ and $q$ if and only if $|p_k|<|r_k|<|q_k|$ for $k=0,1,...,n$.

If $p$ and $q$ have only real zeroes does there exist $r$ that is between $p$ and $q$ and also has only real zeroes?

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  • $\begingroup$ What motivates the between-ness definition for polynomials? It looks very strange to me (for example, if $r$ is between $p$ and $q$, then it need not be that $r(z + 1)$ is between $p(z + 1)$ and $q(z + 1)$). $\endgroup$
    – LSpice
    Mar 24, 2018 at 18:23

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At least, this is true if all roots of $p$ are pairwise distinct (and you do not need to assume anything about $q$ at all); this is why.

Writing $p(z)=C(z-z_1)\dotsb(z-z_n)$ with $z_1,\dotsc,z_n$ real shows that $p$ is a constant multiple of a polynomial with all its coefficients real. Normalizing, we can in fact assume that $C=1$, and so all coefficients of $p$ are real. Now choose a polynomial $s=\sum_{k=0}^n s_kz^k$ with all coefficients real, non-zero, and satisfying $s_kp_k\ge 0$. If, in addition, the coefficients $s_k$ are sufficiently small in absolute value, then $r(z):=p(z)+s(z)$ will have $n$ pairwise distinct real roots (as so has $p$), and the coefficients of $r$ will be between those of $p$ and of $q$ in the absolute value.


BTW, if all coefficients of $p$ are non-zero, then you can simply take $r=(1+\varepsilon)p$ with a sufficiently small $\varepsilon>0$. So, I suppose your question is essentially about the situation where some coefficients of $p$ vanish?

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