approximation of products of polynomials

Question

I am wondering whether the following can be proved:

Suppose $p(z)$ and $q(z)$ are polynomials of degree $n$ with real coefficients and leading coefficient 1. Moreover, they have quite different roots: for example, $p$ only have roots inside circle $\{z: |z| = 0.9\}$, and $q$ only have roots outside circle $\{z: |z| = 1.1\}$. Moreover, they satisfy that for any$z$ on unit circle, $c < |p(z)| < C$ and $c < |q(z)| < C$ for constant $c$ and $C$. (Ideally I it would be good to weaken these assumptions of the separation of the roots).

Suppose there are other two degree $n$ polynomials $\hat{p}$ and $\hat{q}$ with leading coefficients 1 such that for any $z$ on unit circle,

$$ |p(z)\hat{q}(z) - q(z)\hat{p}(z)| \le \epsilon.$$

There exists $\epsilon' \le \epsilon^{c_0} n^{c_0} C^{c_0}c^{c_0}$ for some constant $c_00$ such that for any $z$ on unit circle

$$|p(z) - \hat{p}(z)| <\epsilon'$$ and $$|q(z) - \hat{q}(z)| <\epsilon'$$

I think I roughly know how to get $\epsilon' \le \epsilon^{c_1} c_2^n$ for constant $c_1$ and $c_2$ but nothing better. Any pointer would also be very appreciated! (I even don't know how to search the literature ..) (Also if there is any technique that can partially solved problem, it would also be great!)

Thanks a lot!

EDIT: added the requirement that $\hat{p}$ and $\hat{q}$ need to have degree $n$. EDIT2: added the upper and lower bound condition for $p$ and $q$ on unit circle

Answer 1

$\let\eps\varepsilon$You hardly can do better. Set $a=0.9$, $b=1.1$, and take $p(z)=(z-a)^n$, $q(z)=(x-b)^n$. we have $\eps=p\bar q-q\bar p$ with some $\bar p$ and $\bar q$ of degree $n-1$ (then one may set $\hat p=p+\bar p$, $\hat q=q+\bar q$). Moreover, the first $n-1$ derivatives of $p\bar q$ at $b$ vanish, which means that $\bar q$ is simply the sum of the first $n$ terms of the Taylor expansion of $1/p$ at $b$ (multiplied by some constant), i.e. $$ \bar q(z)=\frac{c}{(b-a)^n}\sum_{i=0}^{n-1}{n+i-1\choose i}\left(\frac{z-b}{a-b}\right)^i. $$ Checking the value at $b$, we get $c=\eps$.

Finally, let us look at $\bar q(-1)$. The absolute values of terms in the corresponding sum grow pretty fast (since $\frac{1+b}{b-a}>2$), so this sum has the order of the last term, i.e. almost $\eps C^n$ for $C=\frac{4(1+b)}{(b-a)^2}$.

If you need distinct roots, just perturb the roots a bit.