I'm trying to prove the following two statements about a polynomial $p$ of degree $n$ with complex coefficients:

If $|p(x)|\le1$ for all real $x$ with $|x|\le1$, then every coefficient of $p$ has modulus $\le(10n)^n$.

If $|p(x)|\le1$ for all complex $x$ with $|x|\le1$, then every coefficient of $p$ has modulus $\le1$.

One way I tried attacking it (in the real case) was by induction: if the leading coefficient has modulus $\le K$, then every other coefficient has modulus $\le\frac{(10n-10)^{n-1}}{K+1}$. So it'd be enough to show that the modulus of the leading coefficient is at most, say, $20n-1$. But I'm not sure whether this is necessarily true. EDIT: no, it isn't true - thanks, @MaMing! So the induction method probably doesn't have much mileage in it.

Another idea (again in the real case): the constant coefficient clearly has modulus $\le1$, so maybe we can prove that the $x^k$-coefficient is $\le(10k)^k$ or $(10n)^k$, e.g. by considering the polynomial $p(nx)$?

The complex case looks like it should be doable using the maximum or minimum modulus principle and roots of unity, but I can't quite get this to work.

EDIT: in the complex case, |($z^k$-coefficient)|$=\frac{1}{2\pi}\Big|\int_{|z|=1}z^{-k-1}p(z)dz\Big|\le \sup_{|z|=1}\big|z^{-k-1}p(z)\big|\le1$. Thanks, @GEdgar!

Many thanks for any help with this!

  • $\begingroup$ You guess the leading coefficient is at most $20n-1$ is not correct. en.wikipedia.org/wiki/Chebyshev_polynomials $\endgroup$ – Ma Ming Apr 17 '13 at 19:55
  • $\begingroup$ Hint: can you compute a given coefficient of the polynomial by doing a certain integral around the circle? $\endgroup$ – GEdgar Apr 17 '13 at 20:24

First assume that $p$ is a real polynomial of degree $n \geq 1$. Without loss of generality we will assume that the leading coefficient of $p$ is positive.

Let $T_n$ be the Chebyshev polynomial of the first kind of degree $n$. The leading coefficient of $T_n$ is $2^{n-1}$. If $p$ is such that $|p(x)| < 1$ for $x \in [-1,1]$ then $\alpha \, T_n - p$ has $n$ roots in $[-1,1]$ for all $\alpha \geq 1$. This implies that the leading coefficient of $p$ is less than $2^{n-1}$.

If all roots of a polynomial $a_nx^n + a_{n-1}x^{n-1}+ \dotsc +a_0$ of degree $n$ lie in the interval $[-1,1]$ then

$$ |a_k| \leq |a_n| {n \choose k} $$

for all $k$ (expand its factored form). This holds therefore for both $T_n$ and $T_n - p$ and since $p = T_n - (T_n - p)$ we get the estimate

$$ |p_k| \leq 2 \cdot 2^{n-1} {n \choose k} < 4^n $$

for the coefficients of $p$.

This inequality also holds if $p$ is a complex polynomial as required. To see this let $\omega \in \mathbb{C}$ be such that $|\omega|=1$. Then $x \mapsto \mathrm{Re}\left(\omega \, p(x) \right)$ for $x \in \mathbb{R}$ is a real polynomial for which the coefficient bound above holds. Since $\omega$ is arbitrary the same bound must hold for the complex coefficients of $p$ itself.


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