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I wonder if there exists a complex polynomial $P(z),z\in \mathbb{C}$ s.t $$\forall |z|\leq 1, P(z)<1.$$

I know that using modulus maximum principle, we only need to find $$P(z)<1, \forall |z|=1.$$

I tried several polynomial (e.g. Chebyshev's polynomial) but did not succeed. Any ideas?

Any help would be appreciated !

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    $\begingroup$ What about $P(z)=cz$ for any $c\in \mathbb{C}$ with $\left|c\right|<1$? $\endgroup$
    – Amitesh Datta
    Commented Dec 23, 2014 at 7:02
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    $\begingroup$ constant polynomials also should work $\endgroup$
    – clark
    Commented Dec 23, 2014 at 7:03
  • $\begingroup$ More generally, you could also take $P(z)=a_0+a_1z+\cdots+a_nz^n$ for $a_0,a_1,\dots,a_n\in\mathbb{C}$ such that $\left|a_0\right|+\left|a_1\right|+\cdots+\left|a_n\right|<1$. $\endgroup$
    – Amitesh Datta
    Commented Dec 23, 2014 at 7:07
  • $\begingroup$ Do you mean $|P(z)| < 1$, or do you want $P(z)$ to be real? $\endgroup$
    – Robert Israel
    Commented Dec 23, 2014 at 7:15
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    $\begingroup$ Thanks for all the post! I forget to add one more condition, $P(z)=z+ \sum_{i\not=1} a_iz^i$. I will ask again in another question. $\endgroup$
    – Brian Ding
    Commented Dec 23, 2014 at 7:43

2 Answers 2

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What about $P(z)=\frac{1}{2}z$. Then $\forall z\in \mathbb{C}$ with $|z|\leq 1$ we have $P(z)=\frac{1}{2}z$ and $|P(z)|=|\frac{1}{2}||z|=\frac{1}{2}|z|\leq\frac{1}{2}$.

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Let $p(z)$ be any polynomial of degree $n$ and $M=\max_{|z|=1}|p(z)|.$ Define $f(z)=\frac{p(z)}{M+\epsilon},$ where $\epsilon>0,$ then $|f(z)|<1$ for all $|z|=1.$

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