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The polynomial is $p(z)=\sum^n_{k=0} a_kz^k$. And I want to prove the following inequality on the unit disk$$\max_{B_1(0)}|p(z)|\geq |a_n|+|a_0|$$

By the maximum modulus principle, the maximum must be on the unit circle and greater than $|a_0|$ by considering $p(0)$. However, I cannot make further conclusions from this, since any attempt of using the triangle inequality will result in the opposite direction of the wanted result.

I have also seen a similar problem, although I can conclude $\max_{|z|=1}|p(z)|$is greater than any of the two on RHS, but since there is no relation of $\max_{k\in\{0,\ldots,n\}}|a_k|\geq|a_0|+|a_k|$, a tighter bound is needed.

I also tried expanding it into trig functions, and consider the roots, but it didn't work as expected.

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2 Answers 2

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This was trickier than I first thought. Perhaps there is a simpler solution, but here is one approach: Let $\omega = \exp(2\pi i/n)$. Then $$ \sum_{j=0}^{n-1} {(\omega^k)}^j = \begin{cases} n, & k \in \{ 0, n \} \\ 0, & 1 \le k \le n-1. \end{cases} $$ (This is a well-known property of roots of unity, or follows from computing the geometric sum if you prefer.)

Hence \begin{align} \frac1n \sum_{j=0}^{n-1} f(\omega^j z) &= \frac1n \sum_{j=0}^{n-1} \sum_{k=0}^{n} a_k {(\omega^j)}^k z^k \\ &= \frac1n \sum_{k=0}^{n} a_k z^k \sum_{j=0}^{n-1} {(\omega^k)}^j = a_0 + a_nz^n. \end{align}

Consequently, we get \begin{align} |a_0| + |a_n| &= \max_{|z|=1} |a_0 + a_nz^n | \\ &= \max_{|z|=1} \Big| \frac1n \sum_{j=0}^{n-1} f(\omega^j z) \Big| \\ &\le \frac1n \sum_{j=0}^{n-1} \max_{|z|=1} \big| f(\omega^j z) \big| = \max_{|z|=1} |f(z)|. \end{align}

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  • $\begingroup$ This is really helpful, thanks a lot $\endgroup$ May 17, 2017 at 2:44
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One gets that $$\max_{B_1(0)}|p(z)|\geq \sqrt{|a_n|^2+|a_{n-1}|^2+ \cdots+|a_0|^2}$$ by calculating $$\mathbb{E} \left[ |p(e^{i \Theta})|^2\right],$$ where $\Theta$ is a uniform random variable on $[0,2\pi]$.

Even though it does not answer precisely the question, it could be of interest.

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  • $\begingroup$ ,how does the greater than or equal to sign arise in your ineqluality $\endgroup$ Apr 4, 2022 at 17:01

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