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While thinking about this question of Clin, I wanted to consider the polynomial:

$P(z) = 1+x_1z+x_2z^2+\cdots+x_nz^n$, satisfying:

(I) $1\geq x_{1}\geq x_2\geq\cdots\geq x_{n}\geq0$ and $\sum\limits_{k=1}^{n}x_{k}=1$.

Then, $P(1) = 2$ and $P$ has no root (real or complex) inside the disc $|z| < 1$,

since, $\displaystyle (1-z)P(z) = 1+(x_1-1)z+(x_2-x_1)z+\cdots+(x_n-x_{n-1})z^n-x_nz^{n+1}$,

Now if $P$ has a root $z_0$ in the region $|z| < 1$, then,

$\displaystyle 1+(x_1-1)z_0+(x_2-x_1)z_0+\cdots+(x_n-x_{n-1})z_0^n - x_nz_0^{n+1} = 0 \implies \begin{align} 1 &\le |x_nz_0^{n+1}|+|(x_1-1)z_0|+|(x_2-x_1)z_0|+\cdots+|(x_n-x_{n-1})z_0^n| \\ &= x_n|z_0^{n+1}| + (1-x_1)|z_0| + \cdots + (x_{n-1}-x_n)|z_0^n| \\ &< x_n + (1-x_1)+\cdots + (x_{n-1}-x_n) = 1\end{align}$

leads to a contradiction! Hence, $P$ (polynomial of degree $n$) has no root in the region $|z| < 1$ flatly satisfies the conditions of Erdos-Lax Theorem, which states: $\displaystyle \max\limits_{|z| = 1}|P'(z)| \le \frac{n}{2}\max\limits_{|z| = 1}|P(z)|$, and equality holds for polynomials of type $P(z) = \alpha+\beta z^n$, where, $|\alpha| = |\beta|$.

Thus, $\displaystyle \sum\limits_{k=1}^{n} kx_k = P'(1) \le \frac{n}{2}P(1) = n$. Of course this bound can be improved for $P$ satisfying (I).

Now my questions are:

(1) What is the best constant $c_n$, for polynomials satisfying (I) such that: $$\displaystyle \max\limits_{|z| = 1}|P'(z)| \le c_n\max\limits_{|z| = 1}|P(z)|$$

and

(2) Does an upper bound exist for: $\displaystyle \max\limits_{|z| = 1} \lVert P\rVert_2^2 \, |P'(z)|$ and $\displaystyle \max\limits_{|z| = 1} (\lVert P\rVert_2^2-1) \, |P'(z)|$

where, $\lVert P\rVert_2 := \sqrt{1+|x_1|^2+\cdots+|x_n|^2}$.

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  • 2
    $\begingroup$ It might be simpler to see that there is no root for $|z|\lt1$ using $$ \begin{align} \left|1+\sum_{k=1}^nx_kz^k\right| &\ge1-\left|\sum_{k=1}^nx_kz^k\right|\\ &\gt1-\sum_{k=1}^nx_k\\ &=0 \end{align} $$ $\endgroup$ – robjohn Nov 16 '14 at 9:28
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This is for the first question about the best constant $c_n$. We find $c_n = \frac{n+1}{4}$.
I have no idea about the second question.


Let $A(z) = \sum\limits_{k=0}^m a_k z^k$ be any polynomial with non-negative coefficients. The coefficients of its derivative $A'(z)$ will be non-negative too. Over the unit circle $|z|= 1$, we have

$$|A(z)| = \left|\sum_{k=0}^m a_k z^k\right| \le \sum_{k=0}^m |a_kz^k| = \sum_{k=0}^m a_k = A(1)$$ This leads to

$$\max_{|z| = 1}|A(z)| = A(1)$$

By a similar argument, we have

$$\max_{|z| = 1} |A'(z)| = A'(1)$$

Apply this to the polynomial $P(z)$ at hand. we obtain

$$\begin{align} \max_{|z| = 1} |P(z)| &= P(1) = 1 + \sum_{k=1}^n x_k = 2\\ \max_{|z| = 1} |P'(z)| &= P'(1) = \sum_{k=1}^n k x_k \end{align}$$

As finite sequences, $(x_1,x_2,\ldots,x_n)$ is decreasing while $(1,2,\ldots,n)$ is increasing. By rearrangment inequality, we have

$$\sum_{k=1}^n kx_k \le \sum_{k=1}^n (n - k + 1) x_k$$ Taking average with the trivial equality $\sum\limits_{k=1}^n kx_k = \sum\limits_{k=1}^n k x_k$, we obtain $$P'(1) = \sum_{k=1}^n kx_k \le \frac12 \sum_{k=1}^n (k + (n - k + 1)) x_k = \frac{n+1}{2} \sum_{k=1}^n x_k = \frac{n+1}{2}$$

As a result,

$$\max_{|z| = 1} |P'(z)| = P'(1) \le \frac{n+1}{2} = \frac{n+1}{4} P(1) = \frac{n+1}{4} \max_{|z| = 1} |P(z)|$$

Notice when $x_1 = x_2 = \cdots = x_n = \frac{1}{n}$, the inequality above becomes an equality. This implies the best constant $c_n$ equals to $\frac{n+1}{4}$.

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  • $\begingroup$ (+1) Thank you very much! Nice answer!! :) $\endgroup$ – r9m May 20 at 0:06

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