Inequalities for polynomials in the complex plane

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Let KK be a compact, connected subset of C\mathbb {C} containing more than one point and let ff be a one-to-one conformal map C∖K\mathbb {C} \setminus K onto Δ={z:∣z∣<1}\Delta = \left\{ z : \left|z\right| < 1\right\} with f(∞)=0f(\infty ) = 0. If pp is a polynomial of degree nn for which ∣p(z)∣≤1\left|p(z)\right| \leq 1 for z∈Kz \in K, prove that

∣p(z)∣≤∣f(z)∣−nfor z∈C∖K. \left|p(z)\right| \leq \left|f(z)\right|^{-n} \quad \text{for } z \in \mathbb {C} \setminus K.
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Prove the theorem: if a polynomial f(z)f(z) of degree nn does not exceed MM in modulus on the interval −1⩽z⩽1-1 \leqslant z \leqslant 1, then for any zz the inequality ∣f(z)∣⩽M(a+b)n\left|f(z)\right| \leqslant M(a+b)^{n} holds, where aa and bb are the semi-axes of the ellipse passing through the point zz and having its foci at the points z=±1z = \pm 1.

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Remark.
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Hint. Set z=12(ζ+1ζ)z = \dfrac {1}{2}\left(\zeta +\dfrac {1}{\zeta }\right) and apply the theorem on the modulus to the function

ζ−nf(ζ2+12ζ). \zeta ^{-n}f\left(\frac{\zeta ^{2}+1}{2\zeta }\right).

Suppose that f(z)f(z) is a polynomial of degree nn and that

∣f(z)∣≤M \left|f(z)\right| \leq M

on the real interval −1≤z≤1-1\leq z\leq 1. Then we have

∣f(z)∣≤M(a+b)n \left|f(z)\right| \leq M(a+b)^{n}

for any zz outside this interval; aa and bb are the semi-axes of the ellipse through zz and with foci −1-1 and 11. What does the proposition imply for z→∞z\to \infty?

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We assume that f(z)f(z) is a polynomial of degree nn, that E1E_{1} and E2E_{2} denote two homofocal ellipses with semi-axes a1,b1a_{1},b_{1} and a2,b2a_{2},b_{2}, a1<a2a_{1}<a_{2}, b1<b2b_{1}<b_{2}. The maximum of ∣f(z)∣\left|f(z)\right| on E1E_{1} and E2E_{2} is denoted by M1M_{1} and M2M_{2} resp.; then

M1(a1+b1)n≥M2(a2+b2)n. \frac{M_{1}}{(a_{1}+b_{1})^{n}} \geq \frac{M_{2}}{(a_{2}+b_{2})^{n}}.

Derive problems 269 and 270 from this proposition.

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Let P(z)P(z) be a polynomial of degree nn, and let M(r)=max⁡∣z∣=r∣P(z)∣M(r) = \max_{\left|z\right| = r}\left|P(z)\right|. Prove that for 0<r1<r20 < r_{1} < r_{2} the inequality

M(r1)r1n⩾M(r2)r2n, \frac{M\left(r_{1}\right)}{r_{1}^{n}} \geqslant \frac{M\left(r_{2}\right)}{r_{2}^{n}},

holds, and equality, for even one pair of values r1r_{1} and r2r_{2}, is possible only for a polynomial of the form P(z)=aznP(z) = a z^{n}.

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Let P(z)=zn+a1zn−1+…+anP(z) = z^{n}+a_{1} z^{n-1}+\ldots +a_{n}. Prove that either the inequality ∣P(z)∣>1\left|P(z)\right| > 1 holds at some point of the circle ∣z∣=1\left|z\right| = 1, or P(z)≡znP(z) \equiv z^{n}.

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Let P(z)P(z) be a polynomial of degree nn satisfying, on the interval (−1,1)(-1,1), the inequality ∣P(z)∣⩽M\left|P(z)\right| \leqslant M. Prove that at every point z0z_{0} lying outside this interval the inequality ∣P(z0)∣⩽M(a+b)n\left|P\left(z_{0}\right)\right| \leqslant M(a+b)^{n} holds, where aa and bb are the semi-axes of the ellipse with foci −1-1 and 11 passing through the point z0z_{0}.

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References

[CJP+11] Chen J., Jiang G., Pan Y., Qin T., Tong Y., Wu Q., Xu S., Li T. Problems and Solutions in Mathematics. 2nd ed., World Scientific, 2011.
[ГюКу03] Гюнтер Н.М., Кузьмин Р.О. Сборник задач по высшей математике. 13-е изд., Издательство «Лань», 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[ШПК15] Шабунин М.И., Половинкин Е.С., Карлов М.И. Сборник задач по теории функций комплексного переменного. 4-е изд., БИНОМ. Лаборатория знаний, 2015.