Hadamard's three circles theorem

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Prove the theorem: if f(z)f(z) is regular for r<∣z∣<Rr < \left|z\right| < R, and M(ρ)M(\rho ) is the maximum modulus of f(ρeφi)f(\rho e^{\varphi i}), then for r<ρ<Rr < \rho < R the quantity M(ρ)M(\rho ) is a convex function of ln⁡ρ\ln \rho (Hadamard's Three-Circles Theorem).

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Remark.
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Hint. It is useful to consider the function znf(z)=F(z)z^{n}f(z) = F(z), choosing nn so that the quantities M(r)M(r) and M(R)M(R) for F(z)F(z) are equal.

Let the function f(z)f(z) be regular in the disk ∣z∣<R\left|z\right|<R. Suppose

0<r1<r2<r3<R. 0 < r_{1} < r_{2} < r_{3} < R.

Then

log⁡M(r2)≤log⁡r2−log⁡r1log⁡r3−log⁡r1log⁡M(r3)+log⁡r3−log⁡r2log⁡r3−log⁡r1log⁡M(r1). \log M(r_{2}) \leq \frac{\log r_{2}-\log r_{1}}{\log r_{3}-\log r_{1}}\log M(r_{3}) + \frac{\log r_{3}-\log r_{2}}{\log r_{3}-\log r_{1}}\log M(r_{1}).

This means that in an orthogonal system of coordinates the graph of log⁡M(r)\log M(r) as a function of log⁡r\log r appears as a convex curve. (Hadamard's three circle theorem.) [Examine zαf(z)z^{\alpha }f(z) with suitably chosen α\alpha.]

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(continued). The function log⁡M(r)\log M(r) is a strictly convex function of log⁡r\log r unless f(z)f(z) is of the form azαaz^{\alpha }, a,αa,\alpha constants, α\alpha real: this is the only type of function for which the inequality in problem 304 becomes an equality.

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Suppose that the function f(z)f(z) is regular for ∣z∣<R\left|z\right|<R but not of the form czncz^{n}, cc constant, nn integer; let

I2(r)=12π∫02π∣f(reiϑ)∣2 dϑ I_{2}(r) = \frac{1}{2\pi }\int _{0}^{2\pi } \left|f(re^{i\vartheta })\right|^{2}\, d\vartheta

denote the arithmetic mean of ∣f(z)∣2\left|f(z)\right|^{2} on the circle ∣z∣=r\left|z\right|=r, r<Rr<R. The function I2(r)I_{2}(r) is monotone increasing with rr and log⁡I2(r)\log I_{2}(r) is a strictly convex function of log⁡r\log r.

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Let f(z)f(z) be regular for ∣z∣<R\left|z\right|<R;

G(r)=e12π∫02πlog⁡∣f(reiϑ)∣ dϑ \mathfrak {G}(r) = e^{\frac{1}{2\pi }\int _{0}^{2\pi } \log \left|f(re^{i\vartheta })\right|\, d\vartheta }

denotes the geometric mean of ∣f(z)∣\left|f(z)\right| on the circle ∣z∣=r\left|z\right|=r, r<Rr<R. The function G(r)\mathfrak {G}(r) is monotone increasing with rr and a convex function of log⁡r\log r (in the wide sense).

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The function f(z)f(z) is supposed to be regular for ∣z∣<R\left|z\right|<R and not a constant. We put

I(r)=12π∫02π∣f(reiϑ)∣ dϑ,r<R. I(r) = \frac{1}{2\pi }\int _{0}^{2\pi } \left|f(re^{i\vartheta })\right|\, d\vartheta , \quad r<R.

The function I(r)I(r) is monotone increasing with rr and log⁡I(r)\log I(r) is a convex function of log⁡r\log r. [Problems 299, 304.]

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We suppose that f(z)f(z) is regular and not constant for ∣z∣<R\left|z\right|<R and that pp is a positive number. We define

Ip(r)=12π∫02π∣f(reiϑ)∣p dϑ,r<R. I_{p}(r) = \frac{1}{2\pi }\int _{0}^{2\pi } \left|f(re^{i\vartheta })\right|^{p}\, d\vartheta , \quad r<R.

The function Ip(r)I_{p}(r) is monotone increasing with rr and log⁡Ip(r)\log I_{p}(r) is a convex function of log⁡r\log r. (Cf.problems 306, 308, 307, 267 and 304, for an analogous case see Part IV, problem 19.)

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Consider a regular harmonic function in the disk ∣z∣<R\left|z\right|<R. We denote by A(r)A(r) its maximum on the circle ∣z∣=r\left|z\right|=r, r<Rr<R. When 0<r1<r2<r3<R0<r_{1}<r_{2}<r_{3}<R we have

A(r2)≤log⁡r2−log⁡r1log⁡r3−log⁡r1A(r3)+log⁡r3−log⁡r2log⁡r3−log⁡r1A(r1), A(r_{2}) \leq \frac{\log r_{2}-\log r_{1}}{\log r_{3}-\log r_{1}}A(r_{3}) + \frac{\log r_{3}-\log r_{2}}{\log r_{3}-\log r_{1}}A(r_{1}),

i.e.A(r)A(r) is a convex function of log⁡r\log r.

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Deduce Hadamard's three circle theorem 304 from problem 320 and vice versa problem 320 from Hadamard's three circle theorem.

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The function η=log⁡M(eξ)\eta = \log M(e^{\xi }) is represented in a rectangular coordinate system ξ,η\xi , \eta by a curve which is always increasing and is strictly convex, apart from certain exceptional polynomials for which the curve degenerates into a straight line.

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Let α\alpha be fixed, 0<α<10 < \alpha < 1. The quotient M(αr)/M(r)M(\alpha r)/M(r) is strictly decreasing with increasing rr, apart from certain polynomials (see problem 20).

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With the aid of Problem 123 of Part II, obtain a new proof for the Hadamard three-circle theorem (Problem 304 of Part III).

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Remark.
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E. Landau.

The modular graph (see Volume I, p. 130) of an entire function g(z)g(z) is everywhere convex from below in just two cases: Either when

g(z)=(az+b)n g(z) = (az+b)^{n}

or when

g(z)=eaz+b, g(z) = e^{az+b},

where aa and bb are arbitrary complex numbers, nn a non-negative integer.

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Remark.
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See the definition preceding Problem 130 of Part III. Cf. G. Polya, problem: Jber. deutsch. Math. Ver., Section 2, 35, p. 48 (1926). Solution by H. Kneser, op. cit. 35, p. 119 (1926), and by N. Tchebotarev, op. cit. 36, pp. 40-41 (1927).

Let the function f(z)f(z) be regular in the annulus ρ<∣z∣<R\rho < \left|z\right| < R. Denote

M(r)=max⁡∣z∣=r∣f(z)∣ M(r) = \max _{\left|z\right| = r}\left|f(z)\right|

Prove that for ρ<r1<r2<r3<R\rho < r_{1} < r_{2} < r_{3} < R the inequality holds

ln⁡M(r2)⩽ln⁡r3−ln⁡r2ln⁡r3−ln⁡r1ln⁡M(r1)+ln⁡r2−ln⁡r1ln⁡r3−ln⁡r1ln⁡M(r3). \ln M\left(r_{2}\right) \leqslant \frac{\ln r_{3}-\ln r_{2}}{\ln r_{3}-\ln r_{1}} \ln M\left(r_{1}\right)+\frac{\ln r_{2}-\ln r_{1}}{\ln r_{3}-\ln r_{1}} \ln M\left(r_{3}\right).

(Hadamard's three-circle theorem.)

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References

[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.