Convexity in the growth of analytic and subharmonic functions

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Problem 4.18 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 1 "Analogy between $\mu(r)$ and $M(r)$, $\nu(r)$ and $N(r)$"

If $f(0) = 0$ , $0 < r_{1} < r_{2} < R$ then

\frac{M(r_{2})}{M(r_{1})} \geq \frac{r_{2}}{r_{1}}.

Problem 4.19 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 1 "Analogy between $\mu(r)$ and $M(r)$, $\nu(r)$ and $N(r)$"

The function $\eta = \log \mu (e^{\xi })$ is represented in a rectangular coordinate system $\xi , \eta$ by a curve which is non-decreasing and convex.

Remark.

Consider the aggregate of the straight lines

\eta = \log \left|a_{0}\right|, \qquad \eta = \xi + \log \left|a_{1}\right|, \qquad \eta = 2\xi + \log \left|a_{2}\right|, \ldots \eta = n\xi + \log \left|a_{n}\right|, \ldots ,

omitting meaningless terms with $a_{n} = 0$ . How can $\mu (r)$ and $\nu (r)$ be interpreted in terms of this configuration?

Problem 4.41 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 2 "Further Results on $\mu(r)$ and $\nu(r)$"

In a rectangular coordinate system mark the points

(0, -\log \left|a_{0}\right|), \quad (1, -\log \left|a_{1}\right|), \quad (2, -\log \left|a_{2}\right|), \ldots , \quad (n, -\log \left|a_{n}\right|), \ldots

(omitting meaningless ones with $a_{n} = 0$ ) and from each point draw a ray directed vertically upwards. The smallest convex region $\mathfrak {R}$ which includes all these rays extends to infinity. Draw the supporting line with direction coefficient $\log r$ (the straight line on which lies at least one boundary point and no interior point of $\mathfrak {R}$ and which forms an angle with the positive real axis whose tangent is $\log r$ ). Interpret in the resulting figure $\log \mu (r)$ , $\nu (r)$ , $\log \rho_{n}$ .

Problem 4.172.1 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Problem 4.172 "[NO TITLE]"

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

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

or when

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

where $a$ and $b$ are arbitrary complex numbers, $n$ a non-negative integer.

Remark.

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).

Problem 10.26 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 10 "The Maximum Principle"

Prove that the function

u(z) = \varphi (\mathrm{Re}\left(z\right))

is subharmonic in the strip $a < \mathrm{Re}\left(z\right) < b$ if and only if the function $\varphi (x)$ is convex downward on the interval $a < x < b$ .

Problem 10.27 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 10 "The Maximum Principle"

Prove that the function $u\left(r e^{i \theta }\right) = \varphi (r)$ is subharmonic in the annulus $\rho < \left|z\right| < R$ if and only if the function $\varphi (r)$ is logarithmically convex on the interval $(\rho , R)$ , i.e., if for any three values $\rho < r_{1} < r_{2} < r_{3} < R$ the inequality holds

\varphi \left(r_{2}\right) \leqslant \varphi \left(r_{1}\right) \frac{\ln r_{3}-\ln r_{2}}{\ln r_{3}-\ln r_{1}}+\varphi \left(r_{3}\right) \frac{\ln r_{2}-\ln r_{1}}{\ln r_{3}-\ln r_{1}}.

Problem 10.28 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 10 "The Maximum Principle"

Prove that the function $u\left(r e^{i \theta }\right) = r^{\rho } \varphi (\theta )$ is subharmonic in the angle $\alpha < \theta < \beta$ if and only if the function $\varphi (\theta )$ is trigonometrically $\rho$ -convex on the interval $(\alpha , \beta )$ , i.e., if for any triple of values $\theta_{1}, \theta_{2}, \theta_{3}$ satisfying the conditions $\alpha < \theta_{1} < \theta_{2} < \theta_{3} < \beta$ and $\theta_{3}-\theta_{1} < \pi / \rho$ , the inequality holds

\varphi \left(\theta _{2}\right) \leqslant \varphi \left(\theta _{1}\right) \frac{\sin \rho \left(\theta _{3}-\theta _{2}\right)}{\sin \rho \left(\theta _{3}-\theta _{1}\right)}+\varphi \left(\theta _{3}\right) \frac{\sin \rho \left(\theta _{2}-\theta _{1}\right)}{\sin \rho \left(\theta _{3}-\theta _{1}\right)}.

Problem 10.29 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 10 "The Maximum Principle"

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

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

Prove that for $\rho < r_{1} < r_{2} < r_{3} < R$ the inequality holds

\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.)

Problem 10.31 from Шабунин М.И., Половинкин Е.С., Карлов М.И. "Сборник задач по теории функций комплексного переменного" [ШПК15]More problems in Section 10 "The Maximum Principle"

Let the function $f(z)$ be regular in the half-plane $\mathrm{Re}\left(z\right) > 0$ and satisfy the inequality

\left|f(z)\right| \leqslant \frac{M\left|z\right|}{(1+\left|z\right|)^{\alpha }}, \quad \alpha > 1 \quad (\mathrm{Re}\left(z\right) > 0)

Prove that the function

I(\varphi ) = \int _{0}^{\infty }\left|f\left(r e^{i \varphi }\right)\right| \frac{d r}{r}

is a downward convex function of $\varphi$ on the interval $-\pi / 2 < \varphi < \pi / 2$ .

References

[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.