MATHEMATICS

N. A. LEBEDEV

ON THE RANGE OF VALUES OF A FUNCTIONAL IN THE PROBLEM OF NON-OVERLAPPING DOMAINS

(Presented by Academician V. I. Smirnov on 18 III 1957)

Let (B) and (B^) be two arbitrary simply connected domains lying in the (w)-plane, having no common points, and such that the domain (B) contains the point (w=0), while the domain (B^) contains the point (w=\infty). Let (w=f(z)), (f(0)=0), be a function regular and univalent in the disk (|z|<1) and mapping the disk (|z|<1) onto (B), and let (w=F(\zeta)), (F(\infty)=\infty), be a function regular and univalent in (1<|\zeta|<+\infty) and mapping (|\zeta|>1) onto (B^*). We shall call the set of pairs of functions ((f(z), F(\zeta))) of the indicated type the class (\mathfrak M).

In this paper we find the domain (E) of values of the quantity (\xi=f(z_0)/F(\zeta_0)), for fixed (z_0) and (\zeta_0), respectively from (0<|z_0|<1) and (1<|\zeta_0|<\infty), in the class (\mathfrak M) (i.e., the set of all those values of the quantity (\xi) which it assumes when the pair of functions ((f(z),F(\xi))) ranges over the whole class (\mathfrak M)).

If ((f(z),F(\zeta))\in\mathfrak M) and (\varphi) and (\psi) are real numbers, then ((f(ze^{i\varphi}),F(\zeta e^{i\psi}))\in\mathfrak M). Hence we conclude that the domain (E) of values of the quantity
[
\xi=f(z_0)/F(\zeta_0)
]
coincides with the domain of values of the quantity (\xi=f(r)/F(\rho)), (r=|z_0|), (\rho=|\zeta_0|). Therefore in what follows we put (z_0=r), (\zeta_0=\rho).

It is easy to prove that the domain (E) of values of the quantity (\xi) is a closed connected bounded set with the point (\xi=0) removed. The determination of the domain (E) reduces ((^1)) to finding the minimum (J_0) of the expression
[
J=\left|\frac{f(r)}{F(\rho)}-a\right|
]
in the class (\mathfrak M), for an arbitrary point (a\in \overline E).

Let ((f(z),F(\zeta))) be a pair of functions from (\mathfrak M) for which this minimum (J_0) of the quantity (J) is attained. We shall call the functions of such a pair extremal. Relying on the variational method of G. M. Goluzin ((^2)), we obtain: if ((f(z),F(\zeta))\in\mathfrak M), then, for (|h|) sufficiently small, the class (\mathfrak M) also contains:

1) the pair of functions
[
f_(z)=f(z)+h\,\frac{f(z)}{f(z)-w_0},\qquad
F_(\zeta)=F(\zeta)+h\,\frac{F(\zeta)}{F(\zeta)-w_0},
\tag{1}
]
if (w_0) is an exterior point simultaneously for the domains (B) and (B^*);

2) the pair of functions
[
f_(z)=f(z)+h\,\frac{f(z)}{f(z)-f(z')}
-h\left(\frac{f(z)}{z'f'(z')^2}\right)\frac{zf'(z)}{z-z'}
]
[
+\bar h\,\frac{\overline{f(z')}}{z'f'(z')^2}\,
\frac{z^2 f'(z)}{1-\overline{z'}z}
+O(h^2),
]
[
F_(\zeta)=F(\zeta)+h\,\frac{F(\zeta)}{F(\zeta)-f(z')},
\tag{2}
]
for (|z'|<1);

3) the pair of functions

[
f_*(z)=f(z)+\frac{f(z)}{f(z)-F(\zeta')},
]

[
F_*(\zeta)=F(\zeta)+h\,\frac{F(\zeta)}{F(\zeta)-F(\zeta')}
-h\left(\frac{F(\zeta')}{\zeta'^2 F'(\zeta')^2}\right)
\frac{\zeta^2 F'(\zeta)}{\zeta-\zeta'}
+
\bar h\left(\frac{F(\zeta')}{\zeta'^2 F'(\zeta')^2}\right)
\frac{\zeta F'(\zeta)}{1-\bar\zeta'\zeta}
+O(h^2)
\tag{3}
]

for (|\zeta'|>1).

Let ((f(z),F(\zeta))) be a pair of extremal functions from (\mathfrak M), corresponding to some point (a\bar\in \bar E), and let ((f_(z),F_(\zeta))) be the varied functions obtained by one of the pairs of formulas (1), (2), or (3). Then

[
J_=\left|\frac{f_(r)}{F_*(\rho)}-a\right|
\geq
\left|\frac{f(r)}{F(\rho)}-a\right|
=J_0.
]

Relying on this inequality and formulas (1), in the usual way ({}^{(2)}) we prove that the domains (B) and (B^) in this case cover the whole (w)-plane, in the sense that there are no points external simultaneously to both domains (B) and (B^). If, however, we use formulas (2) and (3), we obtain the following two differential equations for the extremal functions:

[
e^{-i\alpha}\frac{z f'(z)^2}{f(z)}
\left(\frac{1}{f(z)-F(\rho)}-\frac{1}{f(z)-f(r)}\right)
=
\frac{A}{(r-z)(1-rz)},
]

[
e^{-i\alpha}\frac{\zeta F'(\zeta)^2}{F(\zeta)}
\left(\frac{1}{F(\zeta)-F(\rho)}-\frac{1}{F(z)-f(r)}\right)
=
\frac{B}{(\rho-\zeta)(1-\rho\zeta)},
\tag{4}
]

where

[
\alpha=\pi-\arg\left{\left(\frac{f(r)}{F(\rho)}-a\right)\frac{f(r)}{F(\rho)}\right},
\qquad
0\leq \alpha<2\pi;
]

[
A=e^{-i\alpha}\frac{r f'(r)}{f(r)}(1-r^2)>0;
\qquad
B=e^{-i\alpha}\frac{\rho F'(\rho)}{F(\rho)}(\rho^2-1)>0.
]

From the fact that the quantity

[
J=\left|\frac{f(re^{i\varphi})}{F(\rho)}-a\right|,
\qquad -\pi\leq \varphi\leq \pi,
]

attains its minimum at (\varphi=0), we conclude that (A) is a real number. From the fact that the quantity

[
J=\left|\frac{f(\psi(r,t))}{F(\rho)}-a\right|,
]

where (\psi(z,t)) is the solution of the Loewner equation

[
\frac{\partial\psi}{\partial t}
=
-\psi\,\frac{1+k\psi}{1-k\psi},
\qquad
k=\text{const},
]

[
|k|=1,\qquad \psi\big|_{t=0}=z,\qquad t\geq 0,
]

attains its minimum at (t=0), we conclude that (A>0). Analogously, we prove that (B>0).

Analyzing the differential equations (4), we conclude that the domains (B) and (B^*) have as a common boundary some closed simple analytic curve.

Extracting the square root from both sides of equations (4) and integrating the first of the resulting equalities with respect to (z) from (0) to (r), and the second with respect to (\zeta) from

(\rho) to (\infty), we have

[
-e^{-i\alpha/2}K\left(\sqrt{\frac{\xi}{\xi-1}}\right)
=
\sqrt{A}\,K(r)
=
\sqrt{B}\,\frac{1}{\rho}K\left(\frac{1}{\rho}\right),
]

[
K(k)=\int\limits_{0}^{\pi/2}\frac{du}{\sqrt{1-k^2\sin^2 u}}
\tag{5}
]

(where by the radical (\sqrt{1-k^2\sin^2 u}) we mean that branch which becomes unity at (u=0)).

Mark in the (w)-plane the point (F(1)). Let this point coincide with the point (f(e^{i\varphi_0})), (\pi\leq\varphi_0<3\pi). We shall now integrate the equalities obtained from (4), after taking the square root, with respect to (z) from (0) to (-1), then along the arc (|z|=1) counterclockwise from (-1) to (e^{i\varphi}), and, finally, with respect to (\zeta) from (1) to (\rho). Adding the equalities thus obtained, we have

[
-e^{-i\alpha/2}K\left(\sqrt{\frac{1}{1-\xi}}\right)
=
\sqrt{A}\,\frac{1}{2}K(\sqrt{1-r^2})+
]

[
+\sqrt{B}\,\frac{1}{2\rho}K\left(\sqrt{1-\frac{1}{\rho^2}}\right)
+\sqrt{A}\,K(r)\lambda i,
]

where (0\leq \lambda<2).

Eliminating from this equality the numbers (\sqrt{A}) and (B), with the aid of equalities (5), we obtain

[
\frac{
K\left(\sqrt{\frac{1}{1-\xi}}\right)
}{
K\left(\sqrt{\frac{\xi}{\xi-1}}\right)
}
=
\frac{1}{2}\left{
\frac{K(\sqrt{1-r^2})}{K(r)}
+
\frac{
K\left(\sqrt{1-\frac{1}{\rho^2}}\right)
}{
K\left(\frac{1}{\rho}\right)
}
\right}
+i\lambda
=\mu .
\tag{6}
]

Now, instead of the parameter (\alpha), one should consider the parameter (\lambda).

From (6), relying on the known formulas ((^3)) (p. 321), we have

[
\xi=\xi(\lambda)=-4^2 q\prod_{n=1}^{\infty}
\left(\frac{1+q^{2n}}{1-q^{2n-1}}\right)^8,
\qquad
q=e^{-\pi\mu}.
\tag{7}
]

Hence we conclude that (\xi(\lambda)) ((0\leq\lambda<2)) is a single-valued function. As (\lambda) varies in the interval ([0;2)), the point (\xi=\xi(\lambda)) describes the boundary of the domain (E) of values of the quantity (\xi=f(r)/F(\rho)) in the class (\mathfrak{M}) (the extremal functions corresponding to a given (\lambda) can also be found). The domain (E) is symmetric with respect to the real axis.

Consequences.

1. From formula (7) we have*

[
\left|\frac{f(r)}{F(\rho)}\right|
\leq
-\xi(0)
\leq
\frac{r}{\sqrt{(1-r^2)(\rho^2-1)}} .
]

In the first of the inequalities the equality sign is realized for all (r), (0<r<1), and (\rho), (1<\rho<+\infty); in the second—only for (\rho=1/r).

1. In the class (\mathfrak{M}) we have the estimate (((^3),) p. 321)

[
\left|\ln\left(1-\frac{f(r)}{F(\rho)}\right)\right|
\leq
\ln(1-\xi(0))
\leq
\ln\left(1+\frac{r}{\sqrt{(1-r^2)(\rho^2-1)}}\right)^{**}.
]

[
\text{* The estimate }\quad
\left|\frac{f(r)}{F(\rho)}\right|
\leq
\frac{r}{\sqrt{(1-r^2)(\rho^2-1)}}
\quad
\text{was obtained by another method by Yu. E. Alenitsyn }(^4).
]

[
\text{** By }\ln(1-w)\text{ is meant that branch which becomes zero at }w=0.
]

Here, in the first of the inequalities the equality sign is attained for all (r) and (\rho), while in the second—only when (\rho=1/r).

1. Relying on formulas (7), we obtain: the range of values of the quantity (f'(0)/F(\rho)) (or (f(r)/F'(\infty))) in the class (\mathfrak M) is the disk

[
\left|\frac{f'(0)}{F(\rho)}\right|
\leqslant
4\exp\left[
-\frac{\pi}{2}\,
\frac{K\left(\sqrt{1-1/\rho}\right)}{K(1/\rho)}
\right]
]

[
\left(
\left|\frac{f(r)}{F'(\infty)}\right|
\leqslant
4\exp\left[
-\frac{\pi}{2}\,
\frac{K\left(\sqrt{1-r^{2}}\right)}{K(r)}
\right]
\right)
\tag{8}
]

with the center removed. Hence we easily obtain the known estimate in the class (\mathfrak M):

[
\left|\frac{f'(0)}{F'(\infty)}\right|\leqslant 1.
]

1. The range of values of the system (\left(|f(r)|,\dfrac{1}{|F(\rho)|}\right)) in the class (\mathfrak M) is determined by the inequalities

[
0<|f(r)|,\frac{1}{|F(\rho)|}\leqslant -\xi(0).
]

1. Let (C) be the class of functions (f(z)), (f(0)=0), regular and univalent in the disk (|z|<1) and satisfying there the condition

[
f(z_1)f(z_2)\ne 1
]

(the Bieberbach–Eilenberg class), and let (C') be the subclass of univalent functions from (C). It follows from (7) that the range of values of the quantity (f(z)) in the class (C') is the closed domain with the point (\xi=0) removed, bounded by the curve

[
\xi=\xi_1(\lambda)=4iq^{1/2}\prod_{n=1}^{\infty}
\left(\frac{1+q^{2n}}{1-q^{2n-1}}\right)^4,
]

[
q=e^{-\pi\mu_1},\qquad
\mu_1=\frac{K\left(\sqrt{1-|z|^2}\right)}{K(|z|)}+i\lambda,
\qquad
0\leqslant \lambda<4.
]

Received
15 III 1957

CITED LITERATURE

¹ N. A. Lebedev, Vestn. Leningradsk. univ., No. 8, 29 (1955).
² G. M. Goluzin, Geometric Theory of Functions of a Complex Variable, Moscow–Leningrad, 1952.
³ I. M. Ryzhik, I. S. Gradshtein, Tables of Integrals, Sums, Series, and Products, Moscow–Leningrad, 1951.
⁴ Yu. E. Alyoshin, DAN, 115, No. 6 (1957).