Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 18, 1993, 343–354

APPROXIMATION OF THE HERSCH–PFLUGER DISTORTION FUNCTION

Dariusz Partyka

Maria Curie–Skłodowska University, Department of Mathematics pl. M. Curie–Skłodowskiej 1, PL–20–031 Lublin, Poland; partyka@plumcs11.bitnet

Abstract. This paper aims at giving a method of approximating the distortion function Φ_{K} by means of elementary functions. As a consequence, new bounds for the function Φ_{K} are established. Moreover, the n-dimensional counterpart Φ_{K,n} of Φ_{K}, n = 2, 3, … is considered.

0. Introduction

Let Φ_{K}(r) = μ^{−1}(μ(r)/K), 0 < r < 1, Φ_{K}(0) = 0, Φ_{K}(1) = 1, K > 0, where μ stands for the module of the Grötzsch extremal domain B^{2} \ [0, r]. Φ_{K} is called the Hersch–Pfluger distortion function, cf. [HP], and it plays an important role in the theory of plane quasiconformal mappings. In the first section we examine the functions φ_{K,t}, φ̃_{K,t}, ψ_{K,t}, ψ̃_{K,t}, depending on a real parameter t ≥ 1, defined by (1.8) and (1.20). We show (Theorem 1.3 and Corollary 1.4) that these sequences are monotonically convergent to the function Φ_{K}. Putting t = 2^{n}, n = 0, 1, 2, …, we are able to approximate Φ_{K} in a simple way by elementary functions with an arbitrarily preassigned accuracy. In Section 2 we derive (Theorems 2.1 and 2.2) new upper and lower bounds for the function Φ_{K}, which improve some recent results obtained by Anderson, Vamanamurthy, Vuorinen [AVV1], [AVV3] and Zaja̧c [Z1], [Z2]. In the last section of this paper we study the approximation problem for the n-dimensional counterpart Φ_{K,n}, cf. [V], [AVV2], of Φ_{K}, n = 2, 3, . . .. We establish Theorem 3.1, the n-dimensional counterpart of Theorem 1.3.

Actually, in view of Theorem 1.5 and Corollary 1.6 the sequences ψ_{K,2^{n}}, ψ̃_{K,2^{n}}, n = 0, 1, 2, … are convergent to the function Φ_{K} very fast. So they may be used conveniently for the calculation of the values of Φ_{K} by a computer. The details will be given in a separate paper [P] in which we shall also estimate the familiar functions μ, μ^{−1} and the distortion function λ, cf. [LV], [L], introduced by Lehto, Virtanen and Väisälä in [LVV], as an application of Theorems 1.3, 1.5 and Corollaries 1.4, 1.6.

1991 Mathematics Subject Classification: Primary 30C60; Secondary 41A25.

1. Basic approximation theorems

The function Φ_{K} was studied by many mathematicians. Recently Anderson, Vamanamurthy, Vuorinen, cf. [AVV1], [AVV3], [VV], and Zaja̧c, cf. [Z1], [Z2] obtained many interesting results concerning the properties of the function Φ_{K} and its estimates. We shall now state some inequalities concerning Φ_{K} which will be used later on.

Lemma 1.1. ^{∗)} For any p > 1 and 0 < x < 1

(1.1)  Φ_{K}(x^{p}) < Φ_{K}^{p}(x),  K > 1

and

(1.2)  Φ_{K}(x^{p}) > Φ_{K}^{p}(x),  0 < K < 1.

Proof. We shall first prove by induction that the inequality (1.1) holds for all p = 1 + 1/n, n = 1, 2, . . .. To this end we apply the fact shown in [AVV1] that for any 0 < c < 1 and K > 1 the function

(1.3)  f(t) = \frac{Φ_{K}(ct)}{Φ_{K}(t)},  0 < t ≤ 1

is strictly increasing. Thus setting c = x and t_{1} = x < t_{2} = 1, we have f(t_{1}) < f(t_{2}) for n = 1 so

(1.4)  Φ_{K}(x^{1+1/n}) < Φ_{K}^{1+1/n}(x).

If now n = 1, 2, … is any positive integer and (1.4) holds then setting c = x^{1/(n+1)} and t_{1} = x < t_{2} = x^{n/(n+1)} we have f(t_{1}) < f(t_{2}) from which

(1.5)  Φ_{K}(x^{1+1/(n+1)}) < Φ_{K}(x)\frac{Φ_{K}(x)}{Φ_{K}(x^{n/(n+1)})}.

But, in view of the induction assumption we get

Φ_{K}(x) = Φ_{K}((x^{n/(n+1)})^{1+1/n}) < Φ_{K}^{1+1/n}(x^{n/(n+1)}).

This and (1.5) yield

Φ_{K}(x^{1+1/(n+1)}) < Φ_{K}^{2}(x)Φ_{K}^{−n/(n+1)}(x) = Φ_{K}^{1+1/(n+1)}(x)

∗) The inequalities (1.1) and (1.2) have been proved independently in a different way in [VV]. so the inequality (1.1) is true for every p = 1 + 1/n, n = 1, 2, . . .. If now m is any positive integer and the inequality (1.1) holds for p = 1 + m/n, where n is any fixed positive integer, then setting c = x^{m/n} and t_{1} = x^{1+1/n} < t_{2} = x we have f(t_{1}) < f(t_{2}) so

Φ_{K}(x^{1+(m+1)/n}) < Φ_{K}(x^{1+1/n})\frac{Φ_{K}(x^{1+m/n})}{Φ_{K}(x)} < Φ_{K}^{1+1/n}(x)\frac{Φ_{K}^{1+m/n}(x)}{Φ_{K}(x)}
= Φ_{K}^{1+(m+1)/n}(x),

because of (1.4). This way the inequality (1.1) holds for any rational number p > 1. It follows from the continuity of the function Φ_{K} that for every p > 1, Φ_{K}(x^{p}) ≤ Φ_{K}^{p}(x) as 0 < x < 1. Then there exists a rational number q between 1 and p which allows us to improve the above inequality because

Φ_{K}(x^{p}) = Φ_{K}((x^{q})^{p/q}) ≤ Φ_{K}^{p/q}(x^{q}) < (Φ_{K}^{q}(x))^{p/q} = Φ_{K}^{p}(x).

A substitution x := Φ_{1/K}(x) in the inequality (1.1) yields the inequality (1.2) so the lemma is proved.

Lemma 1.2. For any p > 1 and 0 < x < 1

(1.6)  Φ_{K}(4^{1−p}x^{p}) > 4^{1−p}Φ_{K}^{p}(x),  K > 1

and

(1.7)  Φ_{K}(4^{1−p}x^{p}) < 4^{1−p}Φ_{K}^{p}(x),  0 < K < 1.

Proof. Applying the inequality Φ_{K}(r) < 4^{1−1/K}r^{1/K}, K ≥ 1, cf. [W], [Hü], [LV], [AVV1], and the inequality

Φ_{K}(ab) ≥ max{b^{1/K}Φ_{K}(a), a^{1/K}Φ_{K}(b)}, 0 ≤ a, b ≤ 1,

showed in [AVV1] we obtain

\begin{aligned}Φ_{K}(4^{1−p}x^{p}) & = Φ_{K}((\frac{x}{4})^{p−1}x)≥(\frac{x}{4})^{(p−1)/K}Φ_{K}(x) \\ & = 4^{1−p}(4^{1−1/K}x^{1/K})^{p−1}Φ_{K}(x) > 4^{1−p}Φ_{K}^{p}(x). \end{aligned}

A substitution x := Φ_{1/K}(x) in the inequality (1.6) yields the inequality (1.7) and this ends the proof.

Now we define the functions φ_{K,t}, ψ_{K,t}, t ≥ 1, important for our further considerations. We put

(1.8)  φ_{K,t} = Φ_{t} ∘ φ_{K,1} ∘ Φ_{1/t}  and  ψ_{K,t} = Φ_{t} ∘ ψ_{K,1} ∘ Φ_{1/t}

for any K > 0 and t ≥ 1 where

φ_{K,1}(x) = x^{1/K}  and  ψ_{K,1}(x) = min{4^{1−1/K}x^{1/K}, 1},  0 ≤ x ≤ 1.
Theorem 1.3. For any 0 ≤ x ≤ 1 and 1 ≤ t_{1} ≤ t_{2}
(1.9)  φ_{K,t_{1}}(x) ≤ φ_{K,t_{2}}(x) ≤ Φ_{K}(x) ≤ ψ_{K,t_{2}}(x) ≤ ψ_{K,t_{1}}(x),  K ≥ 1

and

(1.10) ψ_{K,t_{1}}(x) ≤ ψ_{K,t_{2}}(x) ≤ Φ_{K}(x) ≤ φ_{K,t_{2}}(x) ≤ φ_{K,t_{1}}(x),  0 < K ≤ 1.

Moreover, for any 0 ≤ x ≤ 1 and K > 0

(1.11)  lim_{t→∞} φ_{K,t}(x) = lim_{t→∞} ψ_{K,t}(x) = Φ_{K}(x).

Proof. Let K ≥ 1 be fixed. Since

(1.12)  φ_{K,1}(x) ≤ Φ_{K}(x) ≤ ψ_{K,1}(x),  0 ≤ x ≤ 1,

cf. [AVV1], [LV], and Φ_{L} is an increasing function for any L > 0 we obtain for every t ≥ 1 the following inequality

(1.13) φ_{K,t}(x) = Φ_{t} ∘ φ_{K,1} ∘ Φ_{1/t}(x) ≤ Φ_{t} ∘ Φ_{K} ∘ Φ_{1/t}(x) = Φ_{K}(x) ≤ Φ_{t} ∘ ψ_{K,1} ∘ Φ_{1/t}(x) = ψ_{K,t}(x).

Moreover, setting r = Φ_{1/t}^{1/K}(x) we get by Lemma 1.1 that

(1.14) φ_{K,1}(x) = Φ_{t}^{1/K}(r^{K}) ≤ Φ_{t}(r) = Φ_{t} ∘ φ_{K,1} ∘ Φ_{1/t}(x) = φ_{K,t}(x), 0 ≤ x ≤ 1.

Let now r = 4^{1−1/K}Φ_{1/t}^{1/K}(x). If r ≤ 1 then by Lemma 1.2 we have

(1.15)  x = Φ_{t}(4^{1−K}r^{K}) > 4^{1−K}Φ_{t}^{K}(r) = 4^{1−K}ψ_{K,t}^{K}(x).

Otherwise 1 < r ≤ 4^{1−1/K}x^{1/K} so ψ_{K,1}(x) = 1 ≥ ψ_{K,t}(x). This and (1.15) lead to the following inequality

(1.16)  ψ_{K,1}(x) ≥ ψ_{K,t}(x),  0 ≤ x ≤ 1.

Suppose 1 ≤ t_{1} ≤ t_{2} are arbitrary. Then t = t_{2}/t_{1} ≥ 1 and in view of (1.14)

φ_{K,t_{1}}(x) = Φ_{t_{1}} ∘ φ_{K,1} ∘ Φ_{1/t_{1}}(x) ≤ Φ_{t_{1}} ∘ φ_{K,t} ∘ Φ_{1/t_{1}}(x)

= Φ_{t_{1}} ∘ Φ_{t} ∘ φ_{K,1} ∘ Φ_{1/t} ∘ Φ_{1/t_{1}}(x) = Φ_{t_{2}} ∘ φ_{K,1} ∘ Φ_{1/t_{2}}(x) = φ_{K,t_{2}}(x)

for 0 ≤ x ≤ 1, and similarly, it follows from (1.16) that

ψ_{K,t_{1}}(x) ≥ ψ_{K,t_{2}}(x), 0 ≤ x ≤ 1.

This proves the inequality (1.9).

Let us consider the sequences φ_{K,2^{n}}, ψ_{K,2^{n}}, n = 0, 1, 2, . . .. Obviously for any 0 < x ≤ 1

(1.17)  \frac{ψ_{K,1}(x)}{φ_{K,1}(x)} ≤ 4^{1−1/K}.

If n = 0, 1, 2, … is arbitrary then by (1.9) and the equality

(1.18)  Φ_{2}(r) = \frac{2\sqrt{r}}{1 + r} ,  0 ≤ r ≤ 1

we obtain that (1.19)

1 ≤ \frac{ψ_{K,2^{n+1}}(x)}{φ_{K,2^{n+1}}(x)} = \frac{Φ_{2}(ψ_{K,2^{n}}(Φ_{1/2}(x)))}{Φ_{2}(φ_{K,2^{n}}(Φ_{1/2}(x)))}

= (\frac{ψ_{K,2^{n}}(Φ_{1/2}(x))}{φ_{K,2^{n}}(Φ_{1/2}(x))})^{1/2}(\frac{1 + φ_{K,2^{n}}(Φ_{1/2}(x))}{1 + ψ_{K,2^{n}}(Φ_{1/2}(x))}) ≤ sup_{0<x≤1}(\frac{ψ_{K,2^{n}}(x)}{φ_{K,2^{n}}(x)})^{1/2}.

Hence and by (1.17) we conclude that

φ_{K,2^{n}}(x) ≤ ψ_{K,2^{n}}(x) ≤ 4^{(1−1/K)2^{−n}}φ_{K,2^{n}}(x), 0 ≤ x ≤ 1,

and this together with the inequality (1.9) implies (1.11). In a similar way we arrive at the inequality (1.10) and derive (1.11) in the case when 0 < K ≤ 1, which ends the proof.

The functions φ_{K,t}, ψ_{K,t}, t ≥ 1, behave nicely near 0 but not so nicely close to 1. To improve this we shall introduce another couple of functions as follows:

(1.20)  φ̃_{K,t} = h ∘ φ_{1/K,t} ∘ h, ψ̃_{K,t} = h ∘ ψ_{1/K,t} ∘ h,  K > 0, t ≥ 1

where h(x) = (1 − x)/(1 + x), 0 ≤ x ≤ 1. As shown in [AVV1]

(1.21)  Φ_{1/K} = h ∘ Φ_{K} ∘ h,  K > 0,

so in view of Theorem 1.3 we immediately obtain the following Corollary 1.4. For any 0 ≤ x ≤ 1 and 1 ≤ t_{1} ≤ t_{2}

(1.22)  φ̃_{K,t_{1}}(x) ≤ φ̃_{K,t_{2}}(x) ≤ Φ_{K}(x) ≤ ψ̃_{K,t_{2}}(x) ≤ ψ̃_{K,t_{1}}(x),  K ≥ 1

and

(1.23) ψ̃_{K,t_{1}}(x) ≤ ψ̃_{K,t_{2}}(x) ≤ Φ_{K}(x) ≤ φ̃_{K,t_{2}}(x) ≤ φ̃_{K,t_{1}}(x),  0 < K ≤ 1.

Moreover, for any 0 < x ≤ 1 and K > 0

(1.24)  lim_{t→∞} φ̃_{K,t}(x) = lim_{t→∞} ψ̃_{K,t}(x) = Φ_{K}(x),  0 ≤ x ≤ 1.

It follows from Theorem 1.3 that the sequences φ_{K,2^{n}}, ψ_{K,2^{n}}, n = 0, 1, 2, … are convergent to Φ_{K} but computer tests show that ψ_{K,2^{n}} approaches much faster Φ_{K} than φ_{K,2^{n}} as n → ∞, K > 0. There is a simple justification of this fact. Namely, it can be explained by the asymptotic behaviour of Φ_{K} near 0 given by

lim_{r→0^{+}} \frac{Φ_{K}(r)}{r^{1/K}} = 4^{1−1/K} = lim_{r→0^{+}} \frac{ψ_{K,1}(r)}{r^{1/K}}.

Actually the sequence ψ_{K,2^{n}}, n = 0, 1, 2, … converges to Φ_{K} very fast. We will prove the following

Theorem 1.5. For any 0 < x < 1

(1.25) (1 − x^{2^{n+1}/K})ψ_{K,2^{n}}(x) ≤ Φ_{K}(x) ≤ ψ_{K,2^{n}}(x),  K ≥ 1, n = 2, 3, 4, …

as well as (1.26)

ψ_{K,2^{n}}(x) ≤ Φ_{K}(x) ≤ (1 − x^{2^{n+1}})^{−1/K2^{n}}ψ_{K,2^{n}}(x), 0 < K ≤ 1, n = 1, 2, 3, ….

Proof. At first we consider the case when K ≥ 1. We put

(1.27)  r_{K,n}(x) = \frac{ψ_{K,2^{n−1}}(x)}{ψ_{K,2^{n}}(x)},  0 < x ≤ 1, n = 1, 2, ….

Similarly as in the proof of Theorem 1.3, see (1.19), we get by the inequality (1.9) and the equality (1.18) that for any 0 < x ≤ 1 and n = 1, 2, …

(1.28)  \begin{aligned}1 ≤ r_{K,n+1}(x) & = \frac{Φ_{2}(ψ_{K,2^{n−1}}(Φ_{1/2}(x)))}{Φ_{2}(ψ_{K,2^{n}}(Φ_{1/2}(x)))} \\ & ≤ (\frac{ψ_{K,2^{n−1}}(Φ_{1/2}(x))}{ψ_{K,2^{n}}(Φ_{1/2}(x))})^{1/2} = r_{K,n}^{1/2}(Φ_{1/2}(x)). \end{aligned}

Hence for any 0 < x ≤ 1 and n = 1, 2, …

(1.29)  r_{K,n+1}(x) ≤ r_{K,1}^{1/2^{n}}(Φ_{1/2^{n}}(x)).

It follows from (1.18) that

r_{K,1}(x) ≤ (\frac{1 + \sqrt{1 − x^{2}}}{2} )^{1/K} + 4^{1−1/K}(\frac{1 − \sqrt{1 − x^{2}}}{2} )^{1/K}

≤ 1 + 4^{1−1/K}(\frac{1 − \sqrt{1 − x^{2}}}{1 + \sqrt{1 − x^{2}}} )^{1/K} = 1 + 4^{1−1/K}(\frac{x}{1 + \sqrt{1 − x^{2}}} )^{2/K}

= 1 + 4^{1−1/K}(Φ_{1/2}(x))^{1/K}.

This together with (1.29) and (1.10) gives

(1.30) \begin{aligned}r_{K,n+1}(x) & ≤ (1 + 4^{1−1/K}(Φ_{1/2^{n+1}}(x))^{1/K})^{1/2^{n}} ≤ 1 + \frac{4^{1−1/K}}{2^{n}} x^{2^{n+1}/K} \\ & ≤ 1 + x^{2^{n+1}/K} \end{aligned}

as n = 2, 3, … and 0 < x ≤ 1. By (1.11) we have for any 0 < x < 1 and n = 2, 3, …

\begin{aligned}\frac{ψ_{K,2^{n}}(x)}{Φ_{K}(x)} & = lim_{m→∞} \frac{ψ_{K,2^{n}}(x)}{ψ_{K,2^{m}}(x)} = lim_{m→∞} ∏_{l=n+1}^{m} r_{K,l}(x) = ∏_{l=n+1}^{∞} r_{K,l}(x) \\ & ≤ ∏_{l=n+1}^{∞}(1 + x^{2^{l}/K}) = \frac{1}{1 − x^{2^{n+1}/K}} , \end{aligned}

because of (1.30). This proves the inequality (1.25).

Let now 0 < K ≤ 1. Similarly as in the first case we put

(1.31)  r_{K,n}(x) = \frac{ψ_{K,2^{n}}(x)}{ψ_{K,2^{n−1}}(x)},  0 < x ≤ 1, n = 1, 2, …,

and by (1.10) and (1.18) we get the inequalities (1.28) and (1.29) true for 0 < K ≤ 1, 0 < x ≤ 1 and n = 1, 2, . . .. Applying the equality (1.18) once again we have

(1.32)  \begin{aligned}r_{K,1}(x) & ≤ \frac{2^{1/K}}{(1 + \sqrt{1 − x^{2}} )^{1/K} + 4^{1−1/K} (1 − \sqrt{1 − x^{2}} )^{1/K}} \\ & ≤ (\frac{2}{1 + \sqrt{1 − x^{2}}} )^{1/K} = (1 + Φ_{1/2}(x))^{1/K} \end{aligned}

which together with (1.29) and (1.10) leads to

(1.33)  r_{K,n+1}(x) ≤ (1 + Φ_{1/2^{n+1}}(x))^{1/K2^{n}} ≤ (1 + x^{2^{n+1}})^{1/K2^{n}}

as 0 < x ≤ 1 and n = 1, 2, . . .. Hence and by (1.11)

\frac{Φ_{K}(x)}{ψ_{K,2^{n}}(x)} = lim_{m→∞} \frac{ψ_{K,2^{m}}(x)}{ψ_{K,2^{n}}(x)} = lim_{m→∞} ∏_{l=n+1}^{m} r_{K,l}(x) ≤ ( lim_{m→∞} ∏_{l=n+1}^{m}(1 + x^{2^{l}}))^{1/K2^{n}}

= ( ∏_{l=n+1}^{∞}(1 + x^{2^{l}}))^{1/K2^{n}} = \frac{1}{(1 − x^{2^{n+1}})^{1/K2^{n}}} , 0 < x < 1, n = 1, 2, ….

This implies the inequality (1.26) and ends the proof.

It follows from the above theorem and the equalities (1.20) and (1.21) that the following counterpart of Theorem 1.5 for ψ̃_{K,t}, t ≥ 1, holds.

Corollary 1.6. For any 0 < x < 1

(1.34)  0 ≤ ψ̃_{K,2^{n}}(x) − Φ_{K}(x) ≤ 2((1 − h(x)^{2^{n+1}})^{−K2^{−n}} − 1)h^{K}(x)

as K ≥ 1 and n = 1, 2, 3, … as well as

(1.35) 0 ≤ Φ_{K}(x) − ψ̃_{K,2^{n}}(x) ≤ 2((1 − h(x)^{K2^{n+1}})^{−1} − 1) min{4^{1−K}h^{K}(x), 1}

as 0 < K ≤ 1 and n = 2, 3, 4, . . ..

Proof. Let K ≥ 1 be any fixed number. By the inequality

|h(x) − h(y)| ≤ 2|x − y|, 0 ≤ x, y ≤ 1,

the equalities (1.20), (1.21) and the inequalities (1.9), (1.10), (1.22) we get for any 0 < x < 1

\begin{aligned}0 ≤ ψ̃_{K,2^{n}}(x) − Φ_{K}(x) & ≤ 2(Φ_{1/K}(h(x)) − ψ_{1/K,2^{n}}(h(x))) \\ & ≤ 2(\frac{Φ_{1/K}(h(x))}{ψ_{1/K,2^{n}}(h(x))} − 1)ψ_{1/K,2^{n}}(h(x)). \end{aligned}

Hence and by the inequalities (1.26) and (1.10) we obtain immediately the inequality (1.34). In a similar way we derive the inequality (1.35) which ends the proof.

The convergence of another couple of sequences φ_{K,2^{n}}, φ_{K,2^{n}}, n = 0, 1, 2, …, K > 0, will be considered in a separate paper [P].

2. Applications

By Theorem 1.3 we obtain upper and lower bounds for the function Φ_{K} of the form

(2.1)  φ_{K,2^{n}}(x) ≤ Φ_{K}(x) ≤ ψ_{K,2^{n}}(x)  as K ≥ 1

and

(2.2)  ψ_{K,2^{n}}(x) ≤ Φ_{K}(x) ≤ φ_{K,2^{n}}(x)  as 0 < K ≤ 1

for 0 ≤ x ≤ 1 and n = 0, 1, 2, . . .. In view of (1.18) the estimates (2.1) and (2.2) are expressed in an explicit form by elementary functions. Moreover, Theorem 1.3 says that they are closer step by step to the function Φ_{K} as n increases to infinity. In particular, if n = 0, we obtain from (2.1) the classical bounds

x^{1/K} ≤ Φ_{K}(x) ≤ 4^{1−1/K}x^{1/K}, 0 ≤ x ≤ 1, K ≥ 1,

cf. [W], [Hü], [LV], [AVV1]. If n = 1 then (2.1) yields for any K ≥ 1 and 0 ≤ x ≤ 1 the estimate

\begin{aligned}\frac{2x^{1/K}}{(1 + \sqrt{1 − x^{2}} )^{1/K} + (1 − \sqrt{1 − x^{2}} )^{1/K}} & ≤ Φ_{K}(x) \\ & ≤ \frac{2^{2−1/K}x^{1/K}}{(1 + \sqrt{1 − x^{2}} )^{1/K} + (1 − \sqrt{1 − x^{2}} )^{1/K}} \end{aligned}

which coincides with that established in [AVV1]. So setting n = 2 we obtain a new improvement of the above mentioned results:

Theorem 2.1. For K ≥ 1 and 0 ≤ x ≤ 1 the following bounds hold

(2.3) Φ_{K}(x) ≥ \frac{2^{3/2}x^{1/K}}{(1 + \sqrt{1 − x^{2}} )^{1/2K}} \frac{((1 + ^{4} \sqrt{1 − x^{2}} )^{2/K} + (1 − ^{4} \sqrt{1 − x^{2}} )^{2/K})^{1/2}}{((1 + ^{4} \sqrt{1 − x^{2}} )^{1/K} + (1 − ^{4} \sqrt{1 − x^{2}} )^{1/K})^{2}}

and (2.4)

Φ_{K}(x) ≤ \frac{2^{2−1/2K}x^{1/K}}{(1 + \sqrt{1 − x^{2}} )^{1/2K}} \frac{((1 + ^{4} \sqrt{1 − x^{2}} )^{2/K} + 4^{1−1/K}(1 − ^{4} \sqrt{1 − x^{2}} )^{2/K})^{1/2}}{((1 + ^{4} \sqrt{1 − x^{2}} )^{1/K} + 2^{1−1/K}(1 − ^{4} \sqrt{1 − x^{2}} )^{1/K})^{2}}

as 0 ≤ x < Φ_{4}(4^{1−K}) and Φ_{K}(x) ≤ 1 as Φ_{4}(4^{1−K}) ≤ x ≤ 1.

From the inequalities (2.2), estimates of the function Φ_{K} for 0 < K ≤ 1 corresponding to (2.3) and (2.4) can be derived.

Contrary to Theorem 1.3, Corollary 1.4 yields upper and lower bounds for the function Φ_{K}, good for x near 1, of the form

(2.5)  φ_{K,2^{n}}(x) ≤ Φ_{K}(x) ≤ ψ̃_{K,2^{n}}(x)  as K ≥ 1

and

(2.6)  ψ̃_{K,2^{n}}(x) ≤ Φ_{K}(x) ≤ φ_{K,2^{n}}(x)  as 0 < K ≤ 1

for 0 ≤ x ≤ 1 and n = 0, 1, 2, . . .. Similarly as (2.1) and (2.2), the estimates (2.5) and (2.6) can be expressed in an explicit form by elementary functions and they are, in view of Corollary 1.4, closer step by step to the function Φ_{K} as n increases to infinity. Setting, for example, n = 0, 1 in (2.5) we obtain the following explicit estimates of Φ_{K} for K ≥ 1 and 0 ≤ x ≤ 1, good for x close to 1, cf. [Z1],

\frac{(1 + x)^{K} − (1 − x)^{K}}{(1 + x)^{K} + (1 − x)^{K}} ≤ Φ_{K}(x) ≤ \frac{(1 + x)^{K} − 4^{1−K}(1 − x)^{K}}{(1 + x)^{K} + 4^{1−K}(1 − x)^{K}}

and a more precise bound

(\frac{(1 + \sqrt{x})^{K} − (1 − \sqrt{x})^{K}}{(1 + \sqrt{x})^{K} + (1 − \sqrt{x})^{K}} )^{2} ≤ Φ_{K}(x) ≤ (\frac{(1 + \sqrt{x})^{K} − 2^{1−K}(1 − \sqrt{x})^{K}}{(1 + \sqrt{x})^{K} + 2^{1−K}(1 − \sqrt{x})^{K}} )^{2},

found in [Z2] and [AVV3]. So, if n = 2 then the inequalities (2.5) yield a new improvement of the above mentioned results.

Theorem 2.2. For K ≥ 1 and 0 ≤ x ≤ 1 the following bounds hold (2.7)

(\frac{1 − p(x)}{\sqrt{1 + p^{2}(x)} + \sqrt{2p(x)}})^{4} ≤ Φ_{K}(x) ≤(\frac{1 − 2^{1−K}p(x)}{\sqrt{1 + 4^{1−K}p^{2}(x)} + \sqrt{2^{2−K}p(x)}})^{4}

where

p(x) = (\frac{\sqrt{1 + x} − ^{4} \sqrt{4x}}{\sqrt{1 + x} + ^{4} \sqrt{4x}} )^{K}, 0 ≤ x ≤ 1.

From the inequalities (2.6), estimates of the function Φ_{K} for 0 < K ≤ 1 corresponding to (2.7) can be derived.

It follows from Theorem 1.3 and Corollary 1.4 that the sequences ψ_{K,2^{n}}, ψ̃_{K,2^{n}}, n = 0, 1, 2, … are convergent to the distortion function Φ_{K} and the convergence is very fast, because of Theorem 1.5 and Corollary 1.6. For example, if 1 ≤ K ≤ 2 and 0 ≤ x ≤ 0.1 then by virtue of Theorem 1.5

0 ≤ ψ_{K,2^{7}}(x) − Φ_{K}(x) ≤ x^{2^{8}/K}ψ_{K,2^{7}}(x) ≤ 10^{−128}ψ_{K,2^{7}}(x) < 10^{−100}

and

0 ≤ ψ_{K,2^{10}}(x) − Φ_{K}(x) ≤ x^{2^{10}/K}ψ_{K,2^{10}}(x) ≤ 10^{−1024}ψ_{K,2^{10}}(x) < 10^{−1000}.

This way we obtained a convenient algorithm for computation of values of the function Φ_{K} with an arbitrary accuracy, which can be easily handled with a computer. Details will be presented in a separate paper [P]. We shall present there some other applications of the results obtained in the first section to the estimates of functions μ, μ^{−1} and λ closely related to Φ_{K} as well as a method of numerical computation of their values.

3. Generalization to the n-dimensional case

Let γ_{n}(r), r > 1 denote the conformal capacity of the Grötzsch extremal ring in R^{n}, n = 2, 3, 4, …, cf. [AVV2]. We define the function μ_{n}(r), 0 < r < 1, by

μ_{n}(r) = (\frac{1}{ω_{n−1}} γ_{n}(\frac{1}{r}))^{1/(1−n)}

where ω_{n−1} is the n−1-dimensional measure of the unit sphere S^{n−1} in R^{n}. For K > 0 the function

(3.1) Φ_{K,n}(r) = μ_{n}^{−1}(K^{1/(1−n)}μ_{n}(r)),  0 < r < 1, Φ_{K,n}(0) = 0, Φ_{K,n}(1) = 1

is called the n-dimensional distortion function. Obviously, the function Φ_{K,2} coincides with the function Φ_{K}.

Theorem 3.1. For every K > 0 and n = 2, 3, 4, …

(3.2)  lim_{t→∞} Φ_{t,n} ∘ φ ∘ Φ_{1/t,n}(x) = Φ_{K,n}(x),  0 ≤ x ≤ 1

where φ: [0, 1] → [0, 1] is any function such that

(3.3)  lim_{r→0^{+}} \frac{log φ(r)}{log r} = K^{1/(1−n)}.

Proof. Let n ≥ 2 be an arbitrary positive integer and x be any point between 0 and 1. As shown in [G1], [G2], the limit

lim_{r→0^{+}}(μ_{n}(r) − log r) = log λ_{n}

exists and λ_{n} is called the Grötzsch ring constant. Hence, by (3.3) and the equality

\frac{μ_{n}(φ(τ))}{μ_{n}(τ)} = \frac{log τ}{μ_{n}(τ)} \frac{μ_{n}(φ(τ))}{log φ(τ)} \frac{log φ(τ)}{log τ}, 0 < τ < 1,

we get

(3.4)  lim_{τ→0^{+}} \frac{μ_{n}(φ(τ))}{μ_{n}(τ)} = K^{1/(1−n)}.

Setting τ = Φ_{1/t,n}(x) we derive from (3.1) that μ_{n}(τ) = t^{1/(n−1)}μ_{n}(x) and further on

Φ_{t,n} ∘ φ ∘ Φ_{1/t,n}(x) = μ_{n}^{−1} (t^{1/(1−n)}μ_{n}(φ(τ))) = μ_{n}^{−1}(\frac{μ_{n}(φ(τ))}{μ_{n}(τ)}μ_{n}(x)).

Hence, by (3.4) and the continuity of the function μ_{n} we finally obtain the convergence statement in (3.2), because of lim_{t→∞} Φ_{1/t,n}(x) = 0. This ends the proof.

354 Dariusz Partyka

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Received 8 May 1992