On asymptotic behavior of fractional Cauchy transform

Abstract

We study non-regularity of growth of the fractional Cauchy transform

$$\begin{aligned} f(z)=\int _{-\pi }^{\pi } \frac{d\psi (t)}{(1-ze^{-it})^\alpha }, \quad \alpha >0, \psi \in BV[-\pi ,\pi ], \end{aligned}$$

in terms of the modulus of continuity of the function $\psi$. Sharp estimates of the lower logarithmic order of f are found. In the case $\alpha \in (0,1)$ the estimates are of different form than that for the logarithmic order.

1 Introduction

Let ${\mathbb {D}}=\{ z\in {\mathbb {C}}:|z|<1\}$ denote the unit disc, $\alpha >0$; $(1-w)^\alpha$, $w\in {\mathbb {D}}$, be the principle branch of the power function, i.e. $(1-w)^{\alpha }\Bigr |_{w=0}=1$.

Our main focus is to study of the following Cauchy type integral

$$\begin{aligned} f(z)=\int _{-\pi }^{\pi } \frac{d\psi (t)}{(1-ze^{-it})^\alpha }, \end{aligned}$$

(1)

where $\psi$ is a complex-valued function of bounded variation, i.e., $\psi \in BV [-\pi , \pi ]$, or a fractional Cauchy transform of the measure $\psi ^*$ on $\partial {\mathbb {D}}$ associated with $\psi$. The family of all functions of the form (1) is denoted by ${\mathcal {F}}_\alpha$. A study of ${\mathcal {F}}_\alpha$ can be found in the book [11], here we mention only few properties. ${\mathcal {F}}_\alpha$ is a Banach space with respect to the norm $\Vert f\Vert _{{\mathcal {F}}_\alpha }=\inf _{\psi ^*} \Vert \psi ^* \Vert$, where $\Vert \psi ^* \Vert$ denotes the total variation of $\psi ^*$. A straightforward consequence of the definition is that for $f\in {\mathcal {F}}_\alpha$ one has

$$\begin{aligned} M(r,f)\le \frac{\Vert \psi ^*\Vert }{(1-r)^\alpha }, \end{aligned}$$

(2)

where $M(r,f)=\max \{ |f(re^{i\theta })|: |z|=r\}$.

Many relations with Hardy, Besov, and Dirichlet-type spaces are obtained in the mentioned monograph.

Theorem A

If $0<\alpha \le 1$, then ${\mathcal {F}}_\alpha \subset H^p$ for $0<p<1/\alpha$. If $0<p\le 1$ then $H^p \subset {\mathcal {F}}_{1/p}$, where $H^p$ is the Hardy space of analytic functions in ${\mathbb {D}}$ satisfying

$$\begin{aligned} \sup _{0<r<1} \int _{-\pi }^{\pi } |f(re^{i\theta })|^p\, d\theta <\infty . \end{aligned}$$

Theorem B

If $0<\alpha <\beta$, then ${\mathcal {B}}_\alpha \subset {\mathcal {F}}_\alpha \subset {\mathcal {B}}_\beta$, where ${\mathcal {B}}_\alpha$ is the Besov space of analytic functions in ${\mathbb {D}}$ satisfying

$$\begin{aligned} \int _{{\mathbb {D}}} |f^{\prime }(z)|(1-|z|)^{\alpha -1}\, dm(z) <\infty , \end{aligned}$$

where m is the planar Lebesgue measure.

Functions from ${\mathcal {F}}_ \alpha$ appear frequently in representation theorems for different classes of functions defined on the unit disc ${\mathbb {D}}$. In particular, Djrbashian introduced [5] classes $A_\alpha$, $\alpha >-1$, of analytic functions in ${\mathbb {D}}$. For $\alpha >0$, an analytic function f belongs to the class $A_\alpha$, $\alpha >0$, if and only if

$$\begin{aligned} \sup _{0<r< 1}\int _0^{2\pi } \biggl (\int _0^r (r-t)^{\alpha -1} \log |f(te^{i\varphi })|\, dt\biggr )^+d\varphi <+\infty . \end{aligned}$$

Note that the union of $\bigcup _{\alpha >0} A_\alpha$ contains the class of analytic functions f in ${\mathbb {D}}$ of finite order of growth, that is $\log \log M(r,f)=O\bigl (\log \frac{1}{1-r}\bigr )$ ($r\rightarrow 1-$). Due to results of Djrbashian [5, Chap. IX] functions $f\in {\mathcal {F}}_{\alpha +1}$ appear in a parametric representation of the class $A_{\alpha }$, $\alpha >-1$.

Radial and non-tangential limits of $f\in {\mathcal {F}}_\alpha$ were investigated in many papers, e.g., Hallenbeck and MacGregor [8, 9], and Sheremeta [13], see also [11, 12]. It turns out that the estimate (2) can be improved in terms of the modulus of continuity for $\psi$. Let

$$\begin{aligned} \omega (\delta , \psi )=\sup \{ |\psi (x)-\psi (y)|: |x-y|<\delta , x,y\in [-\pi ,\pi ]\} \end{aligned}$$

be the modulus of continuity of $\psi$.

Denote by ${\varLambda }_\gamma$ [15] the class of functions $\psi$ such that $\omega (\delta , \psi )=O(\delta ^\gamma )$, $\gamma \in (0,1]$. It comes from Hardy and Littlewood [10], see also [6, 14], that there is a strong interplay between smoothness of a measure (or associated function $\psi$ of bounded variation) and the growth of integral (1).

The next result in this spirit is a consequence of Theorem 3.4 [4] (see also Theorem 3.1 [1]).

Theorem C

Let f be an analytic function in ${\mathbb {D}}$ of the form (1). Let $\alpha > \gamma$ and $0<\gamma \le 1$. Then $\psi \in {\varLambda }_\gamma$, if and only if

$$\begin{aligned} M(r,f) =O((1-r)^{\gamma -\alpha }), \quad r\rightarrow 1- . \end{aligned}$$

Remark 1

More general results using the concept of a proximate order are obtained in [4]. There are counterparts for harmonic functions represented by Poisson-type integrals as well, see [1, 3].

Remark 2

Growth of pth means of functions $f\in {\mathcal {F}}_\alpha$ is described in [2].

A common feature of the above mentioned results is that the growth of f depends on which Hölder class ${\varLambda }_\gamma$ the modulus of continuity of $\psi$ belongs to. Let

$$\begin{aligned} p[f]:=\limsup _{r\rightarrow 1-} \frac{\log M(r,f)}{\log \frac{1}{1-r}}, \quad \lambda [f]:=\liminf _{r\rightarrow 1-} \frac{\log M(r,f)}{\log \frac{1}{1-r}} \end{aligned}$$

(3)

denote the logarithmic order and the lower logarithmic order, respectively. It follows from Theorem C that for an arbitrary $f\in {\mathcal {F}}_\alpha$ and appropriate $\alpha$, $\gamma$ we have $p[f]\le \alpha -\gamma$ if and only if $\psi \in {\varLambda }_\gamma$. The following question appears naturally. Does a smoothness of the modulus of continuity$\omega (\delta , \psi )$on a sequence$(\delta _n)$of values$\delta$tending to zero imply an upper estimate for the lower logarithmic order similar to that for the logarithmic order?

We shall see that the answer is in the positive provided that $\alpha > 1$, but in general, the situation is more complicated.

Theorem 3

Let $\alpha >0$, $0\le \gamma<\lambda <1$.

Suppose that $\omega (\delta _n, \psi )=O(\delta _n^\lambda )$, $\delta _n \rightarrow 0+$, while $\omega (\delta , \psi )=O(\delta ^\gamma )$, $\delta \rightarrow 0+$, and that f is of the form (1).

• (i) If $0<\alpha <\gamma$, then $M(r,f)=O(1)$.

• (ii) If $0<\alpha =\gamma$, then $M(r,f)=O(\log \frac{1}{1-r})$.

• (iii) If $\gamma < \alpha$, then

  $$\begin{aligned} M(r_n,f_\alpha )=O\Bigl (\Bigl (\frac{1}{1-r_n}\Bigr )^{{\varDelta }} \Bigr ), \quad n\rightarrow \infty , \end{aligned}$$

  on a sequence $r_n\rightarrow 1-$ ($n\rightarrow \infty$), where

  $$\begin{aligned} {\varDelta }= {\left\{ \begin{array}{ll} \frac{\alpha (\alpha -\gamma )(1-\lambda )}{\alpha (1-\lambda )+\lambda -\gamma },&{} \alpha <1 \\ \alpha -\lambda , &{} \alpha > 1, \end{array}\right. } \end{aligned}$$

  (4)

  and, for $\alpha =1$

  $$\begin{aligned} M(r_n,f_\alpha )=O\Bigl (\Bigl (\frac{1}{1-r_n}\Bigr )^{1-\varkappa _2}\log \frac{1}{1-r_n} \Bigr ), \quad n\rightarrow \infty . \end{aligned}$$

Remark 4

The quantity ${\varDelta }$ in (4) is continuous at 1 as a function of $\alpha$. Moreover, ${\varDelta }\rightarrow \alpha -\lambda$ as $\gamma \rightarrow \lambda$.

Remark 5

Any modulus of continuity $\omega$ satisfies [7, Chap. 3]

$$\begin{aligned} \frac{\omega (t_2)}{t_2}\le \frac{2\omega (t_1)}{t_1}, \quad 0<t_1< t_2. \end{aligned}$$

(5)

If $\omega (\delta _n, \psi )\le C \delta _n$, $\delta _n \rightarrow 0+$, i.e., $\lambda =1$, then

$$\begin{aligned} \omega (\delta , \psi )\le 2C \delta ,\quad \delta >\delta _n, \end{aligned}$$

so $\psi \in {\varLambda }_1$, and $\gamma =\lambda =1$. This is why we exclude this case in the assumptions of Theorem 3.

Corollary 1

Let $\alpha >0$. Suppose that for arbitrary $\varepsilon >0$ there exists a sequence $\delta _n \rightarrow 0+$$(n\rightarrow \infty )$ such that $\omega (\delta _n, \psi )=O(\delta _n^{\lambda -\varepsilon })$$(n\rightarrow \infty )$, while $\psi \in {\varLambda }_\gamma$, $0\le \gamma<\lambda <1$, $\gamma < \alpha$. Then

$$\begin{aligned} \lambda [f] \le {\left\{ \begin{array}{ll} \frac{\alpha (\alpha -\gamma )(1-\lambda )}{\alpha (1-\lambda )+\lambda -\gamma },&{} \alpha <1 \\ \alpha -\lambda , &{} \alpha \ge 1. \end{array}\right. } \end{aligned}$$

(6)

Proof of Theorem 3

Let $\varphi$ be arbitrary on $[-\pi , \pi ]$. We extend $\psi$ on ${\mathbb {R}}$ by the formula $\psi (t+2\pi )-\psi (t)= \psi (\pi )-\psi (-\pi )$. Since the kernel of the integral (1) is a $2\pi$-periodic function in t, integrating by parts gives us

$$\begin{aligned} f_\alpha (re^{i\varphi })&= \int \limits _{-\pi +\varphi }^{\pi +\varphi } \frac{ d(\psi (t) - \psi (\varphi ))}{(1-re^{i(\varphi -t)})^{\alpha }} \\&=\frac{\psi (\pi )-\psi (-\pi )}{(1+r)^\alpha } + ir\alpha \int \limits _{-\pi }^{\pi } \frac{e^{-i\tau }(\psi (\tau +\varphi ) - \psi (\varphi ))d\tau }{(1-re^{-i\tau })^{\alpha +1}}. \end{aligned}$$

Therefore, uniformly in $\varphi$, $\omega (\delta ):=\omega (\delta ; \psi )$

$$\begin{aligned} |f_\alpha (re^{i\varphi })|\le 2\alpha \int _{0}^\pi \frac{\omega (x)\, dx}{|1-re^{ix}|^{\alpha +1}}+O(1)\le 2\alpha \int _{0}^1 \frac{\omega (x)\, dx}{|1-re^{ix}|^{\alpha +1}}+O(1) . \end{aligned}$$

(7)

We start with the simple cases (i) and (ii). Suppose that $0<\alpha \le \gamma$. Then $\omega (\delta )=O(\delta ^\gamma )$, and standard estimates yield

$$\begin{aligned} \int _{0}^1 \frac{\omega (x)\, dx}{|1-re^{ix}|^{\alpha +1}}\le \int _0^{1-r} \frac{Cx^\gamma \, dx}{(1-r)^{\alpha +1}} +\int _{1-r}^{1} \frac{dx}{Cx^{\alpha +1-\gamma }}, \end{aligned}$$

where C is a positive constant. Hence, the statements (i) and (ii) follow.

Let now $\alpha >\gamma \ge 0$. Consider two cases. First, we suppose that $\alpha < 1$. We then choose

$$\begin{aligned} {{\tilde{\delta }}}_n:= \delta _n^ {\frac{\alpha (1-\lambda )+\lambda -\gamma }{\alpha (1-\gamma )}} =\delta _n^{1+ \frac{(\lambda -\gamma )(1-\alpha )}{1-\gamma }}. \end{aligned}$$

Note that ${{\tilde{\delta }}}_n\le \delta _n$, because $\alpha \le 1$.

In what follows the inequality $g_1(n)\lesssim g_2(n)$ means that there exists a constant C independent of n such that $g_1(n)\le C g_2(n)$$(n\rightarrow \infty )$.

We put $r_n=1-{{\tilde{\delta }}}_n$. Since a modulus of continuity is nondecreasing, we have

$$\begin{aligned} \int _{0}^{{{\tilde{\delta }}}_n} \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \le \omega ( \delta _n) \int _{0}^{{{\tilde{\delta }}}_n} \frac{dx}{(1-r_n)^{\alpha +1}} \lesssim \frac{\delta _n^\lambda }{{{\tilde{\delta }}}_n^\alpha }. \end{aligned}$$

(8)

Then

$$\begin{aligned} \int _{{{\tilde{\delta }}}_n}^{\delta _n} \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \lesssim \delta _n^\lambda \int _{{{\tilde{\delta }}}_n} ^{\delta _n} \frac{dx}{x^{\alpha +1}} \lesssim \frac{\delta _n^\lambda }{{{\tilde{\delta }}}_n^\alpha }. \end{aligned}$$

(9)

Taking into account (5), and setting $\hat{\delta }_n=\delta _n ^{\frac{1-\lambda }{1-\gamma }}$, we obtain

$$\begin{aligned} \int _{\delta _n}^{{\hat{\delta }}_n} \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \lesssim \int _{\delta _n}^{{\hat{\delta }}_n} \frac{\delta _n^\lambda x\, dx}{\delta _n x^{\alpha +1}} \lesssim \frac{\hat{\delta }_n^{1-\alpha }}{\delta _n^{1-\lambda }}=\frac{1}{\delta _n^{\frac{(1-\lambda )(\alpha -\gamma )}{1-\gamma }}}. \end{aligned}$$

(10)

Similarly

$$\begin{aligned} \int _{{\hat{\delta }}_n}^1 \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \lesssim \int _{{\hat{\delta }}_n} ^1 \frac{x^\gamma dx}{x^{\alpha +1}} \lesssim \frac{1}{{\hat{\delta }}_n^{\alpha -\gamma }}=\frac{1}{\delta _n^{\frac{(1-\lambda )(\alpha -\gamma )}{1-\gamma }}}. \end{aligned}$$

(11)

Finally, taking into account that ${{\tilde{\delta }}}_n \lesssim \delta _n$, from the above estimates we deduce

$$\begin{aligned} \int _{0}^1 \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}}&\lesssim \frac{\delta _n^\lambda }{{{\tilde{\delta }}}_n^\alpha } +\frac{1}{\delta _n^{\frac{(1-\lambda )(\alpha -\gamma )}{1-\gamma }}}\le \frac{{\tilde{\delta }}_n^{\frac{\lambda \alpha (1-\gamma )}{\alpha (1-\lambda )+\lambda -\gamma }}}{\tilde{\delta }_n^\alpha }+ \frac{1}{{\tilde{\delta }}_n^{\frac{\alpha (1-\gamma )}{\alpha (1-\lambda )+\lambda -\gamma }\frac{(1-\lambda ) (\alpha -\gamma )}{1-\gamma }}} \nonumber \\&\le \frac{1}{{\tilde{\delta }}_n^\frac{\alpha (\alpha -\gamma ) (1-\lambda )}{\alpha (1-\lambda )+\lambda -\gamma }}+\frac{1}{\tilde{\delta }_n^\frac{\alpha (\alpha -\gamma )(1-\lambda )}{\alpha (1-\lambda )+ \lambda -\gamma }} \lesssim \frac{1}{{\tilde{\delta }}_n^\frac{\alpha (\alpha -\gamma ) (1-\lambda )}{\alpha (1-\lambda )+\lambda -\gamma }}. \end{aligned}$$

(12)

This finishes the proof in the first case.

Let now $\alpha >1$. We put $r_n:=1-\delta _n$ in this case. Arguing similarly as in the first case, we obtain

$$\begin{aligned} \int _{0}^{\delta _n} \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \lesssim \delta _n^\lambda \int _{0} ^{\delta _n} \frac{dx}{(1-r_n)^{\alpha +1}} \lesssim \frac{1}{\delta _n^{\alpha -\lambda }}. \end{aligned}$$

(13)

Then

$$\begin{aligned} \int _{\delta _n}^{{\hat{\delta }}_n} \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \lesssim \int _{\delta _n}^{{\hat{\delta }}_n} \frac{\delta _n^\lambda x\, dx}{\delta _n x^{\alpha +1}} \lesssim \frac{1}{\delta _n^{\alpha -\lambda }}. \end{aligned}$$

(14)

Similarly, as in the first case,

$$\begin{aligned} \int _{{\hat{\delta }}_n}^1 \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \lesssim \int _{{\hat{\delta }}_n} ^1 \frac{x^\gamma dx}{x^{\alpha +1}} \lesssim \frac{1}{{\hat{\delta }}_n^{\alpha -\gamma }}. \end{aligned}$$

(15)

Thus, using the definition of ${\hat{\delta }}_n$ and the latter estimates we deduce

$$\begin{aligned} \int _{0}^1 \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{\alpha +1}} \lesssim \frac{1}{ \delta _n^{\alpha -\lambda }} + \frac{1}{{\hat{\delta }}_n^{\alpha -\gamma }}= \frac{1}{ \delta _n^{\alpha -\lambda }} + \frac{1}{\delta _n^{\frac{(1-\lambda )(\alpha -\gamma )}{1-\gamma }}} \lesssim \frac{1}{ \delta _n^{\alpha -\lambda }}, \end{aligned}$$

because the inequality $\frac{(1-\lambda )(\alpha -\gamma )}{1-\gamma }< \alpha -\lambda$ is equivalent to $\alpha >1$ provided that $\alpha >\gamma$ and $\gamma <1$.

In the case $\alpha =1$, all the estimates for the case $\alpha >1$ except (14) remain true. Instead of (14) we have

$$\begin{aligned} \int _{\delta _n}^{{\hat{\delta }}_n} \frac{\omega (x)\, dx}{|1-r_n e^{ix}|^{2}} \lesssim \int _{\delta _n}^{{\hat{\delta }}_n} \frac{\delta _n^\lambda x\, dx}{\delta _n x^{2}} \lesssim \frac{1}{\delta _n^{1-\lambda }} \log \frac{1}{\delta _n}. \end{aligned}$$

(16)

The theorem is proved. $\square$

2 Example of a function of non-regular growth

The following theorem shows the sharpness of the corollary.

Theorem 6

Let $0<\varkappa _1<\varkappa _2< 1$, $\varkappa _1<\alpha$, and suppose that $\alpha$ is not an odd number. There exists an analytic function $f_\alpha$ in ${\mathbb {D}}$ of the form (1) such that $\omega (\delta _n, \psi )=O(\delta _n^{\varkappa _2})$, $\delta _n \rightarrow 0+$, while $\psi \in {\varLambda }_{\varkappa _1}$, $p[f_\alpha ]=\alpha -\varkappa _1$, and

$$\begin{aligned} \lambda [f_\alpha ]= {\left\{ \begin{array}{ll} \frac{\alpha (\alpha -\varkappa _1)(1-\varkappa _2)}{\alpha (1-\varkappa _2) +\varkappa _2-\varkappa _1},&{} \alpha <1 \\ \alpha -\varkappa _2 , &{} \alpha > 1. \end{array}\right. } \end{aligned}$$

(17)

Proof

Given two numbers $\varkappa _1$, $\varkappa _2$, $0<\varkappa _1<\varkappa _2<1$ we define a nondecreasing function $\psi :[0,1/2]\rightarrow {\mathbb {R}}$ recursively. Let $\psi (1/2)=2^{-\varkappa _1}$, $t_1=\frac{1}{2}$. Set

$$\begin{aligned} t_n^{\prime }=(t_n)^{\frac{1-\varkappa _1}{1-\varkappa _2}}, \quad t_{n+1}=(t_n^{\prime })^{\frac{\varkappa _2}{\varkappa _1}}, \quad n\in {\mathbb {N}}. \end{aligned}$$

(18)

One can write these sequences explicitly: $t_n=2^{-\beta ^{n-1}}$, where $\beta =\frac{1-\varkappa _1}{1-\varkappa _2} \frac{\varkappa _2}{\varkappa _1}$, and $t_n^{\prime }=2^{-\frac{1-\varkappa _1}{1-\varkappa _2}\beta ^{n-1}}$, $n\in {\mathbb {N}}$. It is clear that $t_{n+1}<t_n^{\prime }<t_n$, $n\in {\mathbb {N}}$, and $t_{n+1}=o(t_n^{\prime })$, $t_{n}^{\prime }=o(t_n)$ ($n\rightarrow \infty$).

We define the function

$$\begin{aligned} \psi (t)={\left\{ \begin{array}{ll} (t_n^{\prime })^{\varkappa _2-1}t, &{} t_n^{\prime }\le t\le t_n;\\ (t_n^{\prime })^{\varkappa _2}, &{} t_{n+1}\le t\le t_{n}^{\prime }. \end{array}\right. } \end{aligned}$$

(19)

It follows from the definition of the sequences $(t_n)$ and $(t_n^{\prime })$ that $\psi$ is a nondecreasing and continuous function on $[0,\frac{1}{2}]$. Moreover, $t^{\varkappa _2}\le \psi (t)\le t^{\varkappa _1}$, $t\in [0,\frac{1}{2}]$, and $\psi (t_n)=t_n^{\varkappa _1}$, $\psi (t_n^{\prime })=(t_n^{\prime })^{\varkappa _2}$, $n\in {\mathbb {N}}$.

It follows directly from the definition of $\psi$ that $\psi (t)/t$ is nonincreasing. By Dzyadyk and Shevchuk [7, Chap. 3] it implies semiadditivity of $\psi$, thus $\psi$ is a modulus of continuity.

Define

$$\begin{aligned} f_\alpha (z)=\int _0^{1/2} \frac{d\psi (t)}{(1-ze^{-it})^\alpha }. \end{aligned}$$

We then have

$$\begin{aligned} f_\alpha (r):= \sum _{n=1}^\infty \int _{t_n'}^{t_n} \frac{(t_n^{\prime })^{\varkappa _2-1}dt}{(1-re^{-it})^\alpha }, \quad r\in [0,1). \end{aligned}$$

Since $\psi$ is a modulus of continuity, $\omega (\delta , \psi )=\psi (\delta )=O(\delta ^{\varkappa _1})$, and, by construction we have $\sup \{ \gamma : \psi \in {\varLambda }_\gamma \}=\varkappa _1$. Therefore, by Theorem C, we have that $p[f_\alpha ]\le \alpha -\varkappa _1$. In fact $|f_\alpha (z)|=O((1-|z|)^{\varkappa _1-\alpha })$$(|z|\rightarrow 1-)$.

Let $\alpha >1$. It follows from the construction of $\psi$ that $\liminf \limits _{\delta \rightarrow 0+}\frac{\omega (\delta , \psi )}{\delta ^{\varkappa _2}}=1$. Hence, by Theorem 3, it follows that $\lambda [f_\alpha ]\le \alpha -\varkappa _2$. We prove that the converse inequality is true as well.

Denote

$$\begin{aligned} I_{n}(r)=(t_n^{\prime })^{\varkappa _2-1} \int _{t_n^{\prime }} ^{t_n} \frac{\cos (\alpha \tau (t))\,dt}{|1-re^{-it}|^\alpha }, \end{aligned}$$

where

$$\begin{aligned} \tau =\tau (t)=\arg \frac{1}{1-re^{-it}}, \quad \tan \tau =\frac{r\sin t}{1-r\cos t}. \end{aligned}$$

(20)

We are going to calculate asymptotics for $I_{n}$ for an appropriate choice of r. Let $A_n=\log \frac{1}{t_n^{\prime }}\rightarrow +\infty$$(n\rightarrow +\infty )$. If $1-r\le t_n^{\prime }/A_n$, then uniformly in r and $t\in [t_n^{\prime }, t_n]$,

$$\begin{aligned} 1-re^{-it}=1-r+r(1-e^{-it})\sim it, \quad n\rightarrow \infty . \end{aligned}$$

Remark 7

In what follows all asymptotic relations as $n\rightarrow \infty$ that depend on parameters r, t, etc. hold uniformly in these parameters at their specified ranges. For simplicity, this fact will no longer be emphasized.

We have $\tau (t)\sim \frac{\pi }{2}$, $\cos \alpha \tau (t)\sim \cos \frac{\alpha \pi }{2}$ as $t\in [t_n^{\prime }, t_n]$, $n\rightarrow +\infty$. Therefore,

$$\begin{aligned} I_{n}(r)\sim \frac{\cos \frac{\alpha \pi }{2}}{(t_n^{\prime })^{1-\varkappa _2}} \int _{t_n^{\prime }}^{t_n} \frac{dt}{t^\alpha } =\frac{C_0(\alpha )+o(1)}{(t_n^{\prime })^{\alpha -\varkappa _2}}, \quad n\rightarrow +\infty , 1-r\le t_n^{\prime }/A_n. \end{aligned}$$

(21)

where $C_0(\alpha )= \frac{\cos \frac{\alpha \pi }{2}}{\alpha -1}$.

If $1-r\ge A_n t_n$, $t\in [t_n^{\prime }, t_n]$, then $\cos \alpha \tau (t)\sim 1$, and $1-re^{-it}\sim 1-r$$(n\rightarrow \infty )$, so we get

$$\begin{aligned} I_{n}(r)\sim \frac{(t_n^{\prime })^{\varkappa _2-1}t_n}{(1-r)^\alpha } = \frac{(t_n)^{\varkappa _1}}{(1-r)^{\alpha }}, \quad n\rightarrow +\infty . \end{aligned}$$

(22)

Finally, let $1-r\in [t_n^{\prime }A_n, t_n/A_n]$. We split $I_n$ into three integrals in this case, namely,

$$\begin{aligned} I_n=\biggl ( \int \limits _{t_n^{\prime }}^{(1-r)/A_n}+\int \limits _{(1-r)/A_n}^{(1-r)A_n}+ \int \limits _{(1-r)A_n}^{t_n}\biggr ) \frac{(t_n^{\prime })^{\varkappa _2-1} \cos (\alpha \tau (t))\,dt}{|1-re^{-it}|^\alpha }=: J_{n1}+J_{n2}+J_{n3}. \end{aligned}$$

We begin with $J_{n2}$. Note that, $\tau (t)=\arctan \frac{r\sin t}{1-r\cos t}\sim \arctan \frac{t}{1-r}$ if $t\in [(1-r)/A_n, (1-r)A_n]$$(n\rightarrow \infty )$. Making the substitution $u=\frac{t}{1-r}$, we obtain

$$\begin{aligned} J_{n2}&\sim (t_n^{\prime })^{\varkappa _2-1} \int _{A_n^{-1}}^{A_n} \frac{\cos (\alpha \arctan u)(1-r)\, du}{\bigl ((1-r)^2+4\sin ^2 \frac{u(1-r)}{2}\bigr )^{\frac{\alpha }{2}}} \nonumber \\&\sim \frac{(t_n^{\prime })^{\varkappa _2-1}}{(1-r)^{\alpha -1}} \int _{A_n^{-1}}^{A_n} \frac{\cos (\alpha \arctan u)\, du}{(u^2+1)^{\frac{\alpha }{2}}}\sim \frac{C_1(\alpha )}{(t_n^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}}, \; n\rightarrow \infty , \end{aligned}$$

(23)

where $C_1(\alpha )=\int _0^\infty \frac{\cos (\alpha \arctan u)\, du}{(u^2+1)^{\frac{\alpha }{2}}}$. Similar arguments give

$$\begin{aligned} J_{n1}&\le (t_n^{\prime })^{\varkappa _2-1} \int _0^{1/A_n} \frac{dt}{(1-r)^{\alpha -1}}=o\Bigl ( \frac{1}{(t_n^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}} \Bigr ),\end{aligned}$$

(24)

$$\begin{aligned} J_{n3}&\le (t_n^{\prime })^{\varkappa _2-1} \int \limits _{(1-r)A_n}^{t_n} \frac{dt}{t^{\alpha }}\lesssim \frac{(t_n^{\prime })^{\varkappa _2-1}}{(A_n(1-r))^{\alpha -1}} =o\Bigl ( \frac{1}{(t_n^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}} \Bigr ) \end{aligned}$$

(25)

Combining (23)–(25), we get

$$\begin{aligned} I_n(r)\sim \frac{C_1(\alpha )}{(t_n^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}}, \quad 1-r\in [t_n^{\prime }A_n, t_n/A_n],\; n\rightarrow \infty . \end{aligned}$$

(26)

We proceed to estimate the growth of $|f_\alpha (r)|$ from below.

Let now $1-r\in [t_{k+1}A_k, \frac{t_k^{\prime }}{A_k}]$. Then, using (21) for $n\le k$, and (22), for $n\ge k+1$, we obtain for some $k_0\in {\mathbb {N}}$ (see Remark 7)

$$\begin{aligned} \mathfrak {R}f_\alpha (r)&=\biggl ( \sum _{n=1}^{k_0} + \sum _{n=k_0+1}^{k}+\sum _{n=k+1}^\infty \biggr ) I_{n} \nonumber \\&=O(1)+ \sum _{n=k_0+1}^k \frac{C_0(\alpha )+o(1)}{(t_n^{\prime })^{\alpha -\varkappa _2}}+ \sum _{n=k+1}^\infty (1+o(1))\frac{(t_n)^{\varkappa _1}}{(1-r)^\alpha }\nonumber \\&=O\Bigl (\frac{1}{(t_k^{\prime })^{\alpha -\varkappa _2}}\Bigr )+ (1+o(1))\frac{(t_{k+1})^{\varkappa _1}}{(1-r)^\alpha } \nonumber \\&=O\Bigl (\frac{1}{(t_k^{\prime })^{\alpha -\varkappa _2}}\Bigr )+ (1+o(1))\frac{(t_{k}^{\prime })^{\varkappa _2}}{(1-r)^\alpha }\sim \frac{(t_{k}^{\prime })^{\varkappa _2}}{(1-r)^\alpha } \nonumber \\&\ge \frac{A_k^{\varkappa _2}}{(1-r)^{\alpha -\varkappa _2}}, \quad k\rightarrow +\infty . \end{aligned}$$

(27)

Further, let $1-r\in [A_kt_k^{\prime }, t_k/A_k]$. In this case, using (21) for $n< k$, (22) for $n\ge k+1$, and (26) for $n=k$, definitions of $(t_n)$ and $(t_n^{\prime })$, we deduce $(k_1\in {\mathbb {N}})$

$$\begin{aligned} \mathfrak {R}f_\alpha (r)&=\biggl ( \sum _{n=1}^{k_1} + \sum _{n=k_1+1}^{k-1}+\sum _{n=k+1}^\infty \biggr ) I_{n}+ I_k \nonumber \\&=O(1) + \sum _{n=k_1+1}^{k-1} \frac{C_0(\alpha )+o(1)}{(t_n^{\prime })^{\alpha -\varkappa _2}}+ \sum _{n=k+1}^\infty \frac{(1+o(1))(t_n)^{\varkappa _1}}{(1-r)^\alpha } \nonumber \\&\quad +\, \frac{C_1(\alpha )+o(1)}{(t_k^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}}\nonumber \\&=O\Bigl (\frac{1}{(t_{k-1}^{\prime })^{\alpha -\varkappa _2}}\Bigr )+ (1+o(1))\frac{(t_{k+1})^{\varkappa _1}}{(1-r)^\alpha } + \frac{C_1(\alpha )+o(1)}{(t_k^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}} \nonumber \\&=O\Bigl (\frac{1}{(t_{k})^{\frac{\varkappa _1}{\varkappa _2} (\alpha -\varkappa _2)}}\Bigr )+ (1+o(1))\frac{(t_{k}^{\prime })^{\varkappa _2}}{(1-r)^\alpha }+ \frac{C_1(\alpha )+o(1)}{(t_k^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}} \nonumber \\&=O\Bigl (\frac{1}{(t_{k}^{\prime })^{\frac{\varkappa _1(1-\varkappa _2)}{\varkappa _2(1-\varkappa _1)}(\alpha -\varkappa _2)}}\Bigr )+ \frac{C_1(\alpha )+o(1)}{(t_k^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}} , \quad k\rightarrow +\infty . \end{aligned}$$

(28)

We show that the second addend in the last correlation dominates, i.e.

$$\begin{aligned} \frac{1}{(t_{k}^{\prime })^{\frac{\varkappa _1(1-\varkappa _2)}{\varkappa _2 (1-\varkappa _1)}(\alpha -\varkappa _2)}} =o\Bigl ( \frac{1}{(t_k^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}}\Bigr ), \quad 1-r\in [A_kt_k^{\prime }, t_k/A_k], k\rightarrow \infty , \end{aligned}$$

or

$$\begin{aligned} (1-r)^{\alpha -1}=o\Bigl ( (t_k^{\prime })^{(1-\varkappa _2) \Bigl (\frac{\varkappa _1(\alpha -\varkappa _2)}{\varkappa _2(1-\varkappa _1)}-1\Bigr )} \Bigr ), \quad 1-r\in [A_kt_k^{\prime }, t_k/A_k], k\rightarrow \infty . \end{aligned}$$

Since $\alpha >1$ and $1-r\le \frac{t_k}{A_k}=\frac{(t_k^{\prime })^{\frac{1-\varkappa _2}{1-\varkappa _1}}}{A_k}$, it is sufficient to check if $\frac{(\alpha -1)(1-\varkappa _2)}{1-\varkappa _1}> (1-\varkappa _2) \Bigl (\frac{\varkappa _1(\alpha -\varkappa _2)}{\varkappa _2(1-\varkappa _1)}-1\Bigr )$. This is equivalent to $\varkappa _2>\varkappa _1$, which holds by assumption. Hence, (28) can be rewritten in the form

$$\begin{aligned} |\mathfrak {R}f_\alpha (r)|&\sim \frac{|C_1(\alpha )|}{(t_k^{\prime })^{1-\varkappa _2}(1-r)^{\alpha -1}} \nonumber \\&\ge \frac{|C_1(\alpha )|}{(\frac{1-r}{A_k})^{1-\varkappa _2} (1-r)^{\alpha -1}} \nonumber \\&=\frac{|C_1(\alpha )| A_k^{1-\varkappa _1}}{(1-r)^{\alpha -\varkappa _2}} , \quad 1-r\in \left[ A_kt_k^{\prime }, \frac{t_k}{A_k}\right] , \; k\rightarrow +\infty . \end{aligned}$$

(29)

Therefore, taking into account (27) and (29), we obtain

$$\begin{aligned} (\alpha -\varkappa _2) \log \frac{1}{1-r} \lesssim \log M(r,f) \lesssim (\alpha -\varkappa _1) \log \frac{1}{1-r} \end{aligned}$$

(30)

as $r\in \bigcup _{k=1}^\infty \Bigl ( [A_k t_{k+1}, \frac{t_k^{\prime }}{A_k}]\cup [A_k t_k^{\prime }, \frac{t_k}{A_k}]\Bigr )$. Since for $r\in [t_{k+1},t_k]$

$$\begin{aligned} \log \Bigl (\frac{A_k^2}{1-r}\Bigr )^{\alpha -\varkappa _2} \le \log \Bigl (\frac{1}{1-r}\Bigr )^{\alpha -\varkappa _2}+ 2(\alpha -\varkappa _2)\log A_k\lesssim \log \Bigl (\frac{1}{1-r}\Bigr )^{\alpha -\varkappa _2}, \end{aligned}$$

(30) holds as $r\rightarrow 1-$ without exceptional sets.

This finishes the proof in the case $\alpha >1$.

Let now $0<\alpha <1$. We use the same function $f_\alpha$. Arguing similarly, we deduce

$$\begin{aligned} I_{n}(r)\sim \frac{\cos \frac{\alpha \pi }{2}}{(t_n^{\prime })^{1-\varkappa _2}} \int _{t_n^{\prime }}^{t_n} \frac{dt}{t^\alpha } \sim \frac{C_2(\alpha )t_n^{1-\alpha }}{(t_n^{\prime })^{1-\varkappa _2}}, \quad n\rightarrow +\infty , 1-r\le t_n^{\prime }/A_n, \end{aligned}$$

(31)

where $C_2(\alpha )=|C_0(\alpha )|=\frac{\cos \frac{\alpha \pi }{2}}{1-\alpha }$. Correlation (22) remains true. For $J_{nm}$, $m\in \{1,2,3\}$ we get

$$\begin{aligned} J_{n2}\sim \frac{(t_n^{\prime })^{\varkappa _2-1}}{(1-r)^{\alpha -1}} \int _{A_n^{-1}}^{A_n} \frac{\cos (\alpha \arctan u)\, du}{(u^2+1)^{\frac{\alpha }{2}}}\sim \frac{C_2(\alpha )A_n^{1-\alpha }(1-r)^{1-\alpha }}{(t_n^{\prime })^{1-\varkappa _2}}, \end{aligned}$$

(32)

while $J_{n1}$ allows the same estimate as for $\alpha >1$,

$$\begin{aligned} J_{n3} \sim \cos \frac{\alpha \pi }{2} (t_n^{\prime })^{\varkappa _2-1} \int _{(1-r)A_n}^{t_n} \frac{dt}{t^{\alpha }}\sim \frac{C_2(\alpha )}{(t_n^{\prime })^{1-\varkappa _2}}(t_n^{1-\alpha }- (A_n(1-r))^{1-\alpha }). \end{aligned}$$

Therefore

$$\begin{aligned} I_n=J_{n1}+J_{n2}+J_{n3}\sim C_2(\alpha ) \frac{t_n^{1-\alpha }}{(t_n^{\prime })^{1-\varkappa _2}}, \quad 1-r\in [t_n^{\prime }A_n, t_n/A_n],\; n\rightarrow \infty . \end{aligned}$$

(33)

For $1-r\in [t_{k+1}A_k, \frac{t_k^{\prime }}{A_k}]$, as in the case $\alpha >1$, using (31) for $n\le k$, and (22), for $n\ge k+1$, we get

$$\begin{aligned} \mathfrak {R}f_\alpha (r)&=O(1)+ \sum _{n=k_0+1}^k \frac{(C_2(\alpha )+o(1)) t_n^{1-\alpha }}{(t_n^{\prime })^{1-\varkappa _2}}+ \sum _{n=k+1}^\infty (1+o(1))\frac{(t_n)^{\varkappa _1}}{(1-r)^\alpha } \\&=\frac{(C_2(\alpha )+o(1)) t_k^{1-\alpha }}{(t_k^{\prime })^{1-\varkappa _2}}+ (1+o(1))\frac{(t_{k+1})^{\varkappa _1}}{(1-r)^\alpha } \\&=\frac{C_2(\alpha )+o(1)}{(t_k^{\prime })^{(1-\varkappa _2) \frac{\alpha -\varkappa _1}{1-\varkappa _1}}}+ (1+o(1))\frac{(t_{k}^{\prime })^{\varkappa _2}}{(1-r)^\alpha }, \quad k\rightarrow +\infty . \end{aligned}$$

We write ${\tilde{t}}_n:=(t_n^{\prime })^{1+ \frac{(\varkappa _2-\varkappa _1)(1-\alpha )}{\alpha (1-\varkappa _1)}}$. We have $t_{n+1}<{\tilde{t}}_n<t_n^{\prime }$, because $\varkappa _1<\alpha <1$. Consider two subcases. If $t_{k+1}A_k\le 1-r \le \tilde{t_k}$, then

$$\begin{aligned} \mathfrak {R}f_\alpha (r)&\ge (1+o(1))\frac{(t_{k}^{\prime })^{\varkappa _2}}{(1-r)^\alpha } \sim \frac{({\tilde{t}}_{k})^{\varkappa _2\frac{\alpha (1-\varkappa _1)}{\alpha (1-\varkappa _2)+\varkappa _2- \varkappa _1}}}{(1-r)^\alpha } \\&=\frac{1}{(1-r)^{\alpha (1- \frac{\varkappa _2(1-\varkappa _1)}{\alpha (1-\varkappa _2)+\varkappa _2- \varkappa _1})}} = \frac{1}{(1-r)^{\frac{\alpha (\alpha -\varkappa _1)(1-\varkappa _2)}{\alpha (1-\varkappa _2)+\varkappa _2- \varkappa _1}}} \end{aligned}$$

as required. If ${\tilde{t}}_k\le 1-r \le \frac{t_k^{\prime }}{A_k}$, then

$$\begin{aligned} \mathfrak {R}f_\alpha (r)\ge \frac{C_2(\alpha )+o(1)}{(t_k^{\prime })^{(1-\varkappa _2) \frac{\alpha -\varkappa _1}{1-\varkappa _1}}}\sim \frac{C_2(\alpha )}{\tilde{t_k}^{\frac{\alpha (1-\varkappa _2) (\alpha -\varkappa _1)}{\alpha (1-\varkappa _2)+\varkappa _2- \varkappa _1}}}\ge \frac{C_2(\alpha )}{(1-r)^{\frac{\alpha (1-\varkappa _2) (\alpha -\varkappa _1)}{\alpha (1-\varkappa _2)+\varkappa _2- \varkappa _1}}}. \end{aligned}$$

Thus, in both subcases we have the required lower estimate.

Finally, let $1-r \in [t_k^{\prime } A_k, \frac{t_k}{A_k}]$. Then using (31) for $n< k$, (22) for $n\ge k+1$, and (33) for $n=k$, and the definitions of $(t_n)$ and $(t_n^{\prime })$, we deduce

$$\begin{aligned} \mathfrak {R}f_\alpha (r)&=O(1)+ \sum _{n=k_0+1}^{k} \frac{(C_2(\alpha )+o(1)) t_n^{1-\alpha }}{(t_n^{\prime })^{1-\varkappa _2}}+ \sum _{n=k+1}^\infty (1+o(1))\frac{(t_n)^{\varkappa _1}}{(1-r)^\alpha } \\&\ge \frac{(C_2(\alpha )+o(1)) t_k^{1-\alpha }}{(t_k^{\prime })^{1-\varkappa _2}} \sim \frac{C_2(\alpha )}{(t_k^{\prime })^{(1-\varkappa _2) \frac{\alpha -\varkappa _1}{1-\varkappa _1}}} \ge \frac{C_2(\alpha )}{(\frac{1-r}{A_k} )^{(1-\varkappa _2)\frac{\alpha -\varkappa _1}{1-\varkappa _1}}} \quad k\rightarrow +\infty . \end{aligned}$$

It is easy to check that $(1-\varkappa _2)\frac{\alpha -\varkappa _1}{1-\varkappa _1} > \frac{\alpha (1-\varkappa _2)(\alpha -\varkappa _1)}{\alpha (1-\varkappa _2)+\varkappa _2- \varkappa _1}$, because $\alpha <1$.

We have proved that

$$\begin{aligned} \frac{\alpha (1-\varkappa _2)(\alpha -\varkappa _1)}{\alpha (1-\varkappa _2)+\varkappa _2- \varkappa _1} \log \frac{1}{1-r} \lesssim \log M(r,f) \lesssim (\alpha -\varkappa _1) \log \frac{1}{1-r} \end{aligned}$$

(34)

as $r\in \bigcup _{k=1}^\infty \Bigl ( [A_k t_{k+1}, \frac{t_k^{\prime }}{A_k}]\cup [A_k t_k^{\prime }, \frac{t_k}{A_k}]\Bigr )$$(k\rightarrow +\infty )$. As in the case $\alpha >1$ we deduce that (34) holds as $r\rightarrow 1-$. $\square$