Karamata's theorem for regularized Cauchy transforms

Abstract

We prove Abelian and Tauberian theorems for regularized Cauchy transforms of positive Borel measures on the real line whose distribution functions grow at most polynomially at infinity. In particular, we relate the asymptotics of the distribution functions to the asymptotics of the regularized Cauchy transform.

1. Introduction

Let $\mu$ be a positive measure on ${\mathbb {R}}$ such that $\int _{{\mathbb {R}}}(1+|t|)^{-1}\mkern 4mu\mathrm {d}\mu (t)<\infty$. The Cauchy transform of $\mu$ is the function

(1.1)\begin{equation} C[\mu](z)\mathrel{\mathop:}=\int_{{\mathbb{R}}}\frac 1{t-z}\mkern4mu\mathrm{d}\mu(t) \end{equation}

defined and analytic in the open upper half-plane ${\mathbb {C}}^+$. It plays an important role in many areas, such as spectral theory, moment problems, complex analysis and random matrix theory. A prominent particular case occurs when $\mu$ is supported on $[0,\,\infty )$. Then we speak of the Stieltjes transform of $\mu$ and write

(1.2)\begin{equation} S[\mu](z)\mathrel{\mathop:}=\int_{[0,\infty)}\frac 1{t-z}\mkern4mu\mathrm{d}\mu(t). \end{equation}

This function is defined and analytic in the slit plane ${\mathbb {C}}{\setminus} [0,\,\infty )$.

A measure $\mu$ can be reconstructed from its Cauchy (or Stieltjes) transform by means of the Stieltjes inversion formula,

(1.3)\begin{equation} \mu\bigl((\alpha,\beta)\bigr)= \lim_{\delta\downarrow0}\lim_{\varepsilon\downarrow 0} \frac{1}{\pi}\int_{\alpha+\delta}^{\beta-\delta}\operatorname{Im} C[\mu](x+i\varepsilon)\mkern4mu\mathrm{d} x \end{equation}

for $-\infty <\alpha <\beta <\infty$. This formula can be seen as relating the local behaviour of $\mu$ at a point or on a bounded interval in ${\mathbb {R}}$ with the local behaviour of $C[\mu ]$ around this point or interval: in order to evaluate the right-hand side of (1.3) only the values of $C[\mu ](z)$ for $z$ in some rectangle $(\alpha,\,\beta )\times (0,\,\varepsilon )\subseteq {\mathbb {C}}^+$ have to be known.

One question that has attracted a lot of attention is the relation between the asymptotics of $\mu$ at $\infty$ and the asymptotics of its transform at $\infty$. For the case of the Stieltjes transform results were obtained already in the early 20$^{^{\rm th}}$ century: G. Valiron [35], E.C. Titchmarsh [34], and G.H. Hardy and J.E. Littlewood [12] proved that, for each $\gamma \in (-1,\,0)$,

(1.4)\begin{equation} S[\mu]({-}x)\sim cx^\gamma\quad\Leftrightarrow\quad\mu([0,t))\sim c't^{\gamma+1}, \end{equation}

where $c,\,c'$ are related by a certain formula. Here the symbol $\sim$ means that the quotient of the left-hand and right-hand sides tends to $1$, and is understood for $x,\,t\to +\infty$. The asymptotics of $S[\mu ]$ along the ray $e^{i\pi }(0,\,\infty )$ could be substituted by the asymptotics along any ray contained in the domain of analyticity ${\mathbb {C}}{\setminus} [0,\,\infty )$ (allowing the constant $c$ to depend on the angle of the ray), or even non-tangentially. This early result about the Stieltjes transform was generalized in several directions, and there is a vast literature on that topic. As examples we mention [16] where J. Karamata generalized (1.4) to growth of regular variation instead of power asymptotics, or [27] where asymptotic expansions with infinitely many terms instead of a monomial on the right-hand side of (1.4) is considered. In the bilateral case, meaning measures that are not semi-bounded, asymptotics have to be taken along the positive imaginary axis, or a ray in ${\mathbb {C}}^+$ or non-tangentially. In this case there is much less known. One of the main difficulties is that contributions from the positive and negative half-axes can cancel each other.

In many applications, e.g. spectral theory of Sturm–Liouville and Schrödinger operators, measures are used that grow faster at infinity: instead of satisfying $\int _{{\mathbb {R}}}(1+|t|)^{-1}\mkern 4mu\mathrm {d}\mu (t)<\infty$ they have only power bounded tails, meaning that $\int _{{\mathbb {R}}}(1+t^2)^{-\kappa -1}\mkern 4mu\mathrm {d}\mu (t)<\infty$ for some $\kappa \in {\mathbb {N}}_0$. For such measures the Cauchy transform (1.1) has to be redefined by including appropriate regularizing summands in the integrand. The most common case is that $\mu$ is Poisson integrable, i.e. $\kappa =0$, and a commonly used regularization in this case is

(1.5)\begin{equation} \widetilde C[\mu](z)\mathrel{\mathop:}=\int_{{\mathbb{R}}}\Big(\frac 1{t-z}-\frac t{1+t^2}\Big)\mkern4mu\mathrm{d}\mu(t), \qquad z\in{\mathbb{C}}^+. \end{equation}

Some Abelian and Tauberian theorems dealing with polynomial asymptotics in the bilateral case are given in [28, 29], a Tauberian theorem for the Cauchy transform (1.1) can be found in [30], and an Abelian theorem of somewhat different type (and formulated for integration on the unit circle instead of the real line) is [33]. A Tauberian theorem for the regularized Cauchy transform (1.5) is stated in [2]; however, the proof given contains a mistake. Fortunately the result itself turns out to be true; see theorem 3.2 and the discussion preceding it.

In the current paper we prove Abelian and Tauberian theorems for higher-order regularized Cauchy transforms and growth of regular variation in Karamata's sense (see Appendix A for this notion). We relate the asymptotics of $\mu ([0,\,t))$ and $\mu ((-t,\,0))$ when $t\to +\infty$ to the asymptotics of the higher-order regularized Cauchy transform when $z\to +i\infty$ radially or non-tangentially. The main result of the paper is theorem 5.1 where we give a full characterization (including explicit formulae for constants) in the generic case. There are some boundary cases, namely when the index of regular variation is an integer, where only one direction is possible: either the Abelian direction where we deduce properties of the regularized Cauchy transform from properties of the measure, or the Tauberian direction, which is the other way round. The phenomenon that more complicated behaviour occurs at integer powers was already observed in [28, 29]. We investigate these exceptional cases more closely in theorem 5.5.

For the proof of our results we follow common lines and consider imaginary and real parts of the integral separately. The imaginary part can be written as a Stieltjes transform, and thus inherits being well behaved; see theorem 4.7. Contrasting this, the real part is the difference of two Stieltjes transforms, and this is the point where cancellation may happen.

Let us give a brief overview of the contents of the paper. In § 2 we define higher-order regularized Cauchy transforms and study basic properties. In particular, we explore the relation with generalized Nevanlinna functions in the sense of M.G. Krein and H. Langer [19], characterize the range of the transform, and prove an analogue of the classical Grommer–Hamburger theorem that relates convergence of a sequence of measures to convergence of their Cauchy transforms. We use the latter theorem to prove a basic Tauberian theorem in § 3. The proofs of Abelian theorems, which are contained in § 4, use different methods: the main ingredients are Karamata's theorems. In § 5 we combine the results from § 3 and 4 to prove our main theorems. We also provide counterexamples for the boundary cases. Finally, in Appendix A we recall and extend some results on regularly varying functions and Stieltjes transforms of measures supported on $[0,\,\infty )$.

Notation

Throughout the paper we use the following conventions and notations.

• $\triangleright\ $ We set ${\mathbb {N}}\mathrel {\mathop :}=\{1,\,2,\,\ldots \}$, ${\mathbb {N}}_0\mathrel {\mathop :}={\mathbb {N}}\cup \{0\}$, and let ${\mathbb {C}}$ be the field of complex numbers.

• $\triangleright{\kern-.1pt}$ We always use the branches of the logarithm and complex powers which are analytic on ${\mathbb {C}}{\setminus} (-\infty,\,0]$ and take the value $0$ or $1$, respectively, at the point $1$.

• $\triangleright{\kern.8pt}$ We set ${\mathbb {C}}^+\mathrel {\mathop :}=\{z\in {\mathbb {C}}\colon \mkern 3mu \operatorname {Im} z>0\}$ and ${\mathbb {C}}^-\mathrel {\mathop :}=\{z\in {\mathbb {C}}\colon \mkern 3mu \operatorname {Im} z<0\}$.

• $\triangleright{\kern.7pt}$ For a domain $\Omega$ denote by $\operatorname {Hol}(\Omega )$ and $\operatorname {Mer}(\Omega )$ the set of holomorphic and of meromorphic functions on $\Omega$ respectively.

• $\triangleright{\kern.7pt}$ We use the notation $f\sim g$ to express that $\frac fg\to 1$, and the notation $f\ll g$ if $\frac fg\to 0$. Further, we write $f\lesssim g$ if there exists a constant $c>0$ such that $f\le cg$, and we write $f\asymp g$ if $f\lesssim g$ and $g\lesssim f$. The domain of validity will be stated or will be clear from the context.

• $\triangleright{\kern.7pt}$ When we speak of a ‘measure’, we always mean positive Borel measure (unless explicitly specified differently).

• $\triangleright{\kern.7pt}$ Throughout the rest of the paper we use the Stieltjes transform ${\mathscr {S}}[\mu ]$ as defined in (A.7), where we use a different sign convention from the one used in (1.2).

2. Regularized Cauchy integrals

2.1. Definition of higher-order regularized Cauchy integrals

To start with, let us recall the characterizations of the ranges of the transforms $C$ and $\widetilde C$ introduced in (1.1) and (1.5). These are classical results going back to F. Riesz, G. Herglotz and R. Nevanlinna; for a comprehensive account see, e.g. [14] or [10].

Proposition 2.1 Let $q\in \operatorname {Hol}({\mathbb {C}}^+)$.

1. (i) The function $q$ can be represented in the form

   (2.1)\begin{equation} q(z)=a+C[\mu](z)=a+\int_{{\mathbb{R}}}\frac 1{t-z}\mkern4mu\mathrm{d}\mu(t), \qquad z\in{\mathbb{C}}^+, \end{equation}
   with some $a\in {\mathbb {R}}$ and a positive measure $\mu$ on ${\mathbb {R}}$ with $\int _{{\mathbb {R}}}(1+|t|)^{-1}\mathrm {d}\mu (t)<\infty$ if and only if
   \[ \forall z\in{\mathbb{C}}^+\kern2pt{\mathrel{\mathop:}\kern5pt} \operatorname{Im} q(z)\ge 0 \quad\text{and}\quad \int_1^\infty\frac{\operatorname{Im} q(iy)}y\mkern4mu\mathrm{d} y < \infty. \]

2. (ii) The function $q$ can be represented in the form

   (2.2)\begin{align} q(z)=a+bz+\widetilde C[\mu](z)=a+bz+\int_{{\mathbb{R}}}\Big(\frac 1{t-z}-\frac t{1+t^2}\Big)\mkern4mu\mathrm{d}\mu(t), \qquad z\in{\mathbb{C}}^+, \end{align}
   with some $a\in {\mathbb {R}}$, $b\ge 0$ and a positive measure $\mu$ on ${\mathbb {R}}$ with $\int _{{\mathbb {R}}}(1+t^2)^{-1}\mathrm {d}\mu (t)<\infty$ if and only if
   \[ \forall z\in{\mathbb{C}}^+\kern2pt{\mathrel{\mathop:}\kern5pt} \operatorname{Im} q(z) \ge 0. \]

Remark 2.2

1. (i) Assume that $q$ is represented in the form (2.2). Then the constants $a,\,b$ are given by

   (2.3)\begin{equation} a=\operatorname{Re} q(i), \qquad b=\lim_{y\to+\infty}\frac 1{iy}q(iy), \end{equation}
   and the measure $\mu$ is given by the Stieltjes inversion formula
   \[ \forall \alpha,\beta\in{\mathbb{R}},\,\alpha<\beta\kern2pt{\mathrel{\mathop:}\kern5pt} \lim_{\delta\downarrow0}\lim_{\varepsilon\downarrow 0} \frac{1}{\pi}\int_{\alpha+\delta}^{\beta-\delta}\operatorname{Im} q(x+i\varepsilon)\mkern4mu\mathrm{d} x = \mu\bigl((\alpha,\beta)\bigr); \]
   see, e.g. [14]. In particular, the map $(a,\,b,\,\mu )\mapsto q$ with $q$ satisfying (2.2) is injective.

2. (ii) Assume that $a=b=0$ in (2.2), i.e. $q(z)=\widetilde C[\mu ](z)$. Then

   (2.4)\begin{equation} |q(iy)| \ll y, \qquad y\to+\infty, \end{equation}
   by (2.3), and
   (2.5)\begin{equation} \lim_{y\to+\infty}y\operatorname{Im} q(iy) = \sup_{y>0}y\operatorname{Im}(iy) = \mu({\mathbb{R}}); \end{equation}
   see again [14].

It is apparent that moving from function (2.1) to function (2.2) is only the first step on a ladder: instead of Poisson-integrable measures one may use measures whose tails have at most power growth, and instead of the term $a+bz$ one may use any polynomial with real coefficients. In the integral higher-order regularization will become necessary.

We work with a scale of higher-order regularized Cauchy transforms which is commonly used in the framework of indefinite inner product spaces; see, e.g. [19].

Definition 2.3 Let $\kappa \in {\mathbb {N}}_0$.

1. (i) We denote by ${\mathbb {E}}_{\leq \kappa }$ the set of all pairs $(\mu,\,p)$ where

   • $\triangleright$ $\mu$ is a measure on ${\mathbb {R}}$ that satisfies

     (2.6)\begin{equation} \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}<\infty; \end{equation}

   • $\triangleright$ $p$ is a polynomial with real coefficients whose degree does not exceed $2\kappa +1$;

   • $\triangleright$ the coefficient of $z^{2\kappa +1}$ in $p$ satisfies

     (2.7)\begin{equation} \frac{1}{(2\kappa+1)!}p^{(2\kappa+1)}(0) \ge \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}. \end{equation}

2. (ii) The $\kappa$-regularized Cauchy transform is the map ${\mathscr {C}}_\kappa \colon {\mathbb {E}}_{\le \kappa }\to \operatorname {Hol}({\mathbb {C}}^+)$ defined by

   (2.8)\begin{equation} {{\mathscr{C}}_\kappa}[\mu,p](z) := p(z)+(1+z^2)^{\kappa+1}\int_{{\mathbb{R}}}\frac 1{t-z}\cdot \frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}, \qquad z\in{\mathbb{C}}^+. \end{equation}

In order to represent polynomials with real coefficients as regularized Cauchy transforms, we have to include pairs such as $(0,\,z^{2\kappa })$ in ${\mathbb {E}}_{\leq \kappa }$. For this reason we cannot speak of ‘the leading coefficient of $p$’ in (2.7).

These maps can indeed be seen as higher-order regularized Cauchy integrals: for $k\in {\mathbb {N}}_0$ we have

(2.9)\begin{align} & \frac{1}{t-z}-(t+z)\sum_{j=0}^k\frac{(1+z^2)^j}{(1+t^2)^{j+1}} = \frac{1}{t-z}-\frac{t+z}{1+t^2} \cdot\frac{1-\bigl(\frac{1+z^2}{1+t^2}\bigr)^{k+1}}{1-\frac{1+z^2}{1+t^2}}\nonumber\\ & \quad= \frac{1}{t-z}-(t+z)\cdot\frac{1-\bigl(\frac{1+z^2}{1+t^2}\bigr)^{k+1}}{t^2-z^2} = \frac{1}{t-z}\cdot\frac{(1+z^2)^{k+1}}{(1+t^2)^{k+1}}; \end{align}

hence we obtain, with $k=\kappa$,

(2.10)\begin{equation} {\mathscr{C}}_\kappa[\mu,p](z) = p(z) + \int_{{\mathbb{R}}} \biggl[\frac{1}{t-z}-(t+z)\sum_{j=0}^\kappa\frac{(1+z^2)^j}{(1+t^2)^{j+1}}\biggr] \mkern4mu\mathrm{d}\mu(t). \end{equation}

Note that the regularizing terms in the integral on the right-hand side of (2.10) are the first $\kappa +1$ terms of an expansion of $\frac {1}{t-z}=(t+z)\cdot \frac {1}{t^2-z^2}$ in terms of powers of $\frac {1}{1+t^2}$. For $\kappa =0$ relation (2.10) reads as

(2.11)\begin{equation} {\mathscr{C}}_0[\mu,p](z) = p(z) + \int_{{\mathbb{R}}}\Bigl(\frac{1}{t-z}-\frac{t+z}{1+t^2}\Bigr)\mkern4mu\mathrm{d}\mu(t), \end{equation}

which yields the following connection with the previously discussed regularized Cauchy-type integral (1.5):

• $\triangleright$ $a+bz+\widetilde C[\mu ](z)={\mathscr {C}}_0[\mu,\,p](z)$ with $\displaystyle p(z)\mathrel {\mathop :}= a+\biggl (b+\int _{{\mathbb {R}}}\frac {\mathrm {d}\mu (t)}{1+t^2}\biggr )z$;

• $\triangleright$ $\displaystyle (a,\,b,\,\mu )\in {\mathbb {R}}\times [0,\,\infty )\times \biggl \{\mu \colon \mkern 3mu\text {positive measure with }\int _{{\mathbb {R}}}\frac {\mathrm {d}\mu (t)}{1+t^2}<\infty \biggr \}$

  $\displaystyle \Longleftrightarrow \quad (\mu,\,p)\in {\mathbb {E}}_{\le 0}$ with $p$ related to $a$, $b$ and $\mu$ as above.

Remark 2.4 Using again (2.9) with $\kappa =0$ we obtain the following representation for ${\mathscr {C}}_\kappa$ for arbitrary $\kappa \in {\mathbb {N}}_0$:

(2.12)\begin{align} {\mathscr{C}}_\kappa[\mu,p](z) & = p(z) + (1+z^2)^\kappa\int_{{\mathbb{R}}}\Bigl(\frac{1}{t-z}-\frac{t+z}{1+t^2}\Bigr)\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa} \nonumber\\ & = \bigg(p(z)-z(1+z^2)^\kappa\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}\bigg) \nonumber\\ & \quad +(1+z^2)^\kappa\int_{{\mathbb{R}}}\Bigl(\frac{1}{t-z}-\frac{t}{1+t^2}\Bigr)\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}. \end{align}

If the stronger integrability condition $\int _{{\mathbb {R}}}(1+|t|)^{-(2\kappa +1)}\mathrm {d}\mu (t)<\infty$ is satisfied, then we can split the second integral on the right-hand side of (2.12) and rewrite it as

(2.13)\begin{align} {\mathscr{C}}_\kappa[\mu,p](z) & = \biggl(p(z)-(1+z^2)^\kappa \biggl[z\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}+ \int_{{\mathbb{R}}}\frac{t}{1+t^2}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}\biggr]\biggr) \nonumber\\ & \quad +(1+z^2)^\kappa\int_{{\mathbb{R}}}\frac{1}{t-z}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}. \end{align}

Before we collect some properties of ${\mathbb {E}}_{\le \kappa }$ and ${\mathscr {C}}_\kappa$, we recall the Stieltjes–Livšic inversion formula; see, e.g. [21, Corollary II.1.2] or [11, Theorem 1.2.4]). Let $\sigma$ be a finite measure on ${\mathbb {R}}$, let $\alpha,\,\beta \in {\mathbb {R}}$ with $\alpha <\beta$, and let $f$ be an analytic function on a neighbourhood of $[\alpha,\,\beta ]$. For $\delta,\,\varepsilon >0$ let $\Gamma _\varepsilon ^\delta$ be the path consisting of the two directed line segments

(2.14)\begin{equation} \alpha+\delta-i\varepsilon\rightsquigarrow \beta-\delta-i\varepsilon \quad\text{and}\quad \beta-\delta+i\varepsilon\rightsquigarrow \alpha+\delta+i\varepsilon. \end{equation}

Then

(2.15)\begin{equation} \lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0} \frac{-1}{2\pi i}\int_{\Gamma_\varepsilon^\delta}f(z)\int_{{\mathbb{R}}}\frac{1}{t-z}\mkern4mu\mathrm{d}\sigma(t)\mkern4mu\mathrm{d} z = \int_{(\alpha,\beta)}f(t)\mkern4mu\mathrm{d}\sigma(t). \end{equation}

Lemma 2.5 Let $\kappa \in {\mathbb {N}}_0$.

1. (i) The set ${\mathbb {E}}_{\le \kappa }$ is a positive cone and ${\mathscr {C}}_\kappa$ is a cone map, i.e. compatible with finite sums and non-negative scalar multiples.

2. (ii) The map ${\mathscr {C}}_\kappa$ is injective. For $q\in \operatorname {ran}{\mathscr {C}}_\kappa$ the element $(\mu,\,p)={\mathscr {C}}_\kappa ^{-1}q$ is obtained as follows: the polynomial $p$ can be recovered from solving the $2\kappa +2$ equations obtained by splitting real and imaginary parts of

   (2.16)\begin{equation} q^{(j)}(i) = p^{(j)}(i), \qquad j\in\{0,\ldots,\kappa\}; \end{equation}
   the measure $\mu$ can be obtained via the Stieltjes inversion formula: for $\alpha,\,\beta \in {\mathbb {R}}$ with $\alpha <\beta$ we have
   (2.17)\begin{equation} \mu\bigl((\alpha,\beta)\bigr) = \lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0} \frac{1}{\pi}\int_{\alpha+\delta}^{\beta-\delta} \operatorname{Im} q(t+i\varepsilon)\mkern4mu\mathrm{d}\mu(t). \end{equation}

3. (iii) Let $\kappa '>\kappa$. Then the inclusion $\operatorname {ran}{\mathscr {C}}_\kappa \subseteq \operatorname {ran}{\mathscr {C}}_{\kappa '}$ holds, and, for $(\mu,\,p)\in {\mathbb {E}}_{\le \kappa }$, we have ${\mathscr {C}}_\kappa [\mu,\,p]={\mathscr {C}}_{\kappa '}[\mu,\,\widetilde p]$ with

   \[ \widetilde p(z) = p(z) + \sum_{j=\kappa+1}^{\kappa'}(1+z^2)^j \biggl[z\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{j+1}} + \int_{{\mathbb{R}}}\frac{t\mkern4mu\mathrm{d}\mu(t)}{(1+t^2)^{j+1}}\biggl]. \]

Proof.

1. (i) The statements are clear from the definitions of ${\mathbb {E}}_{\le \kappa }$ and ${\mathscr {C}}_\kappa$.

2. (ii) Let $(\mu,\,p)\in {\mathbb {E}}_{\le \kappa }$ and set $q={\mathscr {C}}_\kappa [\mu,\,p]$. It follows from the definition of ${\mathscr {C}}_\kappa$ that (2.16) holds, which implies that $p$ is uniquely determined by $q$. To show (2.17), let us first extend $q$ to ${\mathbb {C}}{\setminus} {\mathbb {R}}$ by symmetry: $q(z)\mathrel {\mathop :}=\overline {q(\overline z)}$ for $z\in {\mathbb {C}}^-$. Moreover, let $\alpha,\,\beta\in {\mathbb {R}}$ with $\alpha< \beta$ and let $\Gamma _\varepsilon ^\delta$ be the path in (2.14). Then (2.15) implies that

   \begin{align*} & \lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0} \frac{1}{\pi}\int_{\alpha+\delta}^{\beta-\delta} \operatorname{Im} q(t+i\varepsilon)\mkern4mu\mathrm{d}\mu(t) = \lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0}\frac{-1}{2\pi i}\int_{\Gamma_\varepsilon^\delta}q(z)\mkern4mu\mathrm{d} z \\ & =\lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0}\frac{-1}{2\pi i}\int_{\Gamma_\varepsilon^\delta}(1+z^2)^{\kappa+1} \bigg[\int_{{\mathbb{R}}}\frac{1}{t-z}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}\bigg] \mkern4mu\mathrm{d} z \\ & =\int_{(\alpha,\beta)}(1+t^2)^{\kappa+1}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}} = \mu\big((\alpha,\beta)\big). \end{align*}
   The unique determination of $\mu$ and $p$ shows that ${\mathscr {C}}_\kappa$ is injective.

3. (iii) Let $(\mu,\,p)\in {\mathbb {E}}_{\le \kappa }$ and let $\kappa '>\kappa$. It follows from (2.9) that

(2.18)\begin{equation} \frac{1}{t-z}\cdot\frac{(1+z^2)^{\kappa+1}}{(1+t^2)^{\kappa+1}} = \frac{1}{t-z}\cdot\frac{(1+z^2)^{\kappa'+1}}{(1+t^2)^{\kappa'+1}} + (t+z)\sum_{j=\kappa+1}^{\kappa'}\frac{(1+z^2)^j}{(1+t^2)^{j+1}}, \end{equation}

which yields the statement in (iii).

Note that the Stieltjes inversion formula (2.17) for the recovery of $\mu$ is independent of $\kappa$.

2.2. Determining the range of ${\mathscr {C}}_\kappa$

An intrinsic characterization of the range of ${\mathscr {C}}_\kappa$ along the lines of proposition 2.1 (ii) can be given. This is based on [19, 20, 22] and related to [25, Theorem 3.9] (a predecessor of the latter is [15, Lemma 3.6]).

Let us first recall the definition of generalized Nevanlinna functions in the sense of [19]. We need, in particular, functions from the subclasses ${\mathcal {N}}_\kappa ^{(\infty )}$, which are characterized by a special behaviour at infinity and which were studied in, e.g. [6–8, 13, 24–26].

Definition 2.6 For $q\in \operatorname {Mer}({\mathbb {C}}^+)$ we denote by $\Omega _q$ its domain of analyticity; for the constant $q\equiv \infty$ we set $\Omega _q=\emptyset$.

1. (i) For $q\in \operatorname {Mer}({\mathbb {C}}^+)\cup \{\infty \}$ we denote by $\kappa _q\in {\mathbb {N}}_0\cup \{\infty \}$ the number of negative squares of the Hermitian kernel

   (2.19)\begin{equation} K_q(w,z) \mathrel{\mathop:}= \frac{q(z)-\overline{q(w)}\,}{z-\overline w}, \qquad z,w\in\Omega_q, \end{equation}
   i.e. the supremum of all numbers of negative squares of the quadratic forms
   \[ \sum_{i,j=1}^m K_q(w_j,w_i)\xi_i\overline{\xi_j} \]
   with $m\in {\mathbb {N}}$ and $w_1,\,\ldots,\,w_m\in \Omega _q$.

2. (ii) Let $\kappa \in {\mathbb {N}}_0$. We denote by ${\mathcal {N}}_\kappa$ the set of all functions $q\in \operatorname {Mer}({\mathbb {C}}^+)$ with $\kappa _q=\kappa$. Moreover, we set ${\mathcal {N}}_{\le \kappa }\mathrel {\mathop :}=\bigcup \limits _{\kappa '=0}^\kappa {\mathcal {N}}_{\kappa '}$ and ${\mathcal {N}}_{<\infty }\mathrel {\mathop :}=\bigcup \limits _{\kappa '=0}^\infty {\mathcal {N}}_{\kappa '}$.

3. (iii) Let $\kappa \in {\mathbb {N}}_0$. We denote by ${\mathcal {N}}_{\kappa }^{(\infty )}$ the set of all functions $q\in {\mathcal {N}}_\kappa$ for which

   \[ \lim_{y\to+\infty}\bigg|\frac{q(iy)}{y^{2\kappa-1}}\bigg| = \infty \quad\text{or}\quad \lim_{y\to+\infty}\frac{q(iy)}{(iy)^{2\kappa-1}}\in(-\infty,0). \]
   Moreover, we set ${\mathcal {N}}_{\le \kappa }^{(\infty )} \mathrel {\mathop :}=\bigcup \limits _{\kappa '=0}^\kappa {\mathcal {N}}_{\kappa '}^{(\infty )}$ and ${\mathcal {N}}_{<\infty }^{(\infty )}\mathrel {\mathop :}=\bigcup \limits _{\kappa '=0}^\infty {\mathcal {N}}_{\kappa '}^{(\infty )}$.

Remark 2.7

1. (i) Note that the classes ${\mathcal {N}}_0$ and ${\mathcal {N}}_0^{(\infty )}$ coincide with the set of all Nevanlinna functions, i.e. those functions $q$ that are analytic on ${\mathbb {C}}^+$ and satisfy $\operatorname {Im} q(z)\ge 0$ for $z\in {\mathbb {C}}^+$. Further, functions in ${\mathcal {N}}_\kappa$ have at most $\kappa$ poles and at most $\kappa$ zeros in ${\mathbb {C}}^+$.

2. (ii) The classes ${\mathcal {N}}_\kappa ^{(\infty )}$, which have also been denoted by ${\mathcal {N}}_\kappa ^\infty$ in the literature, can also be characterized differently, namely, for $q\in \operatorname {Mer}({\mathbb {C}}^+)$ the following conditions are equivalent (see [13], or also [6, 15]):

   1. (a) $q\in {\mathcal {N}}_{<\infty }^{(\infty )}$;

   2. (b) $\infty$ is the only (generalized) pole not of positive type (in the sense of [19, §3]), i.e. $\infty$ is the only (generalized) eigenvalue with a non-positive eigenvector of a representing relation in a Pontryagin space (see also [22] for an analytic characterization of generalized poles not of positive type);

   3. (c) there exist $m\in {\mathbb {N}}_0$, a real polynomial $p$ and a measure $\sigma$ on ${\mathbb {R}}$ such that $\int _{{\mathbb {R}}}(1+t^2)^{-1}\mathrm {d}\sigma (t)<\infty$ and

      (2.20)\begin{equation} q(z) = (1+z^2)^m\int_{{\mathbb{R}}}\Bigl(\frac{1}{t-z}-\frac{t}{1+t^2}\Bigr)\mkern4mu\mathrm{d}\sigma(t) + p(z); \end{equation}

   4. (d) there exist $n\in {\mathbb {N}}_0$, $\beta _1,\,\ldots,\,\beta _n\in {\mathbb {C}}^+\cup {\mathbb {R}}$, $\rho _j\in {\mathbb {N}}$ and $q_0\in {\mathcal {N}}_0$ such that

      \[ q(z) = q_0(z)\prod\limits_{j=1}^n(z-\beta_j)^{\rho_j}(z-\overline{\beta_j})^{\rho_j}. \]

   If $q$ is as in (2.20) and $\deg p=l$ with leading coefficient $c_l$, then $q\in {\mathcal {N}}_\kappa ^{(\infty )}$ with

   (2.21)\begin{equation} \kappa \le \max\{m,\kappa_p\} \qquad\text{where}\quad \kappa_p = \begin{cases} \frac{l}{2}, & l \ \text{even}, \\ \frac{l-\operatorname{sgn}(c_l)}{2}, & l \ \text{odd}; \end{cases} \end{equation}
   equality holds in (2.21) if $m=0$ or $\sigma$ is an infinite measure; see, e.g. [6, (1.16)]. Note that for a real polynomial $p$ one has $\kappa _p\le \kappa '$ if and only if $\deg p\le 2\kappa '+1$ and $p^{(2\kappa '+1)}(0)\ge 0$.

3. (iii) Representation (2.10) is a special case of the integral representation of ${\mathcal {N}}_{<\infty }$-functions given in [19, Satz 3.1].

4. (iv) In [25] representations of functions in ${\mathcal {N}}_{<\infty }^{(\infty )}$ were constructed with distributions (more precisely, distributional densities) on the one-point compactification ${\mathbb {R}}\cup \{\infty \}$ of ${\mathbb {R}}$ which act like measures on ${\mathbb {R}}$.

5. (v) Functions in ${\mathcal {N}}_{<\infty }^{(\infty )}$ are analytic in ${\mathbb {C}}^+$.

Theorem 2.8 For every $\kappa \in {\mathbb {N}}_0$ the equality $\operatorname {ran}{\mathscr {C}}_\kappa ={\mathcal {N}}_{\le \kappa }^{(\infty )}$ holds.

Proof. Let $\kappa \in {\mathbb {N}}_0$. It follows from (2.12) that $q\in \operatorname {ran}{\mathscr {C}}_\kappa$ if and only if it can be written as

(2.22)\begin{equation} q(z) = \widetilde p(z) + (1+z^2)^\kappa\int_{{\mathbb{R}}}\Bigl(\frac{1}{t-z}-\frac{t}{1+t^2}\Bigr)\mkern4mu\mathrm{d}\sigma(t) \end{equation}

with a real polynomial $\widetilde p$ of degree at most $2\kappa +1$ with $\widetilde p^{(2\kappa +1)}(0)\ge 0$ and a measure $\sigma$ such that $\int _{{\mathbb {R}}}(1+t^2)^{-1}\mathrm {d}\sigma (t)<\infty$.

First assume that $q\in \operatorname {ran}{\mathscr {C}}_\kappa$. Then (2.22) holds with $\sigma$ and $\widetilde p$ as above. By remark 2.7 (ii) we obtain that $q\in {\mathcal {N}}_{\le \kappa }^{(\infty )}$.

Conversely, assume that $q\in {\mathcal {N}}_{\le \kappa }^{(\infty )}$, say $q\in {\mathcal {N}}_{\kappa '}^{(\infty )}$ with $\kappa '\in \{0,\,\ldots,\,\kappa \}$. Then there exists a representation of $q$ as in (2.20) such that $\kappa '=\max \{m,\,\kappa _p\}$, where $\kappa _p$ is as in (2.21); in particular $m\le \kappa$ and $\kappa _p\le \kappa$. The function

\[ \widehat q(z) \mathrel{\mathop:}= (1+z^2)^m \int_{{\mathbb{R}}}\Bigl(\frac{1}{t-z}-\frac{t}{1+t^2}\Bigr)\mkern4mu\mathrm{d}\sigma(t) \]

belongs to $\operatorname {ran}{\mathscr {C}}_m$ by the first paragraph of this proof. Since $m\le \kappa$, we obtain from lemma 2.5 (iii) that $\widehat q\in \operatorname {ran}{\mathscr {C}}_\kappa$. The relations $\deg p\le 2\kappa +1$ and ${p^{(2\kappa +1)}(0)\ge 0}$ show that $(0,\,p)\in {\mathbb {E}}_{\le \kappa }$, and hence $p={\mathscr {C}}_\kappa [0,\,p]\in \operatorname {ran}{\mathscr {C}}_\kappa$. Now lemma 2.5 (i) implies that $q=\widehat q+p\in \operatorname {ran}{\mathscr {C}}_\kappa$.

2.3. ${\mathscr {C}}_\kappa$ as a homeomorphism: the Grommer–Hamburger theorem

Next we discuss a continuity property of ${\mathscr {C}}_\kappa$; see theorem 2.12 below. This result is a variant of a classical theorem of J. Grommer and H. Hamburger; see the discussion in remark 2.13.

Before being able to formulate a result, we have to make clear which topologies we use. On the set $\operatorname {Hol}({\mathbb {C}}^+)$ we always use the topology of locally uniform convergence. Topologizing ${\mathbb {E}}_{\le \kappa }$ is slightly more subtle. We proceed as follows. Fix $\kappa \in {\mathbb {N}}_0$. The set of all positive measures $\mu$ that satisfies (2.6) is a subset of the dual space of the weighted $C_0$-space

\[ C_0({\mathbb{R}},\omega_\kappa) \mathrel{\mathop:}= \Bigl\{f\in C({\mathbb{R}})\colon\mkern3mu \lim_{|x|\to\infty}|f(x)|\omega_\kappa(x)=0\Bigr\}, \qquad \|f\|\mathrel{\mathop:}=\sup_{x\in{\mathbb{R}}}|f(x)|\omega_\kappa(x), \]

where $\omega _\kappa$ is the weight function $\omega _\kappa (x)\mathrel {\mathop :}=(1+x^2)^{\kappa +1}$; note that

(2.23)\begin{equation} \|\mu\|_{C_0({\mathbb{R}},\omega_k)'} = \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}} \end{equation}

for a positive measure $\mu$ that satisfies (2.6). We endow the set of all positive measures $\mu$ that satisfy (2.6) with the subspace topology of the $w^*$-topology in $C_0({\mathbb {R}},\,\omega _\kappa )'$. The set of all polynomials of degree at most $2\kappa +1$ is isomorphic to ${\mathbb {R}}^{2\kappa +2}$ mapping a polynomial to its coefficients, and we endow polynomials with the Euclidean norm transported via this isomorphism. The set ${\mathbb {E}}_{\le \kappa }$ is now topologized as a subspace of the product.

This topology has some very nice properties, which are summarized in the following lemma.

Lemma 2.9 Let $\kappa \in {\mathbb {N}}_0$.

1. (i) ${\mathbb {E}}_{\le \kappa }$ is a closed subset of $C_0({\mathbb {R}},\,\omega _\kappa )'\times {\mathbb {R}}^{2\kappa +2}$.

2. (ii) A subset ${\mathcal {E}}\subseteq {\mathbb {E}}_{\leq \kappa }$ is relatively compact if and only if

   (2.24)\begin{equation} \sup\big\{\|p\|\colon\mkern3mu(\mu,p)\in{\mathcal{E}}\big\} < \infty. \end{equation}

3. (iii) The sets

   \[ {\mathcal{E}}_N\mathrel{\mathop:}=\big\{(\mu,p)\in{\mathbb{E}}_{\le\kappa}\colon\mkern3mu\|p\|\leq N\big\} \]
   are compact. We have ${\mathcal {E}}_N\subseteq \operatorname {Int}{\mathcal {E}}_{N+1}$, and $\bigcup _{N\in {\mathbb {N}}}{\mathcal {E}}_N={\mathbb {E}}_{\leq \kappa }$, where the interior $\operatorname {Int}{\mathcal {E}}_{N+1}$ of ${\mathcal {E}}_{N+1}$ is understood within ${\mathbb {E}}_{\leq \kappa }$.

Proof.

1. (i) Assume that $((\mu _i,\,p_i))_{i\in I}$ is a net in ${\mathbb {E}}_{\leq \kappa }$ that converges to some element $(\mu,\,p)\in C_0({\mathbb {R}},\,\omega _\kappa )'\times {\mathbb {R}}^{2\kappa +2}$. Then $\mu$ is again a positive measure, and, by (2.23),

   \begin{align*} \frac 1{(2\kappa+1)!}p^{(2\kappa+1)}(0) & = \lim_{i\in I}\frac 1{(2\kappa+1)!}p_i^{(2\kappa+1)}(0) \\ & \ge \limsup_{i\in I}\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu_i(t)}{(1+t^2)^{\kappa+1}}\ge \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}. \end{align*}
   Thus $(\mu,\,p)\in {\mathbb {E}}_{\leq \kappa }$, and we see that ${\mathbb {E}}_{\leq \kappa }$ is indeed closed.

2. (ii) Let $\pi _2$ be the projection onto the second component of $C_0({\mathbb {R}},\,\omega _\kappa )'\times {\mathbb {R}}^{2\kappa +2}$. Then $\pi _2$ is continuous. This already shows the implication ‘$\Rightarrow$’ in (ii). Conversely, assume that (2.24) holds. It follows from (2.23) that

   \[ \sup\Bigl\{\|\mu\|_{C_0({\mathbb{R}},\omega_\kappa)'}\colon\mkern3mu (\mu,p)\in{\mathcal{E}}\Bigr\} \le \sup\bigl\{\|p\|\colon\mkern3mu(\mu,p)\in{\mathcal{E}}\bigr\} =\mathrel{\mathop:} c < \infty. \]
   We see that
   \[ {\mathcal{E}} \subseteq \{\mu\in C_0({\mathbb{R}},\omega_\kappa)'\colon\mkern3mu\|\mu\|_{C_0({\mathbb{R}},\omega_\kappa)'}\le c\} \times \{p\in{\mathbb{R}}^{2\kappa+2}\colon\mkern3mu\|p\|\le c\}, \]
   and hence ${\mathcal {E}}$ is relatively compact in $C_0({\mathbb {R}},\,\omega _\kappa )'\times {\mathbb {R}}^{2\kappa +2}$ by the Banach–Alaoglu Theorem. Since ${\mathbb {E}}_{\le \kappa }$ is closed in this product space, ${\mathcal {E}}$ is also relatively compact in ${\mathbb {E}}_{\leq \kappa }$.

3. (iii) The statement follows from what we have shown so far and from the fact that the continuity of $\pi _2$ implies that ${\mathcal {E}}_N$ is closed and that

   \[ \operatorname{Int}{\mathcal{E}}_{N+1}=\big\{(\mu,p)\in{\mathbb{E}}_{{\leq}\kappa}\colon\mkern3mu\|p\|< N+1\big\}. \]

Proposition 2.10 The range of ${\mathscr {C}}_\kappa$ is closed in $\operatorname {Hol}({\mathbb {C}}^+)$, and ${\mathscr {C}}_\kappa$ is a homeomorphism onto its range.

Proof. Continuity of ${\mathscr {C}}_\kappa$ is clear from our choice of topology. Let $(q_i)_{i\in I}$ be a net in $\operatorname {ran}{\mathscr {C}}_\kappa$, and assume that $\lim _{i\in I}q_i=q$ in $\operatorname {Hol}({\mathbb {C}}^+)$. Remembering (2.16) we find $i_0\in I$ and $N\in {\mathbb {N}}$ such that ${\mathscr {C}}_\kappa ^{-1}(q_i)\in {\mathcal {E}}_N$ for all $i\ge i_0$. Since ${\mathcal {E}}_N$ is compact (and ${\mathscr {C}}_\kappa$ is continuous and injective), it follows that the limit $\lim _{i\in I}{\mathscr {C}}_\kappa ^{-1}(q_i)$ exists in ${\mathcal {E}}_N\subseteq {\mathbb {E}}_{\leq \kappa }$.

The following result is used in the proofs of theorems 2.12 and 3.1.

Proposition 2.11 Let $\kappa \in {\mathbb {N}}_0$ and $q_n\in {\mathcal {N}}_{\le \kappa }$, $n\in {\mathbb {N}}_0$. Assume that

1. (i) for each compact $K\subseteq {\mathbb {C}}^+$ with non-empty interior $O$ there exists $m_K\in {\mathbb {N}}$ such that $q_n$ is analytic on $O$ for all $n\ge m_K$;

2. (ii) there exists $M\subseteq {\mathbb {C}}^+$ with accumulation point in ${\mathbb {C}}^+$ such that $\lim \limits _{n\to \infty }q_n(z)$ exists for all $z\in M$.

Then there exists $\mathring q\in {\mathcal {N}}_{\le \kappa }\cap \operatorname {Hol}({\mathbb {C}}^+)$ such that $\lim \limits _{n\to \infty }q_n=\mathring q$ locally uniformly in ${\mathbb {C}}^+$. Here we understand locally uniform convergence in the space of meromorphic functions considered as analytic functions into the Riemann sphere.

Proof. Assumptions (i) and (ii) imply, in particular, that there exist $\kappa +1$ points $z_0,\,\ldots,\,z_\kappa \in {\mathbb {C}}^+$ such that $|q_n(z_i)|\le c$ for all $n\in {\mathbb {N}}$ and $i\in \{0,\,\ldots,\,\kappa \}$ and some $c>0$. By [23, Theorem 3.2] there exist a subsequence $(q_{n_k})_{k\in {\mathbb {N}}}$, a set $P\subseteq {\mathbb {C}}^+$ with $|P|\le \kappa$, and $\mathring q\in {\mathcal {N}}_{\le \kappa }$ such that

\[ \lim_{k\to\infty}q_{n_k} = \mathring q \qquad\text{locally uniformly on} \ {\mathbb{C}}^+{{\setminus}} P. \]

Note that $\mathring q$ is meromorphic on ${\mathbb {C}}^+$ because $\mathring q\in {\mathcal {N}}_{\le \kappa }$. Let $w\in P$. There exists a closed disc $K\subseteq {\mathbb {C}}^+$ around $w$ with interior $O$ such that $\mathring q$ is zero-free on $O{\setminus} \{w\}$. By assumption (i), $q_{n_k}$ is analytic on $O$ for all $k$ with $n_k\ge m_K$. The convergence of the logarithmic residue implies that $\mathring q$ is analytic at $w$, and hence $\lim _{k\to \infty }q_{n_k}=\mathring q$ locally uniformly on $O$. Since $w$ was arbitrary in $P$, this shows that $\mathring q$ is analytic on ${\mathbb {C}}^+$ and that $\lim _{k\to \infty }q_{n_k}=\mathring q$ locally uniformly on ${\mathbb {C}}^+$.

The above considerations can be done for every subsequence of $(q_n)$ instead of $(q_n)$ itself. Now assumption (ii) implies that $\lim _{n\to \infty }q_n=\mathring q$ locally uniformly on ${\mathbb {C}}^+$.

We can now prove an analogue of the classical Grommer–Hamburger theorem for regularized Cauchy transforms.

Theorem 2.12 Let $\kappa \in {\mathbb {N}}_0$, let $(\mu _n,\,p_n)\in {\mathbb {E}}_{\le \kappa }$ for $n\in {\mathbb {N}}$, set $q_n\mathrel {\mathop :}={\mathscr {C}}_\kappa [\mu _n,\,p_n]$, and let $\mathring q\in \operatorname {Hol}({\mathbb {C}}^+)$. Then the following three statements are equivalent:

1. (i) $\exists M\subseteq {\mathbb {C}}^+$ such that $M$ has an accumulation point in ${\mathbb {C}}^+$ and that $\lim \limits _{n\to \infty }q_n(z)=\mathring q(z)$ for all $z\in M$;

2. (ii) $\lim \limits _{n\to \infty }q_n=\mathring q$ locally uniformly on ${\mathbb {C}}^+$;

3. (iii) $\exists (\mathring \mu,\,\mathring p)\in {\mathbb {E}}_{\le \kappa }$ such that $\mathring q={\mathscr {C}}_\kappa [\mathring \mu,\,\mathring p]$, $\lim \limits _{n\to \infty }p_n=\mathring p$ and

   (2.25)\begin{equation} \forall a,b\in{\mathbb{R}}\kern2pt{\mathrel{\mathop:}\kern5pt} a< b,\, \mathring\mu(\{a\})=\mathring\mu(\{b\})=0 \quad\implies\quad \lim_{n\to\infty}\mu_n((a,b)) = \mathring\mu((a,b)). \end{equation}

Proof. The equivalence of (i) and (ii) follows directly from theorem 2.8 and proposition 2.11 since $q_n$ is analytic on ${\mathbb {C}}^+$ for every $n\in {\mathbb {N}}$.

Let us now prove the equivalence of (ii) and (iii). It follows from proposition 2.10 that (ii) is equivalent to

(2.26)\begin{equation} \exists(\mathring\mu,\mathring p)\in{\mathbb{E}}_{\le\kappa}\kern2pt{\mathrel{\mathop:}\kern5pt} \mathring q = {\mathscr{C}}_\kappa[\mathring\mu,\mathring p],\,\lim_{n\to\infty}p_n=\mathring p \end{equation}

together with $\lim _{n\to \infty }\mu _n=\mathring \mu$ w.r.t. $w^*$ in $C_0({\mathbb {R}},\,\omega _\kappa )'$. Now assume that (2.26) holds. Then the convergence of $(p_n)$ implies that

\[ \|\mu_n\|_{C_0({\mathbb{R}},\omega_\kappa)'} = \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu_n(t)}{(1+t^2)^{\kappa+1}} \le \frac{1}{(2\kappa+1)!}p_n^{(2\kappa+1)}(0) \le c \]

for some $c>0$. Since $C_{00}({\mathbb {R}})$, which denotes the set of compactly supported continuous functions on ${\mathbb {R}}$, is dense in $C_0({\mathbb {R}},\,\omega _\kappa )$, the relation $\mu _n\to \mathring \mu$ w.r.t. $w^*$ in $C_0({\mathbb {R}},\,\omega _\kappa )'$ is equivalent to $\mu _n\to \mathring \mu$ w.r.t. $w^*$ in $C_{00}({\mathbb {R}})'$. The latter relation is equivalent to (2.25) by the portmanteau-type theorem [1, Theorem 1].

Remark 2.13

1. (i) Let us make the connection with the original formulation of the Grommer–Hamburger theorem; see, e.g. [36, §48]. The latter is about Cauchy transforms, (1.1), of finite measures, and states the following: let $(\mu _n)_{n\in {\mathbb {N}}}$ be a sequence of finite measures on ${\mathbb {R}}$ whose total variations are uniformly bounded and let $\mathring q\in \operatorname {Hol}({\mathbb {C}}^+)$; then the following statements are equivalent:

   1. (a) $\lim \limits _{n\to \infty }C[\mu _n](z) = \mathring q(z)$ for all $z\in {\mathbb {C}}^+$;

   2. (b) there exists a finite measure $\mathring \mu$ such that $\mathring q=C[\mathring \mu ]$ and (2.25) holds.

2. (ii) There has been some confusion about the formulation of the Grommer–Hamburger theorem. The condition in (2.25) says that $(\mu _n)_{n\in {\mathbb {N}}}$ converges vaguely, i.e. w.r.t. $w^*$ in $C_{00}({\mathbb {R}})'$. However, at some places in the literature it is claimed that (a) in item (i) implies that $(\mu _n)_{n\in {\mathbb {N}}}$ converges weakly, i.e. w.r.t. $w^*$ in $C_{\rm b}({\mathbb {R}})'$, where $C_{\rm b}({\mathbb {R}})$ is the space of bounded continuous functions. The example $\mu _n=\delta _n$, where $\delta _n$ denotes the Dirac measure at $n$, shows that this is not true: $\lim _{n\to \infty }C[\delta _n]=0$ locally uniformly, but $\lim _{n\to \infty }\delta _n=0$ only vaguely and not weakly; in particular, mass is lost. Note that, by the portmanteau theorem (see, e.g. [18, Theorem 13.16]), a sequence of uniformly bounded measures $(\mu _n)$ on ${\mathbb {R}}$ converges weakly to a measure $\mathring \mu$ if and only if it converges vaguely to $\mathring \mu$ and $\lim _{n\to \infty }\mu _n({\mathbb {R}})=\mathring \mu ({\mathbb {R}})$. See also the discussion in [9].

3. (iii) In the original Grommer–Hamburger theorem one needs the a priori assumption that the total variations are uniformly bounded. On the other hand, in theorem 2.12 the integrals $\int _{{\mathbb {R}}}(1+t^2)^{-(\kappa +1)}\mathrm {d}\mu _n(t)$ are automatically bounded by (2.7) and the convergence of the polynomials $(p_n)$. Consider also the following example: let $\mu _n=n^2\delta _n$ and $p_n(z)=\frac {n^2}{1+n^2}z$. By (2.11) we have

   \begin{align*} {\mathscr{C}}_0[\mu_n,p_n](z) & = p_n(z) + \int_{{\mathbb{R}}}\Bigl(\frac{1}{t-z}-\frac{t+z}{1+t^2}\Bigr) \mkern4mu\mathrm{d}\mu_n(t) \\ & = \widetilde C[\mu_n](z) = n^2\Bigl(\frac{1}{n-z}-\frac{n}{1+n^2}\Bigr)\to z \end{align*}
   locally uniformly as $n\to \infty$. Note that the limit function belongs to $\operatorname {ran}{\mathscr {C}}_0$ but is not the Cauchy transform of a finite measure.

4. (iv) It follows from [1, Theorem 1] that (2.25) is equivalent to the following condition:

   \[ \text{for every bounded Borel set $A$ with} \ \mathring\mu(\partial A)=0:\quad \lim_{n\to\infty}\mu_n(A) = \mathring\mu(A). \]

5. (v) Under the additional assumption that $\mathring q\in {\mathcal {N}}_\kappa ^{(\infty )}$, i.e. $\mathring q$ has the same number of negative squares as $q_n$, the implication (ii) $\Rightarrow$ (2.25) in theorem 2.12 can also be deduced from [23, Corollary 3.1]; see also [26, Lemma 3.7].

3. A Tauberian theorem

In this section we prove a Tauberian theorem for the transform ${\mathscr {C}}_\kappa$ where the asymptotic behaviour of the measure $\mu$ towards infinity can be derived from the asymptotic behaviour of the function $q={\mathscr {C}}_\kappa [\mu,\,p]$ at infinity. As mentioned in the Introduction, there is a wide range of Tauberian theorems for Stieltjes transforms, where the measure is only supported on the positive half-line. Surprisingly, it seems there is much less known for Cauchy integrals, where the measure is allowed to be supported on the whole real line. One result, which has been frequently cited, is claimed in [2, Theorem 7.5]. The proof given in that paper contains a mistake¹. Fortunately, the result itself turns out to be true. In this section we provide a simple and conclusive argument which allows us, at the same time, to drop one assumption made in [2] and to generalize it to higher-order regularized Cauchy transforms.

Theorem 3.2 contains the above mentioned Tauberian theorem for the transform $q={\mathscr {C}}_\kappa [\mu,\,p]$. In most cases the asymptotic behaviour of the measure $\mu$ can be determined independently for the positive and the negative real axis; see (3.6) and (3.7). The assumption about the asymptotic behaviour of $q$ at infinity can be formulated in different ways. Theorem 3.1 shows the equivalence of these assumptions, where conditions (i) and (ii) are relatively minimal assumptions. In particular, (ii) says that, along one ray towards infinity, $q$ behaves like a constant times a regularly varying function; for the latter notion see Appendix A. Note that we prove theorem 3.1 for the larger class ${\mathcal {N}}_{\le \kappa }$ instead of the class ${\mathcal {N}}_{\le \kappa }^{(\infty )}=\operatorname {ran}{\mathscr {C}}_\kappa$.

The following functions play an important role in the current section, namely as limiting functions of rescalings of a given $q$:

(3.1)\begin{equation} Q_{\alpha,\omega}(z) \mathrel{\mathop:}= i\omega\Bigl(\frac zi\Bigr)^\alpha, \qquad z\in{\mathbb{C}}^+, \end{equation}

where $\alpha \in {\mathbb {R}}$ and $\omega \in {\mathbb {C}}{\setminus} \{0\}$. Note that, by the normalization of the power function at the end of the Introduction, we have $Q_{\alpha,\omega }(i)=i\omega$.

Theorem 3.1 Let $\kappa \in {\mathbb {N}}_0$ and $q\in {\mathcal {N}}_{\le \kappa }$, and let $f:[x_0,\,\infty )\to (0,\,\infty )$ with $x_0>0$ be measurable. Then the following statements are equivalent:

1. (i) there exists $M\subseteq {\mathbb {C}}^+$ with an accumulation point in ${\mathbb {C}}^+$ such that

   (3.2)\begin{equation} \forall z\in M\kern2pt{\mathrel{\mathop:}\kern5pt} \lim_{r\to\infty}\frac{q(rz)}{f(r)} \quad\text{exists and is non-zero;} \end{equation}

2. (ii) $f$ is regularly varying and there exists $z_0\in {\mathbb {C}}^+$ such that

   (3.3)\begin{equation} \lim_{r\to\infty}\frac{q(rz_0)}{f(r)} \quad\text{exists and is non-zero;} \end{equation}

3. (iii) $f$ is regularly varying with index $\alpha \in [-2\kappa -1,\,2\kappa +1]$ and there exists $\omega \in {\mathbb {C}}{\setminus} \{0\}$ such that

   \[ \lim_{r\to\infty}\frac{q(rz)}{f(r)} = Q_{\alpha,\omega}(z) \qquad\text{locally uniformly for} \ z\in{\mathbb{C}}^+; \]
   moreover, there exists $\kappa '\in \{0,\,\ldots,\,\kappa \}$ such that $Q_{\alpha,\omega }\in {\mathcal {N}}_{\kappa '}$ and hence
   (3.4)\begin{equation} \big|\arg\bigl(({-}1)^{\kappa'}\omega\bigr)\big| \le \frac\pi 2\Bigl(1-\big||\alpha|-2\kappa'\big|\Bigr). \end{equation}

If (i)–(iii) are satisfied and $\xi$ denotes the limit in (3.3), then $\omega =\frac {\xi }{i}\bigl (\frac {z_0}{i}\bigr )^{-\alpha }$.

Note that in the case when $\kappa =0$ we must have $\kappa '=0$ in (iii) and hence (3.4) reduces to $|\arg \omega |\le \frac {\pi }{2}(1-|\alpha |)$.

Theorem 3.2 Let $\kappa \in {\mathbb {N}}_0$ and $(\mu,\,p)\in {\mathbb {E}}_{\le \kappa }$, set $q\mathrel {\mathop :}={\mathscr {C}}_\kappa [\mu,\,p]$, and let $f:[x_0,\,\infty )\to (0,\,\infty )$ with $x_0>0$ be measurable. Assume that the equivalent conditions (i)–(iii) in theorem 3.1 hold. Then $\alpha \in [-1,\,2\kappa +1]$ and $Q_{\alpha,\omega }\in \operatorname {ran}{\mathscr {C}}_\kappa$. Let $\mu _{\alpha,\omega }$ be the measure component of ${\mathscr {C}}_\kappa ^{-1}Q_{\alpha,\omega }$. For all $a,\,b\in {\mathbb {R}}$ with $a< b$ and $\mu _{\alpha,\omega }(\{a\})=\mu _{\alpha,\omega }(\{b\})=0$, we have

(3.5)\begin{equation} \lim_{r\to\infty}\frac{1}{rf(r)}\mu\bigl((ra,rb)\bigr) = \mu_{\alpha,\omega}\bigl((a,b)\bigr). \end{equation}

In particular, if $\alpha >-1$, then

(3.6)\begin{align} \lim_{r\to\infty}\frac{1}{rf(r)}\mu\bigl((0,r)\bigr) & = \frac{1}{\pi}\cdot\frac{|\omega|}{\alpha+1} \cos\Bigl(\frac{\alpha\pi}{2}-\arg\omega\Bigr), \end{align}

(3.7)\begin{align} \lim_{r\to\infty}\frac{1}{rf(r)}\mu\bigl(({-}r,0)\bigr) & = \frac{1}{\pi}\cdot\frac{|\omega|}{\alpha+1} \cos\Bigl(\frac{\alpha\pi}{2}+\arg\omega\Bigr); \end{align}

if $\alpha =-1$, then $\omega >0$ and

(3.8)\begin{equation} \lim_{r\to\infty}\frac{1}{rf(r)}\mu\bigl(({-}r,r)\bigr) = \omega. \end{equation}

Remark 3.3

1. (i) Any asymmetry of the limits on the right-hand sides of (3.6) and (3.7) can be seen from $\omega$. To this end, assume that $\alpha >-1$ and write $\alpha =2m+\alpha _0$ with $m\in {\mathbb {N}}_0$ and $|\alpha _0|\le 1$; if $\alpha \in 2{\mathbb {N}}_0+1$, choose $m$ such that $(-1)^m\omega >0$. The limits on the right-hand sides of (3.6) and (3.7) can be rewritten as follows:

   \begin{align*} c_\pm & \mathrel{\mathop:}= \frac{1}{\pi}\cdot\frac{|\omega|}{\alpha+1}\cos\Bigl(\frac{\alpha\pi}{2}\mp\arg\omega\Bigr) \\ & = \frac{1}{\pi}\cdot\frac{|\omega|}{\alpha+1} \begin{cases} \displaystyle \cos\Bigl(\frac{\alpha_0\pi}{2}\mp\arg\omega\Bigr), & m \ \text{even}, \\ \displaystyle \cos\Bigl(\frac{\alpha_0\pi}{2}+\pi\mp\arg\omega\Bigr), & m \ \text{odd}, \end{cases} \\ & = \frac{1}{\pi}\cdot\frac{|\omega|}{\alpha+1} \cos\Bigl(\frac{\alpha_0\pi}{2}\mp\arg\bigl(({-}1)^m\omega\bigr)\Bigr). \end{align*}
   It follows from lemma 3.4 below and its proof that $m=\kappa '$ and hence, by (3.4), that $\big |\frac {\alpha _0\pi }{2}\mp \arg \bigl ((-1)^m\omega \bigr )\big |\le \frac {\pi }{2}$. From this the following equivalences follow easily, where we set $\psi \mathrel {\mathop :}=\arg ((-1)^m\omega )$,
   \begin{align*} c_+ & > c_- \quad & \iff\quad & \alpha_0,\psi\ne0 \;\;\wedge\;\; \operatorname{sgn}\alpha_0 = \operatorname{sgn}\psi, \\ c_+ & = c_- \quad & \iff\quad & \alpha_0=0 \;\;\vee\;\; \psi=0, \\ c_+ & < c_- \quad & \iff\quad & \alpha_0,\psi\ne0 \;\;\wedge\;\; \operatorname{sgn}\alpha_0 ={-}\operatorname{sgn}\psi, \\ c_\pm & = 0 \quad & \iff\quad & \big|\tfrac{\alpha_0\pi}{2}\mp\psi\big| = \tfrac{\pi}{2}, \\ c_+ & = c_-{=} 0 \quad & \iff\quad & |\alpha_0|=1 \;\;\vee\;\; \bigl(\alpha_0=0 \;\wedge\; |\psi|=\tfrac{\pi}{2}\bigr). \end{align*}
   If $c_+\ne 0$, then the function $r\mapsto \mu ((0,\,r))$ is regularly varying with index $\alpha +1$; if $c_-\ne 0$, then $r\mapsto \mu ((-r,\,0))$ is regularly varying with index $\alpha +1$.

2. (ii) Example 4.9 below shows that there are situations where (i)–(iii) in theorem 3.1 are satisfied but none of $t\mapsto \mu ((-t,\,t))$, $t\mapsto \mu ((0,\,t))$, $t\mapsto \mu ((-t,\,0))$ is regularly varying. See also example 5.3.

Before we prove theorems 3.1 and 3.2, we need some lemmas.

Lemma 3.4 Let $\alpha \in {\mathbb {R}}$ and $\omega \in {\mathbb {C}}{\setminus} \{0\}$ and let $Q_{\alpha,\omega }$ be as in (3.1).

1. (i) We have $Q_{\alpha,\omega }\in {\mathcal {N}}_\kappa$ if and only if

   (3.9)\begin{equation} \big|\arg\bigl(({-}1)^\kappa\omega\bigr)\big| \le \frac\pi 2\Bigl(1-\big||\alpha|-2\kappa\big|\Bigr). \end{equation}
   This is possible only when
   (3.10)\begin{equation} \kappa = \begin{cases} \bigl\lfloor\frac{|\alpha|+1}{2}\bigr\rfloor & \text{if} \ \alpha\notin 2{\mathbb{Z}}+1, \\ \frac{|\alpha|+1}{2} & \text{if} \ \alpha\in 2{\mathbb{Z}}+1 \ \text{and} \ ({-}1)^{\frac{|\alpha|+1}{2}}\omega>0, \\ \frac{|\alpha|-1}{2} & \text{if} \ \alpha\in 2{\mathbb{Z}}+1 \ \text{and} \ ({-}1)^{\frac{|\alpha|+1}{2}}\omega<0. \end{cases} \end{equation}
   In particular, $\frac {|\alpha |-1}{2}\le \kappa \le \frac {|\alpha |+1}{2}$.

2. (ii) Assume that $Q_{\alpha,\omega }\in {\mathcal {N}}_\kappa$, i.e. that (3.9) is satisfied. Then $Q_{\alpha,\omega }\in {\mathcal {N}}_\kappa ^{(\infty )}$ if and only if $\alpha \ge -1$ and, in addition, $\omega >0$ in the case when $\alpha =-1$.

3. (iii) Assume that $Q_{\alpha,\omega }\in {\mathcal {N}}_\kappa ^{(\infty )}$ and let $\mu _{\alpha,\omega }$ be the measure component of ${\mathscr {C}}_\kappa ^{-1} Q_{\alpha,\omega }$. If $\alpha =-1$, then $\mu _{\alpha,\omega }=\omega \delta _0$, where $\delta _0$ is the Dirac measure at $0$. If $\alpha >-1$, then $\mu _{\alpha,\omega }$ is absolutely continuous w.r.t. the Lebesgue measure and has density

   \[ \frac{\mathrm{d}\mu_{\alpha,\omega}}{\mathrm{d} t}(t) = \frac{|\omega|}{\pi}|t|^\alpha \cos\Bigl(\frac{\alpha\pi}{2}-(\operatorname{sgn} t)\arg\omega\Bigr), \qquad \text{a.e.} \ t\in{\mathbb{R}}. \]

Proof.

1. (i) Write $\alpha =2m+\alpha _0$ with $m\in {\mathbb {Z}}$ and $|\alpha _0|\le 1$ (note that if $\alpha$ is an odd integer, then $m$ and $\alpha _0$ are not unique). Then

   (3.11)\begin{equation} Q_{\alpha,\omega}(z) = i\omega\Bigl(\frac{z}{i}\Bigr)^{2m}\Bigl(\frac{z}{i}\Bigr)^{\alpha_0} = z^{2m}q_0(z) \end{equation}
   with
   \[ q_0(z) = i({-}1)^m\omega\Bigl(\frac{z}{i}\Bigr)^{\alpha_0}. \]
   Since the only generalized poles and zeros not of positive type (in the sense of [19]; see also [22]) can be $0$ and $\infty$, it follows from [5, Corollary] or [4, Proposition 3.2 and Theorem 3.3] that $Q_{\alpha,\omega }\in {\mathcal {N}}_\kappa$ if and only if $|m|=\kappa$ and $q_0\in {\mathcal {N}}_0$. Determining the sector onto which ${\mathbb {C}}^+$ is mapped under $q_0$ one can easily show that $q_0\in {\mathcal {N}}_0$ if and only if
   \[ \big|\arg\bigl(({-}1)^m\omega\bigr)\big| \le \frac{\pi}{2}\bigl(1-|\alpha_0|\bigr). \]
   Since $|\alpha _0|=\big ||\alpha |-|2m|\big |=\big ||\alpha |-2\kappa \big |$, the equivalence of $Q_{\alpha,\omega }\in {\mathcal {N}}_\kappa$ and (3.9) follows. The formula for $\kappa$ can be derived easily from (3.9).

2. (ii) It follows from the factorization (3.11) and [5, Corollary] that $q\in {\mathcal {N}}_\kappa ^{(\infty )}$ if and only if $m\ge 0$.

3. (iii) If $\alpha =-1$ and $\omega >0$, then $Q_{\alpha,\omega }(z)=-\frac {\omega }{z}$ and hence $Q_{\alpha,\omega }\in {\mathcal {N}}_0={\mathcal {N}}_0^{(\infty )}$ and $\mu _{\alpha,\omega }=\omega \delta _0$. Assume that $\alpha >-1$. The function $Q_{\alpha,\omega }$ can be extended to a continuous function on $({\mathbb {C}}^+\cup {\mathbb {R}}){\setminus} \{0\}$, and, for $t>0$, we have

   \begin{align*} \operatorname{Im} Q_{\alpha,\omega}({\pm} t) & = \operatorname{Im}\bigl[i\omega({\mp} it)^\alpha\bigr] = \operatorname{Im}\Bigl[|\omega|t^\alpha e^{i(\frac{\pi}{2}+\arg\omega\mp\alpha\frac{\pi}{2})}\Bigr] \\ & = |\omega|t^\alpha\sin\Bigl(\frac{\pi}{2}+\arg\omega\mp\frac{\alpha\pi}{2}\Bigr) = |\omega|t^\alpha\cos\Bigl(\arg\omega\mp\frac{\alpha\pi}{2}\Bigr). \end{align*}
   Now the assertion follows from the Stieltjes inversion formula (2.17). Note that there is no point mass at $0$ if $\alpha >-1$.

Remark 3.5 It follows from (3.9) and (3.10) that $Q_{\alpha,\omega }\in {\mathcal {N}}_{<\infty }$ if and only if

(3.12)\begin{align} & \quad \alpha\notin 2{\mathbb{Z}}+1 \quad\text{and}\quad & & \big|\arg\bigl((-1)^{\lfloor\frac{|\alpha|+1}{2}\rfloor}\omega\bigr)\big| \le \frac{\pi}{2}\operatorname{dist}(\alpha,2{\mathbb{Z}}+1) \\ & \text{or} \nonumber \\ & \quad \alpha\in 2{\mathbb{Z}}+1 \quad\text{and}\quad & & \omega\in{\mathbb{R}}. \end{align}

The significance of regular variation is that having a regularly varying asymptotics for $q(rz)$ for one single point $z$ already suffices to get locally uniform asymptotics depending on $z$ as a power. The reason for this is the multiplicative nature of the argument in $q(rz)$. In the following lemma we use the standard notation $z_0M=\{z_0z\colon \mkern 3mu z\in M\}$ with $M\subseteq {\mathbb {R}}$.

Lemma 3.6 Let $z_0\in {\mathbb {C}}^+$, $r_0>0$ and let $q:z_0(r_0,\,\infty )\to {\mathbb {C}}{\setminus} \{0\}$ be a continuous function. Further, let $f:[x_0,\,\infty )\to (0,\,\infty )$ with $x_0>0$ be a measurable function and let $B\subseteq (r_0,\,\infty )$ be a set with positive Lebesgue measure. Assume that

\[ \forall z\in z_0B\kern2pt{\mathrel{\mathop:}\kern5pt} \lim_{r\to\infty}\frac{q(rz)}{f(r)} \quad\text{exists and is non-zero}. \]

Then there exist $\alpha \in {\mathbb {R}}$ and $\omega \in {\mathbb {C}}{\setminus} \{0\}$ such that $f$ is regularly varying with index $\alpha$ and

(3.14)\begin{equation} \forall z\in z_0(0,\infty)\kern2pt{\mathrel{\mathop:}\kern5pt} \lim_{r\to\infty}\frac{q(rz)}{f(r)} = Q_{\alpha,\omega}(z), \end{equation}

where $Q_{\alpha,\omega }$ is as in (3.1).

Proof. Set $\mathring q(z)\mathrel {\mathop :}=\lim _{r\to \infty }\frac {q(rz)}{f(r)}$, $z\in z_0B$. Choose $s_0\in B$. For every $\lambda \in \frac {1}{s_0}B$ we have $\lambda s_0z_0\in z_0B$ and $s_0z_0\in z_0B$ and hence

\[ \lim_{r\to\infty}\frac{q(r\cdot\lambda s_0z_0)}{f(r)} = \mathring q(\lambda s_0z_0), \quad \lim_{r\to\infty}\frac{q(\lambda r\cdot s_0z_0)}{f(\lambda r)} = \mathring q(s_0z_0). \]

Taking quotients of these equations we obtain

\[ \lim_{r\to\infty}\frac{f(\lambda r)}{f(r)} = \frac{\mathring q(\lambda s_0z_0)}{\mathring q(s_0z_0)}. \]

Since the set $\frac {1}{s_0}B$ has positive measure, the Characterization Theorem [3, Theorem 1.4.1] yields that $f$ is regularly varying with index, say, $\alpha \in {\mathbb {R}}$. Hence, for every $\lambda >0$ we have

\[ \lim_{r\to\infty}\frac{q(r\cdot\lambda s_0z_0)}{f(r)} = \lim_{r\to\infty}\frac{q(\lambda r\cdot s_0z_0)}{f(\lambda r)} \cdot\lim_{r\to\infty}\frac{f(\lambda r)}{f(r)} = \mathring q(s_0z_0)\lambda^\alpha. \]

Replacing $\lambda s_0z_0$ by $z$ we obtain that, for every $z\in z_0(0,\,\infty )$,

\[ \lim_{r\to\infty}\frac{q(rz)}{f(r)} = \mathring q(s_0z_0)\Bigl(\frac{z}{s_0z_0}\Bigr)^\alpha, \]

which implies (3.14).

Lemma 3.7 Let $q\in \operatorname {ran}{\mathscr {C}}_\kappa$, $r>0$, and consider the function $q_{\langle r\rangle }(z)\mathrel {\mathop :}= q(rz)$. Then the following statements hold.

1. (i) $q_{\langle r\rangle }\in \operatorname {ran}{\mathscr {C}}_\kappa$.

2. (ii) The measure components $\mu$ and $\mu _{\langle r\rangle }$ of ${\mathscr {C}}_\kappa ^{-1}(q)$ and ${\mathscr {C}}_\kappa ^{-1}(q_{\langle r\rangle })$, respectively, are related by

   \[ \mu_{\langle r\rangle}=\tfrac{1}{r}\Sigma^r_*\mu, \]
   where $\Sigma ^r_*\mu$ is the push-forward of $\mu$ under the map $\Sigma ^r\colon t\mapsto \frac {1}{r}t$, i.e. $\mu _{\langle r\rangle }(M)=\frac {1}{r}\mu (rM)$ for a measurable set $M\subseteq {\mathbb {R}}$.

Proof. The statement in (i) is obvious from theorem 2.8. Write $q={\mathscr {C}}_\kappa [\mu,\,p]$ and $q_{\langle r\rangle }={\mathscr {C}}_\kappa [\mu _{\langle r\rangle },\,p_{\langle r\rangle }]$. Making a change of variable ($t=rs$) we obtain

\begin{align*} q_{\langle r\rangle}(z) & = q(rz) = p(rz)+\bigl(1+(rz)^2\bigr)^{\kappa+1} \int_{{\mathbb{R}}}\frac{1}{t-rz}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}} \\ & = p(rz)+\bigl(1+(rz)^2\bigr)^{\kappa+1}\cdot\int_{{\mathbb{R}}}\frac{1}{s-z} \cdot\frac{\frac{1}{r}\mkern4mu\mathrm{d}(\Sigma^r_*\mu)(s)}{(1+(rs)^2)^{\kappa+1}}. \end{align*}

Extend $q_{\langle r\rangle }$ to ${\mathbb {C}}^+\cup {\mathbb {C}}^-$ by symmetry: $q_{\langle r\rangle }(z)\mathrel {\mathop :}=\overline {q_{\langle r\rangle }(\overline z)}$, $z\in {\mathbb {C}}^-$, and let $\Gamma _{\varepsilon }^\delta$ be the path in (2.14). The Stieltjes inversion formula (2.17) and the Stieltjes–Livšic inversion formula (2.15) yield that, for all $\alpha,\,\beta \in {\mathbb {R}}$ with $\alpha <\beta$, we have

\begin{align*} \mu_{\langle r\rangle}\bigl((\alpha,\beta)\bigr) & = \lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0} \frac{1}{\pi}\int_{\alpha+\delta}^{\beta-\delta} \operatorname{Im} q_{\langle r\rangle}(t+i\varepsilon)\mkern4mu\mathrm{d} t = \lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0}\frac{-1}{2\pi i} \int_{\Gamma_\varepsilon^\delta}q_{\langle r\rangle}(z)\mkern4mu\mathrm{d} z \\ & = \lim_{\delta\downarrow 0}\lim_{\varepsilon\downarrow 0}\frac{-1}{2\pi i} \int_{\Gamma_\varepsilon^\delta}\bigl(1+(rz)^2\bigr)^{\kappa+1} \biggl[\int_{{\mathbb{R}}}\frac 1{s-z}\cdot \frac{\frac 1r\mkern4mu\mathrm{d}(\Sigma^r_*\mu)(s)}{(1+(rs)^2)^{\kappa+1}}\biggr]\mkern4mu\mathrm{d} z \\ & = \int_{(\alpha,\beta)}\big(1+(rs)^2\big)^{\kappa+1}\cdot \frac{\frac 1r\mkern4mu\mathrm{d}(\Sigma^r_*\mu)(s)}{(1+(rs)^2)^{\kappa+1}} = \tfrac 1r(\Sigma^r_*\mu)\bigl((a,b)\bigr). \end{align*}

Proof of theorem 3.1 The implication (iii) $\Rightarrow$ (ii) is trivial.

Next let us show the implication (ii) $\Rightarrow$ (i). For each $s>0$ the limit

\[ \lim_{r\to\infty}\frac{q(r\cdot sz_0)}{f(r)} = \lim_{r\to\infty}\biggl(\frac{q(rs\cdot z_0)}{f(rs)}\cdot\frac{f(rs)}{f(r)}\biggr) = \lim_{t\to\infty}\frac{q(tz_0)}{f(t)}\cdot\lim_{r\to\infty}\frac{f(rs)}{f(r)} \]

exists and is non-zero. Hence (i) is satisfied with $M=\{sz_0\colon \mkern 3mu s\in (0,\,\infty )\}$.

Finally, we prove the implication (i) $\Rightarrow$ (iii). Let $(r_n)_{n\in {\mathbb {N}}}$ be an arbitrary sequence of positive numbers with $r_n\to \infty$, and set

(3.15)\begin{equation} q_n(z) \mathrel{\mathop:}= \frac{q(r_nz)}{f(r_n)}, \qquad z\in{\mathbb{C}}^+. \end{equation}

It is easy to see that $q_n\in {\mathcal {N}}_{\le \kappa }$. Since $q$, as a function from ${\mathcal {N}}_{\le \kappa }$, has only finitely many poles in ${\mathbb {C}}^+$, assumption (i) in proposition 2.11 is satisfied. By (3.2) also assumption (ii) in proposition 2.11 is fulfilled. Hence, the latter proposition implies that there exists $\mathring q\in {\mathcal {N}}_{\le \kappa }\cap \operatorname {Hol}({\mathbb {C}}^+)$ such that $\lim _{n\to \infty }q_n=\mathring q$ locally uniformly in ${\mathbb {C}}^+$. Since the sequence $(r_n)_{n\in {\mathbb {N}}}$ was arbitrary, it follows again from (3.2) that

\[ \lim_{r\to\infty}\frac{q(rz)}{f(r)} = \mathring q(z) \]

locally uniformly for $z\in {\mathbb {C}}^+$. Now lemma 3.6 implies that there exist $\alpha \in {\mathbb {R}}$ and $\omega \in {\mathbb {C}}{\setminus} \{0\}$ such that $f$ is regularly varying with index $\alpha$ and that $\mathring q=Q_{\alpha,\omega }$. Since ${\mathcal {N}}_{\le \kappa }$ is closed under locally uniform convergence, we have $Q_{\alpha,\omega }\in {\mathcal {N}}_{\kappa '}$ with some $\kappa '\le \kappa$. By lemma 3.4 this shows that $|\alpha |\le 2\kappa '+1\le 2\kappa +1$ and that (3.4) holds.

Proof of theorem 3.2 Let $(r_n)_{n\in {\mathbb {N}}}$ be an arbitrary sequence of positive numbers with $r_n\to \infty$, and define $q_n$ as in (3.15). It follows from lemma 3.7 that $q_n\in \operatorname {ran}{\mathscr {C}}_\kappa$ and that the measure component $\mu _n$ of ${\mathscr {C}}_\kappa ^{-1}q_n$ satisfies

(3.16)\begin{equation} \mu_n\bigl((a,b)\bigr) = \frac{1}{r_nf(r_n)}\mu\bigl((r_na,r_nb)\bigr) \end{equation}

for all $a,\,b\in {\mathbb {R}}$ with $a< b$. Theorem 2.12 implies that $Q_{\alpha,\omega }\in \operatorname {ran}{\mathscr {C}}_\kappa ={\mathcal {N}}_{\le \kappa }^{(\infty )}$, which, by lemma 3.4 (ii) shows that $\alpha \ge -1$. Let $\mu _{\alpha,\omega }$ be the measure component of ${\mathscr {C}}_\kappa ^{-1}Q_{\alpha,\omega }$. Further, let $a,\,b\in {\mathbb {R}}$ be such that $a< b$ and $\mu _{\alpha,\omega }(\{a\})=\mu _{\alpha,\omega }(\{b\})=0$. It follows from theorem 2.12 that

(3.17)\begin{equation} \lim_{n\to\infty}\mu_n\bigl((a,b)\bigr) = \mu_{\alpha,\omega}\bigl((a,b)\bigr). \end{equation}

Since the sequence $(r_n)_{n\in {\mathbb {N}}}$ was arbitrary, relations (3.16) and (3.17) imply (3.5). We obtain from lemma 3.4 (iii) that

\[ \mu_{\alpha,\omega}\bigl({\pm}(0,1)\bigr) = \frac{1}{\pi}\cdot\frac{|\omega|}{\alpha+1} \cos\Bigl(\frac{\alpha\pi}{2}\mp\arg\omega\Bigr) \]

if $\alpha >-1$ and $\mu _{\alpha,\omega }((-1,\,1))=\omega$ if $\alpha =-1$. This, combined with (3.5), yields (3.6)–(3.8).

4. The Abelian direction

In this section we consider Abelian theorems, i.e. we study the asymptotic behaviour of $q$ at infinity using some knowledge about the asymptotic behaviour of the distribution functions $t\mapsto \mu ([0,\,t))$ and $t\mapsto \mu ((-t,\,0))$. Some of the theorems contain also a Tauberian direction, which complement the results in § 3. The main tool is Karamata's theorem about Stieltjes transforms of measures that are supported on the half-axis $[0,\,\infty )$. We follow the common lines to pass from unilateral to bilateral theorems, e.g. [28, 30], and represent imaginary and real parts as Stieltjes transforms. To this end we use the push-forward measure $\mu _*$ of $\mu$ under the map $t\mapsto t^2$, which satisfies

(4.1)\begin{equation} \mu_*((-\infty,0))=0 \qquad\text{and}\qquad \mu_*([0,t^2))=\mu(({-}t,t)), \quad t>0. \end{equation}

We shall often use a substitution to change between $\mu$ and $\mu _*$; let us note that, for a non-negative, measurable function $h$ on $[0,\,\infty )$, we have $\int _{[0,\infty )}h(s)\mkern 4mu\mathrm {d}\mu _*(s)=\int _{{\mathbb {R}}}h(t^2)\mkern 4mu\mathrm {d}\mu (t)$.

In order to apply Karamata's theorem in an effective way, we need a finer classification of the growth properties of the positive measure $\mu$, namely, let us set

(4.2)\begin{equation} {\mathscr{p}}(\mu) \mathrel{\mathop:}= \inf\biggl\{n\in{\mathbb{N}}\colon\mkern3mu \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^n}<\infty\biggr\} \in {\mathbb{N}}\cup\{\infty\}. \end{equation}

In the Abelian theorems we often assume that the symmetrized distribution function $t\mapsto \mu ((-t,\,t))$ is regularly varying. The following lemma can be used to give a different characterization of ${\mathscr {p}}(\mu )$ and its relation to the index of the regularly varying distribution function.

Lemma 4.1 Let $\mu$ be a measure on ${\mathbb {R}}$. For $\gamma >0$ we have

\[ \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\gamma} < \infty \quad\Leftrightarrow\quad \int_1^\infty\frac{\mu(({-}t,t))}{t^{\gamma+1}}\mkern4mu\mathrm{d} t < \infty. \]

If $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$, then ${\mathscr {p}}(\mu )$ is finite and $\beta \in [{\mathscr {p}}(\mu )-1,\,{\mathscr {p}}(\mu )]$.

Proof. Let the measure $\mu _*$ be defined as in (4.1), let $\gamma >0$ and define the measure $\nu$ on $[1,\,\infty )$ such that $\nu ((t,\,\infty ))=t^{-\frac {\gamma }{2}}$. It follows from lemma A.3 that the following equivalences hold:

(4.3)\begin{align} \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\gamma} < \infty \quad& \Leftrightarrow\quad \int_{[0,\infty)}\frac{\mathrm{d}\mu_*(s)}{(1+\sqrt{s})^\gamma} < \infty \quad\Leftrightarrow\quad \int_{[1,\infty)}s^{-\frac{\gamma}{2}}\mkern4mu\mathrm{d}\mu_*(s) < \infty \nonumber\\ & \Leftrightarrow\quad \int_1^\infty s^{-\frac{\gamma}{2}-1}\mu_*([1,s))\mkern4mu\mathrm{d} s < \infty \\ & \Leftrightarrow\quad \int_1^\infty\frac{\mu(({-}t,t))}{t^{\gamma+1}}\mkern4mu\mathrm{d} t < \infty, \nonumber \end{align}

where in the last step we used the substitution $s=t^2$.

Now assume that $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$. Then $s\mapsto \mu _*{[1,\,s)}$ is regularly varying with index ${\beta }/{2}$. It is clear that the integral in (4.3) is finite if $\gamma$ is large enough, which shows that ${\mathscr {p}}(\mu )$ is finite. Further, the fact that the integral in (4.3) is finite for $\gamma ={\mathscr {p}}(\mu )$ and infinite for $\gamma ={\mathscr {p}}(\mu )-1$ (unless $\mu$ itself is finite) implies that ${\beta }/{2}-{{\mathscr {p}}(\mu )}/{2}\le 0$ and ${\beta }/{2}-{{\mathscr {p}}(\mu )-1}/{2}\ge 0$, which finishes the proof; cf. proposition A.4. Note that, when $\mu$ is finite, then $\beta =0$ and ${\mathscr {p}}(\mu )=1$.

For the Abelian theorems we treat real and imaginary parts of $q(iy)$ separately as they have different representations in terms of Stieltjes transforms. This is done in the following two subsections.

4.1. The imaginary part

The imaginary part of $q(iy)$ is relatively well behaved as it can be written in terms of one Stieltjes transform. In order to apply Karamata's theorem, we choose $\kappa$ minimal in (2.6) for a given measure $\mu$. To this end, let us define

(4.4)\begin{equation} \kappa(\mu) \mathrel{\mathop:}= \inf\biggl\{n\in{\mathbb{N}}_0\colon\mkern3mu \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^{2n+2}}<\infty\biggr\} < \infty. \end{equation}

Comparing (4.4) with (4.2) we can easily deduce that

(4.5)\begin{equation} \kappa(\mu) = \Bigl\lfloor\frac{{\mathscr{p}}(\mu)-1}{2}\Bigr\rfloor. \end{equation}

Throughout this section we suppose that the following assumption is satisfied.

Assumption 4.2 Let $\mu$ be a measure on ${\mathbb {R}}$ such that ${\mathscr {p}}(\mu )<\infty$ and set $\kappa \mathrel {\mathop :}=\kappa (\mu )$. Further, let $p\in {\mathbb {R}}[z]$ with $p(z)=c_{2\kappa +1}z^{2\kappa +1}+\ldots +c_0$ such that $(\mu,\,p)\in {\mathbb {E}}_{\le \kappa }$, and set $q\mathrel {\mathop :}={\mathscr {C}}_\kappa [\mu,\,p]$.

Note that, by definition 2.3 (i), we have $c_{2\kappa +1}\ge \int _{{\mathbb {R}}}(1+t^2)^{-(\kappa +1)}\mathrm {d}\mu (t)$. The two cases, equality and strict inequality, lead to different asymptotic behaviour of $\operatorname {Im} q(iy)$ as $y\to \infty$, as the next proposition shows. Naturally, the case when the polynomial dominates the integral is the simpler one.

Proposition 4.3 Let $\mu$, $\kappa$, $p$, and $q$ be as in assumption 4.2.

1. (i) If $c_{2\kappa +1}>\int _{{\mathbb {R}}}(1+t^2)^{-(\kappa +1)}\mathrm {d}\mu (t)$, then

   \[ q(iy) \sim i({-}1)^\kappa\biggl(c_{2\kappa+1}-\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}\biggr)y^{2\kappa+1}, \qquad y\to\infty. \]

2. (ii) If $c_{2\kappa +1}=\int _{{\mathbb {R}}}(1+t^2)^{-(\kappa +1)}\mathrm {d}\mu (t)$, then

   (4.6)\begin{equation} |q(iy)| \ll y^{2\kappa+1} \qquad\text{and}\qquad \operatorname{Im} q(iy) \sim ({-}1)^\kappa y^{2\kappa+1} \int_{{\mathbb{R}}}\frac{1}{t^2+y^2}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}, \end{equation}
   as $y\to \infty$.

3. (iii) If $\mu$ is an infinite measure, then

   (4.7)\begin{equation} ({-}1)^\kappa\operatorname{Im} q(iy) \gg y^{2\kappa-1}, \qquad y\to\infty. \end{equation}

Proof.

• ${\bigcirc{\kern-6pt 1}}$ From (2.12) we obtain

  (4.8)\begin{align} q(iy) & = p(iy) + iy(1-y^2)^\kappa\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}} \nonumber\\& \quad + (1-y^2)^\kappa\int_{{\mathbb{R}}}\biggl(\frac{1}{t-iy}-\frac{t}{1+t^2}\biggr)\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}\nonumber\\ & = i({-}1)^\kappa\biggl(c_{2\kappa+1}-\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}\biggr)y^{2\kappa+1} + c_{2\kappa}({-}1)^\kappa y^{2\kappa} + \mathrm{O}\bigl(y^{2\kappa-1}\bigr)\nonumber\\ & \quad + ({-}1)^\kappa\Bigl(y^{2\kappa}+\mathrm{O}\bigl(y^{2\kappa-2}\bigr)\Bigr) \int_{{\mathbb{R}}}\biggl(\frac{1}{t-iy}-\frac{t}{1+t^2}\biggr)\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}. \end{align}
  Together with (2.4), this proves the assertion in (i), relation (4.7) when $c_{2\kappa +1}>\int _{{\mathbb {R}}}(1+t^2)^{-(\kappa +1)}\mathrm {d}\mu (t)$, and the first relation in (4.6) when $c_{2\kappa +1}=\int _{{\mathbb {R}}}(1+t^2)^{-(\kappa +1)}\mathrm {d}\mu (t)$.

• ${\bigcirc{\kern-6pt 2}}$ For the rest of the proof assume that $c_{2\kappa +1}=\int _{{\mathbb {R}}}(1+t^2)^{-(\kappa +1)}\mathrm {d}\mu (t)$. If the measure $\mu$ is finite, then $\kappa =0$ and

  \[ \operatorname{Im} q(iy) = \operatorname{Im}\int_{{\mathbb{R}}}\frac{1}{t-iy}\mkern4mu\mathrm{d}\mu(t) = y\int_{{\mathbb{R}}}\frac{1}{t^2+y^2}\mkern4mu\mathrm{d}\mu(t). \]

• ${\bigcirc{\kern-6pt 3}}$ Let us now consider the case when $\mu$ is infinite. We can use (4.8) to write the imaginary part as

  \begin{align*} \operatorname{Im} q(iy) & = ({-}1)^\kappa\Bigl(y^{2\kappa}+\mathrm{O}\bigl(y^{2\kappa-2}\bigr)\Bigr) \operatorname{Im}\biggl[\int_{{\mathbb{R}}}\biggl(\frac{1}{t-iy}-\frac{t}{1+t^2}\biggr) \frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}\biggr]\\ & \quad + \mathrm{O}\bigl(y^{2\kappa-1}\bigr). \end{align*}
  By the definition of $\kappa$ we have $\int _{{\mathbb {R}}}(1+t^2)^{-\kappa }\mathrm {d}\mu (t)=\infty$. Hence (2.5) implies that
  \[ \operatorname{Im}\biggl[\int_{{\mathbb{R}}}\biggl(\frac{1}{t-iy}-\frac{t}{1+t^2}\biggr) \frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}\biggr] \gg \frac{1}{y}, \]
  from which the second relation in (4.6) follows.

In the following we assume that the leading asymptotics of $q(iy)$ is not given by a polynomial term but the measure $\mu$. More precisely, we suppose that the following assumption is satisfied.

Assumption 4.4 Let $\mu$, $\kappa$, $p$, and $q$ satisfy the conditions in assumption 4.2. Further, assume that

(4.9)\begin{equation} c_{2\kappa+1} = \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}. \end{equation}

Remark 4.5 Suppose that assumption 4.2 holds and that $\kappa =\kappa (\mu )=0$, with $\kappa$ defined in (4.4). According to (2.11) we can write

(4.10)\begin{equation} q(z) = {\mathscr{C}}_0[\mu,p] = \biggl(c_1-\int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{1+t^2}\biggr)z + c_0 + \int_{{\mathbb{R}}}\biggl(\frac{1}{t-z}-\frac{t}{1+t^2}\biggr)\mkern4mu\mathrm{d}\mu(t). \end{equation}

Assumption 4.4 is equivalent to the coefficient of $z$ in (4.10) vanishing. Hence with assumption 4.4 being satisfied we have

(4.11)\begin{equation} q(z) = c_0 + \int_{{\mathbb{R}}}\biggl(\frac{1}{t-z}-\frac{t}{1+t^2}\biggr)\mkern4mu\mathrm{d}\mu(t) = c_0 + \widetilde C[\mu](z), \end{equation}

where $\widetilde C[\mu ]$ is defined in (1.5).

The next lemma shows that the imaginary part of $q(iy)$ can be written—at least asymptotically—in terms of a Stieltjes transform. For the definition of the Stieltjes transform see (A.7).

Lemma 4.6 Let $p$, $\kappa$, $p$ and $q$ satisfy the conditions in assumptions 4.2 and 4.4. Further, let $\mu _*$ be the push-forward measure of $\mu$ as in (4.1) and define the measure $\tau _\kappa$ on $[0,\,\infty )$ by

(4.12)\begin{equation} \mathrm{d}\tau_\kappa(s) = \frac{\mathrm{d}\mu_*(s)}{(1+s)^\kappa}, \qquad s\in[0,\infty). \end{equation}

Then the Stieltjes transform ${\mathscr {S}}[\tau _\kappa ]$ is well defined and

(4.13)\begin{equation} \operatorname{Im} q(iy) \sim ({-}1)^\kappa y^{2\kappa+1}{\mathscr{S}}[\tau_\kappa](y^2), \qquad y\to\infty. \end{equation}

Proof. Since

\[ \int_{[0,\infty)}\frac{\mathrm{d}\tau_\kappa(s)}{1+s} = \int_{[0,\infty)}\frac{\mathrm{d}\mu_*(s)}{(1+s)^{\kappa+1}} = \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}} < \infty, \]

the Stieltjes transform ${\mathscr {S}}[\tau _\kappa ]$ is well defined. It follows from proposition 4.3 (ii) that

\begin{align*} \operatorname{Im} q(iy) & \sim ({-}1)^\kappa y^{2\kappa+1} \int_{{\mathbb{R}}}\frac{1}{t^2+y^2}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa} \\ & = ({-}1)^\kappa y^{2\kappa+1} \int_{[0,\infty)}\frac{1}{s+y^2}\cdot\frac{\mathrm{d}\mu_*(s)}{(1+s)^\kappa} \\ & = ({-}1)^\kappa y^{2\kappa+1} \int_{[0,\infty)}\frac{1}{s+y^2}\mkern4mu\mathrm{d}\tau_\kappa(s) = ({-}1)^\kappa y^{2\kappa+1}{\mathscr{S}}[\tau_\kappa](y^2), \end{align*}

which proves (4.13).

In the following theorem we prove that the imaginary part of $q(iy)$ is related to the symmetrised distribution function $\mu ((-t,\,t))$ of the measure $\mu$. In most cases $|\operatorname {Im} q(iy)|$ is regularly varying if and only if $t\mapsto \mu ((-t,\,t))$ is regularly varying.

Theorem 4.7 Let $\mu$, $\kappa$, $p$, and $q$ satisfy the conditions in assumptions 4.2 and 4.4. Further, let $\beta \ge 0$ and consider the following two statements:

1. (a) the symmetrized distribution function $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$;

2. (b) the function $y\mapsto (-1)^\kappa \operatorname {Im} q(iy)$ is regularly varying with index $\beta -1$.

Then we have the following relations.

1. (i) The implication (a) $\Rightarrow$ (b) holds.

2. (ii) Unless $\kappa >0$ and $\beta =2\kappa$, also (b) $\Rightarrow$ (a) holds.

3. (iii) Assume that (a) and (b) are satisfied. Then $\beta \in [2\kappa,\,2\kappa +2]$ and

   (4.14)\begin{equation} \operatorname{Im} q(iy) \sim \begin{cases} \displaystyle \frac{\frac{\pi\beta}{2}}{\sin\frac{\pi\beta}{2}}\cdot\frac{\mu(({-}y,y))}{y}, & \beta\in[0,\infty){\setminus}\{2,4,\ldots\}, \\ \displaystyle ({-}1)^\kappa\beta y^{\beta-1}\int_1^y\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t, & \beta=2\kappa \;\wedge\; \kappa>0, \\ \displaystyle ({-}1)^\kappa\beta y^{\beta-1}\int_y^\infty\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t, & \beta=2\kappa+2, \end{cases} \end{equation}
   as $y\to \infty$, where the first fraction in the first case on the right-hand side is understood as $1$ when $\beta =0$. In particular, if $\beta \in 2{\mathbb {N}}$, then
   (4.15)\begin{equation} ({-}1)^\kappa\operatorname{Im} q(iy) \gg \frac{\mu(({-}y,y))}{y}. \end{equation}

Remark 4.8 In the situation of theorem 4.7 assume that (a) is satisfied and that $\beta \in 2{\mathbb {N}}$. It follows from the definition of $\kappa$ in (4.4) and from lemma 4.1 that

\begin{align*} \beta & = 2\kappa \quad & \Leftrightarrow\quad & \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\beta} = \infty \quad & \Leftrightarrow\quad & \int_1^\infty\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t & = \infty, \\ \beta & = 2\kappa+2 \quad & \Leftrightarrow\quad & \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\beta} < \infty \quad & \Leftrightarrow\quad & \int_1^\infty\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t & < \infty. \end{align*}

Proof of theorem 4.7 Let $\mu _*$ be the push-forward measure of $\mu$ as in (4.1) and define $\tau _\kappa$ as in (4.12). Then (4.13) holds. We prove the theorem in several steps.

• ${\bigcirc{\kern-6pt 1}}$ Let us first consider the case when $\kappa =0$. Then $\tau _0=\mu _*$ and, by theorem A.7, we have the following equivalences:

  \begin{align*} \text{(a)} \quad & \Leftrightarrow\quad s\mapsto\mu_*([0,s)) \ \text{is regularly varying with index} \ \tfrac{\beta}{2} \\ & \Leftrightarrow\quad {\mathscr{S}}[\mu_*] \ \text{is regularly varying with index} \ \tfrac{\beta}{2}-1 \\ & \Leftrightarrow\quad \text{(b)}. \end{align*}
  Assume now that (a) and (b) hold. Another application of theorem A.7 implies that $\frac {\beta }{2}\in [0,\,1]$. When $\beta \in [0,\,2)$, remark A.8 yields
  \[ \operatorname{Im} q(iy) = y\,{\mathscr{S}}[\mu_*](y^2) \sim y\frac{\frac{\pi\beta}{2}}{\sin\frac{\pi\beta}{2}}\cdot\frac{\mu_*([0,y^2))}{y^2} \sim \frac{\frac{\pi\beta}{2}}{\sin\frac{\pi\beta}{2}}\cdot\frac{\mu(({-}y,y))}{y}. \]
  When $\beta = 2$, we obtain from theorem A.7 and the substitution $s=t^2$ that
  \[ \operatorname{Im} q(iy) = y\,{\mathscr{S}}[\mu_*](y^2) \sim y\int_{y^2}^\infty\frac{\mu_*([0,s))}{s^2}\mkern4mu\mathrm{d} s = 2y\int_y^\infty\frac{\mu(({-}t,t))}{t^3}\mkern4mu\mathrm{d} t, \]
  which proves (4.14) when $\kappa =0$.

• ${\bigcirc{\kern-6pt 2}}$ Now let us consider the case when $\kappa >0$. Set $h(s)\mathrel {\mathop :}=\frac {1}{(1+s)^\kappa }\sim s^{-\kappa }$, $s\to \infty$. By the definition of $\kappa$ we have

  \begin{align*} \int_{[0,\infty)}h(s)\mkern4mu\mathrm{d}\mu_*(s) & = \int_{[0,\infty)}\frac{\mathrm{d}\mu_*(s)}{(1+s)^\kappa} = \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa} = \infty, \\ \int_{[0,\infty)}\frac{h(s)}{1+s}\mkern4mu\mathrm{d}\mu_*(s) & = \int_{[0,\infty)}\frac{\mathrm{d}\mu_*(s)}{(1+s)^{\kappa+1}} = \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}} < \infty, \end{align*}
  which shows that (A.20) with $\nu =\mu _*$ is satisfied.

• ${\bigcirc{\kern-6pt 3}}$ Assume that $\kappa >0$ and (a) holds. Then $s\mapsto \mu _*([0,\,s))$ is regularly varying with index $\frac {\beta }{2}$. By proposition A.11 (i) with $\alpha =\frac {\beta }{2}$ and $\gamma =-\kappa$ we have $\alpha +\gamma = \frac {\beta }{2}-\kappa \in [0,\,1]$ and hence $\beta >0$. Therefore we can apply proposition A.11 (i) again to obtain that $y\mapsto y^{2\kappa +1}{\mathscr {S}}[\tau _\kappa ](y^2)$ is regularly varying with index $2\kappa + 1+2\bigl (\frac {\beta }{2}-\kappa -1\bigr )=\beta -1$, i.e. (b) holds.

  If $\beta \notin 2{\mathbb {N}}$, then $\alpha +\gamma =\frac {\beta }{2}-\kappa \in (0,\,1)$ and (A.22) implies that

  \begin{align*} \operatorname{Im} q(iy) & \sim ({-}1)^\kappa y^{2\kappa+1}{\mathscr{S}}[\tau_\kappa](y^2) \\ & \sim ({-}1)^\kappa y^{2\kappa+1} \frac{\pi\frac{\beta}{2}}{\sin\bigl(\pi\bigl(\frac{\beta}{2}-\kappa\bigr)\bigr)} \cdot\bigl(y^2\bigr)^{-\kappa+1}\mu_*([0,y)) \\ & = \frac{\frac{\pi\beta}{2}}{\sin\frac{\pi\beta}{2}}\cdot\frac{\mu(({-}y,y))}{y}. \end{align*}

  Now assume that $\beta \in 2{\mathbb {N}}$. The relation $\frac {\beta }{2}-\kappa \in [0,\,1]$ implies that either $\kappa =\frac {\beta }{2}$ or $\kappa =\frac {\beta }{2}-1$. Let us consider the former case; the other case is similar. It follows again from (A.22) that, with the substitution $s=t^2$,

  \begin{align*} \operatorname{Im} q(iy) & \sim ({-}1)^\kappa y^{2\kappa+1}{\mathscr{S}}[\tau_\kappa](y^2) \\ & \sim ({-}1)^\kappa y^{2\kappa+1}\cdot\frac{\beta}{2}\cdot\frac{1}{y^2} \int_1^{y^2} s^{-\kappa-1}\mu_*([0,s))\mkern4mu\mathrm{d} s \\ & = ({-}1)^\kappa \beta y^{2\kappa-1}\int_1^y \frac{\mu_*([0,t^2))}{t^{2\kappa+1}}\mkern4mu\mathrm{d} t = ({-}1)^\kappa \beta y^{\beta-1}\int_1^y \frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t, \end{align*}
  which proves (4.14) also in the case $\kappa >0$.

• ${\bigcirc{\kern-6pt 4}}$ Now assume that $\kappa >0$, $\beta \ne 2\kappa$ and (b) holds. It follows from (4.13) that ${\mathscr {S}}[\tau _\kappa ]$ is regularly varying. Since $\alpha +\gamma =\frac {\beta }{2}-\kappa >0$, we can use proposition A.11 (ii) to deduce that $s\mapsto \mu _*([0,\,s))$ is regularly varying, which implies that (a) holds.

• ${\bigcirc{\kern-6pt 5}}$ Finally, assume that $\beta \in 2{\mathbb {N}}$. Relation (4.15) follows from (4.14) and theorem A.2.

The following example shows that the implication (b) $\Rightarrow$ (a) is, in general, not valid when $\beta =2\kappa >0$.

Example 4.9 Let $\kappa \in {\mathbb {N}}$ and set $h(s)=\frac {1}{(1+s)^\kappa }$, $s\in [0,\,\infty )$. Choose the measures $\sigma$ and $\nu$ as in example A.12, and let $\mu$ be the symmetric measure on ${\mathbb {R}}$ such that $\mu _*=\nu$, i.e. $\mu$ is the discrete measure with point masses

\[ \mu(\{e^{n/2}\}) = \mu(\{{-}e^{n/2}\}) = \nu(\{e^n\}) = (1+e^n)^\kappa, \qquad n\in{\mathbb{N}}. \]

According to example A.12 the distribution function $t\mapsto \mu ((-t,\,t))=\nu ([0,\,t^2))$ is not regularly varying, which means that (a) does not hold. On the other hand, $t\mapsto \tau _\kappa ([0,\,t))=\sigma ([0,\,t))$ is slowly varying, again by example A.12. It follows from theorem A.7 that ${\mathscr {S}}[\tau _\kappa ]$ is regularly varying with index $-1$, and hence (b) holds with $\beta =2\kappa$; see (4.13). Note that we have

\[ \operatorname{Im} q(iy) \sim 2({-}1)^\kappa y^{2\kappa-1}\log y, \qquad y\to\infty, \]

by (A.27).

Since $\mu$ is symmetric, we have $\operatorname {Re} q(iy)=0$ for $y>0$. Hence (3.3) is satisfied with $z_0=i$ and $f(r)=r^{2\kappa -1}\log r$. Theorem 3.2 implies that also (3.6) and (3.7) are satisfied. However, since $\omega =2(-1)^\kappa \in {\mathbb {R}}$ and $\alpha =2\kappa -1$ is odd, the right-hand sides of (3.6) and (3.7) vanish. This shows that, in some cases, one cannot use theorem 3.2 to deduce from the validity of (i)–(iii) in theorem 3.1 that $t\mapsto \mu ((-t,\,t))$ is regularly varying.

Example 4.10 Let $a,\,b\ge 0$ with $a\ne b$, and consider the function

\[ q(z) = a\log z-b\log({-}z), \]

which belongs to the Nevanlinna class ${\mathcal {N}}_0$. Since $q(ri)=(a-b)\log r + i(a+b)\frac {\pi }{2}$, conditions (i)–(iii) in theorem 3.1 are satisfied with $f(r)=\log r$, $\alpha =0$ and $\omega =i(b-a)$. Let $\mu$ be the measure in the representation $q={\mathscr {C}}_0[\mu,\,p]$. Theorem 3.2 only yields that

\[ \lim_{r\to\infty}\frac{\mu((0,r))}{r\log r} = 0, \qquad \lim_{r\to\infty}\frac{\mu(({-}r,0))}{r\log r} = 0. \]

On the other hand, since $\operatorname {Im} q(ri)=(a+b)\frac {\pi }{2}$, we can apply theorem 4.7 to obtain that $r\mapsto \mu ((-r,\,r))$ is regularly varying and that

\[ \mu(({-}r,r)) \sim \frac{2}{\pi}r\operatorname{Im} q(ir) = (a+b)r. \]

Note that, actually, $\mu ((0,\,r))=br$ and $\mu ((-r,\,0))=ar$.

The next proposition shows that, in the case $\kappa =0$, the validity of the first asymptotic relation in (4.14) implies already (a) and (b) in theorem 4.7.

Proposition 4.11 Let $\mu$ be a measure on ${\mathbb {R}}$ such that $\int _{{\mathbb {R}}}(1+t^2)^{-1}\mathrm {d}\mu (t)<\infty$, let $c_0\in {\mathbb {R}}$ and set $q\mathrel {\mathop :}=c_0+\widetilde C[\mu ]$. Assume that the limit

\[ \lim_{y\to\infty}\biggl({\rm{Im} q(iy)}\bigg/ {\frac{\mu(({-}y,\,y))}{y}}\biggr) \]

exists and is positive. Then (a) and (b) in theorem 4.7 are satisfied.

Proof. Let $\mu _*$ be as in (4.1). By (4.13), the following limit

\[ \lim_{t\to\infty}\frac{\mu_*([0,t))}{t{\mathscr{S}}[\mu_*](t)} = \lim_{y\to\infty}\frac{\mu_*([0,y^2))}{y^2{\mathscr{S}}[\mu_*](y^2)} = \lim_{y\to\infty}\frac{\mu(({-}y,y))}{y\operatorname{Im} q(iy)} \]

exists and is positive. Hence [32, Theorem B] implies that $t\mapsto \mu _*([0,\,t))$ is regularly varying, and hence (a) and (b) hold.

4.2. The real part

The real part of $q(iy)$ is more subtle since it can be written only in terms of a difference of two Stieltjes transforms and cancellations can arise. We introduce the following notation for the main part of $\operatorname {Re} q(iy)$. Let $\mu$ be a measure on ${\mathbb {R}}$ such that there exists $\ell \in {\mathbb {N}}_0$ with

(4.16)\begin{equation} \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^{2\ell+1}}<\infty, \end{equation}

and define

(4.17)\begin{equation} \mathsf{RC}_\ell[\mu](y) \mathrel{\mathop:}= (1-y^2)^\ell\int_{{\mathbb{R}}}\frac{t}{t^2+y^2}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^\ell}, \qquad y>0. \end{equation}

Lemma 4.12 Let $\mu$, $\kappa$, $p$, and $q$ be as in assumption 4.2, and let ${\mathscr {p}}(\mu )$ and $\mathsf {RC}_\ell$ be as in (4.2) and (4.17) respectively.

1. (i) If ${\mathscr {p}}(\mu )$ is odd, then there exists a real, even polynomial $\widetilde p$ of degree at most $2\kappa -2$ such that

   (4.18)\begin{equation} \operatorname{Re} q(iy) = ({-}1)^\kappa \biggl(c_{2\kappa}-\int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\mu(t)\biggr)y^{2\kappa} + \mathsf{RC}_\kappa[\mu](y) + \widetilde p(y). \end{equation}

2. (ii) If ${\mathscr {p}}(\mu )$ is even, then there exists a real, even polynomial $\widetilde p$ of degree at most $2\kappa$ such that

   (4.19)\begin{equation} \operatorname{Re} q(iy) = \mathsf{RC}_{\kappa+1}[\mu](y) + \widetilde p(y) \end{equation}
   with $\widetilde p(y) = (-1)^\kappa c_{2\kappa }y^{2\kappa }+\mathrm {O}(y^{2\kappa -2})$.

Proof.

1. (i) When ${\mathscr {p}}(\mu )$ is odd, then $\int _{{\mathbb {R}}}(1+|t|)^{-(2\kappa +1)}\mathrm {d}\mu (t)<\infty$. Hence from (2.13) we obtain

   \begin{align*} \operatorname{Re} q(iy) & = \operatorname{Re}(p(iy)) -(1-y^2)^\kappa\int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\mu(t)\\ & \quad +(1-y^2)^\kappa\int_{{\mathbb{R}}}\frac{t}{t^2+y^2}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^\kappa}, \end{align*}
   which yields (4.18).

2. (ii) When ${\mathscr {p}}(\mu )$ is even, then we use (2.8) to write

   \[ \operatorname{Re} q(iy) = \operatorname{Re}(p(iy)) + (1-y^2)^{\kappa+1}\int_{{\mathbb{R}}}\frac{t}{t^2+y^2}\cdot\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}, \]
   which gives (4.19).

Assumption 4.13 Let $\mu$, $\kappa$, $p$, and $q$ be as in assumption 4.2. Assume that, if ${\mathscr {p}}(\mu )$ is odd, then

\[ c_{2\kappa} = \int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\mu(t). \]

Remark 4.14 Assume that ${\mathscr {p}}(\mu )=1$ and that assumptions 4.2 and 4.4 are satisfied. We then have $\kappa =\kappa (\mu )=0$ and, by remark 4.5, we can write $q$ as in (4.11), i.e. $q(z)=c_0+\widetilde C[\mu ](z)$. If, in addition, assumption 4.13 is satisfied, then $c_0=\int _{{\mathbb {R}}}\frac {t}{1+t^2}\mkern 4mu\mathrm {d}\mu (t)$ and hence $q$ is the Cauchy transform of $\mu$, i.e.

\[ q(z) = \int_{{\mathbb{R}}}\frac{1}{t-z}\mkern4mu\mathrm{d}\mu(t) = C[\mu](z), \]

where $C[\mu ]$ is defined in (1.1).

To investigate the behaviour of the real part, we often choose $\ell$ minimal so that (4.17) makes sense, i.e. let us set

(4.20)\begin{equation} \ell(\mu) \mathrel{\mathop:}= \inf\biggl\{n\in{\mathbb{N}}_0\colon\mkern3mu \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^{2n+1}}<\infty\biggr\}. \end{equation}

It follows easily that $\ell (\mu )=\bigl \lfloor \frac {{\mathscr {p}}(\mu )}{2}\bigr \rfloor$, where ${\mathscr {p}}(\mu )$ is as in (4.2).

Remark 4.15 Let $\mu$ be a measure on ${\mathbb {R}}$ and define ${\mathscr {p}}(\mu )$, $\kappa (\mu )$ and $\ell (\mu )$ as in (4.2), (4.4) and (4.20) respectively. If ${\mathscr {p}}(\mu )$ is odd, then $\kappa (\mu )=\ell (\mu )=\frac {{\mathscr {p}}(\mu )-1}{2}$. If ${\mathscr {p}}(\mu )$ is even, then $\kappa (\mu )+1=\ell (\mu )=\frac {{\mathscr {p}}(\mu )}{2}$. In both cases we have $\kappa (\mu )+\ell (\mu )={\mathscr {p}}(\mu )-1$.

Lemma 4.16 Let $\mu$, $\kappa$, $p$, and $q$ be as in assumption 4.2 and $\ell =\ell (\mu )$. Then there exists a real, even polynomial $\widetilde p$ of degree at most $2\ell -2$ such that

\[ \operatorname{Re} q(iy) = \mathsf{RC}_\ell[\mu](y) + \widetilde p(y). \]

Proof. If ${\mathscr {p}}(\mu )$ is odd, then $\ell (\mu )=\kappa (\mu )$ and the statement follows from lemma 4.12 (i). If ${\mathscr {p}}(\mu )$ is even, then $\ell (\mu )=\kappa (\mu )+1$ and we can apply lemma 4.12 (ii).

Definition 4.17 Let $\mu$ be a measure on ${\mathbb {R}}$, let $\mu _*$ be the push-forward measure as in (4.1), and let $\ell \in {\mathbb {N}}_0$ be such that (4.16) is satisfied, i.e. $\ell \ge \ell (\mu )$. Define the measure $\sigma _\ell$ on $[0,\,\infty )$ by

(4.21)\begin{equation} \mathrm{d}\sigma_\ell(s) = \frac{\sqrt{s}}{(1+s)^\ell}\mkern4mu\mathrm{d}\mu_*(s), \qquad s\in[0,\infty), \end{equation}

and set

\[ {\mathcal{F}}_\ell[\mu](y) \mathrel{\mathop:}= (1-y^2)^\ell{\mathscr{S}}[\sigma_\ell](y^2), \qquad y>0, \]

where ${\mathscr {S}}$ is the Stieltjes transform defined in (A.7).

Lemma 4.18 Let $\mu$ be a measure on ${\mathbb {R}}$ and define the measures $\mu ^+$ and $\mu ^-$ on ${\mathbb {R}}$ by

(4.22)\begin{equation} \mathrm{d}\mu^+(t) \mathrel{\mathop:}= {\mathbb 1}_{[0,\infty)}(t)\mkern4mu\mathrm{d}\mu(t), \qquad \mathrm{d}\mu^-(t) \mathrel{\mathop:}= {\mathbb 1}_{(-\infty,0)}(t)\mkern4mu\mathrm{d}\mu(t), \qquad t\in{\mathbb{R}}. \end{equation}

Let $\ell \in {\mathbb {N}}_0$ such that (4.16) holds and let ${\mathcal {F}}_\ell$ be as in definition 4.17. Then

(4.23)\begin{align} \mathsf{RC}_\ell[\mu](y) & = {\mathcal{F}}_\ell[\mu^+](y) - {\mathcal{F}}_\ell[\mu^-](y), \end{align}

(4.24)\begin{align} \big|\mathsf{RC}_\ell[\mu](y)\big| & \le \big|{\mathcal{F}}_\ell[\mu](y)\big| \end{align}

for $y>0$.

Proof. Let $\mu _*$ and $\mu _*^\pm$ be the push-forward measures of $\mu$ and $\mu ^\pm$ respectively as in (4.1), and let $\sigma _\ell$, $\sigma _\ell ^\pm$ be as in (4.21).

The definition of $\ell$ in (4.20) implies that

\[ \int_{[0,\infty)}\frac{\mathrm{d}\sigma_\ell^\pm(s)}{1+s} = \int_{[0,\infty)}\frac{\sqrt{s}}{(1+s)^{\ell+1}}\mkern4mu\mathrm{d}\mu_*^\pm(s) = \int_{{\mathbb{R}}}\frac{|t|}{(1+t^2)^{\ell+1}}\mkern4mu\mathrm{d}\mu^\pm(t) < \infty, \]

and hence ${\mathscr {S}}[\sigma _\ell ^+]$ and ${\mathscr {S}}[\sigma _\ell ^-]$ are well defined. Moreover,

\begin{align*} \mathsf{RC}_\ell[\mu](y) & = (1-y^2)^\ell\biggl[\int_{{\mathbb{R}}}\frac{1}{t^2+y^2}\cdot\frac{|t|}{(1+t^2)^\ell}\mkern4mu\mathrm{d}\mu^+(t)\\ & \quad -\int_{{\mathbb{R}}}\frac{1}{t^2+y^2}\cdot\frac{|t|}{(1+t^2)^\ell}\mkern4mu\mathrm{d}\mu^-(t)\biggr] \\ & = (1-y^2)^\ell \biggl[\int_{[0,\infty)}\frac{1}{s+y^2}\cdot\frac{\sqrt{s}}{(1+s)^\ell}\mkern4mu\mathrm{d}\mu_*^+(s) \\ & \quad -\int_{[0,\infty)}\frac{1}{s+y^2}\cdot\frac{\sqrt{s}}{(1+s)^\ell}\mkern4mu\mathrm{d}\mu_*^-(s)\biggr] \\ & = (1-y^2)^\ell\Bigl[{\mathscr{S}}[\sigma_\ell^+](y^2)-{\mathscr{S}}[\sigma_\ell^-](y^2)\Bigr], \end{align*}

which yields (4.23). In a similar way as above one shows that ${\mathscr {S}}[\sigma _\ell ]$ is well defined. Further, we have

\begin{align*} \big|\mathsf{RC}_\ell[\mu](y)\big| & \le \big|(1-y^2)^\ell\big|\int_{{\mathbb{R}}}\frac{1}{t^2+y^2}\cdot\frac{|t|}{(1+t^2)^\ell}\mathrm{d}\mu(t) \\ & = \big|(1-y^2)^\ell\big|\int_{[0,\infty)}\frac{1}{s+y^2}\cdot\frac{\sqrt{s}}{(1+s)^\ell}\mkern4mu\mathrm{d}\mu_*(s) = \big|{\mathscr{S}}[\sigma_\ell](y^2)\big|, \end{align*}

which proves (4.24).

In the following key proposition the asymptotic behaviour of ${\mathcal {F}}_\ell [\mu ]$ is determined. It is used in proposition 4.22 and in the Abelian implications in theorems 5.1 and 5.5.

Proposition 4.19 Let $\mu$ be a measure on ${\mathbb {R}}$, set $\ell \mathrel {\mathop :}=\ell (\mu )$ and define ${\mathcal {F}}_\ell [\mu ]$ as in definition 4.17. Further, assume that $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$. Then $\beta \in [2\ell -1,\,2\ell +1]$. Moreover, if $\beta =0$, then

(4.25)\begin{equation} {\mathcal{F}}_\ell[\mu](y) \ll \frac{\mu(({-}y,y))}{y}, \qquad y\to\infty. \end{equation}

If $\beta >0$, then $(-1)^\ell {\mathcal {F}}_\ell [\mu ]$ is regularly varying with index $\beta -1$ and satisfies

(4.26)\begin{equation} {\mathcal{F}}_\ell[\mu](y) \sim \begin{cases} \displaystyle \frac{\frac{\pi\beta}{2}}{\cos\frac{\pi\beta}{2}}\cdot\frac{\mu(({-}y,y))}{y}, & \beta\notin2{\mathbb{N}}-1, \\ \displaystyle ({-}1)^\ell\beta y^{\beta-1}\int_1^y\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t, & \beta=2\ell-1, \\ \displaystyle ({-}1)^\ell\beta y^{\beta-1}\int_y^\infty\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t, & \beta=2\ell+1, \end{cases} \end{equation}

and

(4.27)\begin{equation} y^{2\ell-2} \ll \big|{\mathcal{F}}_\ell[\mu](y)\big| \ll y^{2\ell} \end{equation}

as $y\to \infty$. In particular, if $\beta \in 2{\mathbb {N}}-1$, then

(4.28)\begin{equation} \big|{\mathcal{F}}_\ell[\mu](y)\big| \gg \frac{\mu(({-}y,y))}{y}. \end{equation}

Remark 4.20

1. (i) In the situation of proposition 4.19 assume that $\beta \in 2{\mathbb {N}}-1$ and that $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$. It follows from the definition of $\ell (\mu )$ in (4.20) and from lemma 4.1 that, with $\ell =\ell (\mu )$,

   \begin{align*} \beta & = 2\ell-1 \quad & \Leftrightarrow\quad & \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\beta}\!\!\!\! & = \infty \quad & \Leftrightarrow\quad & \int_1^\infty\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t & = \infty, \\ \beta & = 2\ell+1 \quad & \Leftrightarrow\quad & \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\beta}\!\!\!\! & < \infty \quad & \Leftrightarrow\quad & \int_1^\infty\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t & < \infty. \end{align*}

2. (ii) Instead of the assumption $\ell =\ell (\mu )$ in the proposition, let us consider the case when $\ell >\ell (\mu )$. Then

   \[ \sigma_\ell([0,\infty)) = \int_{[0,\infty)}\frac{\sqrt{s}}{(1+s)^\ell}\mkern4mu\mathrm{d} s = \int_{{\mathbb{R}}}\frac{|t|}{(1+t^2)^\ell}\mkern4mu\mathrm{d}\mu(t) < \infty, \]
   and, by remark A.6, we have
   \[ {\mathcal{F}}_\ell[\mu](y) \sim ({-}1)^\ell y^2{\mathscr{S}}[\sigma_\ell](y^2) \sim ({-}1)^\ell\sigma_\ell([0,\infty))y^{2\ell-2} \]
   as $y\to \infty$.

Proof of theorem 4.19 Let $\mu _*$ be the push-forward measure of $\mu$ as (4.1), and let $\sigma _\ell$ be as in (4.21). Then $s\mapsto \mu _*([0,\,s))=\mu ((-\sqrt {s},\,\sqrt {s}))$ is regularly varying with index $\frac \beta 2$. Set $h(s) \mathrel {\mathop :}= \frac {\sqrt {s}}{(1+s)^\ell } \sim s^{\frac 12-\ell }$ as $s\to \infty$. Then

(4.29)\begin{equation} \int_{[0,\infty)}h(s)\mkern4mu\mathrm{d}\mu_*(s) = \int_{{\mathbb{R}}}\frac{|t|}{(1+t^2)^\ell}\mkern4mu\mathrm{d}\mu(t). \end{equation}

Let us first consider the case when the integrals in (4.29) are finite. Then $\ell =0$ by the definition of $\ell$, and $\sigma _0$ and $\mu$ are finite measures, which implies that $\beta =0$. From remark A.6 we obtain that

\[ {\mathcal{F}}_0[\mu](y) = {\mathscr{S}}[\sigma_0](y^2) \sim \frac{\sigma_0([0,\infty))}{y^2} \ll \frac{\mu({\mathbb{R}})}{y} \sim \frac{\mu(({-}y,y))}{y}, \]

which proves (4.25) in this case.

For the rest of the proof assume that the integrals in (4.29) are infinite. The definition of $\ell$ also implies that

\[ \int_{[0,\infty)}\frac{h(s)}{1+s}\mkern4mu\mathrm{d}\mu_*(s) = \int_{{\mathbb{R}}}\frac{|t|}{(1+t^2)^{\ell+1}}\mkern4mu\mathrm{d}\mu(t) < \infty, \]

which shows that (A.20) with $\nu =\mu _*$ is satisfied. Hence we can apply proposition A.11 with $\alpha =\frac {\beta }{2}$ and $\gamma =\frac 12-\ell$, which yields $\alpha +\gamma =\frac \beta 2+\frac 12-\ell \in [0,\,1]$, i.e. $\beta \in [2\ell -1,\,2\ell +1]$. Proposition A.11 also implies that, if $\beta >0$, then $(-1)^\ell {\mathcal {F}}_\ell [\mu ](y) \sim y^{2\ell }{\mathscr {S}}[\mu _\ell ](y^2)$ is regularly varying with index $2\ell +2\bigl (\frac {\beta }{2}+\frac 12- \ell -1\bigr )=\beta -1$.

Let us first consider the case when $\beta \notin 2{\mathbb {N}}-1$ and $\beta >0$. Then $\alpha +\gamma =\frac {\beta }{2}+\frac 12-\ell \in (0,\,1)$, and from proposition A.11 we obtain

\begin{align*} {\mathcal{F}}_\ell[\mu](y) & \sim ({-}1)^\ell y^{2\ell}{\mathscr{S}}[\sigma_\ell](y^2) \sim ({-}1)^\ell y^{2\ell}\frac{\pi\frac{\beta}{2}}{\sin\bigl(\pi\bigl(\frac{\beta}{2}+\frac12-\ell\bigr)\bigr)} \bigl(y^2\bigr)^{-\frac12-\ell}\mu_*([0,y^2)) \\ & = \frac{\frac{\pi\beta}{2}}{\sin\bigl(\frac{\pi\beta}{2}+\frac{\pi}{2}\bigr)} \cdot\frac{\mu(({-}y,y))}{y} = \frac{\frac{\pi\beta}{2}}{\cos\frac{\pi\beta}{2}}\cdot\frac{\mu(({-}y,y))}{y} \end{align*}

as $y\to \infty$.

Next assume that $\beta =2\ell -1$. Then $\alpha +\gamma =\frac {\beta }{2}+\frac 12-\ell =0$, and proposition A.11 and the substitution $s=t^2$ yield

\begin{align*} {\mathcal{F}}_\ell[\mu](y) & \sim ({-}1)^\ell y^{2\ell}{\mathscr{S}}[\sigma_\ell](y^2) \sim ({-}1)^\ell y^{2\ell}\frac{\beta}{2}\cdot\frac{1}{y^2} \int_1^{y^2} s^{-\frac12-\ell}\mu_*([0,s))\mkern4mu\mathrm{d} s \\ & = ({-}1)^\ell\frac{\beta}{2} y^{2\ell-2}\int_1^y t^{{-}1-2\ell}\mu_*([0,t^2))\,2t\mkern4mu\mathrm{d} t = ({-}1)^\ell\beta y^{2\ell-2}\int_1^y \frac{\mu(({-}t,t))}{t^{2\ell}}\mkern4mu\mathrm{d} t, \end{align*}

which shows (4.26) in this case.

The proofs of (4.25) when $\beta =0$ and the integrals in (4.29) are infinite and of (4.26) in the remaining case $\beta =2\ell +1$ are similar.

We can use (A.21) to obtain

\[ \big|{\mathcal{F}}_\ell[\mu](y)\big| \sim y^{2\ell}{\mathscr{S}}[\sigma_\ell](y^2) \qquad\text{and}\qquad \frac{1}{y^2} \ll {\mathscr{S}}[\sigma_\ell](y^2) \ll 1, \]

which yields (4.27).

Finally, the relation in (4.28) follows easily from (4.26) and theorem A.2.

We also need the following comparison result.

Lemma 4.21 Let $\mu _1$ and $\mu _2$ be measures on ${\mathbb {R}}$ and set $\ell \mathrel {\mathop :}=\ell (\mu _2)$. Further, assume that $t\mapsto \mu _2((-t,\,t))$ is regularly varying with index $\beta >0$ and that the limit

\[ \lim_{t\to\infty}\frac{\mu_1(({-}t,t))}{\mu_2(({-}t,t))} \]

exists in $[0,\,\infty )$. Then ${\mathcal {F}}_\ell [\mu _1]$ is well defined, i.e. $\ell \ge \ell (\mu _1)$, and

\[ \lim_{y\to\infty}\frac{{\mathcal{F}}_\ell[\mu_1](y)}{{\mathcal{F}}_\ell[\mu_2](y)} = \lim_{t\to\infty}\frac{\mu_1(({-}t,t))}{\mu_2(({-}t,t))}. \]

Proof. Since $\beta >0$, it follows as in the proof of proposition 4.19 that (A.28) with $\nu _2=(\mu _2)_*$ and $h(s)=\frac {\sqrt {s}}{(1+s)^\ell }$ is satisfied. Now the claim follows from lemma A.13.

When $t\mapsto \mu ((-t,\,t))$ is regularly varying with an index that is not an odd integer, the real part of $q(iy)$ is dominated by the imaginary part, as the following proposition shows.

Proposition 4.22 Suppose that $\mu$, $p$ and $q$ satisfy assumptions 4.2, 4.4 and 4.13. Further, assume that $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$ such that $\beta \notin 2{\mathbb {N}}-1$. Then

(4.30)\begin{equation} \limsup_{y\to\infty}\bigg|\frac{\operatorname{Re} q(iy)}{\operatorname{Im} q(iy)}\bigg| \le \Big|\tan\frac{\pi\beta}{2}\Big|. \end{equation}

Proof. Set $\ell \mathrel {\mathop :}=\ell (\mu )$. It follows from lemmas 4.16 and 4.18 that

(4.31)\begin{equation} \bigg|\frac{\operatorname{Re} q(iy)}{\operatorname{Im} q(iy)}\bigg| = \bigg|\frac{\mathsf{RC}_\ell[\mu](y)+\widetilde p(y)}{\operatorname{Im} q(iy)}\bigg| \le \frac{|{\mathcal{F}}_\ell[\mu](y)|}{|\operatorname{Im} q(iy)|} + \mathrm{O}\biggl(\frac{y^{2\ell-2}}{|\operatorname{Im} q(iy)|}\biggr). \end{equation}

Theorem 4.7 implies that $y\mapsto |\operatorname {Im} q(iy)|$ is regularly varying with index $\beta -1$. Further, we obtain from proposition 4.19 that $\beta \in [2\ell -1,\,2\ell +1]$. Since, by assumption, $\beta \notin 2{\mathbb {N}}-1$, we have $\beta -1>2\ell -2$ and hence $|\operatorname {Im} q(iy)|\gg y^{2\ell -2}$, i.e. the $\mathrm {O}$-term on the right-hand side of (4.31) converges to $0$ as $y\to \infty$.

From proposition 4.19 and theorem 4.7 we obtain

\begin{align*} |{\mathcal{F}}_\ell[\mu](y)|\ & \begin{cases} \displaystyle \ll \frac{\mu(({-}y,y))}{y}, & \beta=0, \\ \displaystyle \sim \frac{\frac{\pi\beta}{2}}{|\cos\frac{\pi\beta}{2}|} \cdot\frac{\mu(({-}y,y))}{y}, & \beta\in(0,\infty){\setminus}(2{\mathbb{N}}-1), \end{cases} \\ |\operatorname{Im} q(iy)|\ & \begin{cases} \displaystyle \sim \frac{\frac{\pi\beta}{2}}{|\sin\frac{\pi\beta}{2}|} \cdot\frac{\mu(({-}y,y))}{y}, & \beta\in[0,\infty){\setminus}2{\mathbb{N}}, \\ \displaystyle \gg \frac{\mu(({-}y,y))}{y}, & \beta\in2{\mathbb{N}}, \end{cases} \end{align*}

which yields

\[ \lim_{y\to\infty}\frac{|{\mathcal{F}}_\ell[\mu](y)|}{|\operatorname{Im} q(iy)|} = \Big|\tan\frac{\pi\beta}{2}\Big| \]

and hence (4.30).

5. Abelian–Tauberian theorems

In the following theorem, one of the main results of this paper, we combine the Tauberian and the Abelian theorems from the previous sections. In most cases we can give a full characterization when the asymptotic behaviour of the regularized Cauchy transform is described by a regularly varying function. Thereby, we assume that the behaviour of $q={\mathscr {C}}_\kappa [\mu,\,p]$ is not governed by the polynomial summand.

5.1. The generic situation

Recall the notation ${\mathscr {p}}(\mu )$ from (4.2) and $\kappa (\mu )$ from (4.4).

Theorem 5.1 Let $\mu$ be a measure on ${\mathbb {R}}$ such that ${\mathscr {p}}(\mu )<\infty$ and set $\kappa \mathrel {\mathop :}=\kappa (\mu )$. Further, let $p\in {\mathbb {R}}[z]$ with $p(z)=c_{2\kappa +1}z^{2\kappa +1}+\ldots +c_0$ such that $(\mu,\,p)\in {\mathbb {E}}_{\le \kappa }$, and assume that

\[ c_{2\kappa+1} = \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+t^2)^{\kappa+1}}. \]

and, if ${\mathscr {p}}(\mu )$ is odd, that

\[ c_{2\kappa} = \int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\mu(t). \]

Set $q\mathrel {\mathop :}={\mathscr {C}}_\kappa [\mu,\,p]$.

Let $\beta \in [0,\,\infty )$ and consider the following statements:

1. (a) the symmetrized distribution function $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$;

2. (a)′ the limit

   (5.1)\begin{equation} \zeta\mathrel{\mathop:}=\lim_{t\to\infty}\frac{\mu(({-}t,0))}{\mu([0,t))} \end{equation}
   exists in $[0,\,\infty ]$;

3. (b) there exist a regularly varying function $f\colon [x_0,\,\infty )\to (0,\,\infty )$ with $x_0>0$ and a constant $\omega \in {\mathbb {C}}{\setminus} \{0\}$ such that

   (5.2)\begin{equation} \lim_{r\to\infty}\frac{q(rz)}{f(r)} = i\omega\Bigl(\frac{z}{i}\Bigr)^{\beta-1} \end{equation}
   holds locally uniformly for $z\in {\mathbb {C}}^+$.

Then the following relations hold.

1. (i) If $\beta \notin {\mathbb {N}}_0$, then

   \[ \text{(a)} \;\wedge\; \text{(a)$'$} \quad\Leftrightarrow\quad \text{(b)}. \]

2. (ii) Assume that $\beta \in 2{\mathbb {N}}_0$. Then

   \[ \text{(a)} \;\;\Rightarrow\;\; \text{(b)}. \]
   If $\beta =0$ or $\int _{{\mathbb {R}}}(1+|t|)^{-\beta }\mathrm {d}\mu (t)<\infty$, then
   \[ \text{(b)} \;\;\Rightarrow\;\; \text{(a)}. \]

3. (iii) If $\beta \in 2{\mathbb {N}}-1$, then

   \[ \text{(a)} \;\wedge\; \text{(a)$'$} \;\wedge\; \zeta\ne1 \quad\Rightarrow\quad \text{(b)}. \]

Further, assume that either (a) and (b) hold and $\beta \in 2{\mathbb {N}}_0$, or that (a), (a) $'$ and (b) hold and $\beta \notin {\mathbb {N}}_0$, or that (a), (a) $'$ with $\zeta \ne 1$ and (b) hold and $\beta \in 2{\mathbb {N}}-1$; then $\omega$ and $f$ can be chosen as

(5.3)\begin{equation} \omega = ({-}1)^{{\mathscr{p}}(\mu)-1}\biggl(\cos\frac{\pi\beta}{2} +i\frac{\zeta-1}{\zeta+1}\cdot\sin\frac{\pi\beta}{2}\biggr), \end{equation}

where $\frac {\zeta -1}{\zeta +1}$ is understood as $1$ when $\zeta =\infty$, where ${\mathscr {p}}(\mu )$ is defined in (4.2), and

(5.4)\begin{equation} f(r) = \begin{cases} \dfrac{\pi\beta}{|\sin(\pi\beta)|}\cdot\dfrac{\mu(({-}r,r))}{r}, & \beta\in[0,\infty){\setminus}{\mathbb{N}}, \\ \displaystyle \beta r^{\beta-1}\int_1^r\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t, & \displaystyle \beta\in{\mathbb{N}} \;\wedge\; \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\beta}=\infty, \\ \displaystyle \beta r^{\beta-1}\int_r^\infty\frac{\mu(({-}t,t))}{t^{\beta+1}}\mkern4mu\mathrm{d} t, \quad & \displaystyle \beta\in{\mathbb{N}} \;\wedge\; \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\beta}<\infty; \end{cases} \end{equation}

here $\frac {\pi \beta }{|\sin (\pi \beta )|}$ is understood as $1$ when $\beta =0$.

Before we prove the theorem, let us add a couple of comments.

Remark 5.2

1. (i) Note that, in the case $\beta \in {\mathbb {N}}$, one has $f(r)\gg \frac {\mu ((-r,\,r))}{r}$ by theorem A.2.

2. (ii) When $\beta \in 2{\mathbb {N}}_0$, then $\omega \in {\mathbb {R}}$ and hence $|\operatorname {Im} q(iy)|\gg |\operatorname {Re} q(iy)|$. When $\beta \in 2{\mathbb {N}}-1$ in the theorem, then $\omega \in i{\mathbb {R}}$ and hence $|\operatorname {Re} q(iy)|\gg |\operatorname {Im} q(iy)|$.

3. (iii) It follows from the proof of theorem 5.1 (see (5.12)) that, if $\beta \in 2{\mathbb {N}}-1$ and (a) and (a)$'$ are satisfied with $\zeta =1$, then $|q(iy)|\ll f(y)$ with $f$ as in (5.4). A more detailed discussion of some cases in this situation are contained in theorem 5.5 below.

4. (iv) In the case when $\beta \notin {\mathbb {N}}_0$, condition (a) is not sufficient to guarantee (b) since one can easily construct measures $\mu$ that satisfy (a) but not (a)$'$, e.g. by distributing the mass in an alternating way on the positive and negative half-axes.

5. (v) In example 5.3 below we show that the converse implication in theorem 5.1 (iii) does not hold. However, see theorem 5.5 (ii) below for a Tauberian implication in the case when $\omega \notin i{\mathbb {R}}$.

6. (vi) Example 5.4 below deals with the situation when $\beta =1$ and $\zeta =1$, where theorem 5.1 is not applicable. This example shows that, in general, the asymptotic behaviour of $q(iy)$ is not determined by the leading asymptotic behaviour of $\mu ([0,\,t))$ and $\mu ((-t,\,0))$. However, in this example statement (b) in theorem 5.1 still holds. We do not know whether there exist a measure $\mu$ and a function $q$ such that $\beta \in 2{\mathbb {N}}-1$ and (a) and (a)$'$ hold with $\zeta =1$ but (b) does not hold.

Proof of theorem 5.1 Let $\kappa =\kappa (\mu )$ and $\ell =\ell (\mu )$.

• ${\bigcirc{\kern-6pt 1}}$ It follows from theorem 3.1 that (b) is equivalent to

  1. (c) there exist a regularly varying function $f\colon [x_0,\,\infty )\to (0,\,\infty )$ with $x_0>0$ and a constant $\omega \in {\mathbb {C}}{\setminus} \{0\}$ such that

     (5.5)\begin{equation} \lim_{y\to\infty}\frac{q(iy)}{f(y)} = i\omega. \end{equation}

• ${\bigcirc{\kern-6pt 2}}$ Let us first consider the case when $\beta \in 2{\mathbb {N}}_0$. It follows from theorem 4.22 that $|\operatorname {Re} q(iy)|\ll |\operatorname {Im} q(iy)|$ and hence, (5.5) is equivalent to

  \[ \lim_{y\to\infty}\frac{\operatorname{Im} q(iy)}{f(y)} = \omega. \]
  From theorem 4.7 and remark 4.8 we therefore obtain the implications in (ii). Assume now that (a) and (b) hold. It follows from theorem 4.19 that $\beta \in [2\ell -1,\,2\ell +1]$ and hence $\ell =\frac {\beta }{2}$. With $f$ as in (5.4) we obtain from (4.14) and theorem 4.8 that
  (5.6)\begin{equation} \lim_{y\to\infty}\frac{\operatorname{Im} q(iy)}{f(y)} = ({-}1)^\kappa. \end{equation}
  On the other hand, by remark 4.15, $\omega$ from (5.3) equals $\omega = (-1)^{{\mathscr {p}}(\mu )-1} \cos \frac {\pi \beta }{2} = (-1)^{\kappa +\ell }(-1)^\ell = (-1)^\kappa$, which coincides with the limit in (5.6).

• ${\bigcirc{\kern-6pt 3}}$ Next we assume that $\beta \notin {\mathbb {N}}_0$ and (b) holds. It follows from theorem 3.1 that the index of the regularly varying function $f$ is $\alpha \mathrel {\mathop :}=\beta -1$, and hence $r\mapsto rf(r)$ is regularly varying with index $\beta$. Now theorem 3.2 implies that (3.6) and (3.7) hold. Since $\alpha \notin {\mathbb {N}}_0$, at least one of the right-hand sides of (3.6), (3.7) is non-zero. Taking the quotient of (3.7) and (3.6) we obtain that the limit in (5.1) exists; further, $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$, i.e. (a) and (a)$'$ hold.

• ${\bigcirc{\kern-6pt 4}}$ For the rest of the proof we assume that $\beta \notin 2{\mathbb {N}}_0$, that (a) and (a)$'$ hold, and if $\beta \in 2{\mathbb {N}}-1$, then also $\zeta \ne 1$. Let $\mu ^+$ and $\mu ^-$ be as in (4.22) and $f$ as in (5.4). From (4.16) and (4.18) we obtain that

  (5.7)\begin{align} \frac{1}{i}\cdot\frac{q(iy)}{f(y)} & = \frac{{\mathcal{F}}_\ell[\mu](y)}{f(y)}\cdot\frac{\operatorname{Im} q(iy)-i\operatorname{Re} q(iy)}{{\mathcal{F}}_\ell[\mu](y)} \nonumber\\ & = \frac{{\mathcal{F}}_\ell[\mu](y)}{f(y)}\biggl[\frac{\operatorname{Im} q(iy)}{{\mathcal{F}}_\ell[\mu](y)} -i\biggl(\frac{{\mathcal{F}}_\ell[\mu^+](y)}{{\mathcal{F}}_\ell[\mu](y)} - \frac{{\mathcal{F}}_\ell[\mu^+](y)}{{\mathcal{F}}_\ell[\mu](y)} + \frac{\widetilde p(y)}{{\mathcal{F}}_\ell[\mu](y)}\biggr)\biggr], \end{align}
  where $\widetilde p$ is a real, even polynomial of degree at most $2\ell -2$. In the next couple of steps we evaluate the limits of parts of this expression.

• ${\bigcirc{\kern-6pt 5}}$ First we show that

  (5.8)\begin{equation} \lim_{y\to\infty}\frac{{\mathcal{F}}_\ell[\mu](y)}{f(y)} = ({-}1)^{{\mathscr{p}}(\mu)-1}\sin\frac{\pi\beta}{2}. \end{equation}
  We start with the case when $\beta \notin 2{\mathbb {N}}-1$. It follows from lemma 4.1 that $\beta \in ({\mathscr {p}}(\mu )-1,\,{\mathscr {p}}(\mu ))$ and hence $\operatorname {sgn}(\sin (\pi \beta ))=(-1)^{{\mathscr {p}}(\mu )-1}$. Further, we obtain from proposition 4.19 and (5.4) that
  \[ \lim_{y\to\infty}\frac{{\mathcal{F}}_\ell[\mu](y)}{f(y)} = \frac{|\sin(\pi\beta)|}{2\cos\frac{\pi\beta}{2}} = \frac{({-}1)^{{\mathscr{p}}(\mu)-1}\sin(\pi\beta)}{2\cos\frac{\pi\beta}{2}} = ({-}1)^{{\mathscr{p}}(\mu)-1}\sin\frac{\pi\beta}{2}. \]
  Now assume that $\beta \in 2{\mathbb {N}}-1$. Then $\lim _{y\to \infty }\frac {{\mathcal {F}}_\ell [\mu ](y)}{f(y)}=(-1)^\ell$. On the other hand, by theorem 4.7 we have $\beta \in [2\kappa,\,2\kappa +2]$ and hence $\beta =2\kappa +1$, which, by remark 4.15, implies that
  \[ ({-}1)^{{\mathscr{p}}(\mu)-1}\sin\frac{\pi\beta}{2} = ({-}1)^{\kappa+\ell}\sin\Bigl(\pi\kappa+\frac{\pi}{2}\Bigr) = ({-}1)^{\kappa+\ell}\cos(\pi\kappa) = ({-}1)^\ell, \]
  which proves (5.8) also in this case.

• ${\bigcirc{\kern-6pt 6}}$ Next we show that

  (5.9)\begin{equation} \lim_{y\to\infty}\frac{\operatorname{Im} q(iy)}{{\mathcal{F}}_\ell[\mu](y)} = \cot\frac{\pi\beta}{2}. \end{equation}
  When $\beta \notin 2{\mathbb {N}}-1$, we can use theorem 4.7 and proposition 4.19 to obtain (5.9). When $\beta \in 2{\mathbb {N}}-1$, theorem 4.7 and (4.28) imply that $\lim _{y\to \infty }\frac {\operatorname {Im} q(iy)}{{\mathcal {F}}_\ell [\mu ](y)}=0$, and again (5.9) holds.

• ${\bigcirc{\kern-6pt 7}}$ Let us consider the expressions within the round brackets on the right-hand side of (5.7). Assume first that $\mu ^+$ is not the zero measure. The relation $\mu =\mu ^++\mu ^-$ implies that

  \[ \frac{\mu(({-}t,t))}{\mu^+(({-}t,t))} = 1+\frac{\mu^-(({-}t,t))}{\mu^+(({-}t,t))} = 1+\frac{\mu(({-}t,0))}{\mu([0,t))} \to 1+\zeta, \qquad t\to\infty, \]
  and hence
  \[ \lim_{t\to\infty}\frac{\mu^+(({-}t,t))}{\mu(({-}t,t))} = \frac{1}{1+\zeta}, \qquad \lim_{t\to\infty}\frac{\mu^-(({-}t,t))}{\mu(({-}t,t))} = \frac{\zeta}{1+\zeta}. \]
  If $\mu ^+$ is the zero measure, then these relations also hold with $\zeta =\infty$. Together with lemma 4.21 this yields
  (5.10)\begin{equation} \lim_{y\to\infty}\frac{{\mathcal{F}}_\ell[\mu^+](y)}{{\mathcal{F}}_\ell[\mu](y)} = \frac{1}{1+\zeta}, \qquad \lim_{y\to\infty}\frac{{\mathcal{F}}_\ell[\mu^-](y)}{{\mathcal{F}}_\ell[\mu](y)} = \frac{\zeta}{1+\zeta}. \end{equation}
  Further, (4.27) implies that
  (5.11)\begin{equation} \lim_{y\to\infty}\frac{\widetilde p(y)}{{\mathcal{F}}_\ell[\mu](y)} = 0. \end{equation}

• ${\bigcirc{\kern-6pt 8}}$ Combining (5.7), (5.8), (5.9), (5.10) and (5.11) we arrive at

  (5.12)\begin{equation} \frac{1}{i}\lim_{y\to\infty}\frac{q(iy)}{f(y)} = ({-}1)^{{\mathscr{p}}(\mu)-1}\sin\frac{\pi\beta}{2} \biggl(\cot\frac{\pi\beta}{2}-i\frac{1-\zeta}{1+\zeta}\biggr), \end{equation}
  which is equal to the expression on the right-hand side of (5.3); this proves (5.5). Since, by assumption, either $\beta \notin 2{\mathbb {N}}-1$ or $\zeta \ne 1$, we have $\omega \ne 0$. It follows from (5.8) that
  \[ f(y)\sim\big|\sin\tfrac{\pi\beta}{2}\big|\cdot\big|{\mathcal{F}}_\ell[\mu](y)\big|. \]
  By proposition 4.19 the right-hand side, and hence also $f$, is regularly varying. This shows that (c) is satisfied, which, in turn, implies (b).

The following example shows that the converse of the implication in theorem 5.1 (iii) does not hold.

Example 5.3 Let $\ell \in {\mathbb {N}}$. With $h(s)=\frac {\sqrt {s}}{(1+s)^\ell }$, $s\in [0,\,\infty )$, choose $\sigma$ and $\nu$ as in example A.12. Moreover, let $\mu$ be the measure on ${\mathbb {R}}$ such that $\mu ((-\infty,\,0))=0$ and $\mu _*=\mu ^+_*=\nu$. Then

\[ \mu\bigl(\bigl\{e^{\frac{k}{2}}\bigr\}\bigr) = \mu_*^+(\{e^k\}) = \frac{(1+e^k)^\ell}{e^{\frac{k}{2}}}, \qquad k\in{\mathbb{N}}. \]

From the relation

\[ \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^n} = \sum_{k=1}^\infty \frac{1}{\bigl(1+e^{\frac{k}{2}}\bigr)^n}\cdot\frac{(1+e^k)^\ell}{e^{\frac{k}{2}}} \]

we can easily deduce that ${\mathscr {p}}(\mu )=2\ell$ and hence $\ell =\ell (\mu )$. Now set $q={\mathscr {C}}_\kappa [\mu,\,p]$ with $\kappa =\kappa (\mu )$ and a polynomial $p$ such that (4.9) holds. Then assumptions 4.2, 4.4, and 4.13 are satisfied since ${\mathscr {p}}(\mu )$ is even. It follows from example A.12 that $t\mapsto \mu ((-t,\,t))=\mu ^+([0,\,t))=\nu ([0,\,t^2))$ is not regularly varying, i.e. (a) in theorem 5.1 is not satisfied.

Let us now consider the behaviour of $q(iy)$ as $y\to \infty$. It is clear that $\sigma _\ell =\sigma$ with $\sigma _\ell$ as in (4.21). From lemma 4.18 and (A.27) we obtain

\[ \mathsf{RC}_\ell[\mu](y) = {\mathcal{F}}_\ell[\mu^+](y) \sim ({-}1)^\ell y^{2\ell}{\mathscr{S}}[\sigma](y^2) \sim 2({-}1)^\ell y^{2\ell-2}\log y, \]

and hence

(5.13)\begin{equation} \operatorname{Re} q(iy) \sim 2({-}1)^\ell y^{2\ell-2}\log y, \qquad y\to\infty, \end{equation}

by lemma 4.16. Next we show that the imaginary part is dominated by the real part. Let $\tau _\kappa$ be as in (4.12). Since $\kappa (\mu )=\ell (\mu )-1$ by remark 4.15, we have

\[ \tau_\kappa(\{e^k\}) = \frac{1}{(1+e^k)^\kappa}\cdot\frac{(1+e^k)^\ell}{e^{\frac{k}{2}}} = \frac{1+e^k}{e^{\frac{k}{2}}}, \quad k\in{\mathbb{N}}. \]

Let $\tilde \tau$ be the measure on $[0,\,\infty )$ with $\tilde \tau ([0,\,s))=\frac {2\sqrt {e}}{\sqrt {e}-1}\bigl (s^{\frac 12}-1\bigr )$ for $s\in (1,\,\infty )$ and $\tilde \tau ([0,\,1])=0$. For $t\in (e^n,\,e^{n+1}]$, $n\in {\mathbb {N}}$, we have

\begin{align*} \tau_\kappa([0,t)) & = \sum_{k=1}^n\tau_k(\{e^k\}) \le 2\sum_{k=1}^n e^{\frac{k}{2}}\\ & = \frac{2\sqrt{e}}{\sqrt{e}-1}\bigl(e^{\frac{n}{2}}-1\bigr) \le \frac{2\sqrt{e}}{\sqrt{e}-1}\bigl(t^{\frac12}-1\bigr) = \tilde\tau([0,t)). \end{align*}

By lemma A.9 (i) and remark A.8 we have

(5.14)\begin{align} |\operatorname{Im} q(iy)| & \sim y^{2\kappa+1}{\mathscr{S}}[\tau_\kappa](y^2)\le y^{2\kappa+1}{\mathscr{S}}[\tilde\tau](y^2) \sim y^{2\kappa+1}\frac{\pi}{2}\cdot\frac{\tilde\tau([0,y^2))}{y^2}\nonumber\\ & \sim \frac{\pi\sqrt{e}}{\sqrt{e}-1}y^{2\kappa} = \frac{\pi\sqrt{e}}{\sqrt{e}-1}y^{2\ell-2}, \end{align}

which shows that $|\operatorname {Re} q(iy)|\gg |\operatorname {Im} q(iy)|$. This, together with (5.13), implies that (3.2) with $f(r)=2r^{2\ell -2}\log r$ is satisfied and therefore also (b) in theorem 5.1 with $\beta =2\ell -1$ and $\omega =i(-1)^{\ell +1}$. Note also that (a)$'$ is satisfied with $\zeta =0$.

For $\ell =1$ we obtain a function $q\in {\mathcal {N}}_0$ since $\kappa =0$ in this case. We can choose $p$ such that

\[ q(z) = \widetilde C[\mu](z) = \sum_{k=1}^\infty\biggl(\frac{1}{e^{\frac{k}{2}}-z}-\frac{e^{\frac{k}{2}}}{1+e^k}\biggr) \cdot\frac{1+e^k}{e^{\frac{k}{2}}}. \]

According to (5.13) and (5.14) it satisfies $q(iy)\sim 2\log y$ as $y\to \infty$.

Example 5.4 Let $\gamma \in (0,\,1)$ and define the measure $\mu =\lambda +\nu$ on ${\mathbb {R}}$ where $\lambda$ is the Lebesgue measure and $\nu$ is the measure such that $\nu ((-\infty,\,0))=0$ and

\[ \nu([0,t)) = \begin{cases} t, & t\in(0,e], \\ \dfrac{t}{(\log t)^\gamma}, & t\in(e,\infty). \end{cases} \]

Further, set $q(z)\mathrel {\mathop :}=\widetilde C[\mu ](z)$. Clearly, $\mu ([0,\,t))\sim t$ and $\mu ((-t,\,0))=t$ and hence $\mu ((-t,\,t))\sim 2t$, which shows that (a) and (a)$'$ in theorem 5.1 are satisfied with $\beta =1$ and $\zeta =1$. Further, lemma 4.7 implies that $\operatorname {Im} q(iy)\to \pi$ as $y\to \infty$. For the real part we obtain from lemma 4.18 and proposition 4.19 that

\begin{align*} \operatorname{Re} q(iy) & = \mathsf{RC}_1[\mu](y) = {\mathcal{F}}_1[\lambda^+{+}\nu](y)-{\mathcal{F}}_1[\lambda^-](y) = {\mathcal{F}}_1[\nu](y) \\ & \sim{-}\int_1^y\frac{\nu(({-}t,t))}{t^2}\mkern4mu\mathrm{d} t\sim{-}\int_1^y\frac{1}{t(\log t)^\gamma}\mkern4mu\mathrm{d} t ={-}\frac{1}{1-\gamma}(\log y)^{1-\gamma}. \end{align*}

This shows that $q(iy)\sim -\frac {1}{1-\gamma }(\log y)^{1-\gamma }$ as $y\to \infty$, and hence (b) in theorem 5.1 is satisfied. However, the asymptotics of $q(iy)$ at infinity is not determined by the leading term of the asymptotics of $\mu ([0,\,t))$ or $\mu ((-t,\,0))$ as the former depends on $\gamma$ whereas the latter does not.

5.2. The exceptional case

In the following theorem we consider certain situations when $\beta \in 2{\mathbb {N}}-1$ and $\zeta =1$, a case that is not covered by theorem 5.1. When $\beta \in 2{\mathbb {N}}-1$ and $\zeta \ne 1$, the real part of $q(iy)$ dominates the imaginary part and is strictly larger than $\frac {\mu ((-y,\,y))}{y}$; see theorem 5.1 and remark 5.2 (ii). In the next theorem we consider cases when there is cancellation between contributions from the measure $\mu$ on the positive and negative axes to the real part and where the growth of $q(iy)$ is determined by the imaginary part.

Theorem 5.5 Let $\mu$, $\kappa$, $p$, and $q$ be as in assumptions 4.2 and 4.4 and let $\beta \in 2{\mathbb {N}}-1$. Then the following statements are true.

1. (i) Assume that $\mu =\mu _0+\mu _1+\mu _2$ with measures $\mu _0$, $\mu _1$ and $\mu _2$ such that the following conditions are satisfied:

   1. (a) $\mu _0$ is symmetric and $t\mapsto \mu _0((-t,\,t))$ is regularly varying with index $\beta$;

   2. (b) either $\mu _1([0,\,\infty ))=0$ or $t\mapsto \mu _1([0,\,t))$ is regularly varying with index $\beta$; either $\mu _1((-\infty,\,0))=0$ or $t\mapsto \mu _1((-t,\,0))$ is regularly varying with index $\beta$;

   3. (c) $\mu _2((-t,\,t))\lesssim t^\gamma$ as $t\to \infty$ with some $\gamma <\beta$;

   4. (d) if $\int _{{\mathbb {R}}}(1+|t|)^{-\beta }\mathrm {d}\mu _1(t)<\infty$, then

      \[ c_{2\kappa} = \int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}(\mu_1+\mu_2)(t); \]

   5. (e) with the notation in (4.22) the following two limits (for $+$ and $-$) exist in ${\mathbb {R}}$:

      \[ \eta_\pm \mathrel{\mathop:}= \begin{cases} \displaystyle \lim_{t\to\infty} \biggl({\int_1^t\frac{\mu_1^\pm(({-}s,\,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s} \bigg/{\frac{\mu_0(({-}t,\,t))}{t^\beta}}\biggr) & \displaystyle \text{if} \ \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu_1(t)}{(1+|t|)^\beta}=\infty, \\ \displaystyle -\lim_{t\to\infty} \biggl({\int_t^\infty\frac{\mu_1^\pm(({-}s,\,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s} \bigg/{\frac{\mu_0(({-}t,\,t))}{t^\beta}}\biggr) & \displaystyle \text{if} \ \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu_1(t)}{(1+|t|)^\beta}<\infty. \end{cases} \]

   Then $\mu ((-t,\,t))\sim \mu _0((-t,\,t))$, $\beta =2\kappa +1$, and (5.2) holds locally uniformly for $z\in {\mathbb {C}}^+$ with

   (5.15)\begin{equation} f(r) = \beta\frac{\mu(({-}r,r))}{r}, \qquad \omega = ({-}1)^\kappa\Bigl(\frac{\pi}{2}+i\bigl(\eta_+{-}\eta_-\bigr)\Bigr). \end{equation}

2. (ii) Assume that there exist a regularly varying function $f:[x_0,\,\infty )\to (0,\,\infty )$ with $x_0>0$ and a constant $\omega \notin i{\mathbb {R}}$ such that (5.2) holds uniformly for $z\in {\mathbb {C}}^+$. Then $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$ and (5.1) holds with $\zeta =1$. One can choose $f$ as in (5.15), in which case $\omega$ satisfies $\operatorname {Re}\omega =(-1)^\kappa \frac {\pi }{2}$.

Remark 5.6 Example 5.3 shows that the condition $\omega \notin i{\mathbb {R}}$ in (ii) is essential.

Proof of theorem 5.5 We split the proof into a couple of steps; the first five steps deal with the proof of (i).

• ${\bigcirc{\kern-6pt 1}}$ First we note that either $\mu _1=0$ or $t\mapsto \mu _1((-t,\,t))$ is regularly varying with index $\beta$. In the latter case we obtain from theorem A.2 (i) and assumptions (b) and (e) that, when $\int _{{\mathbb {R}}}(1+|t|)^{-\beta }\mathrm {d}\mu _1(t)=\infty$,

  \begin{align*} \frac{\mu_1(({-}t,t))}{t^\beta} & \ll \int_1^t\frac{\mu_1(({-}s,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s\\ & = \int_1^t\frac{\mu_1^+(({-}s,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s + \int_1^t\frac{\mu_1^-(({-}s,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s \lesssim \frac{\mu_0(({-}t,t))}{t^\beta} \end{align*}
  as $t\to \infty$. A similar calculation and the use of theorem A.2 (ii) show that also inthe case when $\int _{[0,\infty )}(1+|t|)^{-\beta }\mathrm {d}\mu _1(t)<\infty$, we have $\mu _1((-t,\,t))\ll \mu _0((-t,\,t))$.Together with assumption (c), this implies that $\mu ((-t,\,t))\sim \mu _0((-t,\,t))$.

  It follows from theorem 4.7 that $\beta =2\kappa +1$ and

  (5.16)\begin{equation} \operatorname{Im} q(iy) \sim \frac{\frac{\pi\beta}{2}}{\sin\bigl(\pi\kappa+\frac{\pi}{2}\bigr)} \cdot\frac{\mu(({-}y,y))}{y} = ({-}1)^\kappa\frac{\pi}{2}f(y). \end{equation}

• ${\bigcirc{\kern-6pt 2}}$ Set $\ell \mathrel {\mathop :}=\ell (\mu )$ and $\widehat \mu \mathrel {\mathop :}=\mu _1+\mu _2$. By remark 4.15 we have either $\ell =\kappa$ or $\ell =\kappa +1$. It follows from assumption (c) and lemma 4.1 that $\int _{{\mathbb {R}}}(1+|t|)^{-\beta }\mathrm {d}\mu _2(t)<\infty$ and hence $\ell (\mu _2)\le \kappa$. If ${\mathscr {p}}(\mu )$ is odd (i.e. $\ell =\kappa$), then

  \[ \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu_1(t)}{(1+|t|)^\beta} \le \int_{{\mathbb{R}}}\frac{\mathrm{d}\mu(t)}{(1+|t|)^\beta} < \infty \]
  and hence, by the symmetry of $\mu _0$ and assumption (d),
  \[ \int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\mu(t) = \int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\widehat\mu(t) = c_{2\kappa}. \]
  This shows that assumption 4.13 is satisfied. It follows from lemma 4.16 and the symmetry of $\mu _0$ that
  (5.17)\begin{equation} \operatorname{Re} q(iy) = \mathsf{RC}_\ell[\mu](y) + \widetilde p(y) = \mathsf{RC}_\ell[\widehat\mu](y) + \widetilde p(y) \end{equation}
  with $\widetilde p$ as in lemma 4.16.

  Set $\hat \ell \mathrel {\mathop :}=\kappa$ when $\mu _1=0$ and $\hat \ell \mathrel {\mathop :}=\ell (\widehat \mu )$ otherwise. It follows in a similar way as above that, by assumption (b), either $\mu _1^+=0$ or $\ell (\mu _1^+)=\kappa$ or $\ell (\mu _1^+)=\kappa +1$, and similarly for $\mu _1^-$. Since $\ell (\mu _2)\le \kappa$, we have $\kappa \le \hat \ell \le \ell \le \kappa +1$.

  Let us rewrite the expression in (5.17) in the case when $\hat \ell =\kappa$ and $\ell =\kappa +1$. We obtain from (2.18) with $\kappa$ and $\kappa '$ there replaced by $\hat \ell -1$ and $\ell -1$ respectively that

  \begin{align*} \frac{t}{t^2+y^2}\cdot\frac{(1-y^2)^\ell}{(1+t^2)^\ell} & = \operatorname{Re}\biggl[\frac{1}{t-iy}\cdot\frac{(1+(iy)^2)^\ell}{(1+t^2)^\ell}\biggr] \\ & = \operatorname{Re}\biggl[\frac{1}{t-iy}\cdot\frac{(1+(iy)^2)^{\hat\ell}}{(1+t^2)^{\hat\ell}}\biggr] - \operatorname{Re}\biggl[(t+iy)\cdot\frac{(1+(iy)^2)^{\hat\ell}}{(1+t^2)^{\hat\ell+1}}\biggr] \\ & = \frac{t}{t^2+y^2}\cdot\frac{(1-y^2)^{\hat\ell}}{(1+t^2)^{\hat\ell}} - (1-y^2)^{\hat\ell}\cdot\frac{t}{(1+t^2)^{\hat\ell+1}}. \end{align*}
  Hence
  \begin{align*} \operatorname{Re} q(iy) & = \mathsf{RC}_{\hat\ell}[\widehat\mu](y) - (1-y^2)^\kappa\int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\widehat\mu(t) + \widetilde p(y) \\ & = \mathsf{RC}_{\hat\ell}[\widehat\mu])(y) + \widehat p(y), \end{align*}
  where the polynomial $\widehat p$ satisfies
  \begin{align*} \widehat p(y) & = ({-}1)^\kappa\biggl[c_{2\kappa} -\int_{{\mathbb{R}}}\frac{t}{(1+t^2)^{\kappa+1}}\mkern4mu\mathrm{d}\widehat\mu(t)\biggr]y^{2\kappa}\\ & \quad + \mathrm{O}\bigl(y^{2\kappa-2}\bigr) = \mathrm{O}\bigl(y^{2\kappa-2}\bigr) = \mathrm{O}\bigl(y^{2\hat\ell-2}\bigr) \end{align*}
  by assumption (d).

  In all cases we can write

  \[ \operatorname{Re} q(iy) = \mathsf{RC}_{\hat\ell}[\widehat\mu](y) + \widehat p(y) = {\mathcal{F}}_{\hat\ell}[\mu_1^+](y) - {\mathcal{F}}_{\hat\ell}[\mu_1^-](y) + \mathsf{RC}_{\hat\ell}[\mu_2](y) + \widehat p(y) \]
  with an even polynomial $\widehat p$ of degree at most $2\hat \ell -2$; note that we have used lemma 4.18 in the last step and that $\widehat p=\widetilde p$ when $\hat \ell =\ell$. We therefore have
  (5.18)\begin{equation} \frac{1}{i}\cdot\frac{q(iy)}{f(y)} = \frac{\operatorname{Im} q(iy)}{f(y)} - i\frac{{\mathcal{F}}_{\hat\ell}[\mu_1^+](y)}{f(y)} + i\frac{{\mathcal{F}}_{\hat\ell}[\mu_1^-](y)}{f(y)} - i\frac{\mathsf{RC}_{\hat\ell}[\mu_2](y)}{f(y)} - i\frac{\widehat p(y)}{f(y)}. \end{equation}

• ${\bigcirc{\kern-6pt 3}}$ We show that

  (5.19)\begin{equation} \lim_{y\to\infty}\frac{{\mathcal{F}}_{\hat\ell}[\mu_1^\pm](y)}{f(y)} ={-}({-}1)^\kappa\eta_\pm. \end{equation}
  Let us start with $\mu _1^+$. When $\mu _1^+=0$, the equality in (5.19) is obvious. Assume now that $\mu _1^+\ne 0$. Let us first consider the case when $\ell (\mu _1^+)=\hat \ell$. Then the assumptions of proposition 4.19 are satisfied for $\mu$ replaced by $\mu _1^+$. Hence that lemma and assumption (e) yield that, when $\int _{{\mathbb {R}}}(1+|t|)^{-\beta }\mathrm {d}\mu _1(t)=\infty$, i.e. $\ell =\kappa +1$,
  \begin{align*} \lim_{y\to\infty}\frac{{\mathcal{F}}_{\hat\ell}[\mu_1^+](y)}{f(y)} & = \lim_{y\to\infty}\biggl[{ ({-}1)^{\hat\ell}\beta y^{\beta-1} \int_1^y\frac{\mu_1^+(({-}s,\,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s} \bigg/{ \Bigl(\beta\frac{\mu_0(({-}y,\,y))}{y}\Bigr)}\biggr]\\ & ={-}({-}1)^\kappa\eta_+; \end{align*}
  a similar calculation proves relation (5.19) also in the case when $\int _{{\mathbb {R}}}(1+|t|)^{-\beta }\mathrm {d}\mu _1 (t)<\infty$. The same considerations can be applied to $\mu _1^-$ when $\mu _1^-=0$ or $\ell (\mu _1^-)=\hat \ell$.

  It remains to prove (5.19) for $\mu _1^+$ when $\ell (\mu _1^+)<\ell (\mu _1^-)=\hat \ell =\kappa +1$ or for $\mu _1^-$ when $\ell (\mu _1^-)<\ell (\mu _1^+)$. We consider only the first case. It follows from remark 4.20 (ii) applied to $\mu _1^+$, the first inequality in (4.27) and the already proved relation (5.19) for $\mu _1^-$ (since $\ell (\mu _1^-)=\hat \ell$) that, with some $c>0$,

  \[ \big|{\mathcal{F}}_{\hat\ell}[\mu_1^+](y)\big| \sim cy^{2\hat\ell-2} = cy^{2\kappa} \ll \big|{\mathcal{F}}_{\hat\ell}[\mu_1^-](y)\big| \lesssim f(y). \]
  On the other hand, the relations $\ell (\mu _1^-)=\kappa +1$, $\ell (\mu _1^+)<\kappa +1$ and lemma 4.1 imply that
  \[ \int_1^\infty\frac{\mu_1^+(({-}s,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s < \infty, \qquad \int_1^\infty\frac{\mu_1^-(({-}s,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s = \infty, \]
  and hence
  \[ \int_1^t\frac{\mu_1^+(({-}s,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s \ll \int_1^t\frac{\mu_1^-(({-}s,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s. \]
  This and the existence of the limit for $\eta _-$ yield $\eta _+=0$, which shows that (5.19) holds also for $\mu _1^+$ in this case.

• ${\bigcirc{\kern-6pt 4}}$ Next let us show that

  (5.20)\begin{equation} \lim_{y\to\infty}\biggl(\frac{\mathsf{RC}_{\hat\ell}[\mu_2](y)}{f(y)} + \frac{\widehat p(y)}{f(y)}\biggr) = 0. \end{equation}
  Let us consider the two possible cases for $\hat \ell$ separately. First assume that $\hat \ell =\kappa$. Choose $\rho \in (\max \{\gamma,\,\beta -1\},\,\beta )$ and define the measure $\nu$ by $\nu ((-\infty,\,0))=0$ and $\nu ([0,\,t))=t^\rho$ for $t>0$. Since $\rho \in (\beta -1,\,\beta )$, we have $\ell (\nu )=\hat \ell$. It follows from lemma 4.21 and assumption (c) that
  \[ \lim_{y\to\infty}\frac{{\mathcal{F}}_{\hat\ell}[\mu_2](y)}{{\mathcal{F}}_{\hat\ell}[\nu](y)} = \lim_{t\to\infty}\frac{\mu_2(({-}t,t))}{\nu(({-}t,t))} = 0. \]
  From (4.24) and (4.26) we obtain
  \[ \big|\mathsf{RC}_{\hat\ell}[\mu_2](y)\big| \le \big|{\mathcal{F}}_{\hat\ell}[\mu_2](y)\big| \ll \big|{\mathcal{F}}_{\hat\ell}[\nu](y)\big| \sim \frac{\frac{\pi\rho}{2}}{\cos\frac{\pi\rho}{2}}y^{\rho-1}. \]
  Since $f$ is regularly varying with index $\beta -1=2\hat \ell$ and $\widehat p$ is a polynomial of degree at most $2\hat \ell -2$, relation (5.20) follows in this case; see (A.1).

  Now let us consider the case when $\hat \ell =\kappa +1$. Then $\ell (\mu _2)\le \kappa <\hat \ell$, and hence lemma 4.18 and remark 4.20 (ii) imply that

  \[ \big|\mathsf{RC}_{\hat\ell}[\mu_2](y)\big| \le \big|{\mathcal{F}}_{\hat\ell}[\mu_2](y)\big| \sim cy^{2\hat\ell-2} \]
  with some $c>0$. Without loss of generality assume that $\ell (\mu _1^+)=\hat \ell$ (the case $\ell (\mu _1^-)=\hat \ell$ is similar). It follows from assumption (e) and proposition 4.19 that
  \[ f(y) \gtrsim y^{\beta-1}\int_1^y\frac{\mu_1^+([0,s))}{s^{\beta+1}}\mkern4mu\mathrm{d} s \sim \frac{1}{\beta}\big|{\mathcal{F}}_{\hat\ell}[\mu_1^+](y)\big| \gg y^{2\hat\ell-2}, \]
  which yields (5.20) also in this case.

• ${\bigcirc{\kern-6pt 5}}$ Combining (5.18), (5.16), (5.19) and (5.20) we arrive at

  \[ \frac{1}{i}\lim_{y\to\infty}\frac{q(iy)}{f(y)} = ({-}1)^\kappa\Bigl(\frac{\pi}{2}+i\bigl(\eta_+{-}\eta_-\bigr)\Bigr), \]
  which, by theorem 3.1, shows the remaining assertions in (i).

• ${\bigcirc{\kern-6pt 6}}$ Let us now prove item (ii). Assume that $f$ and $\omega \notin i{\mathbb {R}}$ are such that (5.2) holds. We obtain from theorem 3.2 and remark 3.3 (i) that

  (5.21)\begin{equation} \lim_{t\to\infty}\frac{\mu((0,t))}{tf(t)} = \lim_{t\to\infty}\frac{\mu(({-}t,0))}{tf(t)} = \frac{\omega}{\pi(\alpha+1)}\cos\bigl(\arg\bigl(({-}1)^m\omega\bigr)\bigr) \end{equation}
  where $\alpha =\beta -1=2m$ with $m\in {\mathbb {N}}_0$ and $\big |\arg \bigl ((-1)^m\omega \bigr )\big |\le \frac {\pi }{2}$. Since, by assumption, $\omega \notin i{\mathbb {R}}$, we have $\operatorname {Re}((-1)^m\omega )>0$ and hence
  \[ |\omega|\cos\bigl(\arg\bigl(({-}1)^m\omega\bigr)\bigr) = \operatorname{Re}\bigl(({-}1)^m\omega)\bigr) = ({-}1)^m\operatorname{Re}\omega. \]
  In particular, the limits in (5.21) are non-zero, and therefore (5.1) holds with $\zeta =1$ and
  (5.22)\begin{equation} \lim_{t\to\infty}\frac{\mu(({-}t,t))}{tf(t)} = ({-}1)^m\frac{2}{\pi\beta}\operatorname{Re}\omega. \end{equation}
  This implies that $t\mapsto \mu ((-t,\,t))$ is regularly varying with index $\beta$, which, in turn, yields that $\beta =2\kappa +1$ and hence $m=\kappa$. Choosing $f$ as in (5.15) we obtain from (5.22) that $\operatorname {Re}\omega =(-1)^\kappa \frac {\pi }{2}$.

To illustrate the Abelian direction of theorem 5.5, let us consider the following example.

Example 5.7 Let $\mu$ be a measure on ${\mathbb {R}}$ that satisfies

(5.23)\begin{equation} \begin{aligned} \mu([0,t)) & = t\log t + a_+t + \mathrm{O}(t^\gamma), \\ \mu(({-}t,0)) & = t\log t + a_-t + \mathrm{O}(t^\gamma) \end{aligned} \end{equation}

as $t\to +\infty$ with $a_+,\,a_-\ge 0$ and $\gamma <1$. We can write $\mu$ as $\mu =\mu _0+\mu _1+\mu _2$ where $\mu _0,\,\mu _1,\,\mu _2$ are measures satisfying

\begin{alignat*}{2} & \mu_0([-1,1]) = 0; \qquad & & \mu_0([1,t)) = \mu_0((-t,-1)) = t\log t, \qquad t>1; \\ & \mu_1([0,t)) = a_+t, \qquad & & \mu_1((-t,0)) = a_-t, \qquad t>0; \\ & \mu_2((-t,t)) = \mathrm{O}(t^\gamma), \qquad & & t\to+\infty. \end{alignat*}

Since $\int _{{\mathbb {R}}}(1+|t|)^{-1}\mathrm {d}\mu (t)=\infty$ (see lemma 4.1), assumptions (a)–(d) in theorem 5.5 are fulfilled with $\beta =1$. For (e) we consider

\[ \eta_\pm{=} \lim_{t\to\infty} \biggl({\int_1^t\frac{a_\pm s}{s^2}\mkern4mu\mathrm{d} s} \bigg/{\frac{2t\log t}{t}}\biggr) = \frac{a_\pm}{2}, \]

which shows that (e) is also satisfied. Now let $q(z)=c_0+\widetilde C[\mu ](z)$ with $c_0\in {\mathbb {R}}$. Then, by theorem 5.5 (i), relation (5.2) holds with

\[ f(r) \sim \frac{\mu_0(({-}r,r))}{r} = 2\log r, \qquad \omega = \frac{\pi}{2}+i\frac{a_+{-}a_-}{2}, \]

i.e.

\[ \lim_{r\to\infty}\frac{q(rz)}{\log r} ={-}a_+{+}a_-{+}i\pi \]

locally uniform for $z\in {\mathbb {C}}^+$. An example of a function $q$ with a measure $\mu$ as in (5.23) is

\[ q(z) = (a_-{-}a_+{+}i\pi)\log(z+i) + \pi(a_+{+}1)i, \qquad z\in{\mathbb{C}}^+, \]

where the $\mathrm {O}(t^\gamma )$ term is actually $\mathrm {O}(\log t)$.