---
author:
- 'Mikhail Sodin[^1]'
title: |
    A question by Alexei Aleksandrov\
    and logarithmic determinants
---

<span>**§1. Analytic functions represented by Schwarz' integrals.**</span>

This note is motivated by the following question: <span>*when is an analytic function $f$ in the unit disc ${\mathbb D}$ represented by the Schwarz integral $$f(z) = \int_{{\mathbb T}} \frac{{\zeta}+z}{{\zeta}-z}\, d\mu ({\zeta})
\eqno (1.1)$$ of a real measure $\mu$ on the unit circle ${\mathbb T}$?*</span> Two classical necessary conditions are:

<span>**(i)**</span> (Smirnov). $f\in {{\mathcal N}_+}$, that is, $$\log|f(z)|\le \int_{{\mathbb T}} \log|f({\zeta})| \,
\hbox{Re} \left(\frac{{\zeta}+z}{{\zeta}-z}\right) \, dm({\zeta})\,,
\qquad z\in{\mathbb D}\,,
\eqno (1.2)$$ where $m$ is the normalized Lebesgue measure on ${\mathbb T}$, $m({\mathbb T})=1$. (We use the same notation $f$ for an analytic function in the unit disc and for its non-tangential boundary values on the unit circle.) In fact, Smirnov proved a stronger result: $f\in \bigcap_{p<1} H^p$.

<span>**(ii)**</span> (Kolmogorov). $f\in L^{1,\infty}({\mathbb T})$, that is, $m_f(t)=O(t^{-1})$, $t\to\infty$, where $m_f(t) = m\{{\zeta}\in{\mathbb T}:\, |f({\zeta})|\ge t \}$. More precisely, if $f$ is represented in the form (1.1), then $$m_f(t) \le \frac{C||\mu||}{t}\,,
\qquad 0<t<\infty\,,$$ where $C$ is a numerical constant.

Another necessary condition is

<span>**(iii)**</span> (Hruschev and Vinogradov @HV) if $f$ us represented in the form (1.1), then there exists a limit $$\lim_{t\to \infty} tm_f(t) = \frac{||\mu_{\hbox{\small sing}}||}{\pi}\,.$$ A weaker result that $\lim_{t\to \infty} tm_f(t) = 0$ if the measure $\mu$ in the representation (1.1) is absolutely continuous, is due to Titchmarsh @T.

Besides the above conditions, there are the trivial restrictions:

<span>** (iv)**</span> $$\hbox{Re} f \left( = \frac{d\mu_{\hbox{\small a.c}}}{dm}\right)
\in L^1({\mathbb T})\,, 
\qquad \hbox{and} \qquad  
f(0) \left( =\int_{\mathbb T}d\mu \right) \in \mathbb R\,.$$

In @A1 and @A2 Aleksandrov proved

<span>**Theorem A1.** </span><span> *The set of conditions (i), (ii), and (iv) is sufficient for representation (1.1), with*</span> $$||\mu|| \le ||\hbox{Re} f||_1 + C ||f||_{1,\infty}\,,$$ <span>*where $||\,.\,||_1=||\,.\,||_{L^1({\mathbb T})}$, $||f||_{1,\infty} = \sup_{t>0} tm_f(t)$, and $C$ is a positive numerical constant.*</span>

That is, an analytic function $f$ is represented by the Schwarz integral (1.1) iff conditions (i), (ii) and (iv) hold. Furthermore, Aleksandrov suggested that in the sufficiency part condition (ii) can be weakened and asked <span>*whether conditions (i), $$\liminf_{t\to\infty} tm_f(t) < \infty\,,$$ and (iv) already guarantee that $f$ is represented by the Schwarz integral (1.1).*</span> Another form of this question is (cf. @A3): <span>*whether there exists a non-constant analytic function $f$ of Smirnov's class ${{\mathcal N}_+}$ such that $\hbox{Re} f = 0$ a.e. on ${\mathbb T}$ and*</span> $$\liminf_{t\to\infty} tm_f(t) = 0\,.
\eqno (1.4)$$ If such a function exists, then in view of the necessary conditions cited above, it cannot be represented by the Schwarz integral (1.1). Here, we answer this question.

<span>**Theorem 1.** </span><span>*There exists a non-constant analytic function $f\in {{\mathcal N}_+}$ satisfying (1.4) and such that $\hbox{Re} f= 0$ a.e. on ${\mathbb T}$.* </span>

On the other hand, condition (ii) can be really weakened:

<span>**Theorem 2.** </span><span>*Let $f$ be an analytic function in ${\mathbb D}$ satisfying conditions (i) and (iv), and let $$\liminf_{R\to\infty} R\int_R^\infty \frac{m_f(t)}{t}\, dt < \infty\,.
\eqno (1.5)$$ Then $f$ is represented by the Schwarz integral (1.1) of a real measure $\mu$, and*</span> $$||\mu|| \le ||\hbox{Re} f||_1
+ C \liminf_{R\to\infty} R\int_R^\infty \frac{m_f(t)}{t}\, dt\,,$$ <span>*where $C$ is a positive numerical constant.*</span>

It is worth to note that after integration by parts condition (1.5) can be written as $$\liminf_{R\to\infty} R\mathcal N\left(\frac{f}{R}\right) < +\infty\,,$$ where $$\mathcal N(f) = \int_{\mathbb T} \log^+|f|\, dm$$ is a "norm" in the Smirnov class $\mathcal N_+
$.

<span>**§2. Proof of Theorem 1.**</span>

We construct $f$ as a universal covering of ${\mathbb C}\setminus E$, where $E$ is a closed subset of $i{\mathbb R}_+$, by the unit disc ${\mathbb D}$. We shall use some classical facts about universal coverings, harmonic measures and Green functions @N.

Let $E=\bigcup_{n\ge 1} [ir_n, 2ir_n]$, where $r_1=1$, $r_{n+1}/r_n\to\infty$, and let $${\omega}(t) = {\omega}(0, E\cap \{|z|\ge t\}, {\mathbb C}\setminus E)$$ be the harmonic measure of $E\cap \{z:\, |z|\ge t\}$ with respect to ${\mathbb C}\setminus E$ evaluated at the origin. We can choose the sequence $\{r_n\}$ increasing so fast that $$\liminf_{t\to\infty} t {\omega}(t) = 0\,.
\eqno (2.1)$$ Indeed, let $h_n(z)$ be a bounded harmonic function in ${\mathbb C}\setminus 
\left( [i,2i]\bigcup [ir_n, i\infty)\,\right)$, vanishing on $[i,2i]$, and having its boundary values equal identically to $1$ on $[ir_n,
i\infty)$. Evidently, $\lim_{n\to\infty} h_n(0) = 0$. Hence, on the $n+1$-st step we can choose a value $r_{n+1}$ sufficiently large that $h_{n+1}(0) \le r_n^{-2}$. Then by the maximum principle $$\begin{aligned}
\liminf_{t\to\infty} t{\omega}(t) &\le& \liminf_{n\to\infty} 2r_n{\omega}(2r_n) \\
&=& 2 \liminf_{n\to\infty} r_n{\omega}(r_{n+1}) \le 2 \liminf_{n\to\infty} r_n
h_{n+1}(0) = 0\,,\end{aligned}$$ proving (2.1).

Now, let $f: {\mathbb D}\to {\mathbb C}\setminus E$ be the universal covering map normalized by $f(0)=0$. Since $E$ has a positive capacity, $f$ is of bounded type in ${\mathbb D}$, and therefore, has non-tangential boundary values a.e. on ${\mathbb T}$. By the invariance of the harmonic measure, $$\int_{\mathbb T}(\varphi \circ f) ({\zeta}) \,\hbox{Re}\left(\frac{{\zeta}+z}{{\zeta}-z}\right) 
dm({\zeta}) = 
\int_E \varphi (\eta)\, {\omega}(f(z), d\eta, {\mathbb C}\setminus E)\,,
\eqno (2.2)$$ where $\varphi$ is an arbitrary continuous function on $E$ (as the boundary of ${\mathbb C}\setminus E$). In particular, $$\int_T (\varphi\circ f)\, dm = \int_E \varphi (\eta)
\, {\omega}(0, d\eta, {\mathbb C}\setminus E)\,.
\eqno (2.2a)$$ After a monotonic limit transitions, relations (2.2) and (2.2a) also hold for semi-continuous functions, and therefore $$m_f(t) = {\omega}(t)\,,
\qquad 0<t<\infty\,.
\eqno (2.3)$$

Now, the function $f$ has pure imaginary boundary values a.e. on ${\mathbb T}$, and $$\liminf_{t\to\infty} tm_f(t) \stackrel{(2.3)}= \liminf_{t\to\infty} t{\omega}(t)
\stackrel{(2.1)}= 0\,.$$ It remains to observe that $f$ is of Smirnov's class ${{\mathcal N}_+}$. Indeed, representing a subharmonic function $\log|w|$ in ${\mathbb C}\setminus E$ as a sum of the Green function and the Poisson integral, we have $$\log|w| = G_{{\mathbb C}\setminus E} (w,0) +
\int_E \log|\eta| \, {\omega}(w, d\eta, {\mathbb C}\setminus E) + \alpha K_E(w)\,,
\eqno (2.4)$$ where $G_{{\mathbb C}\setminus E} (w,0)$ is Green's function for ${\mathbb C}\setminus E$ with pole at $w=0$, $K_E(w)$ is a positive harmonic function in ${\mathbb C}\setminus E$ vanishing on $E$ (so-called Martin function), and $\alpha\in \mathbb R$.

First, we shall show that $\alpha \le 0$ (in fact, a minor modification of the following argument shows that $\alpha =0$, though it is not needed for our purposes). Indeed, if $\alpha > 0$, then as follows from (2.4), the function $K_E$ has a logarithmic cjohnson@example.org. Setting $K_E(z)=0$ for $z\in E$ we obtain a subharmonic function in ${\mathbb C}$ of logarithmic growth, so that the set $E$ must be <span>*thin*</span> at infinity @AH and according to the Wiener criterion $$\sum_{n=1}^\infty \frac{n}{\log \frac{2}{c(E_n)}} < \infty\,,$$ where $E_n=E\cap \{z:\, 2^n\le |z| <2^{n+1}\}$ and $c(\,.\,)$ is the logarithmic capacity (cf. @Hayman [Chapter 7] or @Ransford [Chapter 5]). Since $E$ contains intervals $[ir_n, 2ir_n]$ and since the logarithmic capacity of the interval equals one quarter of its length, we see that the series must diverge and therefore $\alpha
\le 0$.

Omitting the positive terms on the RHS of (2.4), we get $$\log|w| \le
\int_E \log|\eta| \, {\omega}(w, d\eta, {\mathbb C}\setminus E) \,.$$ Setting here $w=f(z)$, $\eta = f({\zeta})$, and making use of (2.2), we obtain $$\log|f(z)| \le
\int_{\mathbb T}\log|f({\zeta})|\, 
\hbox{Re}\left(\frac{{\zeta}+z}{{\zeta}-z}\right) dm({\zeta})\,,
\qquad z\in{\mathbb D}\,,$$ which completes the proof. $\Box$

<span>**§3. Proof of Theorem 2.**</span>

We may assume that $\hbox{Re} f = 0$ a.e. on ${\mathbb T}$. Otherwise, we decompose $$f(z) = f_1(z) + \int_{\mathbb T}\frac{{\zeta}+z}{{\zeta}-z} 
\left( \hbox{Re} f\right)({\zeta})\, dm({\zeta}) = f_1 + f_2\,,$$ where $\hbox{Re} f_1 = 0$ a.e. on ${\mathbb T}$, and $\lim_{t\to\infty} tm_{f_2}(t) = 0$.

Following @MOS, @MS1, and @MS2, we introduce a "logarithmic determinant" $$u_f(w) := \int_{\mathbb T}\log|1-wf({\zeta})| \, dm({\zeta})\,.
\eqno (3.1)$$ The function $u_f$ is subharmonic in ${\mathbb C}$, harmonic in the right and left half-planes $\Pi_\pm$, and $u(0)=0$. Furthermore, $$u_f(w) \le O(\log |w|)\,, 
\qquad w\to\infty\,,$$ so that, $u_f$ is represented by the Poisson integrals in $\Pi_\pm$. The function $u_f$ has a lower bound $$u_f(it) \ge \log|1-it f(0)| \ge 0\,,
\qquad t\in \mathbb R.$$ By the maximum principle applied to $u_f$ in $\Pi_\pm$, the function $u_f$ is non-negative everywhere in ${\mathbb C}$.

Our goal is to estimate from above the integral $${{\mathcal I}_f}:= \frac{1}{\pi} \int_{\mathbb R} \frac{u_f(it)}{t^2}\, dt\,.$$ But first, we shall show that this integral controls the norm $||f||_{1,\infty}$ (cf. the proof of Theorem 2 in @MS1). Indeed, estimating the Poisson integrals in $\Pi_\pm$ we have $$u(re^{i\theta}) = \frac{r|\cos\theta|}{\pi} \int_{\mathbb R}\frac{u_f(it)}{t^2} \, \frac{t^2}{|it-re^{i\theta}|^2}\, dt
\le \frac{r}{|\cos\theta|} \, {{\mathcal I}_f}\,,
\qquad 0<r<\infty\,.$$ In particular, $u_f(re^{i\theta})\le r\sqrt{2}\, {{\mathcal I}_f}$ within the angles $\{|\theta|\le \pi/4\}$ and $\{|\theta|\ge 3\pi/4 \}$. Applying the maximum principle to the harmonic function $u_f(z) - 2{{\mathcal I}_f}|\hbox{Im} z|$ within the complementary angles $\{\pi/4 \le |\theta| \le 3\pi/4\}$, we get $$M(r, u_f) := \max_{\theta\in [-\pi, \pi]} u_f(re^{i\theta}) 
\le 2{{\mathcal I}_f}r\,, 
\qquad 0<r<\infty\,.$$ Let $\mu_f$ be the Riesz measure of the function $u_f$, and let $\mu_f(r) = \mu_f\{|w|\le r\}$ be its counting function. Then by the Jensen formula $$\mu_f(r) \le M(er, u_f) \le 2e{{\mathcal I}_f}r\,, \qquad  0<r<\infty\,.$$ It remains to observe that $$m_f(\tau) \stackrel{(3.1)}= \mu_f(1/\tau)\,,
\eqno (3.2)$$ so that $$||f||_{1,\infty} \le 2e {{\mathcal I}_f}\,. 
\eqno (3.3)$$

Now, we estimate the integral ${{\mathcal I}_f}$ using an integral formula which has been used previously in a similar situation (cf. the proof of Theorem 3 in @MOS). For $|\theta| < \pi/2$ we have $$u_f(re^{i\theta}) = \frac{r\cos\theta}{\pi}
\int_{\mathbb R}\frac{u_f(it) dt}{|re^{i\theta} - it|^2}\,.$$ Integrating this against $\cos\theta$, we get $$\begin{aligned}
\int_{-\pi/2}^{\pi/2} u_f(re^{i\theta}) \cos\theta\, d\theta
&=& \frac{r}{\pi} \int_{\mathbb R}u_f(it) \, dt 
\int_{-\pi/2}^{\pi/2} \frac{\cos^2\theta d\theta}{|re^{i\theta} - it|^2}
\\ \\ 
&=& \frac{r}{2} \int_{\mathbb R}u_f(it) \min\left( \frac{1}{t^2},
\frac{1}{r^2}\right) dt\,,\end{aligned}$$ since $$\int_{-\pi/2}^{\pi/2} \frac{\cos^2\theta d\theta}{|re^{i\theta} - it|^2}
= \frac{\pi}{2} \min\left( \frac{1}{t^2}, \frac{1}{r^2}\right) \,.$$ Using a similar relation in the left half-plane, we have $$\begin{aligned}
\int_{\mathbb R}u_f(it) \min\left( \frac{1}{t^2}, \frac{1}{r^2}\right) dt
&= & \frac{2}{r} \int_{-\pi}^\pi u_f(re^{i\theta}) |\cos\theta|\, d\theta
\\ \\ 
&< & \frac{2}{r} \int_{-\pi}^\pi u_f(re^{i\theta}) \, d\theta \\ \\
&= & \frac{4\pi}{r} \int_0^r \frac{\mu_f (s)}{s}\, ds\,.\end{aligned}$$ At last, making the monotonic limit transition as $r\to 0$, and setting $R=1/r$, we get $${{\mathcal I}_f}\le  4\pi \liminf_{r\to 0} \frac{1}{r} \int_0^r \frac{\mu_f(s)}{s}\, ds 
\stackrel{(3.2)}\le 4\pi \liminf_{R\to\infty} R\int_R^\infty
\frac{m_f(\tau)}{\tau}\, d\tau\,.
\eqno (3.4)$$ Combining inequalities (3.3) and (3.4) with Theorem A1, we complete the proof. $\Box$

<span>**§4. Concluding remarks.** </span>

<span>**4.1** </span> The technique of logarithmic determinants used in the proof of Theorem 2 also allows us to prove another theorem of Aleksandrov @A1, @A2:

<span>**Theorem A2.** </span><span>*Let $f$ be an analytic function in the unit disc satisfying condition (iv), let $f\in
\bigcap_{p<1} H^p$, and let $$\liminf_{p\uparrow 1} (1-p)||f||_{H^p} < \infty\,.$$ Then $f$ is represented by the Schwarz integral (1.1) of a real measure $\mu$, and*</span> $$||\mu|| \le ||\hbox{Re} f||_1 + C\liminf_{p\uparrow 1}
(1-p)||f||_{H^p}\,. 
\eqno (4.1)$$

(In fact, Aleksandrov proved this estimate with $C=\frac{\pi}{2}$ on the RHS.)

Our proof follows the same lines as that of Theorem 2, only in the last step we use other integral formulas to estimate the integral ${{\mathcal I}_f}$: $$\int_{\mathbb R}\log|1-it \lambda|\, \frac{dt}{|t|^{1+p}}
= |\lambda|^p\, \frac{\pi}{p} \cot \frac{\pi p}{2}\,,
\qquad \lambda\in {\mathbb C}\,, \quad 0<p<1\,,$$ and $$\int_{\mathbb R}\frac{u_f(it)}{|t|^{1+p}}\, dt 
= \frac{\pi}{p} \cot \frac{\pi p}{2} \int_{\mathbb T}|f({\zeta})|^p\, dm({\zeta})\,,
\qquad 0<p<1\,.$$

On the other hand, Theorem A2 follows directly from our Theorem 2 since the $p$-th power of the $L^p$ norm can be written as the integral $$p\int_0^\infty t^p\, \frac{m_f(t)}{t}\, dt
= p^2 \int_0^\infty t^{p-1} 
\left( \int_t^\infty \frac{m_f(s)}{s}\, ds \right) dt\,.$$ This yields that if $$\lim_{t\to \infty} t \int_t^\infty \frac{m_f(s)}{s}\, ds = \infty\,,$$ then $\lim_{p\uparrow 1} (1-p)||f||_p = \infty$.

We also note that it is not difficult to construct a function $h$ on $[-\pi,\pi]$ such that $$\liminf_{t\to\infty} t\int_t^\infty
\frac{m_h(s)}{s}\, ds = 0$$ while $h$ does not belong to any of the $L^p$ spaces for $p>0$. This shows that the assumptions of Theorem 2 are really weaker than those of Theorem A2.

<span>**4.2** </span> The counter-example provided by Theorem 1 is related to the logarithmic determinants by the following formula (in the notation of $\S 2$): $$\int_{\mathbb T}\log|1-wf({\zeta})|\, dm({\zeta}) \stackrel{(2.2a)}= 
G_{{\mathbb C}\setminus E^*}(w,\infty)\,,$$ where $E^*=\{\lambda:\, 1/\lambda\in E\}$.

<span>**4.3** </span> Theorems 1 and 2 can be easily reformulated for analytic functions represented by Cauchy-type integrals $$f(z) = \int_{\mathbb T}\frac{d\mu(\zeta)}{\zeta-z}$$ of complex-valued measures $\mu$ of finite variations. In this case, condition (iv) must be replaced by

<span>**(iv')**</span> $Tf\in L^1({\mathbb T})$, where $Tf(\zeta) = \lim_{r\uparrow 1} \left(f(r\zeta) - f(r^{-1}\zeta)
\right)$.

We leave the details to the reader (cf. @A1, @A2).

<span>**Acknowledgements.**</span> I thank A. Aleksandrov, I. Ostrovskii and P. Yuditskii for their numerous useful remarks and suggestions.

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*School of Mathematical Sciences, Tel-Aviv University, Ramat-Aviv, 69978, Israel*

cjohnson@example.org

[^1]: Supported by the Israel Science Foundation of the Israel Academy of Sciences and Humanities under Grant No. 37/00-1.

{'source': 'arXiv_out/math0108187'}