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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
     {\large Siberian Electronic Mathematical Reports} \\[1mm]
     {\LARGE\tt{http://semr.math.nsc.ru}}\\[0.5mm]
%     {\small Институт математики им. С.Л.Соболева СО РАН}\\[-1mm]
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     {\flushleft\it Том 16, стр. 144--144 (2019) \hspace{65mm}{\rm\small УДК \udcs}}
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     	{\rm\small DOI~10.33048/semi.2019.16.xxx}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}{\rm\small MSC\ \ \mscs }
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\title[On the dissymmetrization theorem]{On the dissymmetrization theorem}
\author{{V.N. DUBININ}}%
\address{Dubinin Vladimir Nikolaevich % Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii}Institute for Applied Mathematics, FEBRAS,
\newline\hphantom{iii} Radio str., 7,
\newline\hphantom{iii} 690041, Vladivostok, Russia}%
\email{dubinin@iam.dvo.ru}%

\thanks{\sc Dubinin, V.N.,
On the dissymmetrization theorem}
\thanks{\copyright \ 2023 Dubinin V.N.}
\thanks{\rm This work was supported by the Russian Science Foundation
under grant 23-21-00056.}
\thanks{\it Поступила ... 2023 г., опубликована ... 2023 г.}%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } A new property of the previously proposed dissymmetrization of functions is established. The conjecture about  the capacity of condensers in a circular ring with plates in the form of circles or radial cuts is discussed. The connection of this conjecture with the well-known Gonchar-Baernstein  problem of a harmonic measure is shown.\medskip

\noindent{\bf Keywords:} dissymmetrization, harmonic measure, Dirichlet integral, condenser capacity.
 \end{quote}
}


\section{Introduction and statement of result}

Let $n\ge2$ be a natural number and let 
$$L^*_j=\{z:\;{\rm arg}\,z=2\pi j/n\},\;\;\;\;j=1,...,n.$$
Denote by $\Phi$ the group of symmetries of $\overline{\mathbb{C}}$ formed by the composites of the reflections in the rays $L^*_j,\; j= 1, . . . , n$, and in the bisectors of the angles formed by these rays. Throughout this paper symmetry means $\Phi$-invariance. We say that a set $A\subset \overline{\mathbb{C}}$ is symmetric if $\varphi(A)=A$ for any isometry $\varphi\in \Phi$. A real function $v$ on a symmetric set $\Omega$ is said to be symmetric if  $v(z)\equiv v(\varphi(z))$ for any $\varphi\in \Phi$. We call a system of closed sectors with vertices at origin a decomposition of the sphere $\overline{\mathbb{C}}$ if no two sectors have common interior points and their union is $\overline{\mathbb{C}}$. It has been shown in [1] (see also [2], [3, Sec. 4.4]) that for any different rays $L_j,\; j= 1, . . . , n,$                               starting from the origin there exists a decomposition  $\{P_j\}_{j=1}^{j_0},\;j_0\ge n,$ of the plane $\overline{\mathbb{C}}$ and a set of rotations $\{\lambda_j\}_{j=1}^{j_0}$              of the form  $\lambda_j(z)=e^{i\theta_j}z,\;j=1,...,j_0$     
(dissymmetrization) such that the set  $\{\lambda_j(P_j)\}_{j=1}^{j_0}$  is also a decomposition  $\overline{\mathbb{C}},$ ${\rm Dis}\,L_j^*=L_j,\;j=1,...,n,$ and for any the symmetric function $v,\;z\in \Omega,$ the function ${\rm Dis} \,v$ is well defined:
$${\rm Dis}\,v(z):=v(\lambda_j^{-1}(z)),\;z\in \lambda_j(P_j)\cap{\rm Dis}\,\Omega,\;\;\;j=1,...,j_0.$$
Here ${\rm Dis}\,A$ means $\bigcup_{j=1}^{j_0}\lambda_j(A\cap P_j).$ Conceived as a method for solving the Gonchar problem on harmonic measure [1], dissymmetrization has found application in other questions of function theory (see, for example, [3]--[8]). In this paper, we consider a new property of this transformation. Let us introduce the notation 
$$B(s,t)=\{z:\;s<|z|<t\},\;\;\;\;0<s<t<\infty,$$
$$T(r)=\{z:\;|z|=r\},\;\;\;\;T=T(1).$$
The following theorem is true. 
\begin{theorem}
Let $E^*$ be a symmetric compact set on $\bigcup_{j=1}^n L_j^*\cap B(s,1)$ and let $f^*$  be a symmetric continuous function on  $\overline{B(s,1)}$, constant on $T$, harmonic in $B(s,1)\setminus E^*$, and $f$ is a continuous function on $\overline{B(s,1)}$, coinciding with the function ${\rm Dis}\,f^*$ on $\partial(B(s,1)\setminus {\rm Dis}\,E^*)$             and harmonic in $B(s,1)\setminus {\rm Dis}\,E^*.$ Then 
\begin{equation}
\int\limits_T \left[\left(\frac{\partial f}{\partial n}\right)^2-\left(\frac{\partial f^*}{\partial n}\right)^2\right]ds\ge \frac{1}{\pi}\int\limits_T\frac{\partial f^*}{\partial n} ds\int\limits_T\left(\frac{\partial f}{\partial n}-\frac{\partial f^*}{\partial n}\right)ds,
\end{equation}
where  $\partial/\partial n$ denotes differentiation along the inward normal. 
\end{theorem}

It is known that the Dirichlet integral does not increase under the dissymmetrization [3, Sec. 4.4]. Inequality [1] gives information about the behavior of the variation of the Dirichlet integral under dissymmetrization (see Lemma 1). In particular, when the right-hand side of (1) is equal to zero, the variation of the Dirichlet integral of a symmetric function does not exceed the variation of such an integral of a function that does not have such symmetry (cf. [8]) This statement, together with [8], leads to a new conjecture about the behavior of the capacities of some condensers during dissymmetrization which is closely related to the well-known Gonchar-Baernstein  problem on the harmonic measure of radial cuts [9]. The next section is auxiliary.

\section{Variation of the Dirichlet integral}
Everywhere below, the notation 
$$I(f,B):=\iint\limits_B |\triangledown f|^2 dxdy$$
is adopted. In the unit disk $|z|<1$, consider a finitely connected domain $B$  whose boundary consists of analytic Jordan curves, including the circle $T$. Let $f$ be a function twice continuously differentiable in   $\overline{B}$, harmonic in $B$ and equal to zero on $T$, and let $f_t$ be a function harmonic in $B_t:=B\cap\{z:\;|z|<t\},\;t<1,$ twice continuously differentiable in $\overline{B}_t$, taking boundary values $f$ on $(\partial B)\setminus T$ and equal to zero on $T(t)$. 
\begin{lemma} The following asymptotic formula holds:
$$I(f_t,B_t)=I(f,B)+(1-t)\int\limits_T\left(\frac{\partial f}{\partial n}\right)^2ds+O((1-t)^2),\;\;\;\;t\to 1.$$
\end{lemma}
\hspace{-0.5cm}{\it Proof.}
On the circle $T(t)$ the uniform estimate 
$$f(z)=f(z)-f_t(z)=(1-t)\frac{\partial f}{\partial n}(z/t)+O((1-t)^2)$$
is satisfied. Applying Green's formula, we obtain the equality 
$$I(f-f_t,B_t)=-\int\limits_{\partial B_t}(f-f_t)\frac{\partial( f-f_t)}{\partial n}ds=-\int\limits_{T(t)}(f-f_t)\frac{\partial (f-f_t)}{\partial n}ds=O((1-t)^2).$$
 On the other hand, again by Green's formula
$$I(f-f_t,B_t)=I(f,B_t)+I(f_t,B_t)+2\int\limits_{ T(t)}(f_t-f+f)\frac{\partial f}{\partial n}ds$$$$=I(f_t,B_t)-I(f,B_t)+2\int\limits_{T(t)}(f_t-f)\frac{\partial f}{\partial n}ds=I(f_t,B_t)-I(f,B)-\int\limits_{T(t)}f\frac{\partial f}{\partial n}ds.$$
Therefore, 
$$I(f_t,B_t)-I(f,B)-\int\limits_{T(t)}f\frac{\partial f}{\partial n}ds=O((1-t)^2),$$
which completes the proof of Lemma 1. 
%\end{proof}
 
 We note that in the case when the function $f$  takes constant values on the connected components of the boundary of the domain  $B$, our formula follows from the classical variational formula [10, (A3.12)].

\section{Proof of the theorem\,1}
It suffices to establish inequality (1) in the new formulation of the problem. Namely, to replace the set $E^*$ in the hypothesis of Theorem 1 by a symmetric set $\mathcal{E}^*$ located sufficiently close to it, bounded by a finite number of analytic Jordan curves, and the function  $f^*$  to be assumed to be three times continuously differentiable on  $\partial(B(s,1)\setminus\mathcal{E}^*)$ and equal to zero on  $T$. We fix $t$ such that $0<t<1,$ $\sup\{|z|:\;z\in \mathcal{E}^*\}<1-2\Delta t$ $(\Delta t=1-t)$                                       and consider functions 
$$b(z):=\log|z|,\;\;\;\;u^*=b+\varepsilon f^*,$$
where $\varepsilon>0$ is sufficiently small. It is easy to see that the Hausdorff  distance  
between the curve     
$$\gamma_t^*:\;u^*=b(t)$$
and the circle $T(t)$ is the quantity$^1$\footnotetext[1]{Here and below, the prime at the symbol $O$ -- large means that the corresponding quantity admits a uniform estimate for all sufficiently small $\Delta t$     at $\varepsilon\to 0.$} $O'(\varepsilon \Delta t),\;\varepsilon\to 0.$                   
Further, the notation $(\gamma,\Gamma)$ means a doubly connected domain on the plane $\mathbb{C}$, bounded by closed curves $\gamma,\;\Gamma$,      and $\mod(\gamma,\Gamma)$                       is the module of the domain $(\gamma,\Gamma)$. Let the quantities $t(\varepsilon)$ and $R(\varepsilon)$ be defined by the relations: 
$$\mod(\gamma^*_\tau,\gamma^*_t)=\hspace{-0.35cm}\mod(\gamma^*_\tau,T(t(\varepsilon))),\;\;\;\;\tau=1-2\Delta t,$$
$$\mod(\gamma^*_t,T)=\hspace{-0.35cm}\mod(T(t(\varepsilon)),T(R(\varepsilon))).$$
Following the arguments in [8] from formula (3.1) to (3.5), where it is necessary to set the $-1/\log r=\varepsilon,$  we see that
\begin{equation}
R(\varepsilon)\ge 1 \;\;\mbox{and}\;\;R(\varepsilon)-1=O'((\varepsilon\Delta t)^2).
\end{equation} 
By definition,   
$$\mod(\gamma_t^*,T)=\frac{1}{2\pi}\log\frac{R(\varepsilon)}{t(\varepsilon)}=\frac{1}{2\pi}\log\frac{1}{t(\varepsilon)}+O'((\varepsilon\Delta t)^2).$$
On the other hand, the Hadamard formula [10 (A3.11)], [8, (2.2)] gives 
$$\mod(\gamma_t^*,T)=\frac{1}{2\pi}\log\frac{1}{t}-\left(\frac{1}{2\pi}\log\frac{1}{t}\right)^2\int\limits_{T(t)}\left(\frac{\partial \omega}{\partial n}\right)^2[-\varepsilon t f^*(t e^{i\theta})]ds+O'((\varepsilon\Delta t)^2),$$
where $\omega(z)=(\log|z|)/\log t.$                                           
Hence 
\begin{equation}
\log\frac{t}{t(\varepsilon)}=\frac{\varepsilon}{2\pi}\int\limits_0^{2\pi}f^*(t e^{i\theta})d\theta+O'((\varepsilon\Delta t)^2).
\end{equation} 

Let $F^{-}$ be some conformal mapping of the domain $(T(t(\varepsilon)),T(R(\varepsilon)))$ onto $(\gamma^*_t,T)$
 and let $F^{+}$ be a conformal mapping of $(\gamma^*_\tau,T(t(\varepsilon)))$ onto $(\gamma^*_\tau,\gamma^*_t)$
in such a way that $F^{-}(T(t(\varepsilon)))=F^{+}(T(t(\varepsilon)))=\gamma^*_t.$
Following the proof of Theorem 1 [8], we set 
$$v^*(z)=\begin{cases}
u^*(F^{-}(z)), & z\in \overline{(T(t(\varepsilon)),T(R(\varepsilon)))},\\
u^*(F^{+}(z)), & z\in \overline{(\gamma^*_\tau,T(t(\varepsilon)))},\\
u^*(z), & z\in \overline{(T(s),\gamma^*_\tau)\setminus \mathcal{E}^*}.\\
\end{cases} $$
Note that in the ring $(T(t(\varepsilon)),T(R(\varepsilon)))$
$$v^*(z)=\frac{\log|z/R(\varepsilon)|}{\log|t(\varepsilon)/R(\varepsilon)|}\log t$$
is fulfilled. 

The conformal invariance of the Dirichlet integral and the Dirichlet principle implies 
$$I(u^*,(T(s),T)\setminus\mathcal{E}^*)=I(v^*,(T(s),T(R(\varepsilon)))\setminus\mathcal{E}^*)$$
$$\ge I(h^*,(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*)+I(v^*,(T(t(\varepsilon))),T(R(\varepsilon))).$$
Here $h^*$ is a harmonic function on the set $(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*$, continuous in the closure of this set and taking the following boundary values: 
$$h^*(z)=\begin{cases}
b(t), & z\in T(t(\varepsilon)),\\
b(z)+\varepsilon f^*(z), & z\in T(s)\cup\partial \mathcal{E}^*.
\end{cases} $$
Note the symmetry of the function $h^*$ and the fact that the circles $T(s),\;T(t(\varepsilon))$ do not change under dissymmetrization  ${\rm Dis}$ (Sec.1). Using the Dirichlet principle again, we obtain 
$$I(h^*,(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*)=I({\rm Dis}\,h^*,(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*)$$
$$\ge I(h,(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*),$$
where $h$ is a harmonic function on the set $(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*$, continuous in the closure of this set and taking boundary values:
$$h(z)=\begin{cases}
b(t), & z\in (T(t(\varepsilon)),\\
b(z)+\varepsilon {\rm Dis}\, f^*(z), & z\in T(s)\cup\partial{\rm Dis}\,\mathcal{E}^*.
\end{cases}$$
Again, the Dirichlet principle gives 
$$I(h,(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*)+I(\tilde{v}^*,(T(t(\varepsilon)),T))\ge I(u,(T(s),T)\setminus{\rm Dis}\,\mathcal{E}^*)$$
for $$\tilde{v}^*(z)=\frac{\log|z|}{\log t(\varepsilon)}\log t$$                                                     and the function $u$, harmonic in $(T(s),T)\setminus{\rm Dis}\,\mathcal{E}^*$ continuous in the closure of this set and taking boundary values: 
$$u(z)=\begin{cases}
0, & z\in T,\\
b(z)+\varepsilon {\rm Dis}\, f^*(z), & z\in T(s)\cup\partial{\rm Dis}\,\mathcal{E}^*.
\end{cases}$$
The main inequality in the proof of Theorem 1 follows from the inequalities written above: 
$$I(u^*,(T(s),T)\setminus\mathcal{E}^*)-I(u,(T(s),T)\setminus{\rm Dis}\,\mathcal{E}^*)\ge I(h^*,(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*)$$
\begin{equation}
-I(h,(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*)+I(v^*,(T(t(\varepsilon)),T(R(\varepsilon))))-I(\tilde{v}^*,(T(t(s)),T)).
\end{equation}
Let us now turn to estimates of the integrals in (4). According to Green's formula, the first integral on the left in (4) is 
$$I(u^*,(T(s),T)\setminus\mathcal{E}^*)=I(b,(T(s),T)\setminus\mathcal{E}^*)+\varepsilon^2 I(f^*,(T(s),T)\setminus\mathcal{E}^*)$$
\begin{equation}
-2\varepsilon\int\limits_{T(s)\cup\partial \mathcal{E}^*}f^* \frac{\partial b}{\partial n}ds.
\end{equation}
By the uniqueness theorem for harmonic functions in                                       $(T(s),T)\setminus {\rm Dis}\,\mathcal{E}^*$, the equality
$$u=b+\varepsilon f$$
is true. Therefore, 
$$I(u,(T(s),T)\setminus{\rm Dis}\,\mathcal{E}^*)=I(b,(T(s),T)\setminus{\rm Dis}\,\mathcal{E}^*)$$
\begin{equation}
+\varepsilon^2 I(f,(T(s),T)\setminus{\rm Dis}\,\mathcal{E}^*)-2\varepsilon\int\limits_{T(s)\cup\partial{\rm Dis}\, \mathcal{E}^*}f \frac{\partial b}{\partial n}ds.
\end{equation}
To calculate the first integral of the right-hand side of (4), we represent the function $h^*$      in the form 
$$h^*=b+\varepsilon f^*_{t(\varepsilon)}+\sigma^*,$$
where the functions $f^*_{t(\varepsilon)},\;\sigma^*$                   are harmonic on the set $(T(s),T(t(\varepsilon)))\setminus \mathcal{E}^*$, continuous in the closure of this set and take boundary values
\begin{center}
 \begin{tabular}{l l}
$f^*_{t(\varepsilon)}(z)=\begin{cases}
0, & z\in T(t(\varepsilon)),\\
f^*(z), & z\in T(s)\cup\partial\mathcal{E}^*,
\end{cases}$&$\sigma^*(z)=\begin{cases}
\log (t/t(\varepsilon)), & z\in T(t(\varepsilon)),\\
0, & z\in T(s)\cup\partial\mathcal{E}^*.
\end{cases}$
\end{tabular}
\end{center}
By successively applying Green's formula, we obtain
$$I(h^*,(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*)=I(b,(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*)+\varepsilon^2 I(f^*_{t(\varepsilon)},(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*)$$
$$+I(\sigma^*,(T(s),T(t(\varepsilon)))\setminus\mathcal{E}^*)-2\varepsilon\int\limits_{T(s)\cup\partial\mathcal{E}^*}f^* \frac{\partial b}{\partial n}ds$$
\begin{equation}
-2\log\frac{t}{t(\varepsilon)}\int\limits_{T(t(\varepsilon))} \frac{\partial b}{\partial n}ds-2\varepsilon\log\frac{t}{t(\varepsilon)}\int\limits_{T(t(\varepsilon))} \frac{\partial f^*_{t(\varepsilon)}}{\partial n}ds.
\end{equation}
Similar to (7)
$$h=b+\varepsilon f_{t(\varepsilon)}+\sigma$$
where the functions  $f_{t(\varepsilon)},\;\sigma$ are harmonic in  $(T(s),T(t(\varepsilon)))\setminus {\rm Dis}\,\mathcal{E}^*$, continuous in the closure of this set, and take the boundary values 
\begin{center}
 \begin{tabular}{l l}
$h_{t(\varepsilon)}(z)=\begin{cases}
0, & z\in T(t(\varepsilon)),\\
{\rm Dis}\,f^*, & z\in T(s)\cup\partial{\rm Dis}\,\mathcal{E}^*,
\end{cases}$&$\sigma(z)=\begin{cases}
\log (t/t(\varepsilon)), & z\in T(t(\varepsilon)),\\
0, & z\in T(s)\cup\partial{\rm Dis}\,\mathcal{E}^*,
\end{cases}$
\end{tabular}
\end{center}
$$I(h,(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*)=I(b,(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*)$$$$+\varepsilon^2 I(f_{t(\varepsilon)},(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*)+I(\sigma,(T(s),T(t(\varepsilon)))\setminus{\rm Dis}\,\mathcal{E}^*)$$
\begin{equation}-2\varepsilon\int\limits_{T(s)\cup\partial{\rm Dis}\,\mathcal{E}^*}f \frac{\partial b}{\partial n}ds-2\log\frac{t}{t(\varepsilon)}\int\limits_{T(t(\varepsilon))} \frac{\partial b}{\partial n}ds-2\varepsilon\log\frac{t}{t(\varepsilon)}\int\limits_{T(t(\varepsilon))} \frac{\partial f_{t(\varepsilon)}}{\partial n}ds.
\end{equation}
Finally, taking into account (2) and (3), we have 
$$I(v^*,(T(t(\varepsilon)),T(R(\varepsilon))))-I(\tilde{v}^*,(T(t(\varepsilon)),T))=\frac{2\pi(\log t)^2}{\log\frac{R(\varepsilon)}{t(\varepsilon)}}-\frac{2\pi(\log t)^2}{\log\frac{1}{t(\varepsilon)}}$$
\begin{equation}
=2\pi(\log t)^2\left[\frac{1}{O'((\varepsilon\Delta t)^2)-\log t(\varepsilon)}-\frac{1}{-\log t(\varepsilon)}\right]=O'((\varepsilon\Delta t)^2).
\end{equation}

Substituting (5) - (9) into inequality (4), after obvious reductions and $\varepsilon\to 0$ we arrive at the relation:
$$I(f^*,(T(s),T)\setminus\mathcal{E}^*)-I(f,(T(s),T)\setminus{\rm Dis}\,\mathcal{E}^*)\ge I(f^*_t,(T(s),T(t))\setminus\mathcal{E}^*)$$
\begin{equation}
-I(f_t,(T(s),T(t))\setminus{\rm Dis}\,\mathcal{E}^*)-\frac{\Delta t}{\pi}\int\limits_{0}^{2\pi} \frac{\partial f^*}{\partial n}(e^{i\theta})d\theta\int\limits_{T(t)} \left(\frac{\partial f^*_t}{\partial n}-\frac{\partial f_t}{\partial n}\right)ds+O((\Delta t)^2).
\end{equation}
Here the function $f^*_t$    is harmonic in                                         $(T(s),T(t))\setminus \mathcal{E}^*$, continuous in the closure of this set and 
$$f^*_t(z)=\begin{cases}
0, & z\in T(t),\\
f^*(z), & z\in T(s)\cup\partial\mathcal{E}^*.
\end{cases}$$
The function $f_t$    is harmonic in                                 $(T(s),T(t))\setminus {\rm Dis}\,\mathcal{E}^*$, continuous in the closure of this set and 
$$f_t(z)=\begin{cases}
0, & z\in T(t),\\
f(z), & z\in T(s)\cup\partial{\rm Dis}\,\mathcal{E}^*.
\end{cases}$$
According to Kellogg's theorem, the functions  $f^*_t$    and         $f_t$ are twice continuously differentiable in the closure of the corresponding domains. By virtue of Lemma 1, inequality (10) can be rewritten in the form of 
$$\Delta t\int\limits_{T}\left(\frac{\partial f}{\partial n}\right)^2ds-\Delta t\int\limits_{T}\left(\frac{\partial f^*}{\partial n}\right)^2ds\ge \frac{\Delta t}{\pi}\int\limits_{T}\frac{\partial f^*}{\partial n}ds\int\limits_{T}\left(\frac{\partial f_t}{\partial n}-\frac{\partial f^*_t}{\partial n}\right)ds+O((\Delta t)^2).$$
It remains to divide both parts of the last inequality by     $\Delta t$ and pass to the limit at $\Delta t\to 0$           Theorem 1 is proved. 
\section{Another look at the Gonchar-Baernstein conjecture} 
Let $a=(a_1,....,a_n)$ be a set of arbitrary distinct points on the circle  $T$, and  $a^*=(a^*_1,....,a^*_n)$ be a set of symmetric points $a_j^*=\exp(i 2\pi j/n),\;j=1,...,n.$
Let  $K$     be an arbitrary fixed compact set on the half-interval  $(0,1]$. Denote by $\omega(z)$  the harmonic measure of the circle  $T$ with respect to the domain of $D(a,K)=\{z:\;|z|<1\}\setminus\bigcup_{j=1}^n\{z=ta_j:\;t\in K\},$   
and let $\omega^*(z)$         be the harmonic measure of  $T$      with respect to $D(a^*,K)$.  Gonchar suggested that the  inequality 
\begin{equation}
\omega(0)\ge \omega^*(0)
\end{equation}
holds for $K=[t,1],\;0<t<1$                                   (see, for example, [11, Problem 7.45]). The solution of the Gonchar problem is given in paper [1]. This solution is implicitly related to numerous unsolved problems of function theory, where the assumed extremal configuration has n-fold symmetry [3] – [5]. In 1987, Baernstein published a statement of the generalized problem of the validity of inequality (11) for any compact   $K\subset(0,1]$                     [9]. Such a natural generalization caused serious difficulties for experts. Solynin [12] achieved the greatest result by proving (11) in the case of $K=[t_1,t_2],\;0<t_1<t_2\le 1$. Theorem 1 in [8] and the proof of Theorem 1 in this article lead us to the following conjecture about the condenser capacity. Let                          $C=(E_0,E_1)$ be the simplest condenser considered as a pair of non-empty non-intersecting closed sets   $E_0$  and $E_1$     of the plane $\mathbb{C}$ [3, Sec. 1.2]. The capacity ${\rm cap}\,C$ of the condenser $C$ is defined as the greatest lower bound of the integrals $I(v,\mathbb{C}\setminus(E_0\cup E_1))$
over all functions  $v$, continuous in $\overline{\mathbb{C}}$, satisfying the Lipschitz condition locally in $\mathbb{C}\setminus (E_0\cup E_1)$, equal to zero on  $E_0$    and to unity on  $E_1$. Under the condition of the existence of a potential function  $u$   of the condenser  $C$, continuous in      $\overline{\mathbb{C}}$, harmonic in $\mathbb{C}\setminus (E_0\cup E_1)$, equal to zero on  $E_0$ and unity on  $E_1$, the equality 
$${\rm cap}\,C=I(u,\mathbb{C}\setminus (E_0\cup E_1))$$
is fulfilled. Let $K$    be an arbitrary compact set on   $(0,1)$, and let $s$  be an arbitrary number, $0<s<\inf K$. In the notation adopted above, we put 
$$C(s)=(T(s)\cup\bigcup_{j=1}^n\{z=t a_j:\;t\in K\},T),$$
$$C^*(s)=(T(s)\cup\bigcup_{j=1}^n\{z=t a^*_j:\;t\in K\},T).$$
{\bf Conjecture.} For any $s_1,\;s_2,\;0\le s_1<s_2<\inf K$
the inequality 
\begin{equation}
{\rm cap}\,C^*(s_1)-{\rm cap}\,C(s_1)\ge{\rm cap}\,C^*(s_2)-{\rm cap}\,C(s_2)
\end{equation}
is valid.\\

Since the capacity of the condenser does not increase with dissymmetrization [3, Sec. 4.4], then both differences in (12) are non-negative, and our Conjecture (just like Theorem 1 [8]) defines a new level of complexity in this matter. The Hadamard formula [10,(A3.11)] (see also Lemma 1) implies the “differential” form of Conjecture (12):
$$\int\limits_{T(s)}\left(\frac{\partial u}{\partial n}\right)^2ds\ge\int\limits_{T(s)}\left(\frac{\partial u^*}{\partial n}\right)^2ds $$
where $0<s<\inf K$  and $u,\;u^*$  are the potential functions of the condensers $C(s),$ and $C^*(c)$ 
respectively.
\begin{theorem}
If Conjecture {\rm (12)} is true, then Conjecture {\rm (11)} is also true.       
\end{theorem}
\hspace{-0.5cm}{\it Proof.}
Consider the functions $u=\omega-g$ and
$$g(z)=-\frac{w(0)}{\log r}\left(1+\frac{\log r(D(a,K),0)}{\log r}\right)g_{D(a,K)}(z,0)$$
where $g_{D(a,K)}(z,0)$ is the Green's function of the domain $D(a,K),$ and  $r(D(a,K),0)$ is the inner radius of this domain with respect to the origin, and the parameter $r>0$    is sufficiently small (cf. [3, proof of Theorem 2.1]).  The set 
$E(r)=\{z:\;|z|<\sqrt{r},\;u(z)\le 0|\}$                                                          is an "almost disk"\, centered at the origin and radius   $r$. Let $\varphi(s)$            be an infinitesimal quantity at                        $s\to 0$ such that the set $E(r),\;r=s(1+\varphi(s))$                         contains the disk $|z|\le s$. In view of the monotonicity of the capacity [3, Theorem 1.15], 
$${\rm cap}\,C(s)\le {\rm cap}\,(\{z:\;u(z)=0\},\{z:\;u(z)=1\}).$$
The capacity of the last condenser is 
$$I(u,D(a,K)\setminus E(r))=I(\omega,D(a,K)\setminus E(r))+I(g,D(a,K)\setminus E(r))+2\int\limits_{\partial E(r)}g\frac{\partial \omega}{\partial n}ds$$
$$=I(\omega,D(a,K)\setminus E(r))-\frac{2\pi \omega^2(0)}{\log(s(1+\varphi(s)))}+O\left(\left(\frac{1}{\log s}\right)^2\right)$$
$$=I(\omega,D(a,K))-\frac{2\pi \omega^2(0)}{\log s}+O\left(\left(\frac{1}{\log s}\right)^2\right),\;\;\;s\to 0.$$
We have used the formula [3, (2.12)]. Therefore, the inequality   
\begin{equation}
{\rm cap}\,C(s)\le I(\omega,D(a,K))-\frac{2\pi \omega^2(0)}{\log s}+O\left(\left(\frac{1}{\log s}\right)^2\right),\;\;\;s\to 0,
\end{equation}
holds.  Similarly, inequality is shown in the other direction. Thus, in (13) there is an equal sign. Writing this equality for the capacities
$${\rm cap}\,C^*(s^2),\;{\rm cap}\,C(s^2),\;{\rm cap}\,C^*(s)\;\mbox{and}\;{\rm cap}\,C(s)$$
and substituting these asymptotics into inequality (12), we obtain (11). Theorem 2 is proved.
%\end{proof}

\bigskip
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\end{document}

