Remarks on the thin obstacle problem and constrained Ginibre ensembles

TL;DR

This paper links the constrained Ginibre eigenvalue problem to a thin obstacle problem, proving well-posedness in a Sobolev space and advancing understanding of eigenvalue distributions with constraints.

Contribution

It establishes the well-posedness of the obstacle problem in H^1(R^2), improving prior results limited to local Sobolev spaces, and relates eigenvalue constraints to potential theory.

Findings

The obstacle problem is well posed in H^1(R^2).

The coincidence set has two well-separated components.

The results improve previous H^1_{loc} results.

Abstract

We consider the problem of constrained Ginibre ensemble with prescribed portion of eigenvalues on a given curve $\Gamma\subset \mathbb R^2$ and relate it to a thin obstacle problem. The key step in the proof is the $H^1$ estimate for the logarithmic potential of the equilibrium measure. The coincidence set has two components: one in $\Gamma$ and another one in $\mathbb R^2\setminus \Gamma$ which are well separated. Our main result here asserts that this obstacle problem is well posed in $H^1(\mathbb R^2)$ which improves previous results in $H^1_{loc}(\mathbb R^2)$.

Equations

$$\mu (\Gamma )\ge a,a\in (0,1).$$

$$|\dot{\Gamma }(t)|=1,t\in \mathbb{R}.$$

$$I[\mu ]=\int \int \log \frac{1}{|x-y|}d\mu (x)d\mu (y)+\int Qd\mu$$

$$K(x-y)=\left\{\begin{array}[]{lll}\log\frac{1}{|x-y|},&d=2,\\ \frac{1}{|x-y|^{d-2}},&d\geq 3,\end{array}\right.$$

$$I[\mu ]=\int \int K(x-y)d\mu (x)d\mu (y)+\int Qd\mu .$$

$$U^{{\mu }_a}=K\ast {\mu }_a$$

$$\psi(x)=\left\{\begin{array}[]{lll}\frac{1}{2}(A_{\Gamma}-|x|^{2})&\hbox{if}\ x\in\Gamma,\\ \frac{1}{2}(A_{0}-|x|^{2})&\hbox{if}\ x\in{\mathbb{R}}^{2}\setminus\Gamma.\end{array}\right.$$

$$I[{\mu }_a]=\underset{\mu \in {\mathcal{M}}_a}{\inf }I[\mu ].$$

$$\mu =a\frac{1}{L}{\mathcal{H}}^1\text{with}(\Gamma \cap \Omega )+(1-a)\frac{1}{|B|}{\chi }_B$$

$${\int }_{\Omega }\log \frac{1}{|x-y|}d\mu (x)=\frac{1}{L}{\int }_0^L\log \frac{1}{|\Gamma (t)-y|}dt+\frac{1}{|B|}{\int }_B\log \frac{1}{|x-y|}d\mu (x).$$

$$\mathcal{L}[\mu ]=\frac{a^2}{L^2}{\int }_0^L{\int }_0^L\log \frac{1}{|\Gamma (t)-\Gamma (s)|}dtds+\frac{2a(1-a)}{L|B|}{\int }_0^L{\int }_B\log \frac{1}{|\Gamma (t)-y|}dtdy+\frac{(1-a{)}^2}{|B{|}^2}{\int }_B{\int }_B\log \frac{1}{|x-y|}dxdy.$$

$${\int }_{B\rho (z)}\log \frac{1}{|x-y|}dx={\int }_{B_{\rho }(z-y)}\log \frac{1}{|\xi |}d\xi \le {\int }_{B_{2\rho }(0)}\log \frac{1}{|\xi |}dx<\infty$$

$${\int }_0^L\log \frac{1}{|\Gamma (t)-\Gamma (s)|}dt$$

$$=$$

$$|I_0|$$

$$|\Gamma (\tau +s)-\Gamma (s)|$$

$$|I_0|\le \frac{4L^2}{C_{\delta }}+2\delta \left(1-{\delta }^{\alpha }\right)<\infty .$$

$${\int }_0^L{\int }_0^L\log \frac{1}{|\Gamma (t)-\Gamma (s)|}dtds$$

$$\mu (\Gamma )>a+[{\mu }_a(\Gamma )-a]-\epsilon \delta .$$

$$I[(1-\epsilon ){\mu }_a+\epsilon {\mu }_{a-\delta }]$$

$$U^{{\mu }_a}(y)=\int K(x-y)d{\mu }_a(x).$$

$$I[{\mu }_{\epsilon }]$$

$$\int (2U^{{\mu }_a}+Q)d(\nu -{\mu }_a)\ge 0,\forall \nu \in {\mathcal{M}}_a.$$

$$\displaystyle 2U^{\mu_{a}}+Q\begin{array}[]{lll}=A_{\Gamma}&\hbox{on}\ \ \Gamma\cap\operatorname{supp}\mu_{a},\\ \geq A_{\Gamma}&\hbox{on}\ \ \Gamma.\end{array}$$

$$\displaystyle 2U^{\mu_{a}}+Q\begin{array}[]{lll}=A_{0}&\hbox{on}\ \ \operatorname{supp}\mu_{a}\setminus\Gamma,\\ \geq A_{0}&\hbox{on}\ \ {\mathbb{R}}^{2}\setminus(\operatorname{supp}\mu_{a}\setminus\Gamma).\end{array}$$

$$A_{\Gamma }>A_0.$$

$$2U^{{\mu }_a}+Q<A_{\Gamma }-\delta q.e.\; onE$$

$$I[{\mu }_{\epsilon }]$$

$$a{A}_{\Gamma }={\int }_{\Gamma }(2U^{{\mu }_a}+Q)d{\mu }_a\ge a{A}_{\Gamma }$$

$$I[\nu ]>I[{\mu }_a]+⟨DI[{\mu }_a],\nu -{\mu }_a⟩$$

$$0\ge I[\nu ]-I[{\mu }_a]>⟨DI[{\mu }_a],\nu -{\mu }_a⟩$$

Full text

2000 Mathematics Subject Classification. Primary 35R35, 31A35, 49K10, 60B20.

Keywords: Obstacle problem, thin obstacle, free boundary, global regularity.

1. Introduction

Let $\Gamma$ be a regular curve in ${\mathbb{R}}^{2}$ with locally finite length and $\mathcal{M}_{a}$ the set of all probability measures such that

[TABLE]

By an abuse of notation we let $\Gamma:{\mathbb{R}}\to{\mathbb{R}}^{2}$ be the arc-length parametrization of the curve such that

[TABLE]

In this paper we consider the minimizers of the energy

[TABLE]

where $Q(x)$ is a given function such that the weight function $w=e^{-Q}$ on ${\mathbb{R}}^{2}$ is admissible (see Definition 1.1 p.26 [8]). This means that $w$ satisfies the following three conditions:

• (H1)

$w$ is upper semi-continuous;

• (H2)

$\{w\in{\mathbb{R}}^{2}\ s.t.\ w(z)>0\}$ has positive capacity;

• (H3)

$|z|w(z)\to 0$ as $|z|\to\infty$.

In higher dimensions ${\mathbb{R}}^{d},d\geq 3$ one can consider more general kernels

[TABLE]

with $\Gamma$ being a Lyapunov surface in ${\mathbb{R}}^{d}$ and define the energy as follows

[TABLE]

In this note we mostly confine ourselves with quadratic potentials $Q(x)=|x|^{2}$ in ${\mathbb{R}}^{2}$, although all our results remain valid for more general $Q$ satisfying $\bf(H1)-(H3)$. Furthermore, our main result on global $L^{2}$ estimate of the gradient of the equilibrium potential with kernel $K(x-y)=|x-y|^{-d}$ remains valid in in ${\mathbb{R}}^{d},d\geq 3$, see Theorem 4.1.

The functional $I[\mu]$, with $Q=|x|^{2},d=2$, arises in the description of the convergence of the spectral measure of square $N\times N$ matrices with complex independent, standard Gaussian entries (i.e., the Ginibre ensemble) as $N\to\infty$. In case when there are no constraints imposed on the eigenvalues, it is well known that the eigenvalues spread evenly in the ball of radius $\sqrt{N}$, and after renormalization by a factor $\frac{1}{\sqrt{N}}$ the normalized spectral measure converges to the characteristic function of the unit disc. This is known as the circular law [4], [2]. In this context the functional $I$ is used to prove large deviation principles for the spectral measure.

If one demands that the eigenvalues are real (i.e. when $a=1,\Gamma={\mathbb{R}}$) we get the so called semicircle law. More generally, one can demand that a portion of eigenvalues is contained in a prescribed set $\Gamma$. This is considered in [2] when a portion of eigenvalues are contained in an open bounded subset of ${\mathbb{R}}^{2}$ and in [4] when $\Gamma$ is a line. These problems can be related to the thin obstacle and obstacle problems respectively. The key step in proving this is to establish $H^{1}_{loc}({\mathbb{R}}^{2})$ estimates for the logarithmic potential

[TABLE]

of the corresponding equilibrium measure. The aim of this note is to show that the thin obstacle problem is well-posed in $H^{1}({\mathbb{R}}^{2})$ by showing that in fact $U^{\mu_{a}}\in H^{1}({\mathbb{R}}^{2})$, see Theorem 4.1. This improves the previous results in [2] and [4].

The paper is organized as follows: In the next section we prove the existence and uniqueness of the equilibrium measure $\mu_{a}$ minimizing the energy $I[\mu]$. In section 3 we discuss some basic properties of $\mu_{a}$. In particular we show that there are two positive constants $A_{\Gamma}$ and $A_{0}$ such that $2U^{\mu_{a}}+Q=A_{\Gamma}$ on $\operatorname{supp}\mu_{a}\cap\Gamma$ and $2U^{\mu_{a}}+Q=A_{0}$ on $\operatorname{supp}\mu_{a}\setminus\Gamma$. Furthermore, $A_{\Gamma}>A_{0}$. This fact will be used later to show that $\operatorname{supp}\mu_{a}\setminus\Gamma$ and $\operatorname{supp}\mu_{a}\cap\Gamma$ are disjoint.

Our main result Theorem 4.1 is contained in section 4. To prove it we study the Fourier transformations of $U^{\mu_{a}}$ and $\mu_{a}$. It leads to some integral identity involving Bessel functions. This approach is based on a method of L. Carleson [3]. Finally, combining the results obtained, in section 5 we show that $U^{\mu_{a}}$ solves the obstacle problem where the obstacle is given by

[TABLE]

2. Existence of minimizers

In this section we show the existence of a unique equilibrium measure.

Theorem 2.1**.**

Suppose $d=2,\Gamma\subset{\mathbb{R}}^{2}$ is a regular $C^{1,\alpha}$ smooth planar curve without self-intersections. There is a unique minimizer $\mu_{a}\in\mathcal{M}_{a}$ of $I[\mu]$ such that

[TABLE]

**Proof. **Observe that the uniqueness follows from the convexity of $\mathcal{M}_{a}$ and can be proved as in [4]. Moreover, $I[\mu]$ is also semicontinuous. Thus, we have to show that $I[\mu]$ is bounded by below for all $\mu\in\mathcal{M}_{a}$ and there is at least one $\mu_{0}$ such that $I[\mu]$ is finite. The lower bound follows as in the proof of Theorem 1.3 (a) p. 27 [8].

It remains show that the $\inf_{\mu\in\mathcal{M}_{a}}I[\mu]<\infty$. Let $\chi_{D}$ denote the characteristic function of the set $D$ and take

[TABLE]

where $B=B_{\rho}(z)=\{x\in{\mathbb{R}}^{2}\ :\ |x-z|<\rho\}$ with small $\rho$ such that $B\subset\Omega$, $\Omega\subset{\mathbb{R}}^{2}$ is a compact, $L=\mathcal{H}^{1}(\Gamma\cap\Omega)>0$, and $\operatorname{dist}(\Gamma,B)>0$. Observe that for this choice of $\mu$ we have

[TABLE]

Assuming that $\Gamma$ is given by arc-length parametrization we have for the logarithmic energy

[TABLE]

Since $\operatorname{dist}(\Gamma,B)>0$ then the second integral is bounded. As for the last integral then after change of variables $x-y=\xi$ we have

[TABLE]

where we used $|z-y|\leq\rho$ and the fact that $\rho$ is small by construction.

It remains to check that the first integral is finite. Let us fix $s\in[0,L]$ Then we have that

[TABLE]

where $I_{0}$ is the last integral. Using the crude estimate

[TABLE]

because $|\Gamma(\tau+s)-\Gamma(s)|\geq C_{\delta}$ if $|\tau|\geq\delta$. Finally, from $C^{1,\alpha}$ regularity of $\Gamma$ we get

[TABLE]

Combining (2) with (2) we get

[TABLE]

Returning to the first integral in (2.1) we infer

[TABLE]

if we choose $\delta>0$ suitably small. This finishes the proof for $d=2$. ∎

Remark 2.2**.**

If $d\geq 3,Q(x)=|x|^{2}$ then clearly $I[\mu]\geq 0$. The upper estimate for $I[\mu]$ follows from a similar argument if we assume that $\Gamma$ is a Lyapunov surface and take $\mu=a\frac{1}{L}\mathcal{H}^{d-1}\with(\Gamma\cap\Omega)+(1-a)\frac{1}{|B|}\chi_{B}$ with $L=\mathcal{H}^{d-1}(\Gamma\cap\Omega)$ and $\operatorname{dist}(B,\Gamma)>0$. Therefore Theorem 2.1 remains valid for $d\geq 3.$

3. Basic properties of minimizers

In this section we prove some basic properties of the equilibrium measure. The arguments are along the line of those in [2]. Therefore, we mostly focus on those aspects of the proofs which are new or differ essentially. The results to follow are valid in ${\mathbb{R}}^{d},d\geq 2$ unless otherwise stated.

Lemma 3.1**.**

Let $\mu_{a}$ be as in Theorem 2.1. Then $\mu_{a}(\Gamma)=a$.

**Proof. **If the claim fails then $\mu_{a}(\Gamma)>a$. Fix $\delta\in(0,a)$ and let $\mu_{a-\delta}$ be the minimizer of $I[\cdot]$ over $\mathcal{M}_{a-\delta}\supset\mathcal{M}_{a}$. Form $\mu=(1-\varepsilon)\mu_{a}+\varepsilon\mu_{a-\delta},\varepsilon\in[0,1]$. Clearly, $\mu\in\mathcal{M}_{a}$ if we choose $\varepsilon\delta$ sufficiently small because

[TABLE]

Consequently, we have from the strict convexity of $I$

[TABLE]

which is in contradiction with the fact that $\mu_{a}$ is a minimizer. ∎

Observe that the Fréchet derivative of $I[\mu]$ is $2U^{\mu_{a}}+Q$ where

[TABLE]

It is convenient to consider variations of the equilibrium measure in terms of affine combinations. More precisely, let $\mu_{\varepsilon}=(1-\varepsilon)\mu_{a}+\varepsilon\nu,\nu\in\mathcal{M}_{a},\varepsilon\in[0,1]$, then by direct computation we have that

[TABLE]

Since $\mu_{a}$ is the minimizer then $I[\mu_{a}]\leq I[\mu]$, and after sending $\varepsilon\to 0$ it follows that

[TABLE]

Lemma 3.2**.**

Let $A_{\Gamma}=\frac{1}{a}\int_{\Gamma}(2U^{\mu_{a}}+Q)d\mu_{a}$ then quasi everywhere

[TABLE]

Similarly, let us denote $A_{0}=\frac{1}{1-a}\int_{{\mathbb{R}}^{2}\setminus\Gamma}(2U^{\mu_{a}}+Q)d\mu_{a}$ then

[TABLE]

Furthermore,

[TABLE]

**Proof. **We first prove (3.5). Suppose that there is a set capacitable $E$ of positive capacity such that $\Gamma\cap E$ has zero capacity and

[TABLE]

for some positive $\delta.$ Let $\mu_{E}$ be the equilibrium measure of $E$ and form $\nu=\mu_{a}\with({\mathbb{R}}^{2}\setminus\Gamma)+a\mu_{E}$. Clearly $\nu\in\mathcal{M}_{a}$. Therefore, in view of (3) for the measure $\mu_{\varepsilon}=\varepsilon\mu_{a}+(1-\varepsilon)\nu\in\mathcal{M}_{a}$ we get

[TABLE]

if $\varepsilon$ and $\delta$ are sufficiently small. This will be in contradiction with the fact that $\mu_{a}$ is the minimizer. Thus we have proved that $2U^{\mu_{a}}+Q\geq A_{\Gamma}$ q.e. on $\Gamma$.

Next we show that on $\operatorname{supp}\mu_{a}\cap\Gamma$ we have $2U^{\mu_{a}}+Q=A_{\Gamma}$ q.e. Indeed, from the definition of $A_{\Gamma}$ it follows

[TABLE]

where the last inequality follows from the first inequality in (3.5). The proof of (3.8) is similar. In order to prove the last claim $A_{\Gamma}>A_{0}$ we first observe that there exists a measure $\nu\in\mathcal{M}_{a}$ such that

• •

$a>\nu(\Gamma)$,

• •

$I[\nu]\leq I[\mu_{a}]$.

First notice that $\mathcal{M}_{a}\subset\mathcal{M}_{a-\delta}$ for $\delta\in(0,a)$. Fix such $\delta>0$ and let $\mu_{a-\delta}$ be the minimizer of $I[\cdot]$ over $\mathcal{M}_{a-\delta}$. Then by Lemma 3.1 $\mu_{a-\delta}(\Gamma)=a-\delta<a$ and $I[\mu_{a-\delta}]=\inf_{\mathcal{M}_{a-\delta}}I[\mu]\leq I[\mu_{a}]=\inf_{\mathcal{M}_{a}}I[\mu]$. Therefore one can take $\nu=\mu_{a-\delta}$.

From the strict convexity of $I$ it follows that

[TABLE]

where $DI[\mu]=2U^{\mu}+Q$ is the Fréchet derivative of $I[\mu]$. Therefore, from the properties of $\nu$ we infer

[TABLE]

or equivalently

[TABLE]

On the other hand

[TABLE]

while

[TABLE]

This together with (3.12), (3.11) yields

[TABLE]

Finally, the property $\nu(\Gamma)<a$ implies that $A_{\Gamma}>A_{0}$. ∎

Corollary 3.3**.**

$\operatorname{supp}\mu_{a}$* is compact.*

**Proof. **If $d\geq 3$ then $K(x-y)\geq 0$, hence by Lemma 3.2 for $x\in\operatorname{supp}\mu_{a}$ we have

[TABLE]

which is a contradiction. If $d=2$ then from the triangle inequality we get that

[TABLE]

Consequently, for $x\in\operatorname{supp}\mu_{a}$

[TABLE]

for sufficiently large $|x|$, where the last inequality follows from (4.12) and $\int Qd\mu_{a}<I[\mu_{a}]<\infty$. Since $Q=|x|^{2}$ (of for the general case from the hypotheses on $Q$ $\bf(H1)-(H3)$) it again follows that $\operatorname{supp}\mu_{a}$ is bounded. ∎

4. Global $L^{2}$ estimates for $U^{\mu_{a}}$ and $\nabla U^{\mu_{a}}$

Our main result is contained in the following

Theorem 4.1**.**

Let $U^{\mu_{a}}(y)=\int K(x-y)d\mu_{a}$, if $d\geq 3$ then $\nabla U^{\mu_{a}}\in L^{2}({\mathbb{R}}^{d})$. If $d=2$ then $U^{\mu_{a}}\in H^{1}({\mathbb{R}}^{2})$. Furthermore, there holds

[TABLE]

Here $\mathcal{E}[\mu]$ is the energy of $\mu$ defined as $\int\int K(x-y)d\mu(x)d\mu(y)$.

Remark 4.2**.**

It is shown in [3] that $\mathcal{E}[\mu]>0$ for any probability measure $\mu$ and $d\geq 2$. In fact, this can be seen from the proof to follow (see also Corollary 4.3).

**Proof. **The case $d\geq 3$ follows from Lemma 1.6 p. 92 [7] (see also Lemma 17 p. 95), which assert that

[TABLE]

almost everywhere and morover

[TABLE]

The case of the logarithmic potential follows from a modification of the argument by L. Carleson [3] Lemma 3 page 22. We begin with computing the Fourier transformation of $K$. Note that since $\operatorname{supp}\mu_{a}$ is compact we can assume that $K(r)=0$ for $r\geq r_{0}$ for some fixed $r_{0}>0$. We have

[TABLE]

Let us denote $K_{0}(y)=K\left(\frac{y}{2\pi|\xi|}\right)$ and define

[TABLE]

From Lemma 2 p. 21 [3] it follows that there is a universal constant $c_{1}$ such that

[TABLE]

where $J$ is the Bessel function

[TABLE]

Therefore $F(\eta)$ can be further simplified as follows

[TABLE]

because from the definition of $K_{0}$ we have $\operatorname{supp}K_{0}\subset[0,2\pi|\xi|r_{0}]$. Moreover, $K_{0}^{\prime}(r)=-\frac{1}{r}$ hence

[TABLE]

Consequently,

[TABLE]

Next we restrict $\mu_{1}=\mu_{a}\with\mathcal{C}$ where $\mathcal{C}\subset\operatorname{supp}\mu_{a}$ is a compact such that $U^{\mu_{1}}$ is continuous. Observe that $\int U^{\mu_{a}}d\mu_{a}$ is finite hence $U^{\mu_{a}}$ is finite $\mu_{a}$ almost everywhere. By Theorem 1.8 p. 70 [7] for every $\varepsilon>0$ small there is a restriction of $\mu_{a}$ such that

[TABLE]

Note that if $\tau=\mu_{a}-\mu_{1}$ then we have

[TABLE]

Let $\phi_{n}(y)=n^{\frac{d}{2}}e^{-n\pi|y|^{2}}$ be the sequence of normalised Gaussian kernels. It is well-known that $\phi_{n}$ is a mollification kernel for every $n\in\mathbb{N}$ and moreover $\widehat{\phi_{n}}=e^{-\frac{\phi|\xi|^{2}}{n}}$. From the Parseval relation

[TABLE]

If we first send $n\to\infty$ and then $\varepsilon\to 0$ to conclude the identity

[TABLE]

On the other hand $\widehat{U^{\mu_{a}}}=\widehat{K}\widehat{\mu_{a}}$, which yields

[TABLE]

Using the expansion $J(t)=\sum_{s=0}^{\infty}\frac{(-1)^{s}}{(s!)^{2}}\left(\frac{t}{2}\right)^{2s}=1-\frac{t^{2}}{4}+\frac{t^{4}}{64}+\dots$ we see that

[TABLE]

hence the first integral is bounded below by $C(\delta)\frac{1}{r_{0}^{2}c_{1}}\int_{|\xi|<\delta}|\widehat{U^{\mu_{a}}}(\xi)|^{2}d\xi$ for sufficiently small $\delta>0$. As for the second integral, we have

[TABLE]

Combining we see that $\widehat{U^{\mu_{a}}}\in L^{2}({\mathbb{R}}^{2})$ which, after we apply Parseval’s relation again, yields $U^{\mu_{a}}\in L^{2}({\mathbb{R}}^{2})$ and

[TABLE]

To finish the proof we use that $4\pi^{2}|\xi|^{2}|\widehat{U^{\mu_{a}}}|^{2}=|\widehat{\nabla U^{\mu_{a}}}|^{2}$ which together with (4) implies that

[TABLE]

which finishes the proof. ∎

Corollary 4.3**.**

Let $\mu_{a}$ be as in Theorem 2.1. Then there holds

[TABLE]

5. The thin obstacle problem

From the $H^{1}({\mathbb{R}}^{2})$ estimate for $U^{\mu_{a}}$ it follows that $U^{\mu_{a}}$ is a solution to some variational inequality, and hence $U^{\mu_{a}}$ can be interpreted as a solution to an obstacle problem with a combination of both thin (on $\Gamma$) and ”thick” obstacles (on ${\mathbb{R}}^{2}\setminus\Gamma$). It is convenient to define the obstacle as follows

[TABLE]

Lemma 5.1**.**

Let $U^{\mu_{a}}$ be the logarithmic potential of $\mu_{a}$ and define

[TABLE]

Then $U^{\mu_{a}}$ solves the following obstacle problem:

[TABLE]

The proof is the same as in [2].

Corollary 5.2**.**

$\operatorname{dist}(\Gamma,\operatorname{supp}(\mu_{a}\setminus\Gamma))>0$.

**Proof. **This follows from the estimate $A_{\Gamma}>A_{0}$. Indeed, let us assume that $x_{0}\in\Gamma\cap\operatorname{supp}\mu_{a}$ and there is a sequence $\{x_{k}\}_{k=1}^{\infty},x_{k}\in\operatorname{supp}\mu_{a}\setminus\Gamma$ such that $\lim_{k\to\infty}x_{k}\to x_{0}$. Using the lower semicontinuity of $U^{\mu_{a}}$ (see Lemma 1 p.15 [3]) we see that

[TABLE]

Let $\rho>0$ be such that $\{x_{k}\}\subset B_{\rho}(x_{0})$. If $\rho$ is small then $\Gamma$ divides $B_{\rho}(x_{0})$ into two parts $D^{+}$ and $D^{-}$. To fix the ideas let us suppose that $D^{+}$ contains a subsequence $\{x_{k}\}$. Let $h$ be the harmonic function in $D^{+}$ such that $h=\psi$ on $\partial D^{+}$. Observe that $h$ is continuous at $x_{0}$ because $\Gamma\in C^{1,\alpha}$. Since $U^{\mu_{a}}$ is superharmonic and on $\partial D^{+}$ we have $U^{\mu_{a}}\geq\psi=h$ then the comparison principle implies that

[TABLE]

Combining (5.2) and (5.3) we see that $A_{0}\geq A_{\Gamma}$ which is a contradiction in view of (3.9). ∎

From Corollary 5.2 it follows that near $\Gamma$ the potential $U^{\mu_{a}}$ is a solution to a thin obstacle problem in the following sense, see [5] p. 108:

[TABLE]

where $n^{\pm}$ are the outward normals on the $\Gamma$ corresponding to the domains that $\Gamma$ separates. In particular, if $\Gamma$ is $C^{3}$ regular then $U^{\mu_{a}}$ is $C^{1,\alpha}$ up to $\Gamma$ from each of its side, see Theorem 11.4 p.111 [5].

A particular case is $\Gamma={\mathbb{R}}$ [4]. Using a simple symmetrization argument (see e.g. [6] p. 119 Theorem 4.6) we can show that the potential $U^{\mu_{a}}$ is symmetric w.r.t. the real line and hence we get the Signorini problem near ${\mathbb{R}}$ [5] p. 111.

One can make the connections with the obstacle problem more explicit by using the $H^{1}({\mathbb{R}}^{2})$ estimate in Theorem 4.1 and transforming the energy $I[\mu_{a}]$. Let $R>0$ be fixed then using the divergence theorem

[TABLE]

For a.e. $R>0$ the last integral can be estimated as follows

[TABLE]

From Theorem 4.1 and Fubini’s theorem it follows that

[TABLE]

Consequently,

[TABLE]

and we infer from (5) that

[TABLE]

Recalling that by Corollary 3.3 $\operatorname{supp}\mu_{a}\subset B_{r_{0}}$ for some $r_{0}>0$ and using the divergence theorem again we conclude

[TABLE]

Combining these we have that the energy can be rewritten in terms of $U^{\mu_{a}}$ in the following form

[TABLE]