Arcsine equilibrium measure and capacity of a segment

Put ${\mu }_1:={\omega }_{D(0,1)}^0$, the normalized arclength measure on the unit circle of [F5], and let $\phi (t):=\cos t$; define $\nu :={\phi }_{\ast }\lambda$ where $\lambda$ is normalized Lebesgue measure on $[0,\pi ]$, that is, $\nu (E)=\lambda (\{t\in [0,\pi ]:\cos t\in E\})$ for Borel $E\subseteq \mathbb{R}$. By [F6] the set function $\nu$ is a Borel probability measure concentrated on $[-1,1]$, and $\int fd\nu =\frac{1}{\pi }{\int }_0^{\pi }f(\cos t)dt$ for every nonnegative Borel $f$.

The Joukowski factorization. Let $z\in \mathbb{C}$ and let $s$ satisfy $s^2=z^2-1$ with $s=0$ when $z=\pm 1$, which exists by [F7]; put $w_{+}:=z+s$ and $w_{-}:=z-s$, so that $w_{+}+w_{-}=2z$, $w_{+}{w}_{-}=z^2-s^2=1$, and $w^2-2zw+1=(w-w_{+})(w-w_{-})$ for all $w$. Interchanging the two roots if necessary, $\mathrm{\mid }{w}_{+}\mathrm{\mid }\ge 1\ge \mathrm{\mid }{w}_{-}\mathrm{\mid }$, because $\mathrm{\mid }{w}_{+}\mathrm{\mid }\mathrm{\mid }{w}_{-}\mathrm{\mid }=1$.

The arcsine density of $\nu$. Define the nonnegative Borel function $\rho (x):=\frac{1}{\pi \sqrt{1-x^2}}$ for $-1<x<1$ and $\rho (x):=0$ otherwise, and let $\eta :=E\mapsto {\int }_E\rho dx$ be the measure of [F8]. For $-1<x<1$ its distribution function is $F_{\eta }(x)=\frac{1}{\pi }{\int }_0^x\frac{dy}{\sqrt{1-y^2}}=\frac{1}{\pi }\arcsin x$ for $x\ge 0$, and $F_{\eta }(x)=-\frac{1}{\pi }{\int }_x^0\frac{dy}{\sqrt{1-y^2}}=\frac{1}{\pi }\arcsin x$ for $x<0$, the primitive being [F8]; hence $F_{\eta }(x)=\frac{1}{\pi }(\frac{\pi }{2}-\arccos x)$ on $(-1,1)$ with $\eta$ a probability because $\arcsin 1-\arcsin (-1)=\pi$.

For real $t$ with $w=e^{it}$ the identity $z-\cos t=(2zw-w^2-1)\mathrm{/}(2w)=-\frac{(w-w_{+})(w-w_{-})}{2w}$ holds, and taking moduli with $\mathrm{\mid }w\mathrm{\mid }=1$ gives $\mathrm{\mid }z-\cos t\mathrm{\mid }=\frac{1}{2}\mathrm{\mid }{e}^{it}-w_{+}\mathrm{\mid }\mathrm{\mid }{e}^{it}-w_{-}\mathrm{\mid }$; both sides vanish simultaneously, and if $z\mathrm{\notin }[-1,1]$ then $w_{+}\mathrm{\ne }{w}_{-}$ and neither root lies on the unit circle.

The distribution function of $\nu$ of step 1.1 is the same: for $0\le x<1$, $F_{\nu }(x)=\frac{1}{\pi }\mathrm{\mid }\{t\in [0,\pi ]:0<\cos t\le x\}\mathrm{\mid }=\frac{1}{\pi }(\frac{\pi }{2}-\arccos x)$ because $\cos$ decreases on $[0,\pi ]$, and for $-1<x<0$ the same computation of $\{t:\cos t\in (x,0]\}=[\frac{\pi }{2},\arccos x)$ gives $F_{\nu }(x)=\frac{1}{\pi }(\frac{\pi }{2}-\arccos x)$; both distributions equal $-\frac{1}{2}$ for $x\le -1$ and $\frac{1}{2}$ for $x\ge 1$.

The potential of $\nu$. By [F6] and step 2.1, applied to the positive and negative parts of the logarithmic kernel separately, the symmetry of $\cos t$ gives $U^{\nu }(z)=\frac{1}{\pi }{\int }_0^{\pi }\log \frac{1}{\mathrm{\mid }z-\cos t\mathrm{\mid }}dt=\frac{1}{2\pi }{\int }_0^{2\pi }\log \frac{1}{\mathrm{\mid }z-\cos t\mathrm{\mid }}dt=\log 2+U^{{\mu }_1}(w_{+})+U^{{\mu }_1}(w_{-})\mathrm{.}$ Indeed, step 2.1 writes $\log \frac{1}{\mathrm{\mid }z-\cos t\mathrm{\mid }}=\log 2-\log \mathrm{\mid }{e}^{it}-w_{+}\mathrm{\mid }-\log \mathrm{\mid }{e}^{it}-w_{-}\mathrm{\mid }$, and each negative logarithm has circle average equal to the corresponding unit-circle potential.

By step 2.2 the two Borel probability measures $\nu$ and $\eta$ on $\mathbb{R}$, both finite on compacts, have equal distribution functions; by [F9] they coincide, so $\nu$ has the density $\rho (x)=\frac{1}{\pi \sqrt{1-x^2}}$ on $(-1,1)$, as claimed for $[a,b]=[-1,1]$.

Since $\mathrm{\mid }{w}_{+}\mathrm{\mid }\ge 1\ge \mathrm{\mid }{w}_{-}\mathrm{\mid }$, [F5] gives $U^{{\mu }_1}(w_{+})=\log \frac{1}{\mathrm{\mid }{w}_{+}\mathrm{\mid }}$ and $U^{{\mu }_1}(w_{-})=0$, so step 3.1 yields $U^{\nu }(z)=\log 2-\log \mathrm{\mid }{w}_{+}\mathrm{\mid }$ for $z\mathrm{\notin }[-1,1]$; if $z\in [-1,1]$ then $z=\cos \tau$ and $s$ is purely imaginary with $\mathrm{\mid }{w}_{+}\mathrm{\mid }=\mathrm{\mid }{w}_{-}\mathrm{\mid }=1$, so the same identity gives $U^{\nu }(z)=\log 2$. In particular $U^{\nu }=\log 2$ on $[-1,1]$.

Scaling. Let $\alpha :=\frac{a+b}{2}$, $\beta :=\frac{b-a}{2}>0$ and $T(x):=\alpha +\beta x$, so $T$ maps $[-1,1]$ bijectively onto $[a,b]$. Put $\mu :=T_{\ast }\nu$; by [F6] the measure $\mu$ is a Borel probability on $[a,b]$, and for finite positive compactly supported measures the identities $k(Tu,Tv)=k(u,v)-\log \beta$, hence, writing $M:=\rho (\mathbb{C})$, $I(T_{\ast }\rho )=I(\rho )-M^2\log \beta$ and $U^{T_{\ast }\rho }(Tx)=U^{\rho }(x)-M\log \beta$, follow by substituting $\mathrm{\mid }Tu-Tv\mathrm{\mid }=\beta \mathrm{\mid }u-v\mathrm{\mid }$ and the shift convention of [F1]. For probabilities $M=1$, which is the case used in steps 7.1 and 8.1; the density transforms by $dx=d(T^{-1}y)=\frac{dy}{\beta }$ and $\sqrt{1-x^2}=\frac{1}{\beta }\sqrt{(y-a)(b-y)}$, so $d\mu =dy\mathrm{/}\left(\pi \sqrt{(y-a)(b-y)}\right)$ on $(a,b)$.

The energy and minimality. Choose $R>2$; by [F1] and step 4.1 with $z\in [-1,1]$, $I(\nu )=\int U^{\nu }d\nu =\log 2$, a finite real number. For any Borel probability $\sigma$ on $[-1,1]$ with $I(\sigma )<+\mathrm{\infty }$, the support of $\sigma$ lies in $[-1,1]$, so $U^{\nu }=\log 2$ on $\mathrm{supp}\sigma$ by step 4.1, and [F1] gives $I(\nu ,\sigma )=\int U^{\nu }d\sigma =\log 2=I(\nu )$.

By [F4], whose Countable Choice hypothesis is supplied by the Axiom of Choice of the statement, the pair $\nu ,\sigma$ of step 5.1 satisfies $I(\sigma -\nu )=I(\sigma )-2I(\nu ,\sigma )+I(\nu )=I(\sigma )-\log 2\ge 0$, with equality if and only if $\sigma =\nu$; hence every $\sigma \in P([-1,1])$ has $I(\sigma )\ge \log 2=I(\nu )$ with equality only for $\sigma =\nu$, so $V_{[-1,1]}=\log 2$, $\mathrm{cap}([-1,1])=\frac{1}{2}$, and $\nu$ is the unique equilibrium measure of $[-1,1]$, with $U^{\nu }=\log 2$ on $[-1,1]$.

Applying steps 6.1 and 4.1 to $\mu =T_{\ast }\nu$ with the identities of step 4.2 gives $I(\mu )=I(\nu )-\log \beta =\log \frac{2}{\beta }=\log \frac{4}{b-a}$, $U^{\mu }(x)=\log 2-\log \beta =\log \frac{4}{b-a}$ for $x\in [a,b]$, and $U^{\mu }(y)=\log \frac{4}{b-a}-\log \mathrm{\mid }{w}_{+}(z)\mathrm{\mid }$ for $y\mathrm{\notin }[a,b]$ with $z=(2y-a-b)\mathrm{/}(b-a)$ and $w_{+}$ the root of modulus at least $1$ of $w^2-2zw+1=0$.

Every Borel probability $\sigma$ on $[a,b]$ is the pushforward $T_{\ast }\rho$ of the Borel probability $\rho :=T_{\ast }^{-1}\sigma$ on $[-1,1]$, and step 4.2 applied to $\rho$ gives $I(\sigma )=I(\rho )-\log \beta \ge \log 2-\log \beta =\log \frac{4}{b-a}=I(\mu )$, with equality if and only if $\rho =\nu$, that is, if and only if $\sigma =\mu$; hence $V_K=\log \frac{4}{b-a}$, $\mathrm{cap}(K)=e^{-V_K}=\frac{b-a}{4}$, and $\mu$ is the unique equilibrium measure of $K=[a,b]$.

Steps 4.2, 7.1 and 8.1 give the stated density, potential, capacity and uniqueness, with the case $[a,b]=[-1,1]$ recovered by $\alpha =0$, $\beta =1$. The Axiom of Choice enters through [F3], supplies Dependent Choice for [F5], and supplies Countable Choice for [F4] and [F9]; the Joukowski factorization and scaling computation are choice-free.