Fekete–Szegő equality of logarithmic capacity, transfinite diameter, and Chebyshev constant

Let $K\subseteq \mathbb{C}$ be compact and assume the Axiom of Choice. If $K=\mathrm{\varnothing }$, then $\mathrm{cap}(K)=\tau (K)=\mathrm{cheb}(K)=0$ by their empty-set conventions [F1, F3, F5], so the capacity equality holds; the norm and Fekete-sequence assertions require $K$ infinite or $\mathrm{cap}(K)>0$, respectively. For the remaining proof assume $K\mathrm{\ne }\mathrm{\varnothing }$, so that $t_n(K)$ and $V_K$ are defined as well as ${\delta }_n(K)$, $\tau (K)$, $\mathrm{cheb}(K)$ and $\mathrm{cap}(K)$.

For every $n\ge 2$ the set of $n$-point Fekete tuples of $K$ is nonempty by [F1]; the Axiom of Choice yields Countable Choice by [F16], so select a sequence $(z^{(n)}{)}_{n\ge 2}$ with $z^{(n)}=(z_1^{(n)},\dots ,z_n^{(n)})$ an $n$-point Fekete tuple of $K$, and write $F_n(Z)={\prod }_{j=1}^n(Z-z_j^{(n)})$, a monic polynomial of degree $n$ by [F1], with $\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K={\max }_{z\in K}\mathrm{\mid }{F}_n(z)\mathrm{\mid }\in [0,\mathrm{\infty })$ by [F3].

For every nonempty compact metric space $X$, points $z_1,\dots ,z_n\in X$ and continuous $f:X\to \mathbb{R}$, the finite weighted sum $\nu :=\frac{1}{n}{\sum }_{j=1}^n{\delta }_{z_j}$ is a Borel probability measure on $X$ by [F7] and satisfies $\int fd\nu =\frac{1}{n}{\sum }_{j=1}^{n}f(z_j)$; consequently, for every continuous $g:K\times K\to \mathbb{R}$, the iterated integral against ${\nu }_n:=\frac{1}{n}{\sum }_{j=1}^n{\delta }_{z_j^{(n)}}$ evaluates as $\iint gd{\nu }_{n}d{\nu }_n=\frac{1}{n^2}{\sum }_{i=1}^n{\sum }_{j=1}^{n}g(z_i^{(n)},z_j^{(n)})$. Indeed, for every nonnegative measurable $h$ on $X$ one has $\int hd\nu =\frac{1}{n}{\sum }_{j=1}^{n}h(z_j)$: for a nonnegative simple measurable $s={\sum }_k{c}_k{\chi }_{E_k}$ in pairwise disjoint representation, the defining formula $\nu (E)=\frac{1}{n}{\sum }_j{\delta }_{z_j}(E)$ of [F7] and the simple integral of [F21] give $\int sd\nu ={\sum }_k{c}_k\nu (E_k)=\frac{1}{n}{\sum }_j{\sum }_k{c}_k{\chi }_{E_k}(z_j)=\frac{1}{n}{\sum }_{j}s(z_j)$, and for general $h\ge 0$ the increasing simple approximations $s_m\uparrow h$ of [F20] together with [F13] and [F15] give $\int hd\nu ={\lim }_m\int s_{m}d\nu ={\lim }_m\frac{1}{n}{\sum }_j{s}_m(z_j)=\frac{1}{n}{\sum }_{j}h(z_j)$; a continuous real $f$ is bounded by [F17] and Borel measurable by [F22], so $\int \mathrm{\mid }f\mathrm{\mid }d\nu \le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\nu (X)=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}<+\mathrm{\infty }$ by [F22] and $f$ is $\nu$-integrable by [F21]; its positive and negative parts are nonnegative bounded Borel by [F21], so $\int fd\nu =\int f^{+}d\nu -\int f^{-}d\nu =\frac{1}{n}{\sum }_j{f}^{+}(z_j)-\frac{1}{n}{\sum }_j{f}^{-}(z_j)=\frac{1}{n}{\sum }_{j}f(z_j)$; finally the slices $w\mapsto g(z_i^{(n)},w)$ and $z\mapsto g(z,z_j^{(n)})$ of the continuous $g$, and the finite sum $z\mapsto \frac{1}{n}{\sum }_{j=1}^{n}g(z,z_j^{(n)})$, are continuous by [F22], so this formula applied on the compact metric space $K$ to the continuous functions $w\mapsto g(z_i^{(n)},w)$ and $z\mapsto \frac{1}{n}{\sum }_{j=1}^{n}g(z,z_j^{(n)})$ gives $\int g(z_i^{(n)},w)d{\nu }_n(w)=\frac{1}{n}{\sum }_{j}g(z_i^{(n)},z_j^{(n)})$ for each $i$ and hence $\iint gd{\nu }_{n}d{\nu }_n=\frac{1}{n}{\sum }_i\frac{1}{n}{\sum }_{j}g(z_i^{(n)},z_j^{(n)})$.

If $K$ is finite, say $K=\{a_1,\dots ,a_m\}$ with $m\ge 1$, then for every $n\ge m$ the polynomial $G_n(Z):=Z^{n-m}{\prod }_{k=1}^m(Z-a_k)$ is monic of degree $n$ and vanishes on $K$, so $t_n(K)=0$ by [F3]; the infimum defining $\mathrm{cheb}(K)$ is therefore $0$, and $\mathrm{cheb}(K)=0\le \tau (K)$ because $\tau (K)\ge 0$ by [F1].

Assume for the rest of this phase that $K$ is infinite, and fix $n\ge 2$. Then ${\delta }_n(K)>0$: since $K$ contains $n$ distinct points, the tuple of those points has $D_n>0$, so the maximum in [F1] is positive.

For every $z\in K$ the $(n+1)$-tuple $(z,z_1^{(n)},\dots ,z_n^{(n)})$ satisfies $D_{n+1}(z,z_1^{(n)},\dots ,z_n^{(n)})=\mathrm{\mid }{F}_n(z)\mathrm{\mid }{D}_n(z_1^{(n)},\dots ,z_n^{(n)})$, because the pairs not involving $z$ reproduce the Vandermonde product of $z^{(n)}$ and the pairs involving $z$ reproduce $\mathrm{\mid }{F}_n(z)\mathrm{\mid }={\prod }_j\mathrm{\mid }z-z_j^{(n)}\mathrm{\mid }$ by the multiplicativity in [F19]; equivalently $D_{n+1}(z,z^{(n)})=\mathrm{\mid }{F}_n(z)\mathrm{\mid }{\delta }_n(K{)}^{\left(\genfrac{}{}{0px}{}{n}{2}\right)}$ by [F1].

Since $F_n$ is monic of degree $n$, [F1] and [F3] give $t_n(K)\le \mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K$ and $\mathrm{cheb}(K)\le t_n(K{)}^{1\mathrm{/}n}\le \mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K^{1\mathrm{/}n}$ for every $n\ge 2$; hence $\mathrm{cheb}(K)\le {\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K^{1\mathrm{/}n}$ by [F15].

If $K=\mathrm{\varnothing }$ then $\mathrm{cap}(K)=0=\mathrm{cheb}(K)$ by the conventions of [F3] and [F5]; if $K\mathrm{\ne }\mathrm{\varnothing }$, [F6] gives $\mathrm{cap}(K)\le \mathrm{cheb}(K)$. Thus $\mathrm{cap}(K)\le \mathrm{cheb}(K)$ for every compact $K$.

The finite sums ${\sum }_i{f}_i(z)h_i(w)$ of continuous real functions on $K$ form a unital subalgebra of $C(K\times K,\mathbb{R})$ that separates points, so they are uniformly dense in $C(K\times K,\mathbb{R})$ by [F11], since $K\times K$ is a nonempty compact metric space by [F17].

Fix $R>\mathrm{diam}K$ and $M>0$ and put $g_M:=\min (M,k_R)$ on $K\times K$. By [F4] the function $k_R=k+\log R$ is nonnegative on $K\times K$ with diagonal value $+\mathrm{\infty }$, so $g_M$ is continuous, $0\le g_M\le M$, $g_M\le k_R$, and $g_M(z,z)=M$ for every $z\in K$.

Let $L\subseteq \mathbb{C}\setminus K$ be nonempty and compact. The kernel $k$ is continuous on $L\times K$ and uniformly continuous there. Indeed $(z,w)\mapsto \mathrm{\mid }z-w\mathrm{\mid }$ is continuous and never vanishes on $L\times K$, because $L\cap K=\mathrm{\varnothing }$, so by [F17] its modulus attains a minimum $m>0$ on the nonempty compact set $L\times K$; the functions $\mathrm{\mid }z-w\mathrm{\mid }\mapsto \frac{1}{\mathrm{\mid }z-w\mathrm{\mid }}$ and $t\mapsto \log t$ are continuous on their domains and composition preserves continuity by [F19], so $k=-\log \mathrm{\mid }z-w\mathrm{\mid }$ is continuous on $L\times K$; uniform continuity follows from [F18].

The tuple $(z,z^{(n)})$ lies in $K^{n+1}$, so [F1] bounds its $D_{n+1}$ by ${\delta }_{n+1}(K{)}^{\left(\genfrac{}{}{0px}{}{n+1}{2}\right)}$; dividing step 1.6 by ${\delta }_n(K{)}^{\left(\genfrac{}{}{0px}{}{n}{2}\right)}>0$ (step 1.5) and taking the supremum over $z\in K$ gives $\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K\le {\delta }_{n+1}(K{)}^{\left(\genfrac{}{}{0px}{}{n+1}{2}\right)}{\delta }_n(K{)}^{-\left(\genfrac{}{}{0px}{}{n}{2}\right)}$; taking nonnegative $n$-th roots by [F14] and using ${\delta }_{n+1}(K)\le {\delta }_n(K)$ from [F2] gives $\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K^{1\mathrm{/}n}\le ({\delta }_{n+1}(K)\mathrm{/}{\delta }_n(K){)}^{n\mathrm{/}2}\sqrt{{\delta }_{n+1}(K){\delta }_n(K)}\le {\delta }_n(K)$.

Assume $\tau (K)>0$ for this phase. Then ${\delta }_n(K)\ge \tau (K)>0$ for every $n\ge 2$ by [F1], so no Fekete tuple of step 1.2 has a repeated entry and the sum $E_n:={\sum }_{i<j}k(z_i^{(n)},z_j^{(n)})=\left(\genfrac{}{}{0px}{}{n}{2}\right)\log \frac{1}{{\delta }_n(K)}$ is finite by [F4]; moreover $\log \frac{1}{{\delta }_n(K)}\to \log \frac{1}{\tau (K)}$, since ${\delta }_n(K)\to \tau (K)>0$ and $t\mapsto \log \frac{1}{t}$ is continuous on $(0,\mathrm{\infty })$ by [F19].

Put ${\nu }_n:=\frac{1}{n}{\sum }_{j=1}^n{\delta }_{z_j^{(n)}}\in P(K)$ for $n\ge 2$; these are Borel probability measures on $K$ by [F7], so by [F9], whose Axiom of Choice hypothesis is part of the Given, there are a strictly increasing sequence $n_k$ and a measure $\widehat{\nu }\in P(K)$ with ${\nu }_{n_k}\Rightarrow \widehat{\nu }$.

Since ${\delta }_n(K)\to \tau (K)$ by [F2], step 2.1 and [F15] give ${\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K^{1\mathrm{/}n}\le \tau (K)$.

For a finite sum $h={\sum }_i{f}_i{h}_i$ and every Borel probability $\nu$ on $K$, [F12] factorizes the iterated integral, $\iint hd\nu d\nu ={\sum }_i(\int f_{i}d\nu )(\int h_{i}d\nu )$; applying this to $\nu ={\nu }_{n_k}$ and to $\nu =\widehat{\nu }$ and using the weak convergence of step 2.3 gives $\iint hd{\nu }_{n_k}d{\nu }_{n_k}\to \iint hd\widehat{\nu }d\widehat{\nu }$.

For every $n\ge 2$, step 1.3 with $g=g_M$ applied to the Fekete tuple $z^{(n)}$, the diagonal values $g_M(z_i^{(n)},z_i^{(n)})=M$ of step 1.10, the bound $g_M\le k_R$ off the diagonal and the Fekete identity ${\sum }_{i<j}{k}_R(z_i^{(n)},z_j^{(n)})=\left(\genfrac{}{}{0px}{}{n}{2}\right)(\log \frac{1}{{\delta }_n(K)}+\log R)$ of step 2.2 give $\iint g_{M}d{\nu }_{n}d{\nu }_n\le \frac{1}{n^2}(2\left(\genfrac{}{}{0px}{}{n}{2}\right)\log \frac{1}{{\delta }_n(K)}+2\left(\genfrac{}{}{0px}{}{n}{2}\right)\log R+nM)=\frac{n-1}{n}(\log \frac{1}{{\delta }_n(K)}+\log R)+\frac{M}{n}$.

Assume $\mathrm{cap}(K)>0$ and let $L\subseteq \mathbb{C}\setminus K$ be nonempty and compact. For $z\mathrm{\notin }K$ and $n\ge 2$ one has $\mathrm{\mid }{F}_n(z)\mathrm{\mid }={\prod }_j\mathrm{\mid }z-z_j^{(n)}\mathrm{\mid }>0$ by [F19], so the potential of ${\nu }_n$ from step 2.3 is $U^{{\nu }_n}(z)=\frac{1}{n}{\sum }_j\log \frac{1}{\mathrm{\mid }z-z_j^{(n)}\mathrm{\mid }}=\frac{1}{n}\log \frac{1}{\mathrm{\mid }{F}_n(z)\mathrm{\mid }}$ by [F4], step 1.3 and the product law for the logarithm in [F19]; hence $\mathrm{\mid }{F}_n(z){\mathrm{\mid }}^{1\mathrm{/}n}=\exp (-U^{{\nu }_n}(z))$ by the identity $a^{1\mathrm{/}n}=\exp (\frac{1}{n}\log a)$ and the inverse relation between $\exp$ and $\log$ recorded in [F19].

Steps 3.1 and 1.7 give $\mathrm{cheb}(K)\le \tau (K)$ for infinite $K$; with step 1.4 this inequality holds for every compact $K$.

Hence for every bounded continuous $g$ on $K\times K$ one has $\iint gd{\nu }_{n_k}d{\nu }_{n_k}\to \iint gd\widehat{\nu }d\widehat{\nu }$: given $\epsilon >0$, step 1.9 supplies a finite sum $h$ with $\mathrm{\mid }g-h\mathrm{\mid }\le \epsilon$ on $K\times K$, step 3.2 gives convergence of the $h$ integrals, and the triangle inequality bounds the difference of the $g$ integrals by $2\epsilon +\mid \iint hd{\nu }_{n_k}d{\nu }_{n_k}-\iint hd\widehat{\nu }d\widehat{\nu }\mid$.

Letting $k\to \mathrm{\infty }$ in steps 4.2 and 3.3 and using step 2.2 gives $\iint g_{M}d\widehat{\nu }d\widehat{\nu }\le \log \frac{1}{\tau (K)}+\log R$.

As $M\to \mathrm{\infty }$ one has $g_M\uparrow k_R$ pointwise, so [F13] applied to the nonnegative functions $g_M$ gives $\iint k_{R}d\widehat{\nu }d\widehat{\nu }={\lim }_{M\to \mathrm{\infty }}\iint g_{M}d\widehat{\nu }d\widehat{\nu }\le \log \frac{1}{\tau (K)}+\log R$; by [F4], whose shift formula applies since $\widehat{\nu }\in P(K)$ has total mass one and support in $K$ with $\mathrm{diam}K<R$, the left side equals $I(\widehat{\nu })+\log R$, so $I(\widehat{\nu })\le \log \frac{1}{\tau (K)}$.

Since $\widehat{\nu }\in P(K)$, the minimality in [F5] gives $V_K\le I(\widehat{\nu })\le \log \frac{1}{\tau (K)}<+\mathrm{\infty }$; hence $V_K<+\mathrm{\infty }$, $\mathrm{cap}(K)=e^{-V_K}>0$ and, by [F19], $\log \frac{1}{\mathrm{cap}(K)}=V_K\le \log \frac{1}{\tau (K)}$, that is $\tau (K)\le \mathrm{cap}(K)$.

If $\tau (K)=0$ then steps 4.1 and 1.8 give $\mathrm{cap}(K)\le \mathrm{cheb}(K)\le \tau (K)=0$, while $\mathrm{cap}(K)\ge 0$ by [F5]; if $\tau (K)>0$ then step 7.1 gives $\tau (K)\le \mathrm{cap}(K)$ and steps 4.1 and 1.8 give $\mathrm{cap}(K)\le \mathrm{cheb}(K)\le \tau (K)$; in both cases $\mathrm{cap}(K)=\tau (K)=\mathrm{cheb}(K)$.

Assume $\mathrm{cap}(K)>0$. By step 8.1, $\tau (K)=\mathrm{cap}(K)>0$, so the hypotheses of phase 4 hold and the argument of steps 4.2, 3.3, 5.1, 6.1 and 7.1 applies to any subsequence of $({\nu }_n)$ in place of $({\nu }_{n_k})$, because only the Fekete property of each $z^{(n)}$ and the limit $\log \frac{1}{{\delta }_n(K)}\to \log \frac{1}{\tau (K)}$ of step 2.2 are used: if $\widehat{\nu }\in P(K)$ is a weak limit of a subsequence of $({\nu }_n)$, then $I(\widehat{\nu })\le \log \frac{1}{\tau (K)}=V_K$ by steps 6.1, 7.1 and 8.1, while $V_K\le I(\widehat{\nu })$ by [F5]; hence $I(\widehat{\nu })=V_K$, and the uniqueness part of [F10] gives $\widehat{\nu }={\mu }_K$.

If $K$ is infinite, then steps 3.1 and 1.7 with step 8.1 give $\mathrm{cheb}(K)\le {\mathrm{lim\hspace{0.17em}inf}}_n\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K^{1\mathrm{/}n}\le {\mathrm{lim\hspace{0.17em}sup}}_n\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K^{1\mathrm{/}n}\le \tau (K)=\mathrm{cheb}(K)$, so ${\lim }_n\mathrm{\parallel }{F}_n{\mathrm{\parallel }}_K^{1\mathrm{/}n}=\mathrm{cheb}(K)=\mathrm{cap}(K)$; since steps 3.1 and 1.7 use only the Fekete property of each $z^{(n)}$ and the definitions, this limit is the same for every sequence of $n$-point Fekete tuples.

Consequently ${\nu }_n\Rightarrow {\mu }_K$. Indeed, for every bounded continuous real $f$ the sequence $a_n:=\int fd{\nu }_n$ is bounded, and every subsequence of $(a_n)$ has a sub-subsequence converging to $\int fd{\mu }_K$: the corresponding subsequence of $({\nu }_n)$ has a weakly convergent sub-subsequence by [F9], and its limit is ${\mu }_K$ by step 9.1, so the integrals converge by [F8]. A bounded real sequence all of whose subsequences have a sub-subsequence with the same limit $L$ converges to $L$; hence $a_n\to \int fd{\mu }_K$, which is weak convergence.

Let $\epsilon >0$ and let $\delta >0$ satisfy the uniform continuity of step 1.11 for the tolerance $\epsilon$. By compactness of $L$ and [F17] there are finitely many $z_1,\dots ,z_r\in L$ with $L\subseteq {\bigcup }_{i\le r}B(z_i,\delta )$; for each $i$ the function $w\mapsto k(z_i,w)$ is bounded and continuous on $K$, so step 10.1 gives $N$ with $\mid \int k(z_i,w)d({\nu }_n-{\mu }_K)(w)\mid <\epsilon$ for all $n\ge N$ and all $i\le r$. For $z\in L$ choose $i$ with $\mathrm{\mid }z-z_i\mathrm{\mid }<\delta$; then $\mathrm{\mid }{U}^{{\nu }_n}(z)-U^{{\mu }_K}(z)\mathrm{\mid }\le \epsilon +\epsilon +\epsilon$, because the two outer terms are bounded by $\epsilon$ from the uniform continuity of step 1.11 and the middle term is the displayed integral.

Both $U^{{\nu }_n}$ and $U^{{\mu }_K}$ take values in the bounded interval $[-C,C]$, where $C:={\max }_{L\times K}\mathrm{\mid }k\mathrm{\mid }<+\mathrm{\infty }$ exists by [F17] and [F19]; the exponential is continuous and therefore uniformly continuous on $[-C,C]$ by [F18] and [F19], so step 11.1 gives ${\sup }_{z\in L}\mid \exp (-U^{{\nu }_n}(z))-\exp (-U^{{\mu }_K}(z))\mid \to 0$.

By step 3.4 the left-hand function is $\mathrm{\mid }{F}_n(z){\mathrm{\mid }}^{1\mathrm{/}n}$ for $z\in L$, so step 12.1 is exactly the asserted uniform exterior limit ${\sup }_{z\in L}\mid \mathrm{\mid }{F}_n(z){\mathrm{\mid }}^{1\mathrm{/}n}-\exp (-U^{{\mu }_K}(z))\mid \to 0$.

Assembly: step 8.1 proves $\mathrm{cap}(K)=\tau (K)=\mathrm{cheb}(K)$ for every compact $K$; step 9.2 proves the Fekete-polynomial norm limit for infinite $K$; step 10.1 proves weak convergence of the empirical measures to ${\mu }_K$ when $\mathrm{cap}(K)>0$; and step 13.1 proves the uniform exterior limit.