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Logarithmic Potential, Capacity, and Riesz Decomposition: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These examples compute the page's central objects in settings where the formulas can be checked explicitly. Normalized arclength on a circle is the unique equilibrium measure of the closed disc, with constant potential $\log (1 / r)$ and capacity $r$ ; pushing the uniform angle measure through $x = \cos θ$ gives the arcsine equilibrium measure of $[- 1, 1]$ and capacity $(b - a) / 4$ . The unit disc also exhibits the exact Fekete configuration: the $n$ -th roots of unity maximize the Vandermonde product, with $δ_{n} = n^{1 / (n - 1)}$ and Fekete polynomial $F_{n} (z) = z^{n} - 1$ , while the Chebyshev columns describe the corresponding extremal nodes and alternating extrema on $[- 1, 1]$ in terms of the arcsine measure.

The remaining examples separate capacity from more familiar notions. Every finite or countable set is capacity-polar, and on the real line two compact null sets can have positive and zero logarithmic capacity; a Cantor measure with energy at most $3 \log 3$ gives capacity at least $1 / 27$ , while a thin Cantor set whose every probability has infinite energy has capacity zero. For a holomorphic function, the Riesz measure of $\log ∣ f ∣$ is the weighted zero divisor. Finally, the Green function with pole at infinity of the exterior of a circular conductor is $\log (∣ z - a ∣ / r)$ in the stated normalization.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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Capacity of a disc and its circular equilibrium measure

Statement

Assume the Axiom of Choice. Let $a \in C$ , $r > 0$ and let $K : = \overline{D (a, r)}$ be the closed disc. Let $μ$ be normalized arclength on the circle $∣ z - a ∣ = r$ , that is, in the parametrization $w = a + r e^{i t}$ , $d μ = d t / (2 π)$ . Then $μ$ is the unique equilibrium measure of $K$ ,

$U^{μ} (z) = \log \frac{1}{r} (∣ z - a ∣ \leq r), U^{μ} (z) = \log \frac{1}{∣ z - a ∣} (∣ z - a ∣ \geq r),$

and $cap (K) = r$ , with Robin constant $V_{K} = \log (1 / r)$ . The same potential, capacity and equilibrium measure hold for the boundary circle $\partial K = {z : ∣ z - a ∣ = r}$ .

The Axiom of Choice is inherited from the equilibrium framework and supplies Countable Choice for the strict positivity of the zero-mass energy; the calculation of the potential itself is choice-free.

Facts & Assumptions

Given: a point $a \in C$ , a radius $r > 0$ , the closed disc $K = \overline{D (a, r)}$ , its boundary circle $\partial K$ , the logarithmic kernel and potential conventions of Logarithmic potential and energy of a positive compactly supported measure, the Robin constant and capacity of Robin constant and logarithmic capacity of a compact set, and the Axiom of Choice (The Axiom of Choice).

[F1]

$U^{ν} (z) = \int \log \frac{1}{∣ z - w ∣} d ν (w) \in (- \infty, + \infty]$ for finite positive Borel $ν$ of compact support; for $R > diam supp ν$ one has $k_{R} = k + \log R \geq 0$ on the product of the support with itself, $k_{R} = k + \log R$ pointwise as extended functions, and $I (ν) = \iint k_{R} d ν d ν - ν (C)^{2} \log R$ , independently of $R$ ; the mixed energy $I (ν, ρ) = \iint k d ν d ρ$ is symmetric (Logarithmic potential and energy of a positive compactly supported measure).

[F2]

For nonempty compact $F$ , $V_{F} = \inf_{ρ \in P (F)} I (ρ)$ and $cap (F) = e^{- V_{F}}$ when $V_{F} < + \infty$ and $0$ otherwise (Robin constant and logarithmic capacity of a compact set); a Borel probability measure on $F$ is a finite positive measure carried by $F$ (Probability measures and probability spaces).

[F3]

Assume the Axiom of Choice. A compact nonpolar $F$ has exactly one equilibrium measure, namely the unique $ρ \in P (F)$ with $I (ρ) = V_{F}$ (Existence and uniqueness of the equilibrium measure).

[F4]

Assume Countable Choice. If $ν, ρ$ are finite positive compactly supported Borel measures with equal total mass and finite energy, then $I (ν, ρ)$ is finite, $I (ν - ρ) = I (ν) - 2 I (ν, ρ) + I (ρ)$ is a real number, $I (ν - ρ) \geq 0$ , and $I (ν - ρ) = 0$ if and only if $ν = ρ$ (Strict positivity of logarithmic energy for a zero-mass signed charge). The Axiom of Choice implies Countable Choice (AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F5]

Assume Dependent Choice, supplied by the Axiom of Choice of the statement (AC implies DC implies countable choice). For $c \in C$ , $R > 0$ the harmonic measure $ω_{D (c, R)}^{c}$ of the disc at its centre has, on Borel $E \subseteq \partial D (c, R)$ , the form $ω_{D (c, R)}^{c} (E) = \frac{1}{2 π} \int_{{t \in [0, 2 π) : c + R e^{i t} \in E}} d t$ , so it is the normalized arclength measure on the circle, a Borel probability measure on $\partial D (c, R)$ , and for every Borel $f \geq 0$ , $\int f d ω_{D (c, R)}^{c} = \frac{1}{2 π} \int_{0}^{2 π} f (c + R e^{i t}) d t$ (Poisson density of harmonic measure on a disc, Harmonic measure on a bounded regular plane domain).

[F6]

Every plane harmonic function satisfies the circle mean-value property (Plane harmonic functions satisfy the mean-value property, The circle and disc mean-value properties); the function $z \mapsto \log ∣ z - c ∣$ is $C^{\infty}$ and harmonic on $C ∖ {c}$ (Logarithmic modulus is harmonic off its centre, Plane harmonic functions).

[F7]

Jensen's formula: if $f$ is holomorphic on a neighbourhood of the closed unit disc, $f (0) \neq 0$ , and $a_{1}, \dots, a_{N}$ are the zeros of $f$ in $∣ z ∣ < 1$ counted with multiplicity, while $f$ has no zero on $∣ z ∣ = 1$ , then $\log ∣ f (0) ∣ = \frac{1}{2 π} \int_{0}^{2 π} \log ∣ f (e^{i t}) ∣ d t - \sum_{k} \log \frac{1}{∣ a_{k} ∣}$ . Only this boundary-zero-free case is used below (Jensen's formula on a disc).

[F8]

The complex exponential satisfies $∣ e^{i t} ∣ = 1$ for real $t$ , so $∣ a + r e^{i t} - a ∣ = r$ and $t \mapsto a + r e^{i t}$ parametrizes $\partial K$ (The complex exponential by its power series, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ); for each $c \in C$ the polynomial $ζ \mapsto r ζ - c$ is entire (Complex polynomials are entire with the power-rule derivative, and rational functions are holomorphic wherever their denominator is nonzero).

Verification

technique · direct

1.1F5F2given

Put $μ : = ω_{D (a, r)}^{a}$ , the harmonic measure of the disc $K^{\circ} = D (a, r)$ at its centre. By [F5] the measure $μ$ is a Borel probability measure carried by $\partial K \subseteq K$ , and for every Borel $f \geq 0$ one has $\int f d μ = \frac{1}{2 π} \int_{0}^{2 π} f (a + r e^{i t}) d t$ ; in particular $μ$ has no atoms, since for a single point $w$ the set ${t \in [0, 2 π) : a + r e^{i t} = w}$ has at most two elements and Lebesgue measure zero, so $μ ({w}) = 0$ .

1.2F6algebra

The circle average of the kernel. For $c \in C$ put $M (c) : = \frac{1}{2 π} \int_{0}^{2 π} \log ∣ c - r e^{i t} ∣ d t \in [- \infty, \infty)$ . If $∣ c ∣ > r$ , then $z \mapsto \log ∣ z - c ∣$ is harmonic on an open set containing the closed disc $\overline{D (0, r)}$ , so the circle mean-value property of [F6] gives $M (c) = \log ∣ 0 - c ∣ = \log ∣ c ∣$ .

1.3F7F8algebra

If $0 < ∣ c ∣ < r$ , apply Jensen's formula [F7] on the unit disc to the entire function $f (ζ) : = r ζ - c$ , which satisfies $f (0) = - c \neq 0$ and has the single zero $ζ_{0} = c / r$ of modulus $< 1$ : $\frac{1}{2 π} \int_{0}^{2 π} \log ∣ r e^{i t} - c ∣ d t = \log ∣ f (0) ∣ + \log \frac{1}{∣ ζ_{0} ∣} = \log ∣ c ∣ + \log \frac{r}{∣ c ∣} = \log r$ , that is, $M (c) = \log r$ .

2.1step 1.3F8algebra

If $c = 0$ , the integrand defining $M (c)$ is constantly $\log r$ , so $M (0) = \log r$ . If $∣ c ∣ = r$ , rotate the angle to write $c = r$ without changing the average. For $1 / 2 \leq s < 1$ , step 1.3 gives $M (s r) = \log r$ , and $∣ e^{i t} - s ∣^{2} = (1 - s)^{2} + 4 s \sin^{2} (t / 2) \geq 2 \sin^{2} (t / 2)$ . The positive part of $\log ∣ r (e^{i t} - s) ∣$ is bounded by $\log^{+} (2 r)$ ; its negative part is bounded by $∣ \log (r \sqrt{2}) ∣ + \log^{-} ∣ \sin (t / 2) ∣$ . This last function is integrable on $[0, 2 π]$ : $(\sin)^{'} (0) = 1$ (The derivatives of sine and cosine are cosine and minus sine) gives $\sin v \geq v / 2$ for small positive $v$ , the same bound applies near $t = 2 π$ , and away from the endpoints the sine has a positive minimum. Thus its only singularities are bounded by constants plus $- \log t$ or $- \log (2 π - t)$ , both integrable. Dominated convergence (Dominated convergence) along $s ↑ 1$ yields $M (r) = \log r$ , and rotation gives $M (c) = \log r$ for every $∣ c ∣ = r$ .

3.1step 1.2step 1.3step 2.1F1F5algebra

Consequently, for $z \in C$ and $c : = z - a$ , the substitution $w = a + r e^{i t}$ and [F5] give $U^{μ} (z) = \int \log \frac{1}{∣ z - w ∣} d μ (w) = \frac{1}{2 π} \int_{0}^{2 π} \log \frac{1}{∣ z - a - r e^{i t} ∣} d t = - M (z - a)$ ; by steps 1.2, 1.3 and 2.1 this is $\log \frac{1}{r}$ when $∣ z - a ∣ \leq r$ and $\log \frac{1}{∣ z - a ∣}$ when $∣ z - a ∣ \geq r$ .

4.1step 1.1step 3.1F1algebra

The energy of $μ$ . Choose $R > 2 r = diam (\partial K)$ ; by [F1], $k_{R} = k + \log R \geq 0$ on $\partial K \times \partial K$ and $k_{R} = k + \log R$ pointwise, so the iterated integral of $k_{R}$ against $μ \otimes μ$ equals $\int U^{μ} d μ + \log R$ ; since $supp μ = \partial K$ and $U^{μ} = \log \frac{1}{r}$ there by step 3.1 with $∣ z - a ∣ = r$ , [F1] gives $I (μ) = \int U^{μ} d μ = \log \frac{1}{r}$ , a finite real number.

5.1step 3.1step 4.1F1F2

Let $σ \in P (K)$ be a Borel probability measure on $K$ with $I (σ) < + \infty$ ; then $σ$ is a finite positive compactly supported measure of total mass $1$ , and since $supp σ \subseteq K$ step 3.1 gives $U^{μ} = \log \frac{1}{r}$ on $supp σ$ , so by [F1] the mixed energy is $I (μ, σ) = \int U^{μ} d σ = \log \frac{1}{r} = I (μ)$ .

6.1step 5.1F2F4given

By [F4], whose Countable Choice hypothesis is supplied by the Axiom of Choice of the statement, the pair $μ, σ$ of step 5.1 satisfies $I (σ - μ) = I (σ) - 2 I (μ, σ) + I (μ) = I (σ) - \log \frac{1}{r} \geq 0$ , with equality if and only if $σ = μ$ ; hence every $σ \in P (K)$ with finite energy has $I (σ) \geq \log \frac{1}{r} = I (μ)$ , with equality only for $σ = μ$ , while $σ \in P (K)$ with $I (σ) = + \infty$ also satisfies $I (σ) \geq I (μ)$ since $I (μ)$ is finite. Therefore $V_{K} = \inf_{σ \in P (K)} I (σ) = I (μ) = \log \frac{1}{r}$ , and $μ$ is the unique minimizer.

7.1step 6.1F2F3

By step 6.1 the unique minimizer of the energy over $P (K)$ is $μ$ , so [F3] identifies $μ$ as the equilibrium measure of $K$ and shows it is the only one; the capacity is $cap (K) = e^{- V_{K}} = e^{- \log (1 / r)} = r$ .

8.1step 5.1step 6.1step 7.1F1F2F3F5

The boundary circle. The circle $\partial K$ is compact and nonempty and $P (\partial K) \subseteq P (K)$ , so its Robin constant satisfies $V_{\partial K} \geq V_{K}$ ; conversely every $σ \in P (\partial K)$ is a probability carried by $K$ with $supp σ \subseteq \partial K \subseteq K$ , so step 3.1 gives $U^{μ} = \log \frac{1}{r}$ on $supp σ$ and the argument of steps 5.1 and 6.1 applies verbatim to yield $I (σ) \geq I (μ)$ with equality only for $σ = μ$ ; hence $V_{\partial K} = V_{K} = \log \frac{1}{r}$ and $cap (\partial K) = r$ , with unique equilibrium measure $μ$ , which is carried by $\partial K$ .

9.1step 3.1step 7.1step 8.1F4F5given∎

Combining steps 3.1, 7.1 and 8.1 gives the displayed potential, the capacity $r$ of both $K$ and $\partial K$ , the Robin constant $V_{K} = \log (1 / r)$ , and the identification of normalized arclength as the unique equilibrium measure of each of the two compact sets. Two choice principles are spent in the calculation, both supplied by the Axiom of Choice of the statement: Countable Choice in step 6.1 through [F4], and Dependent Choice in steps 1.1, 3.1 and 8.1 through [F5].

Remarks

Where the disc enters. Steps 1.3 and 2.1 are the only places where the specific geometry is used: Jensen's formula computes the circle average of $\log ∣ c - r e^{i t} ∣$ exactly when the singular point stays inside or on the circle, and the mean-value property computes it when the singular point is outside. The two formulas agree on $∣ c ∣ = r$ , which is why the potential is continuous across the boundary of $K$ .

Uniqueness is strict convexity. Step 6.1 does not merely bound $I (σ)$ below by $I (μ)$ : the strict positivity of the zero-mass charge $σ - μ$ gives equality only for $σ = μ$ , which is what makes the equilibrium measure unique rather than merely minimal.

Choice. The statement assumes the Axiom of Choice; it is used only to supply Countable Choice for Strict positivity of logarithmic energy for a zero-mass signed charge and to supply, through AC implies DC implies countable choice, the Dependent Choice hypothesis of the harmonic-measure interface [F5] used in the proof at steps 1.1, 3.1 and 8.1. The circle computation, the atom argument and the energy comparison are otherwise choice-free.

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Arcsine equilibrium measure and capacity of a segment

Statement

Assume the Axiom of Choice. Let $a < b$ and let $K = [a, b]$ . The unique equilibrium measure of $K$ has the density

$d μ_{K} (x) = \frac{d x}{π \sqrt{(x - a) (b - x)}} (a < x < b),$

that is, $μ_{K}$ is the pushforward of $d x / (π \sqrt{1 - x^{2}})$ on $(- 1, 1)$ under $x \mapsto \frac{a + b}{2} + \frac{b - a}{2} x$ , and its potential is

$U^{μ_{K}} (x) = \log \frac{4}{b - a} (a \leq x \leq b), U^{μ_{K}} (z) = \log \frac{4}{b - a} - \log ∣ w_{+} (z) ∣ (z \notin K),$

where $w_{+} (z)$ is the root of $w^{2} - 2 \frac{2 z - a - b}{b - a} w + 1 = 0$ of modulus at least $1$ . In particular $cap (K) = (b - a) / 4$ and $V_{K} = \log \frac{4}{b - a}$ . For $[a, b] = [- 1, 1]$ this says $d μ = \frac{d x}{π \sqrt{1 - x^{2}}}$ , $U^{μ} = \log 2$ on $[- 1, 1]$ and $cap ([- 1, 1]) = \frac{1}{2}$ .

The Axiom of Choice enters through the equilibrium identification, through Dependent Choice for the normalized-arclength harmonic-measure interface [F5], and through Countable Choice for strict energy positivity [F4] and the Lebesgue–Stieltjes identification [F9]. The Joukowski and scaling calculations are choice-free.

Facts & Assumptions

Given: real numbers $a < b$ , the segment $K = [a, b]$ , the logarithmic kernel and potential conventions of Logarithmic potential and energy of a positive compactly supported measure, the Robin constant and capacity of Robin constant and logarithmic capacity of a compact set, and the Axiom of Choice (The Axiom of Choice).

[F1]

$U^{ν} (z) = \int \log \frac{1}{∣ z - w ∣} d ν (w) \in (- \infty, + \infty]$ for finite positive Borel $ν$ of compact support, and for $R > diam supp ν$ one has $I (ν) = \iint (k + \log R) d ν d ν - ν (C)^{2} \log R$ , independently of $R$ ; the mixed energy $I (ν, ρ) = \iint k d ν d ρ$ is symmetric (Logarithmic potential and energy of a positive compactly supported measure).

[F2]

For nonempty compact $F$ , $V_{F} = \inf_{ρ \in P (F)} I (ρ)$ and $cap (F) = e^{- V_{F}}$ when $V_{F} < + \infty$ and $0$ otherwise; a Borel probability measure on $F$ is a finite positive measure carried by $F$ (Robin constant and logarithmic capacity of a compact set, Probability measures and probability spaces).

[F3]

Assume the Axiom of Choice: a compact nonpolar $F$ has exactly one equilibrium measure, the unique minimizer of $I$ over $P (F)$ (Existence and uniqueness of the equilibrium measure).

[F4]

Assume Countable Choice: for finite positive compactly supported $ν, ρ$ with equal mass and finite energy, $I (ν, ρ)$ is finite, $I (ν - ρ) = I (ν) - 2 I (ν, ρ) + I (ρ)$ is real, $I (ν - ρ) \geq 0$ , and $I (ν - ρ) = 0$ only for $ν = ρ$ (Strict positivity of logarithmic energy for a zero-mass signed charge); the Axiom of Choice implies Countable Choice (AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F5]

With $μ_{1}$ the normalized arclength measure on the unit circle, i.e. the harmonic measure at the centre of the unit disc, $U^{μ_{1}} (c) = \log \frac{1}{∣ c ∣}$ for $∣ c ∣ \geq 1$ and $U^{μ_{1}} (c) = 0$ for $∣ c ∣ \leq 1$ , and $I (μ_{1}) = 0$ , with $cap \overline{D (0, 1)} = 1$ (Capacity of a disc and its circular equilibrium measure).

[F6]

Image measures. If $ν$ is a Borel probability measure on a measurable space $X$ and $φ : X \to Y$ is Borel measurable, then $φ_{*} ν (E) : = ν (φ^{- 1} (E))$ is a Borel probability measure on $Y$ and $\int f d φ_{*} ν = \int f \circ φ d ν$ for every nonnegative Borel $f$ ; for continuous $φ$ on a metric space the preimages of Borel sets are Borel, indicators give the identity by definition, simple functions by linearity, and general $f$ by monotone convergence (Measures on sigma-algebras, Probability measures and probability spaces, Monotone convergence for the integral).

[F7]

The complex exponential has $∣ e^{i t} ∣ = 1$ , $(e^{i t} + e^{- i t}) / 2 = \cos t$ , and the modulus is multiplicative: $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ and $∣ u / v ∣ = ∣ u ∣ / ∣ v ∣$ for $v \neq 0$ (The complex exponential by its power series, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive); every complex number has a square root, and more generally every nonzero complex number has an $n$ -th root (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ ).

[F8]

The arcsine density defines a measure: for a nonnegative Borel function $h$ on $R$ , the set function $E \mapsto \int_{E} h (x) d x$ is a measure (The indefinite integral of a nonnegative measurable function is a measure); the derivatives $(\arcsin y)^{'} = 1 / \sqrt{1 - y^{2}}$ and $(\arccos y)^{'} = - 1 / \sqrt{1 - y^{2}}$ hold for $- 1 < y < 1$ (For $- 1 < y < 1$ , $(\arcsin y)^{'} = 1 / \sqrt{1 - y^{2}}$ and $(\arccos y)^{'} = - 1 / \sqrt{1 - y^{2}}$ , Principal inverse sine and inverse cosine), and a function with a continuous derivative on $[u, v]$ is the derivative of its primitive there (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[F9]

Assume Countable Choice: a Borel measure on $R$ that is finite on compacts is determined by its distribution function $F_{μ} (x) = μ ((0, x])$ for $x \geq 0$ , $F_{μ} (x) = - μ ((x, 0])$ for $x < 0$ : two such measures with equal distribution functions coincide, and $F_{μ} (b) - F_{μ} (a) = μ ((a, b])$ for $a < b$ (The distribution function of a Borel measure on $R$ , normalized at $0$ , Assuming countable choice, finite-on-compacts Borel measures on $R$ correspond to nondecreasing right-continuous functions modulo constants).

Verification

technique · direct

1.1F6given

Put $μ_{1} : = ω_{D (0, 1)}^{0}$ , the normalized arclength measure on the unit circle of [F5], and let $φ (t) : = \cos t$ ; define $ν : = φ_{*} λ$ where $λ$ is normalized Lebesgue measure on $[0, π]$ , that is, $ν (E) = λ ({t \in [0, π] : \cos t \in E})$ for Borel $E \subseteq R$ . By [F6] the set function $ν$ is a Borel probability measure concentrated on $[- 1, 1]$ , and $\int f d ν = \frac{1}{π} \int_{0}^{π} f (\cos t) d t$ for every nonnegative Borel $f$ .

1.2F7algebra

The Joukowski factorization. Let $z \in C$ 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 $w_{+} + w_{-} = 2 z$ , $w_{+} w_{-} = z^{2} - s^{2} = 1$ , and $w^{2} - 2 z w + 1 = (w - w_{+}) (w - w_{-})$ for all $w$ . Interchanging the two roots if necessary, $∣ w_{+} ∣ \geq 1 \geq ∣ w_{-} ∣$ , because $∣ w_{+} ∣ ∣ w_{-} ∣ = 1$ .

1.3F8algebra

The arcsine density of $ν$ . Define the nonnegative Borel function $ρ (x) : = \frac{1}{π \sqrt{1 - x^{2}}}$ for $- 1 < x < 1$ and $ρ (x) : = 0$ otherwise, and let $η : = E \mapsto \int_{E} ρ d x$ be the measure of [F8]. For $- 1 < x < 1$ its distribution function is $F_{η} (x) = \frac{1}{π} \int_{0}^{x} \frac{d y}{\sqrt{1 - y^{2}}} = \frac{1}{π} \arcsin x$ for $x \geq 0$ , and $F_{η} (x) = - \frac{1}{π} \int_{x}^{0} \frac{d y}{\sqrt{1 - y^{2}}} = \frac{1}{π} \arcsin x$ for $x < 0$ , the primitive being [F8]; hence $F_{η} (x) = \frac{1}{π} (\frac{π}{2} - \arccos x)$ on $(- 1, 1)$ with $η$ a probability because $\arcsin 1 - \arcsin (- 1) = π$ .

2.1step 1.2F7algebra

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

2.2step 1.1step 1.3F9algebra

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

3.1step 2.1F1F5F6algebra

The potential of $ν$ . By [F6] and step 2.1, applied to the positive and negative parts of the logarithmic kernel separately, the symmetry of $\cos t$ gives $U^{ν} (z) = \frac{1}{π} \int_{0}^{π} \log \frac{1}{∣ z - \cos t ∣} d t = \frac{1}{2 π} \int_{0}^{2 π} \log \frac{1}{∣ z - \cos t ∣} d t = \log 2 + U^{μ_{1}} (w_{+}) + U^{μ_{1}} (w_{-}) .$ Indeed, step 2.1 writes $\log \frac{1}{∣ z - \cos t ∣} = \log 2 - \log ∣ e^{i t} - w_{+} ∣ - \log ∣ e^{i t} - w_{-} ∣$ , and each negative logarithm has circle average equal to the corresponding unit-circle potential.

3.2step 1.3step 2.2F9

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

4.1step 3.1F1F5F7algebra

Since $∣ w_{+} ∣ \geq 1 \geq ∣ w_{-} ∣$ , [F5] gives $U^{μ_{1}} (w_{+}) = \log \frac{1}{∣ w_{+} ∣}$ and $U^{μ_{1}} (w_{-}) = 0$ , so step 3.1 yields $U^{ν} (z) = \log 2 - \log ∣ w_{+} ∣$ for $z \notin [- 1, 1]$ ; if $z \in [- 1, 1]$ then $z = \cos τ$ and $s$ is purely imaginary with $∣ w_{+} ∣ = ∣ w_{-} ∣ = 1$ , so the same identity gives $U^{ν} (z) = \log 2$ . In particular $U^{ν} = \log 2$ on $[- 1, 1]$ .

4.2step 3.2F1F6algebra

Scaling. Let $α : = \frac{a + b}{2}$ , $β : = \frac{b - a}{2} > 0$ and $T (x) : = α + β x$ , so $T$ maps $[- 1, 1]$ bijectively onto $[a, b]$ . Put $μ : = T_{*} ν$ ; by [F6] the measure $μ$ is a Borel probability on $[a, b]$ , and for finite positive compactly supported measures the identities $k (T u, T v) = k (u, v) - \log β$ , hence, writing $M : = ρ (C)$ , $I (T_{*} ρ) = I (ρ) - M^{2} \log β$ and $U^{T_{*} ρ} (T x) = U^{ρ} (x) - M \log β$ , follow by substituting $∣ T u - T v ∣ = β ∣ u - v ∣$ 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 $d x = d (T^{- 1} y) = \frac{d y}{β}$ and $\sqrt{1 - x^{2}} = \frac{1}{β} \sqrt{(y - a) (b - y)}$ , so $d μ = d y / (π \sqrt{(y - a) (b - y)})$ on $(a, b)$ .

5.1step 4.1F1F2given

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

6.1step 5.1F2F3F4given

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

7.1step 4.1step 6.1step 4.2algebra

Applying steps 6.1 and 4.1 to $μ = T_{*} ν$ with the identities of step 4.2 gives $I (μ) = I (ν) - \log β = \log \frac{2}{β} = \log \frac{4}{b - a}$ , $U^{μ} (x) = \log 2 - \log β = \log \frac{4}{b - a}$ for $x \in [a, b]$ , and $U^{μ} (y) = \log \frac{4}{b - a} - \log ∣ w_{+} (z) ∣$ for $y \notin [a, b]$ with $z = (2 y - a - b) / (b - a)$ and $w_{+}$ the root of modulus at least $1$ of $w^{2} - 2 z w + 1 = 0$ .

8.1step 6.1step 4.2step 7.1F2F3

Every Borel probability $σ$ on $[a, b]$ is the pushforward $T_{*} ρ$ of the Borel probability $ρ : = T_{*}^{- 1} σ$ on $[- 1, 1]$ , and step 4.2 applied to $ρ$ gives $I (σ) = I (ρ) - \log β \geq \log 2 - \log β = \log \frac{4}{b - a} = I (μ)$ , with equality if and only if $ρ = ν$ , that is, if and only if $σ = μ$ ; hence $V_{K} = \log \frac{4}{b - a}$ , $cap (K) = e^{- V_{K}} = \frac{b - a}{4}$ , and $μ$ is the unique equilibrium measure of $K = [a, b]$ .

9.1step 4.2step 7.1step 8.1F4given∎

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 $α = 0$ , $β = 1$ . 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.

Remarks

Why the Joukowski variable appears. The quadratic $w^{2} - 2 z w + 1$ has the two roots $w_{\pm}$ with $w_{+} w_{-} = 1$ , so one lies inside and one outside the unit circle (both on it when $z \in [- 1, 1]$ ). The identity of step 2.1 turns the logarithm of $∣ z - \cos t ∣$ into a sum of two logarithms of the form $\log ∣ e^{i t} - c ∣$ , whose circle average is known from the disc example; that is exactly the point at which the interval computation uses Capacity of a disc and its circular equilibrium measure.

The density is identified, not assumed. Step 1.3 defines the arcsine density measure independently, and step 3.2 identifies it with the pushforward $ν$ through the Lebesgue–Stieltjes correspondence, using the arcsin primitive; no change-of-variables formula with a vanishing derivative at the endpoints is invoked.

Choice. The Axiom of Choice is used in [F3] to identify the energy minimizer as the equilibrium measure, and it supplies Dependent Choice for [F5] and Countable Choice for [F4] and [F9]. The image-measure construction, Joukowski factorization, and scaling computation are choice-free.

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Chebyshev extremals and the exact disk Fekete polynomial

Statement

Assume the Axiom of Choice. Let $K = \overline{D (a, R)}$ be a closed disc with $R > 0$ , and let $[- 1, 1] \subseteq R \subseteq C$ be the real unit interval, with the extremal norms $t_{n}$ and Chebyshev constant $cheb$ of Chebyshev constant of a compact planar set and the capacity of Robin constant and logarithmic capacity of a compact set.

For the disc, $t_{n} (K) = R^{n}$ and $cheb (K) = R = cap (K)$ , attained by the monic polynomial $(z - a)^{n}$ .
For the interval, $t_{n} ([- 1, 1]) = 2^{1 - n}$ and $cheb ([- 1, 1]) = \frac{1}{2} = cap ([- 1, 1])$ , attained by the monic Chebyshev polynomial $2^{1 - n} T_{n}$ .
For each $n \geq 2$ the $n$ -th roots of unity form an $n$ -point Fekete tuple of the closed unit disc, with Fekete polynomial $F_{n} (z) = z^{n} - 1$ ; and $δ_{n} (\overline{D (0, 1)}) = n^{1 / (n - 1)}$ .
The empirical probability measures of these tuples converge weakly to normalized arclength on the unit circle, and $∥ z^{n} - 1 ∥_{\overline{D (0, 1)}} = 2$ , so the $n$ -th root of its norm tends to $1 = cheb (\overline{D (0, 1)})$ .

The Axiom of Choice is inherited from the equilibrium theory that supplies the two capacity values and the weak convergence; the Cauchy estimate, the minimax comparison, the Vandermonde determinant and the Gram–Schmidt bound are choice-free.

Facts & Assumptions

Given: a closed disc $K = \overline{D (a, R)}$ with $R > 0$ , the interval $[- 1, 1]$ , the closed unit disc $\overline{D (0, 1)}$ , and the conventions of Chebyshev constant of a compact planar set, Fekete points and the transfinite diameter of a compact set and Robin constant and logarithmic capacity of a compact set.

[F1]

For nonempty compact $K$ : $∥ p ∥_{K} = \sup_{z \in K} ∣ p (z) ∣$ is finite and attained, $t_{n} (K) = \inf {∥ p ∥_{K} : p monic of degree n}$ is a real number, $cheb (K) = \inf_{n \geq 1} t_{n} (K)^{1 / n}$ , and for the disc and the interval the capacities are $cap (\overline{D (a, R)}) = R$ and $cap ([- 1, 1]) = \frac{1}{2}$ (Chebyshev constant of a compact planar set, Capacity of a disc and its circular equilibrium measure, Arcsine equilibrium measure and capacity of a segment).

[F2]

For nonempty compact $K$ and $n \geq 2$ the $n$ -th Fekete diameter is $δ_{n} (K) = (\max_{z \in K^{n}} \prod_{i < j} ∣ z_{i} - z_{j} ∣)^{2 / [n (n - 1)]}$ , tuples attaining the maximum are Fekete tuples, and the associated polynomial $F_{n} (Z) = \prod_{j} (Z - z_{j})$ is monic of degree $n$ (Fekete points and the transfinite diameter of a compact set, Formal complex polynomials, evaluation, degree, leading coefficient, and monic polynomials).

[F3]

Assume the Axiom of Choice. For every compact $K$ one has $cap (K) = τ (K) = cheb (K)$ ; if $cap (K) > 0$ , then the empirical probability measures of any sequence of $n$ -point Fekete tuples of $K$ converge weakly to the unique equilibrium measure $μ_{K}$ (Fekete–Szegő equality of logarithmic capacity, transfinite diameter, and Chebyshev constant, Weak convergence of borel probability measures).

[F4]

Let $f$ be holomorphic on $D (a, R)$ and let $0 < r < R$ with $∣ f (ζ) ∣ \leq M$ on $∣ ζ - a ∣ = r$ ; then $∣ f^{(n)} (a) ∣ \leq n! M / r^{n}$ (Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle). A complex polynomial $P (z) = \sum_{k = 0}^{n} a_{k} z^{k}$ is entire with $P^{'} (z) = \sum_{k = 1}^{n} k a_{k} z^{k - 1}$ , so by iteration the $n$ -th derivative of a monic polynomial of degree $n$ is the constant $n!$ (Complex polynomials are entire with the power-rule derivative, and rational functions are holomorphic wherever their denominator is nonzero).

[F5]

For $n \geq 1$ the polynomial $P_{n} = 2^{1 - n} T_{n}$ is monic of degree $n$ , and for every monic real polynomial $q$ of degree $n$ one has $\max_{x \in [- 1, 1]} ∣ q (x) ∣ \geq 2^{1 - n} = \max_{x \in [- 1, 1]} ∣ P_{n} (x) ∣$ (For $n \geq 1$ , $2^{1 - n} T_{n}$ is the minimax monic polynomial of degree $n$ on $[- 1, 1]$ , Chebyshev polynomials of the first and second kinds by their three-term recurrences, $T_{n} (\cos θ) = \cos (n θ)$ and $U_{n} (\cos θ) \sin θ = \sin ((n + 1) θ)$ for every $n \in N$ ).

[F6]

A complex polynomial $p$ of degree $n$ has real part $Re p$ , a real polynomial whose $x^{n}$ coefficient is the real part of the $x^{n}$ coefficient of $p$ ; hence $Re p$ is monic of degree $n$ when $p$ is monic, and $∣ Re p (x) ∣ \leq ∣ p (x) ∣$ for every real $x$ (Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F7]

Determinant conventions and rules: the determinant is the Leibniz sum $\det (A) = \sum_{σ} sgn (σ) \prod_{i} a_{σ (i), i}$ (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix); it is alternating and multilinear in the rows and columns (The Leibniz determinant is column-multilinear, alternating and normalized over every commutative ring); $\det (A B) = \det A \det B$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ); and the determinant of an upper triangular matrix is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

[F8]

Inner products, norms and orthogonalisation: $∥ v ∥^{2} = ⟨ v, v ⟩$ for the induced norm (Real and complex inner product spaces, with the inner product linear in the first argument, The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product); a finite linearly independent list has an orthonormal list spanning the same successive spans (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans); for orthogonal vectors $∥ u + v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2}$ (Pythagoras, the parallelogram identity, and the real and complex polarisation identities); and a linear isometry carrying an orthonormal basis to an orthonormal basis satisfies $∣ \det T ∣ = 1$ (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{*} T = I$ are equivalent, Orthogonal and unitary operators form groups, and their determinants have modulus one).

[F9]

The $n$ -th roots of unity $ω_{j} = \exp (2 π i j / n)$ , $j = 0, \dots, n - 1$ , are $n$ distinct complex numbers of modulus one, and $z^{n} - 1$ has exactly these $n$ roots, each of multiplicity one (The $n$ -th roots of a complex number and the $n$ distinct roots of unity for every $n \geq 1$ , The complex exponential by its power series, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ , Integer powers in the complex field); a monic polynomial of degree $n$ with this root list is $\prod_{j} (z - ω_{j})$ , by uniqueness of the root factorisation (A complex polynomial of degree $n$ has exactly $n$ roots counted with multiplicity). The Vandermonde polynomial is $Δ_{n} (x_{1}, \dots, x_{n}) = \prod_{i < j} (x_{i} - x_{j})$ (The Vandermonde polynomial $Δ_{n} = \prod_{i < j} (x_{i} - x_{j})$ ).

[F10]

Limits and roots: $t^{1 / n} \to 1$ for every fixed $t > 0$ (For every $a > 0$ , $a^{1 / n} \to 1$ ); sums, products and quotients of convergent sequences converge to the corresponding combination (Algebra of limits: sums, scalar multiples, products and quotients); nonnegative $n$ -th roots exist and are monotone (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ ).

Verification

technique · direct

1.1F1F4algebra

Disc: Cauchy lower bound. Let $p$ be a monic complex polynomial of degree $n$ and $K = \overline{D (a, R)}$ . By [F4] the polynomial $p$ is entire, so it is holomorphic on $D (a, R^{'})$ for every $R^{'} > R$ , and its $n$ -th derivative is the constant $p^{(n)} = n!$ ; applying Cauchy's inequality [F4] with $r = R$ and $M = ∥ p ∥_{K}$ , and noting that $∣ p ∣ \leq ∥ p ∥_{K}$ holds on the circle $∣ ζ - a ∣ = R$ , gives $n! = ∣ p^{(n)} (a) ∣ \leq n! ∥ p ∥_{K} / R^{n}$ , hence $∥ p ∥_{K} \geq R^{n}$ .

1.2F5F6algebra

Interval: complex-to-real reduction. Let $p$ be a monic complex polynomial of degree $n$ and put $q : = Re p$ . By [F6] the polynomial $q$ is a monic real polynomial of degree $n$ with $∣ q (x) ∣ \leq ∣ p (x) ∣$ for all real $x$ , so $\max_{x \in [- 1, 1]} ∣ q (x) ∣ \leq ∥ p ∥_{[- 1, 1]}$ ; the minimax theorem [F5] gives $2^{1 - n} \leq \max_{x \in [- 1, 1]} ∣ q (x) ∣$ , so $∥ p ∥_{[- 1, 1]} \geq 2^{1 - n}$ .

1.3F7algebra

Vandermonde determinant. For $m \geq 1$ and $w_{0}, \dots, w_{m - 1} \in C$ the matrix $V_{m} = (w_{j}^{k})_{j, k = 0}^{m - 1}$ satisfies $\det V_{m} = \prod_{i < j} (w_{j} - w_{i})$ . Indeed for $m = 1$ both sides are $1$ ; for $m \geq 2$ , replacing the $j$ -th row of $V_{m}$ by itself minus the first row does not change the determinant by [F7], because the determinant is multilinear and alternating and the subtracted term has two equal rows; the new $j$ -th row is $(w_{j} - w_{0}) (0, q_{1} (w_{j}), \dots, q_{m - 1} (w_{j}))$ with $q_{k} (z) = z^{k - 1} + z^{k - 2} w_{0} + \dots + w_{0}^{k - 1}$ , so multilinearity [F7] factors $\prod_{j \geq 1} (w_{j} - w_{0})$ out of the last $m - 1$ rows; the remaining matrix has first row $(1, w_{0}, \dots, w_{0}^{m - 1})$ and, below it, zeros in the first column, which Leibniz's formula [F7] evaluates as the determinant of the $(m - 1) \times (m - 1)$ matrix of the polynomials $q_{k}$ at $w_{1}, \dots, w_{m - 1}$ ; since $q_{k} (z) = z^{k - 1} + (lower order terms)$ , the basis change from $(1, z, \dots, z^{m - 2})$ to $(q_{1}, \dots, q_{m - 1})$ is unitriangular and hence a determinant-preserving sequence of column operations [F7], so that last determinant equals $\det V_{m - 1} (w_{1}, \dots, w_{m - 1})$ .

1.4F7F8algebra

Gram–Schmidt (Hadamard) bound. Let $c_{0}, \dots, c_{m - 1} \in C^{m}$ be the columns of a matrix $A$ . If they are linearly dependent then $\det A = 0$ by the alternating multilinearity in [F7]; otherwise [F8] supplies an orthonormal list $e_{0}, \dots, e_{m - 1}$ with $span (e_{0}, \dots, e_{k}) = span (c_{0}, \dots, c_{k})$ for every $k$ , and writing $u_{k} : = c_{k} - \sum_{j < k} ⟨ c_{k}, e_{j} ⟩ e_{j}$ one has $c_{k} = \sum_{j \leq k} ⟨ c_{k}, e_{j} ⟩ e_{j}$ , so $A = Q R$ with $Q$ the matrix of the $e_{j}$ and $R$ upper triangular with diagonal entries $R_{k k} = ⟨ c_{k}, e_{k} ⟩ = ∥ u_{k} ∥$ ; moreover $∥ u_{k} ∥ \leq ∥ c_{k} ∥$ , because $u_{k}$ is orthogonal to $\sum_{j < k} ⟨ c_{k}, e_{j} ⟩ e_{j}$ and Pythagoras [F8] gives $∥ c_{k} ∥^{2} = ∥ u_{k} ∥^{2} + ∥ \sum_{j < k} ⟨ c_{k}, e_{j} ⟩ e_{j} ∥^{2}$ . The matrix $Q$ has orthonormal columns, so it is a linear isometry of $C^{m}$ and $∣ \det Q ∣ = 1$ by [F8]; hence by [F7] $∣ \det A ∣ = ∣ \det Q ∣ ∣ \det R ∣ = \prod_{k} R_{k k} = \prod_{k} ∥ u_{k} ∥ \leq \prod_{k} ∥ c_{k} ∥$ .

2.1step 1.1F1algebra

Disc: extremal value. The monic polynomial $(z - a)^{n}$ has $∥ (z - a)^{n} ∥_{K} = R^{n}$ , so step 1.1 gives $t_{n} (K) = R^{n}$ for every $n \geq 1$ ; hence $cheb (K) = \inf_{n \geq 1} t_{n} (K)^{1 / n} = \inf_{n \geq 1} R = R$ .

2.2step 1.2F5F10algebra

Interval: extremal value and Chebyshev constant. By [F5] the monic polynomial $P_{n} = 2^{1 - n} T_{n}$ has $\max_{x \in [- 1, 1]} ∣ P_{n} (x) ∣ = 2^{1 - n}$ , so step 1.2 gives $t_{n} ([- 1, 1]) = 2^{1 - n}$ ; therefore $cheb ([- 1, 1]) = \inf_{n \geq 1} 2^{(1 - n) / n}$ , and since $2^{(1 - n) / n} = 2^{1 / n} / 2$ with $2^{1 / n} \to 1$ by [F10], that infimum is $\frac{1}{2}$ .

2.3step 1.3F9algebra

Vandermonde identity by induction. Induction on $m$ in the recursion of step 1.3 gives $\det V_{m} = \prod_{j \geq 1} (w_{j} - w_{0}) \cdot \prod_{1 \leq i < j \leq m - 1} (w_{j} - w_{i}) = \prod_{i < j} (w_{j} - w_{i})$ for every $m \geq 1$ ; in particular $∣ Δ_{n} (z_{0}, \dots, z_{n - 1}) ∣ = ∣ \det V_{n} ∣$ by [F9].

3.1step 2.1F1

Disc: capacity. By [F1], $cap (K) = R$ , so $cheb (K) = R = cap (K)$ .

3.2step 2.2F1

Interval: capacity. By [F1], $cap ([- 1, 1]) = \frac{1}{2}$ , so $cheb ([- 1, 1]) = \frac{1}{2} = cap ([- 1, 1])$ .

3.3step 2.3F2F8F9algebra

Unit disc: universal bound. Let $z_{0}, \dots, z_{n - 1} \in \overline{D (0, 1)}$ and $A = (z_{j}^{k})_{j, k = 0}^{n - 1}$ . Its columns are $c_{k} = (z_{j}^{k})_{j = 0}^{n - 1}$ with $∥ c_{k} ∥^{2} = \sum_{j} ∣ z_{j} ∣^{2 k} \leq n$ by [F8], [F9] and $∣ z_{j} ∣ \leq 1$ ; and $∣ \det A ∣ = ∣ Δ_{n} (z) ∣$ by step 2.3, so [F2] gives the Fekete bound $\prod_{i < j} ∣ z_{i} - z_{j} ∣ \leq n^{n / 2}$ .

4.1step 1.4step 3.3F2F9algebra

Unit disc: the roots of unity are Fekete. Let $ω_{j} = \exp (2 π i j / n)$ , $j = 0, \dots, n - 1$ , which are $n$ distinct points of the unit circle by [F9]. For $k \neq ℓ$ put $d : = k - ℓ$ and $r : = ω_{1}^{d}$ ; then $r^{n} = (ω_{1}^{n})^{d} = 1$ and $r \neq 1$ , since $r = 1$ would give $\exp (2 π i d / n) = 1$ , forcing $n ∣ d$ by the kernel statement of $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , contrary to $0 < ∣ d ∣ < n$ . The cyclic-shift computation of For $n \geq 2$ , the sum of all $n$ -th roots of unity is zero, applied to the list $1, r, \dots, r^{n - 1}$ in place of $1, ζ, \dots, ζ^{n - 1}$ (multiply $S = \sum_{j < n} r^{j}$ by $r$ and compare $r S$ with $S$ using $r^{n} = 1$ ), gives $(r - 1) S = 0$ , hence $S = 0$ ; since $ω_{j}^{d} = r^{j}$ by the integer power laws of [F9], the inner product $⟨ c_{k}, c_{ℓ} ⟩ = \sum_{j} ω_{j}^{k} \overline{ω_{j}^{ℓ}} = \sum_{j} ω_{j}^{k - ℓ}$ vanishes, the conjugate being the inverse because $∣ ω_{j} ∣ = 1$ . Each column has norm $\sqrt{n}$ , since $∣ ω_{j}^{2 k} ∣ = 1$ . The Gram–Schmidt residuals of an orthogonal list are the columns themselves, so step 1.4 gives $∣ \det A ∣ = n^{n / 2}$ ; by step 3.3 this is the maximum of $\prod_{i < j} ∣ z_{i} - z_{j} ∣$ over the closed unit disc, so the roots of unity form an $n$ -point Fekete tuple and $δ_{n} (\overline{D (0, 1)}) = n^{1 / (n - 1)}$ . The associated monic polynomial is $F_{n} (z) = \prod_{j} (z - ω_{j}) = z^{n} - 1$ by [F9].

5.1step 4.1F1F3

Weak convergence of the empirical measures. The closed unit disc is compact with $cap (\overline{D (0, 1)}) = 1 > 0$ by [F1], and its equilibrium measure is normalized arclength on the unit circle, which we denote $μ$ (Capacity of a disc and its circular equilibrium measure); since the tuple of step 4.1 is a Fekete tuple for every $n \geq 2$ , [F3] gives $\frac{1}{n} \sum_{j} δ_{ω_{j}} \Rightarrow μ$ .

5.2step 2.1step 3.1step 4.1F10algebra

The norm limit. On the closed unit disc, $∣ z^{n} - 1 ∣ \leq ∣ z ∣^{n} + 1 \leq 2$ by the modulus laws, with equality exactly at the points with $z^{n} = - 1$ , which exist on the unit circle; hence $∥ z^{n} - 1 ∥_{\overline{D (0, 1)}} = 2$ and $∥ F_{n} ∥^{1 / n} = 2^{1 / n} \to 1$ , and this limit is $cheb (\overline{D (0, 1)}) = cap (\overline{D (0, 1)}) = 1$ by steps 2.1 and 3.1 with $a = 0$ and $R = 1$ .

6.1step 2.1step 3.1step 2.2step 3.2step 4.1step 5.1step 5.2F1F3∎

Assembly. Steps 2.1, 3.1, 2.2, 3.2 prove the disc and interval identities $t_{n} = R^{n}$ , $t_{n} = 2^{1 - n}$ and the matching Chebyshev constants and capacities; step 4.1 proves the Fekete property of the roots of unity with $F_{n} (z) = z^{n} - 1$ and the value $δ_{n} = n^{1 / (n - 1)}$ ; step 5.1 proves weak convergence to normalized arclength; and step 5.2 proves the norm value $2$ and the limit of its $n$ -th roots. The Axiom of Choice is used only through [F3] and the capacity values of [F1].

Remarks

Where the two halves of the calculation meet. The first six steps compute the monic extremal norms and combine them with the known capacities of the disc and the interval; steps 1.3, 1.4, 2.3, 3.3 and 4.1 compute the Fekete diameters of the unit disc exactly, with the Vandermonde determinant reducing the Fekete problem to the classical Gram–Schmidt bound for column norms, and with the Fourier columns of the roots-of-unity matrix attaining equality. Step 5.1 then identifies the limit of the empirical measures with the equilibrium measure supplied by Capacity of a disc and its circular equilibrium measure, and step 5.2 confirms the consistency of the Fekete-polynomial norms with the general norm limit of Fekete–Szegő equality of logarithmic capacity, transfinite diameter, and Chebyshev constant.

Why the interval reduction is legitimate. For a monic complex polynomial $p$ on a real interval the real part $Re p$ is again monic — its leading coefficient is $Re (1) = 1$ — and it never exceeds $p$ in modulus on real arguments, so a minimax bound for monic real polynomials applies without loss. This is the only place where the real interval differs from the disc, where Cauchy's estimate handles complex coefficients directly.

Choice. The example states the Axiom of Choice because it quotes the two capacity values from the equilibrium examples and the weak convergence from Fekete–Szegő equality of logarithmic capacity, transfinite diameter, and Chebyshev constant; the determinant, Gram–Schmidt and minimax computations are choice-free.

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Chebyshev extremal nodes converge to the arcsine equilibrium measure

Statement

Assume the Axiom of Choice. For $n \geq 1$ let $x_{k, n} : = \cos \frac{k π}{n}$ for $0 \leq k \leq n$ and let $ν_{n} : = \frac{1}{n + 1} \sum_{k = 0}^{n} δ_{x_{k, n}}$ . Then $ν_{n}$ converges weakly, as $n \to \infty$ , to the arcsine equilibrium measure $d x / (π \sqrt{1 - x^{2}})$ of $[- 1, 1]$ , and the points $x_{0, n}, \dots, x_{n, n}$ are exactly the points of $[- 1, 1]$ at which the Chebyshev polynomial $T_{n}$ attains its extreme values $\pm 1$ , alternately signed.

Facts & Assumptions

Given: the nodes $x_{k, n} = \cos \frac{k π}{n}$ , the empirical measures $ν_{n}$ , the Chebyshev polynomials $T_{n}$ of Chebyshev polynomials of the first and second kinds by their three-term recurrences, and the Axiom of Choice.

[F1]

The multiple-angle identity $T_{n} (\cos θ) = \cos (n θ)$ holds for all real $θ$ ( $T_{n} (\cos θ) = \cos (n θ)$ and $U_{n} (\cos θ) \sin θ = \sin ((n + 1) θ)$ for every $n \in N$ ), and $∣ \cos u ∣ \leq 1$ for real $u$ , with $\cos (k π) = (- 1)^{k}$ (Parity and the Pythagorean identity for sine and cosine).

[F2]

The arcsine measure $μ$ of $[- 1, 1]$ is the unique equilibrium measure of $[- 1, 1]$ , and for every continuous $f$ on $[- 1, 1]$ one has $\int f d μ = \frac{1}{π} \int_{0}^{π} f (\cos θ) d θ$ (Arcsine equilibrium measure and capacity of a segment).

[F3]

Dirac measures are probability measures, finite nonnegative weighted sums of measures are measures, and $ν_{n}$ is therefore a Borel probability measure on $[- 1, 1]$ (The Dirac set function at a point, A Dirac set function is a probability measure, Nonnegative scalar multiples and countable weighted sums of measures are measures).

[F4]

Weak convergence $ν_{n} \Rightarrow μ$ means $\int f d ν_{n} \to \int f d μ$ for every bounded continuous real $f$ (Weak convergence of borel probability measures).

[F5]

A continuous real function on the closed bounded interval $[0, π]$ is Riemann integrable, and its uniform-mesh Riemann sums converge to the integral: $\frac{1}{m} \sum_{j = 0}^{m - 1} g (\frac{j π}{m}) \to \frac{1}{π} \int_{0}^{π} g (θ) d θ$ (apply Every continuous function on a closed nondegenerate rectangle in $R^{m}$ is Riemann integrable in dimension $1$ , then The Darboux and Riemann definitions agree: a bounded $f$ on $[a, b]$ is Darboux integrable with integral $I$ if and only if for every real $ε > 0$ there is a real $δ > 0$ such that $∣ S (f, P, ξ) - I ∣ < ε$ for every tagged partition of mesh below $δ$ ). Its Riemann integral equals its Lebesgue integral by A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral, whose Countable Choice hypothesis is supplied by the assumed Axiom of Choice through AC implies DC implies countable choice.

Verification

technique · direct

1.1F3givenalgebra

For each $n \geq 1$ the points $x_{k, n} = \cos \frac{k π}{n}$ lie in $[- 1, 1]$ ; since $\cos$ is strictly decreasing on $[0, π]$ and $\frac{k π}{n}$ are strictly increasing for $k = 0, \dots, n$ , the $n + 1$ points are pairwise distinct, and by [F3] each $ν_{n}$ is a Borel probability measure on $[- 1, 1]$ .

1.2F1givenalgebra

By the multiple-angle identity [F1], $T_{n} (x_{k, n}) = \cos (k π) = (- 1)^{k}$ ; moreover every $x \in [- 1, 1]$ is $x = \cos θ$ for some $θ \in [0, π]$ , so $∣ T_{n} (x) ∣ = ∣ \cos (n θ) ∣ \leq 1$ for every $x \in [- 1, 1]$ , with equality exactly at the points where $∣ \cos n θ ∣ = 1$ , that is, where $n θ$ is an integer multiple of $π$ .

1.3F5given

Weak convergence. Let $f$ be a continuous real function on $[- 1, 1]$ and put $g (θ) : = f (\cos θ)$ ; then $g$ is continuous on $[0, π]$ , so [F5] applied with $m = n$ gives $\frac{1}{n} \sum_{k = 0}^{n - 1} g (\frac{k π}{n}) \to \frac{1}{π} \int_{0}^{π} g (θ) d θ$ .

2.1step 1.2

Consequently the points $x_{k, n}$ are precisely the points of $[- 1, 1]$ at which $T_{n}$ attains $\pm 1$ , with alternating signs, which is the second assertion.

2.2step 1.3algebra

The empirical sum $\frac{1}{n + 1} \sum_{k = 0}^{n} g (\frac{k π}{n})$ differs from $\frac{1}{n} \sum_{k = 0}^{n - 1} g (\frac{k π}{n})$ by at most $\frac{2}{n} \sup_{[0, π]} ∣ g ∣$ , which tends to $0$ , so it has the same limit $\frac{1}{π} \int_{0}^{π} g d θ$ .

3.1step 2.2F2F3algebra

By the definition of $ν_{n}$ and [F3], $\int f d ν_{n} = \frac{1}{n + 1} \sum_{k = 0}^{n} f (x_{k, n}) = \frac{1}{n + 1} \sum_{k = 0}^{n} g (\frac{k π}{n})$ , and by [F2] the limit $\frac{1}{π} \int_{0}^{π} g d θ$ equals $\int f d μ$ ; hence $\int f d ν_{n} \to \int f d μ$ for every continuous $f$ on $[- 1, 1]$ .

4.1step 2.1step 3.1F2F4given∎

Since $[- 1, 1]$ is compact, every continuous real $f$ on it is bounded; step 3.1 therefore gives convergence of integrals for every bounded continuous test function on the metric space $[- 1, 1]$ . By [F4] this is $ν_{n} \Rightarrow μ$ , which together with step 2.1 proves both assertions.

Remarks

Why the weights are $\frac{1}{n + 1}$ . The extremal points of $T_{n}$ are the $n + 1$ points $\cos \frac{k π}{n}$ ; the Riemann sum of step 1.3 is naturally indexed by $k = 0, \dots, n - 1$ , and step 2.2 records that replacing $n$ weights by $n + 1$ weights and adjoining the endpoint $θ = π$ changes the average by $O (1 / n)$ only. The endpoint contribution vanishes in the limit and does not affect the weak limit.

The limit is the equilibrium measure. The identification of the limit with the arcsine measure is exactly the equilibrium computation of Arcsine equilibrium measure and capacity of a segment; this example supplies the discrete approximation of that measure by Chebyshev nodes.

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Finite and countable planar sets have zero logarithmic capacity

Statement

Assume the Axiom of Countable Choice. Every finite or countable set $E \subseteq C$ is capacity-polar in the compact/local sense of Capacity-polar sets, quasi-everywhere, and subharmonic polar sets: every compact $F \subseteq E$ satisfies $cap (F) = 0$ for the logarithmic capacity of Robin constant and logarithmic capacity of a compact set. Moreover $E$ is contained in the $- \infty$ locus of an explicitly constructed subharmonic function on $C$ that is not identically $- \infty$ , so $E$ is also subharmonically polar.

The Axiom of Countable Choice is used through the local integrability and subharmonicity of compactly supported logarithmic potentials (Distributional Laplacian of a compact logarithmic potential), which enters the construction of the witness; the diagonal $+ \infty$ of the logarithmic kernel and the countable atom computation of the energy are choice-free.

Facts & Assumptions

Given: an at most countable set $E \subseteq C$ , the logarithmic kernel $k (z, w) = \log (1 / ∣ z - w ∣)$ with diagonal value $+ \infty$ , the potentials $U^{μ}$ , $p_{μ} = - U^{μ}$ and the energy $I (μ)$ of Logarithmic potential and energy of a positive compactly supported measure, the Robin constant $V_{F}$ and capacity $cap (F)$ of Robin constant and logarithmic capacity of a compact set, and the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

$k (z, w) = \log \frac{1}{∣ z - w ∣} \in (- \infty, + \infty]$ is Borel, equals $+ \infty$ exactly when $z = w$ , and $p_{μ} (z) = \int \log ∣ z - w ∣ d μ (w) \in [- \infty, + \infty)$ for every finite positive Borel measure $μ$ of compact support; for $R > diam (supp μ)$ one has $k_{R} = k + \log R \geq 0$ on the product of the support with itself and $I (μ) = \iint k_{R} d μ d μ - μ (C)^{2} \log R$ , independently of $R$ (Logarithmic potential and energy of a positive compactly supported measure).

[F2]

For nonempty compact $F$ , $V_{F} = \inf_{ν \in P (F)} I (ν)$ and $cap (F) = e^{- V_{F}}$ when $V_{F} < + \infty$ , $cap (F) = 0$ when $V_{F} = + \infty$ ; $cap (\emptyset) = 0$ ; and $cap (F) = 0$ holds exactly when $I (ν) = + \infty$ for every Borel probability $ν$ on $F$ (Robin constant and logarithmic capacity of a compact set).

[F3]

Capacity-polar means that every compact subset has capacity zero, and subharmonically polar means that every point of the set lies in a complex domain carrying a subharmonic function that is $- \infty$ on the part of the set lying in that domain (Capacity-polar sets, quasi-everywhere, and subharmonic polar sets).

[F4]

Assume ${A C}_{ω}$ : for a finite positive Borel measure $μ$ with nonempty compact support, $p_{μ}$ is locally integrable on $C$ and subharmonic on the domain $C$ , and harmonic on $C ∖ supp μ$ (Distributional Laplacian of a compact logarithmic potential, A complex domain is a nonempty connected open subset of $C$ ).

[F5]

For every $a \in C$ the function $z \mapsto \log ∣ z - a ∣$ is subharmonic on $C$ (The logarithm of the modulus of a holomorphic function is subharmonic) and is $C^{\infty}$ and harmonic on $C ∖ {a}$ (Logarithmic modulus is harmonic off its centre); a real function is harmonic when it is $C^{2}$ and $u_{x x} + u_{y y} = 0$ (Plane harmonic functions), and a $C^{2}$ function with $u_{x x} + u_{y y} \geq 0$ is subharmonic (A C^2 function is subharmonic exactly when its Laplacian is nonnegative).

[F6]

Nonnegative linear combinations of finitely many subharmonic functions are subharmonic (Positive linear combinations and finite maxima preserve subharmonicity); in particular, by [F5] a sum of a subharmonic function and a harmonic function is subharmonic.

[F8]

Differentiation under the integral sign: if $f : X \times I \to C$ has $x \mapsto f (x, t)$ integrable for every $t$ in an open interval $I$ , is differentiable in $t$ for almost every $x$ , has measurable $t$ -derivative, and the $t$ -derivative is dominated in modulus by an integrable $g$ independent of $t$ , then $t \mapsto \int f (x, t) d μ (x)$ is differentiable on $I$ with derivative $\int \partial_{t} f d μ$ (Differentiation under the integral sign).

[F9]

Dirac measures are probability measures, and finite or countable nonnegative weighted sums of measures are measures, with the integral identity $\int g d (\sum_{j} c_{j} μ_{j}) = \sum_{j} c_{j} \int g d μ_{j}$ for nonnegative Borel $g$ , by the pointwise definition and monotone convergence (The Dirac set function at a point, Probability measures and probability spaces, Nonnegative scalar multiples and countable weighted sums of measures, Nonnegative scalar multiples and countable weighted sums of measures are measures, Monotone convergence for the integral).

[F11]

Subharmonicity on a complex domain means: upper semicontinuity, no connected component carrying the value $- \infty$ identically, and the circle mean inequality at every closed disc contained in the domain (Subharmonic functions on plane domains); every subharmonic function on a plane domain is locally integrable (Plane subharmonic functions are locally integrable).

[F12]

A subset of an at most countable set is at most countable, a set is countably infinite when it is in bijection with $N$ , and a finite or countably infinite set can be listed without repetitions (Finite, countably infinite, countable, uncountable).

Verification

technique · direct

1.1givenF3

The statements to prove are the capacity-polarity of $E$ and the existence of a subharmonic witness with $E$ in its $- \infty$ locus; two elementary cases come first, and the countably infinite case occupies the rest of the proof.

1.2F2F3F5given

The empty case. If $E = \emptyset$ there is no compact subset to test, so $E$ is capacity-polar by [F3] and [F2], and the zero function $u \equiv 0$ is of class $C^{2}$ with vanishing Laplacian, hence harmonic and therefore subharmonic on the domain $C$ by [F5], while its $- \infty$ locus is empty; so the statement holds for $E = \emptyset$ .

1.3F12given

A compact at most countable set has capacity zero. Let $F \subseteq E$ be compact and nonempty; by [F12] the set $F$ is at most countable, so it can be listed without repetitions as $F = {b_{1}, b_{2}, \dots}$ (the list is finite when $F$ is finite and otherwise is a bijection with $N$ ; for $F \subseteq E$ with $E$ in bijection with $N$ the listing comes from ordering the corresponding subset of $N$ , which uses no choice). Let $ν$ be a Borel probability measure on $F$ ; by countable additivity over the disjoint singletons $1 = ν (F) = \sum_{i} ν ({b_{i}})$ , so some index $i_{0}$ has $m : = ν ({b_{i_{0}}}) > 0$ , since otherwise the sum would be $0$ .

1.4F5F6F9F12algebra

The finite nonempty case and its witness. Let $E = {a_{1}, \dots, a_{m}}$ be finite and nonempty, listed without repetitions by [F12], and put $c_{j} : = 2^{- j} \in (0, \infty)$ and $σ : = \sum_{j = 1}^{m} c_{j} δ_{a_{j}}$ ; by [F9] the set function $σ$ is a finite positive Borel measure carried by $E$ , and for every nonnegative Borel $g$ one has $\int g d σ = \sum_{j = 1}^{m} c_{j} g (a_{j})$ . Put $u (z) : = p_{σ} (z) = \sum_{j = 1}^{m} c_{j} \log ∣ z - a_{j} ∣$ ; since the sum is finite and each $z \mapsto \log ∣ z - a_{j} ∣$ is subharmonic by [F5], [F6] makes $u$ subharmonic on $C$ . At $z = a_{j}$ every summand with index $k \neq j$ is the finite number $\log ∣ a_{j} - a_{k} ∣$ because the points are distinct, while the $j$ -th summand is $- \infty$ , so $u (a_{j}) = - \infty$ ; thus $E$ lies in the $- \infty$ locus of the subharmonic function $u$ , which is not identically $- \infty$ because it is finite at every point outside the finite set $E$ .

1.5F9givenalgebra

The countably infinite case: the measure and the potential. Let $E = {a_{1}, a_{2}, \dots}$ be a listing without repetitions of a countably infinite set, and put $c_{j} : = 2^{- j} / (1 + \log (1 + ∣ a_{j} ∣)) > 0$ and $σ : = \sum_{j = 1}^{\infty} c_{j} δ_{a_{j}}$ . Since $c_{j} (1 + \log (1 + ∣ a_{j} ∣)) = 2^{- j}$ , one has $\sum_{j} c_{j} (1 + \log (1 + ∣ a_{j} ∣)) = 1 < + \infty$ and in particular $σ (C) = \sum_{j} c_{j} \leq 1$ ; by [F9] the weighted sum $σ$ is a finite positive Borel measure with $\int g d σ = \sum_{j} c_{j} g (a_{j})$ for every nonnegative Borel $g$ , so applying this to $g (w) = \log (1 + ∣ w ∣)$ gives the finite logarithmic moment $\int \log (1 + ∣ w ∣) d σ (w) = \sum_{j} c_{j} \log (1 + ∣ a_{j} ∣) \leq 1 < + \infty$ .

2.1F1F9givenalgebra

With $ν$ , $F$ and $b_{i_{0}}$ as in step 1.3, choose $R > diam (F)$ and put $k_{R} (z, w) = k (z, w) + \log R \geq 0$ on $F \times F$ ; at the diagonal point $(b_{i_{0}}, b_{i_{0}})$ one has $k_{R} = + \infty$ , and the atom ${b_{i_{0}}}$ carries $ν$ -mass $m > 0$ . The inner integral at $z = b_{i_{0}}$ is $+ \infty$ : for every real $M$ the nonnegative function $w \mapsto k_{R} (b_{i_{0}}, w)$ satisfies $k_{R} (b_{i_{0}}, w) \geq M 1_{{b_{i_{0}}}} (w)$ , so by monotonicity of the integral this inner integral is at least $M m$ for every real $M$ and hence equals $+ \infty$ . Therefore the iterated double integral of the nonnegative function $k_{R}$ against $ν \otimes ν$ is infinite, and [F1] gives $I (ν) = + \infty$ .

2.2step 1.5F9given

Put $u (z) : = p_{σ} (z) = \int \log ∣ z - w ∣ d σ (w) \in [- \infty, + \infty)$ . At $z = a_{j_{0}}$ the positive part is finite because $\log^{+} ∣ a_{j_{0}} - w ∣ \leq \log (1 + ∣ a_{j_{0}} ∣) + \log (1 + ∣ w ∣)$ has finite $σ$ -integral by step 1.5, while the negative part satisfies $\log^{-} ∣ a_{j_{0}} - w ∣ \geq M 1_{{a_{j_{0}}}} (w)$ for every real $M$ and $σ ({a_{j_{0}}}) = c_{j_{0}} > 0$ , so $\int \log^{-} ∣ a_{j_{0}} - w ∣ d σ (w) = + \infty$ by monotonicity of the integral; therefore $u (a_{j_{0}}) = \int \log^{+} - \int \log^{-} = - \infty$ , that is, $u = - \infty$ on $E$ .

2.3step 1.5givenalgebra

Local decomposition of the potential. Fix $N \geq 1$ with $N \geq ∣ a_{1} ∣$ , so that the disc below meets $E$ , and split $σ$ into the finite positive Borel measures $σ_{N} : = σ ⁣ ↾_{{∣ w ∣ \leq 2 N + 1}}$ and $σ^{N} : = σ ⁣ ↾_{{∣ w ∣ > 2 N + 1}}$ , whose pointwise sum is $σ$ . For $z \in D (0, N)$ and $w$ with $∣ w ∣ > 2 N + 1$ one has $∣ z - w ∣ \geq ∣ w ∣ - N > N + 1 > 1$ , so the two extended integrals $\int_{{∣ w ∣ \leq 2 N + 1}} \log ∣ z - w ∣ d σ (w)$ and $\int_{{∣ w ∣ > 2 N + 1}} \log ∣ z - w ∣ d σ (w)$ have finite positive parts by the logarithmic moment in step 1.5; the compact part may have infinite negative part, while the tail has zero negative part and finite integral. Thus their sum is a well-defined extended integral and $u (z) = p_{σ_{N}} (z) + p_{σ^{N}} (z)$ for $p_{σ^{N}} (z) : = \int_{{∣ w ∣ > 2 N + 1}} \log ∣ z - w ∣ d σ (w) \in R$ .

3.1step 1.3step 2.1F2F3

Since every Borel probability $ν$ on $F$ has $I (ν) = + \infty$ by step 2.1, the characterization [F2] gives $V_{F} = + \infty$ and $cap (F) = 0$ , and the empty compact set also has $cap (\emptyset) = 0$ by [F2]; as $F \subseteq E$ was an arbitrary compact subset, $E$ is capacity-polar. This proves the first assertion for every at most countable $E$ , including the finite case.

3.2step 2.3F4given

The first summand in step 2.3 is subharmonic on $C$ : $σ_{N}$ is a finite positive Borel measure with nonempty compact support $supp σ_{N} \subseteq {∣ w ∣ \leq 2 N + 1}$ , so [F4], whose hypothesis ${A C}_{ω}$ is the standing assumption, gives that $p_{σ_{N}}$ is locally integrable and subharmonic on the domain $C$ .

3.3step 1.5step 2.3F5F8algebra

The second summand of step 2.3 is harmonic on $D (0, N)$ . Fix $w$ with $∣ w ∣ > 2 N + 1$ and write $z = x + i y$ ; on $D (0, N)$ the function $z \mapsto \log ∣ z - w ∣$ is smooth with $∣ z - w ∣ > N + 1$ , and it is harmonic off $w$ by [F5]. The differentiation theorem [F8] applies to the two real parameters: the integrand is $σ^{N}$ -integrable for every $z \in D (0, N)$ since $0 < \log ∣ z - w ∣ \leq \log 2 + \log (1 + ∣ w ∣)$ and $\int \log (1 + ∣ w ∣) d σ^{N} (w) < + \infty$ by step 1.5, and the partial derivatives of order one and two in $x$ and $y$ are bounded on $D (0, N) \times {∣ w ∣ > 2 N + 1}$ by constants $(N + 1)^{- 1}$ and $(N + 1)^{- 2}$ , which are $σ^{N}$ -integrable because $σ^{N} (C) \leq 1$ . Applying [F8] to the $x$ -parameter and to the $y$ -parameter, and then to the resulting first partial derivatives, shows that $p_{σ^{N}}$ is twice continuously differentiable on $D (0, N)$ with second partial derivatives obtained by differentiating under the integral (continuity of these derivatives follows from their pointwise continuity and the same integrable bounds by Dominated convergence); since $\partial_{x}^{2} \log ∣ z - w ∣ + \partial_{y}^{2} \log ∣ z - w ∣ = 0$ for $z \neq w$ by [F5], summing gives $(\partial_{x}^{2} + \partial_{y}^{2}) p_{σ^{N}} (z) = \int (\partial_{x}^{2} + \partial_{y}^{2}) \log ∣ z - w ∣ d σ^{N} (w) = 0$ on $D (0, N)$ , so $p_{σ^{N}}$ is harmonic there by [F5].

4.1step 2.2step 3.2step 3.3F3F5F6F11

On $D (0, N)$ the function $u = p_{σ_{N}} + p_{σ^{N}}$ is subharmonic: the first summand is subharmonic on $C$ , hence on $D (0, N)$ , by step 3.2, the second is harmonic, hence subharmonic by the $C^{2}$ criterion of [F5], and a sum of two subharmonic functions is subharmonic by [F6]. Since every $z_{0} \in C$ lies in some such disc $D (0, N)$ with $N \geq ∣ a_{1} ∣$ and $N > ∣ z_{0} ∣$ , the function $u$ meets the defining conditions of [F11] on the domain $C$ : it is upper semicontinuous because upper semicontinuity is local and holds on each $D (0, N)$ by subharmonicity there, it is not identically $- \infty$ on any $D (0, N)$ because it is subharmonic there, and the circle mean inequality holds at every closed disc of $C$ because each such disc is contained in some $D (0, N)$ on which $u$ is subharmonic. Therefore $u$ is subharmonic on $C$ and not identically $- \infty$ ; by step 2.2 it is $- \infty$ on $E$ , so $E$ lies in the $- \infty$ locus of the explicitly constructed subharmonic function $u$ , and for every $x \in E$ the single neighbourhood $U_{x} : = C$ with witness $u$ exhibits the local condition of [F3], so $E$ is subharmonically polar.

5.1step 1.2step 3.1step 1.4step 4.1F4F12given∎

The first assertion of the statement was proved in step 3.1 for every at most countable $E$ without further choice, the listings being supplied by countability itself ([F12]) and the atom computation being choice-free; steps 1.4 and 4.1 construct the witness in the finite and countably infinite cases, and step 1.2 covers the empty set. The only choice principle spent anywhere is the Countable Choice of [F4] used in step 3.2, which is the standing hypothesis ${A C}_{ω}$ . This proves both assertions.

Remarks

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Two Cantor sets with different logarithmic capacities

Statement

Assume the Axiom of Choice. Let $C \subseteq [0, 1]$ be the middle-thirds Cantor set with Cantor measure $μ_{c}$ (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds, The Cantor measure). Then

$I (μ_{c}) \leq 3 \log 3 < + \infty, hence cap (C) \geq e^{- 3 \log 3} = \frac{1}{27} > 0.$

Let $ℓ_{n} : = e^{- 4^{n}}$ for $n \geq 1$ and let $K \subseteq [0, 1]$ be the nested binary Cantor set constructed as follows: $K_{0} = [0, 1]$ is one closed cell, and each level- $(n - 1)$ cell $[a, a + s]$ is replaced by the two disjoint closed cells $[a, a + ℓ_{n}]$ and $[a + s - ℓ_{n}, a + s]$ , with $K_{n}$ the union of the resulting $2^{n}$ cells and $K : = ⋂_{n \geq 0} K_{n}$ . Then every Borel probability $ν$ on $K$ has $I (ν) = + \infty$ , so $V_{K} = + \infty$ and $cap (K) = 0$ .

Both $C$ and $K$ are uncountable compact Lebesgue-null subsets of $[0, 1]$ . Thus Lebesgue measure and cardinality alone do not determine logarithmic capacity.

Facts & Assumptions

Given: the Cantor set $C$ and its Cantor measure $μ_{c}$ , the number $ℓ_{n} = e^{- 4^{n}}$ , the Axiom of Choice, and the capacity and energy conventions of Robin constant and logarithmic capacity of a compact set and Logarithmic potential and energy of a positive compactly supported measure.

[F1]

For a finite positive Borel measure $ν$ of compact support and $R > diam (supp ν)$ one has $I (ν) = \iint k_{R} d ν d ν - ν (C)^{2} \log R$ with $k_{R} (z, w) = \log (R / ∣ z - w ∣)$ , and $V_{K} = \inf_{ν \in P (K)} I (ν)$ with $cap (K) = e^{- V_{K}}$ when $V_{K} < + \infty$ and $0$ otherwise (Logarithmic potential and energy of a positive compactly supported measure, Robin constant and logarithmic capacity of a compact set).

[F2]

The Cantor set $C = ⋂_{n} C_{n}$ is compact, uncountable and $λ_{1} (C) = 0$ ; every $x \in C$ is $Φ (a) = \sum_{k \geq 0} a_{k} 3^{- k - 1}$ for a unique sequence $a$ with values in ${0, 2}$ , the first $m$ digits determine the level- $m$ basic interval $I_{b} = [\sum_{j = 1}^{m} 2 b_{j} 3^{- j}, \sum_{j = 1}^{m} 2 b_{j} 3^{- j} + 3^{- m}]$ , and $b \mapsto Φ ((2 b_{k})_{k \geq 0})$ is a bijection from ${0, 1}^{N}$ onto $C$ (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds, The Cantor set is exactly the set of $\sum_{k \geq 1} a_{k} 3^{- k}$ with every $a_{k} \in {0, 2}$ , and this gives a bijection with ${0, 1}^{N}$ , The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points).

[F3]

Assume Countable Choice. The Cantor measure $μ_{c}$ is a Borel probability measure with $μ_{c} (R ∖ C) = 0$ , it is atomless, and $μ_{c} (I_{b}) = 2^{- m}$ for every level- $m$ basic interval $I_{b}$ (The Cantor measure, The Cantor measure is a singular atomless probability measure concentrated on the Cantor set, Cantor basic intervals have their expected masses).

[F4]

Layer cake for $p = 1$ : for a measure space $(X, A, ρ)$ and a measurable $f \geq 0$ one has $\int_{X} f d ρ = \int_{0}^{\infty} ρ ({f > t}) d t$ , both sides allowed to be $+ \infty$ (For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function).

[F5]

The product measure $μ_{c} \otimes μ_{c}$ is a measure on $R^{2}$ with $(μ_{c} \otimes μ_{c}) (A \times B) = μ_{c} (A) μ_{c} (B)$ , and Tonelli's theorem computes integrals of nonnegative product-measurable integrands as iterated integrals (The product measure of two sigma-finite measure spaces, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[F6]

Finite Cauchy–Schwarz: $(\sum_{k < n} a_{k})^{2} \leq n \sum_{k < n} a_{k}^{2}$ for reals $a_{k}$ (The Cauchy-Schwarz inequality for finite sums).

[F7]

A nested sequence of closed bounded intervals whose lengths tend to $0$ has an intersection that is exactly one point (A nested sequence of nonempty closed bounded intervals has nonempty intersection, and the intersection is a single point exactly when the lengths tend to $0$ ), and the recursive construction of the families $(K_{n})$ is licensed by the recursion theorem (The recursion theorem).

[F8]

Under Countable Choice, the Axiom of Choice yields Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), AC implies DC implies countable choice); a subset of a countable set is countable (Finite, countably infinite, countable, uncountable); and a set is Lebesgue-null when it is contained in the union of countably many intervals of arbitrarily small total length (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

Verification

technique · direct

1.1F2F3F8

By [F3] and [F8] the Cantor measure $μ_{c}$ is a Borel probability with $μ_{c} (C) = 1$ and no atoms, so $μ_{c} \otimes μ_{c} ({(x, x) : x \in C}) = \int μ_{c} ({x}) d μ_{c} (x) = 0$ ; by [F2] the set $C$ is compact and uncountable with $λ_{1} (C) = 0$ .

1.2F2F3

For every $m \geq 0$ and every word $b \in {0, 1}^{m}$ the level- $m$ basic interval $I_{b}$ has $μ_{c} (I_{b}) = 2^{- m}$ , and by [F2] the intervals $I_{b}$ , $b \in {0, 1}^{m}$ , are the digit cylinders and are pairwise disjoint.

2.1step 1.2F2F5

If $x = Φ (a)$ and $y = Φ (a^{'})$ have different first $m$ digits, let $k < m$ be the first index with $a_{k} \neq a_{k}^{'}$ ; then $∣ x - y ∣ \geq 2 \cdot 3^{- k - 1} - \sum_{j > k} 2 \cdot 3^{- j - 1} = 3^{- k - 1} \geq 3^{- m}$ , so ${∣ x - y ∣ < 3^{- m}}$ is contained in the set where the first $m$ digits agree, that is, in $⋃_{b} I_{b} \times I_{b}$ with the $I_{b}$ of step 1.2. Hence $(μ_{c} \otimes μ_{c}) ({∣ x - y ∣ < 3^{- m}}) \leq \sum_{b \in {0, 1}^{m}} μ_{c} (I_{b})^{2} = 2^{m} \cdot 2^{- 2 m} = 2^{- m}$ .

2.2step 1.1F7F8

The construction of the cells is licensed by [F7]. For $n = 1$ , $2 ℓ_{1} = 2 e^{- 4} < 1 = ℓ_{0}$ ; for $n \geq 2$ , $ℓ_{n} = ℓ_{n - 1}^{4}$ and $ℓ_{n - 1} \leq e^{- 4}$ , so $2 ℓ_{n} / ℓ_{n - 1} = 2 ℓ_{n - 1}^{3} \leq 2 e^{- 12} < 1$ . Thus the two children of each level- $(n - 1)$ cell are disjoint closed intervals of length $ℓ_{n}$ contained in it, and $(K_{n})$ is a nested sequence of nonempty compact sets with $2^{n}$ cells of length $ℓ_{n}$ at level $n$ . Therefore $K = ⋂_{n} K_{n}$ is compact and nonempty, and $λ_{1} (K) = 0$ because the level- $n$ cells cover $K$ and their total length $2^{n} ℓ_{n} = 2^{n} e^{- 4^{n}} \to 0$ .

3.1step 2.1F1F4F5

Since $μ_{c}$ has total mass $1$ and support in $[0, 1]$ , [F1] gives $I (μ_{c}) = \iint \log \frac{1}{∣ x - y ∣} d μ_{c} (x) d μ_{c} (y)$ ; applying [F4] on the product measure of [F5] and splitting the integral at $t = m \log 3$ , $I (μ_{c}) = \int_{0}^{\infty} (μ_{c} \otimes μ_{c}) ({∣ x - y ∣ < e^{- t}}) d t \leq \log 3 + \sum_{m \geq 0} (\log 3) (μ_{c} \otimes μ_{c}) ({∣ x - y ∣ < 3^{- m}}) \leq \log 3 + (\log 3) \sum_{m \geq 0} 2^{- m} = 3 \log 3 < + \infty$ , where step 2.1 bounds each dyadic piece. Therefore $V_{C} \leq I (μ_{c}) < + \infty$ and $cap (C) = e^{- V_{C}} \geq e^{- 3 \log 3} = \frac{1}{27} > 0$ .

3.2step 2.2F6

For a Borel probability $ν$ on $K$ put $m_{n, k} : = ν (I_{n, k})$ for the $2^{n}$ level- $n$ cells $I_{n, k}$ of step 2.2; since $ν$ is carried by $K \subseteq ⋃_{k} I_{n, k}$ and the cells are pairwise disjoint, $\sum_{k} m_{n, k} = 1$ , so by [F6] with the constant list $1$ one has $1 = (\sum_{k} m_{n, k})^{2} \leq 2^{n} \sum_{k} m_{n, k}^{2}$ , that is, $\sum_{k} m_{n, k}^{2} \geq 2^{- n}$ .

4.1step 3.2F1F4F5

With $ν$ , the cells $I_{n, k}$ and the masses $m_{n, k}$ of step 3.2, [F1] gives $I (ν) = \iint \log \frac{1}{∣ x - y ∣} d ν (x) d ν (y)$ because $ν$ is a probability on $[0, 1]$ ; two points of one level- $n$ cell satisfy $∣ x - y ∣ \leq ℓ_{n}$ , so for $t < 4^{n}$ the event of lying in the same level- $n$ cell is contained in ${∣ x - y ∣ < e^{- t}}$ and hence $(ν \otimes ν) ({∣ x - y ∣ < e^{- t}}) \geq \sum_{k} m_{n, k}^{2} \geq 2^{- n}$ ; by [F4] and [F5], $I (ν) \geq \sum_{n \geq 1} \int_{4^{n - 1}}^{4^{n}} 2^{- n} d t = \sum_{n \geq 1} (4^{n} - 4^{n - 1}) 2^{- n} = \frac{3}{4} \sum_{n \geq 1} 2^{n} = + \infty$ .

5.1step 4.1F1

Every $ν \in P (K)$ has $I (ν) = + \infty$ by step 4.1, so the infimum $V_{K}$ is $+ \infty$ and $cap (K) = 0$ by [F1].

6.1step 1.1step 3.1step 2.2step 5.1F2F7F8∎

For each $b \in {0, 1}^{N}$ the cells $I_{n, b ↾ n}$ form a nested family of closed intervals with lengths $ℓ_{n} \to 0$ , so by [F7] their intersection contains exactly one point $φ (b) \in K$ ; distinct infinite words differ at some level $n$ , where their cells are disjoint, so $φ$ is injective. If $K$ were countable then its subset $φ ({0, 1}^{N})$ would be countable by [F8], and since ${0, 1}^{N}$ is in bijection with $C$ by [F2] and $C$ is uncountable, that is impossible; hence $K$ is uncountable. Thus $C$ has positive capacity and $K$ has zero capacity although both are uncountable compact Lebesgue-null sets.

Remarks

Where the thin geometric decay is used. In step 4.1 the level- $n$ cells have length $ℓ_{n} = e^{- 4^{n}}$ , so the time window $(4^{n - 1}, 4^{n})$ in the layer-cake formula sees the whole level- $n$ cell mass; the divergent series $\sum_{n} 3 \cdot 2^{n - 1}$ is what forces $I (ν) = + \infty$ . The zero-capacity conclusion here uses the divergent weighted logarithmic windows, not merely summability of $\sum_{n} 2^{n} ℓ_{n}$ or decay faster than every exponential. For example, lengths $ℓ_{n} = e^{- n^{2}}$ also decay faster than every exponential, but the equal-branch probability (the pushforward of $μ_{c}$ under the binary coding map) has finite energy: pairs first separated at level $n$ have distance at least $ℓ_{n - 1} - 2 ℓ_{n}$ , with $ℓ_{0} = 1$ , and have probability $2^{- n}$ , while the diagonal has probability zero because the probability of agreeing through level $m$ is $2^{- m} \to 0$ . Thus their energy contribution is bounded by $2^{- n} ((n - 1)^{2} + C)$ for a fixed constant $C$ , a summable series. The middle-thirds scaling likewise gives finite energy by step 3.1.

Choice. The statement assumes the Axiom of Choice, but the proof uses only Countable Choice, through the Cantor measure and cylinder-mass suppliers [F3] and the general conversion [F8]; with those suppliers granted, the construction of $K$ , the layer-cake computations and the cardinality argument are choice-free.

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Riesz measure of a log modulus records the holomorphic zeros

Statement

Assume Dependent Choice. Let $Ω \subseteq C$ be a complex domain and let $f$ be holomorphic on $Ω$ , not identically zero on any connected component of $Ω$ . Put $u : = \log ∣ f ∣$ , subharmonic on $Ω$ (The logarithm of the modulus of a holomorphic function is subharmonic), with Riesz measure $μ_{u} = (2 π)^{- 1} Δ u$ (Distributional Riesz measure of a plane subharmonic function). Then

$μ_{u} = \sum_{a \in Z (f)} {ord}_{a} (f) δ_{a},$

where $Z (f) = {a \in Ω : f (a) = 0}$ , the integers ${ord}_{a} (f) \geq 1$ are the vanishing orders (The order of a zero is the exponent in its local holomorphic factorization) and $δ_{a}$ is the unit Dirac measure at $a$ (The Dirac set function at a point); the sum is a locally finite positive measure on $Ω$ . In particular $\log ∣ f ∣$ has no Riesz mass on $Ω ∖ Z (f)$ .

Facts & Assumptions

Given: a complex domain $Ω$ , a holomorphic $f$ on $Ω$ not identically zero on any component, the function $u = \log ∣ f ∣$ , and Dependent Choice.

[F1]

A holomorphic function on a complex domain that vanishes on a neighbourhood of a point vanishes identically: that neighbourhood supplies an accumulating set of zeros for Identity theorem for holomorphic functions. Thus the given nonzero $f$ cannot have infinite order anywhere, since infinite order is equivalent to local vanishing by The order of a zero is the exponent in its local holomorphic factorization. For $a \in Ω$ , a holomorphic function has finite order $m = {ord}_{a} (f)$ at $a$ exactly when $f (z) = (z - a)^{m} g (z)$ on a neighbourhood of $a$ with $g$ holomorphic and $g (a) \neq 0$ ; $f (a) = 0$ holds exactly when $m \geq 1$ , and if $m = 0$ then $f \neq 0$ near $a$ (The order of a zero is the exponent in its local holomorphic factorization).

[F2]

The function $u = \log ∣ f ∣$ is subharmonic on $Ω$ (The logarithm of the modulus of a holomorphic function is subharmonic); a zero-free holomorphic $h$ on a disc admits $h = \exp L$ with $L$ holomorphic (A nonvanishing holomorphic function on a disc has a holomorphic logarithm); a holomorphic function on an open set is smooth, and its real part is $C^{2}$ with $Δ (Re L) = 0$ wherever $L$ is holomorphic (Holomorphic functions are real analytic and smooth in their two real coordinates, The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair).

[F3]

Under Dependent Choice the Riesz functional $μ_{u} (φ) = (2 π)^{- 1} \int_{Ω} u Δ φ d A$ is a positive Radon measure, and it is the unique positive Radon measure representing $μ_{u}$ on $C_{c}^{\infty} (Ω)$ (The distributional Riesz functional of a subharmonic function is a positive Radon measure, Distributional Riesz measure of a plane subharmonic function); moreover $(2 π)^{- 1} \int \log ∣ z - a ∣ Δ φ d A = φ (a)$ for every $φ \in C_{c}^{\infty} (C)$ , that is, $μ_{\log ∣ \cdot - a ∣} = δ_{a}$ (Distributional Laplacian of a compact logarithmic potential); Dependent Choice yields Countable Choice (Dependent choice implies countable choice).

[F4]

For $h \in C^{2}$ on an open set, $Δ T_{h} = T_{Δ h}$ ; in particular if $Δ h = 0$ pointwise then $\int h Δ φ d A = 0$ for every compactly supported smooth test function $φ$ (Distributional differentiation is continuous and commutes).

[F5]

A $δ_{a}$ is a probability measure concentrated at $a$ , finite or countable nonnegative sums of measures are measures, and every bounded infinite subset of $R^{2} ≅ C$ has an accumulation point (The Dirac set function at a point, Nonnegative scalar multiples and countable weighted sums of measures are measures, For $n \geq 1$ every bounded sequence in $R^{n}$ has a convergent subsequence).

[F6]

If a compact $C$ lies in an open set $U \subseteq R^{2}$ , there is a smooth compactly supported cutoff in $U$ equal to $1$ on a neighbourhood of $C$ (Test function cutoffs and euclidean localization).

[F7]

Under Countable Choice, every Borel measure finite on compact sets on a second-countable locally compact Hausdorff space is Radon (Locally finite Borel measures on second-countable LCH spaces are regular); Dependent Choice supplies Countable Choice by Dependent choice implies countable choice.

[F8]

The plane is second-countable, locally compact and Hausdorff, and each open subset inherits these properties ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis, $R^{n}$ is locally compact and $σ$ -compact, Distinct points of a metric space have disjoint balls around them, In a locally compact Hausdorff space every point has a neighbourhood base of compact sets, every open subspace and every closed subspace is locally compact, every open set around a point contains an open set with compact closure inside it, and every compact set sits inside an open set with compact closure, $T_{0}$ , $T_{1}$ , and Hausdorffness are hereditary).

Verification

technique · direct

1.1F1given

Let $a \in Ω$ with $f (a) = 0$ and let $m : = {ord}_{a} (f) \geq 1$ by [F1]. By [F1] there are a disc $D (a, r) \subseteq Ω$ and a holomorphic zero-free $g$ on it with $f (z) = (z - a)^{m} g (z)$ , so $f (z) \neq 0$ for $0 < ∣ z - a ∣ < r$ : every zero of $f$ is isolated.

1.2F6given

Fix $φ \in C_{c}^{\infty} (Ω)$ . If $φ = 0$ the identity is immediate; otherwise put $K_{0} : = supp φ$ . For each $x \in K_{0}$ there are concentric relatively compact discs $V ⋐ D ⋐ Ω$ with $x \in V$ and $\overline{D}$ containing at most one zero of $f$ , since zeros are isolated. These smaller discs form an open cover of $K_{0}$ ; compactness gives finitely many pairs $(V_{i}, D_{i})$ with the $V_{i}$ covering $K_{0}$ . By [F6], choose $β_{i} \in C_{c}^{\infty} (D_{i})$ with $β_{i} \geq 0$ and $β_{i} = 1$ on a neighbourhood of $\overline{V_{i}}$ . On $W : = ⋃_{i} V_{i}$ the sum $S : = \sum_{i} β_{i}$ is positive. Define $φ_{i} : = φ β_{i} / S$ on $W$ and zero off $W$ ; since $K_{0} ⋐ W$ , each $φ_{i} \in C_{c}^{\infty} (D_{i})$ , and $\sum_{i} φ_{i} = φ$ .

2.1step 1.1F5F7F8

Let $K \subseteq Ω$ be compact and suppose it contained infinitely many distinct zeros of $f$ . By [F5] the infinite bounded set $Z (f) \cap K$ has an accumulation point $z_{*} \in K \subseteq Ω$ ; continuity gives $f (z_{*}) = 0$ , contradicting the isolation of zeros from step 1.1. Thus each compact subset meets $Z (f)$ finitely. Since $Ω$ is second-countable and every zero is isolated, $Z (f)$ is at most countable: assign each zero the least element of a fixed enumerated basis that contains it and no other zero; distinct zeros receive distinct basis elements. The countable sum $λ : = \sum_{a \in Z (f)} {ord}_{a} (f) δ_{a}$ is a positive Borel measure by [F5], locally finite by the compact finiteness just proved. The open set $Ω$ is second-countable and locally compact Hausdorff by [F8]; Dependent Choice supplies the Countable Choice of [F7], so $λ$ is Radon.

2.2step 1.2F1F2F3F4

For each $D_{i}$ from step 1.2, either $f$ has no zero, or it has exactly one zero $a_{i}$ of order $m_{i}$ . In the latter case the local factorization [F1] extends $F_{i} (z) : = f (z) / (z - a_{i})^{m_{i}}$ holomorphically and without zeros throughout $D_{i}$ ; in the zero-free case set $F_{i} : = f$ and $m_{i} : = 0$ . Then $u = m_{i} \log ∣ z - a_{i} ∣ + \log ∣ F_{i} ∣$ on $D_{i}$ when $m_{i} > 0$ , and $u = \log ∣ F_{i} ∣$ when $m_{i} = 0$ . By [F2], $\log ∣ F_{i} ∣$ is harmonic, so its Riesz functional vanishes by [F4]; the point-mass normalization [F3] therefore gives $μ_{u} (φ_{i}) = m_{i} φ_{i} (a_{i}) = \int φ_{i} d λ$ when $m_{i} > 0$ , and both sides are zero when $m_{i} = 0$ .

3.1step 1.2step 2.2

Summing the local identities of step 2.2 over the finite partition $φ = \sum_{i} φ_{i}$ from step 1.2 gives $μ_{u} (φ) = \sum_{i} μ_{u} (φ_{i}) = \sum_{i} \int φ_{i} d λ = \int φ d λ$ for every $φ \in C_{c}^{\infty} (Ω)$ .

4.1step 2.1step 3.1F3∎

Since $λ$ is a positive Radon measure by step 2.1 that represents $μ_{u}$ on all smooth compactly supported tests, the uniqueness clause of [F3] gives $μ_{u} = λ = \sum_{a \in Z (f)} {ord}_{a} (f) δ_{a}$ , which is the asserted formula; in particular every compact subset of $Ω ∖ Z (f)$ carries no $λ$ -mass, so $\log ∣ f ∣$ has no Riesz mass off the zero set.

Remarks

The vanishing order is exactly the Riesz mass. Step 3.1 shows the mass at a zero $a$ is the order $m = {ord}_{a} (f)$ , not merely a positive integer: the factor $m \log ∣ z - a ∣$ contributes $m δ_{a}$ through the normalization $Δ \log ∣ z - a ∣ = 2 π δ_{a}$ , while the zero-free factor $F$ contributes nothing.

Finite local cover. Each compact test support is covered by finitely many discs on which the zero divisor has at most one point. A finite smooth partition subordinate to this cover reduces the distributional identity to the local factorization at each zero.

Choice. Dependent Choice is used through Countable Choice in the Riesz representation, kernel-normalization and local-regularity suppliers, and through the Bolzano–Weierstrass accumulation step. The finite cover, cutoffs, factorization and harmonicity calculations use no further choice.

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Infinity-pole Green function recovered from a circular conductor

Statement

Assume the Axiom of Choice. Let $a \in C$ , $r > 0$ , let $K : = \overline{D (a, r)}$ be the closed disc, and let $Ω : = {z \in C : ∣ z - a ∣ > r}$ be its exterior. Then the normalized infinity-pole Green function of $Ω$ is

$g_{Ω} (z, \infty) = g (z) : = \log \frac{∣ z - a ∣}{r} (z \in Ω),$

so that $g_{Ω} (z, \infty) = V_{K} - U^{μ_{K}} (z)$ on $Ω$ . Moreover $g$ has boundary limit $0$ at every point of the boundary circle ${z : ∣ z - a ∣ = r}$ , with no exceptional set, and $g (z) - \log ∣ z ∣ \to - \log r$ as $∣ z ∣ \to \infty$ .

The Axiom of Choice is spent through the equilibrium-measure input (Capacity of a disc and its circular equilibrium measure and Green function at infinity from the equilibrium potential); the explicit radial computations for $g$ are choice-free.

Facts & Assumptions

Given: $a \in C$ , $r > 0$ , the closed disc $K : = \overline{D (a, r)}$ , its exterior $Ω : = {z : ∣ z - a ∣ > r}$ , the Axiom of Choice, and the conventions of Logarithmic potential and energy of a positive compactly supported measure, Robin constant and logarithmic capacity of a compact set, Capacity-polar sets, quasi-everywhere, and subharmonic polar sets and Green function with a pole at infinity.

[F1]

Assume the Axiom of Choice. With $K : = \overline{D (a, r)}$ and $μ$ the normalized arclength measure $d μ = d t / (2 π)$ on the circle $w = a + r e^{i t}$ , $μ$ is the unique equilibrium measure of $K$ , $U^{μ} (z) = \log \frac{1}{r} (∣ z - a ∣ \leq r), U^{μ} (z) = \log \frac{1}{∣ z - a ∣} (∣ z - a ∣ \geq r),$ and $cap (K) = r$ with Robin constant $V_{K} = \log (1 / r)$ ; the same potential, capacity and equilibrium measure hold for the boundary circle (Capacity of a disc and its circular equilibrium measure).

[F2]

A Green function of $Ω$ with pole at infinity and Robin constant $V_{K}$ is a function $g : Ω \to R$ that is positive and harmonic on $Ω$ , satisfies $g (z) - \log ∣ z ∣ \to V_{K}$ as $∣ z ∣ \to \infty$ , is locally bounded near every point of $\partial Ω$ , and has boundary limit $0$ outside a Borel capacity-polar subset of $\partial Ω$ ; if existence and uniqueness hold the function is written $g_{Ω} (\cdot, \infty)$ (Green function with a pole at infinity).

[F3]

A compact set $K$ with $cap (K) > 0$ is nonempty, $Ω : = Ω (K)$ denotes the unbounded connected component of $C ∖ K$ and is a complex domain with compact boundary $\partial Ω \subseteq K$ , and $V_{K} = \inf_{μ \in P (K)} I (μ)$ is a real number (Green function with a pole at infinity, Robin constant and logarithmic capacity of a compact set).

[F4]

Assume the Axiom of Choice. For $K \subseteq C$ compact with $cap (K) > 0$ , the unbounded component $Ω$ of $C ∖ K$ , the equilibrium measure $μ_{K}$ and $V_{K} = \log \frac{1}{cap (K)}$ , the function $g (z) = V_{K} - U^{μ_{K}} (z)$ satisfies properties 1-4 of [F2], every function satisfying properties 1-4 equals it, and $g_{Ω} (z, \infty) = V_{K} - U^{μ_{K}} (z)$ on $Ω$ ; hence $K$ has exactly one Green function with pole at infinity (Green function at infinity from the equilibrium potential).

[F5]

For every $a \in C$ the function $u_{a} (z) = \log ∣ z - a ∣$ is smooth and harmonic on $C ∖ {a}$ ; no choice principle is required (Logarithmic modulus is harmonic off its centre, Plane harmonic functions).

[F6]

If $K \subseteq C$ is compact, the complement $C ∖ K$ has exactly one unbounded connected component and every other component is bounded; for $K = \overline{D (a, r)}$ that component is ${z : ∣ z - a ∣ > r}$ , which is therefore a complex domain (The complement of a compact plane set has exactly one unbounded connected component, A complex domain is a nonempty connected open subset of $C$ ).

[F7]

A set is capacity-polar when every compact subset of it has capacity zero; $\emptyset$ is capacity-polar, and a subset of a capacity-polar set is capacity-polar (Capacity-polar sets, quasi-everywhere, and subharmonic polar sets).

[F8]

The Axiom of Choice implies Dependent Choice, which implies Countable Choice (AC implies DC implies countable choice).

Verification

technique · direct

1.1F1F3F6given

Setup. With $K = \overline{D (a, r)}$ and $Ω = {z : ∣ z - a ∣ > r}$ , [F6] identifies $Ω$ with the unbounded connected component of $C ∖ K$ and makes it a complex domain with $\partial Ω = {z : ∣ z - a ∣ = r}$ ; by [F1] $cap (K) = r > 0$ , $V_{K} = \log \frac{1}{r}$ , and the equilibrium measure $μ_{K}$ has potential $U^{μ_{K}} (z) = \log \frac{1}{∣ z - a ∣}$ for $∣ z - a ∣ \geq r$ and $U^{μ_{K}} (z) = \log \frac{1}{r}$ for $∣ z - a ∣ \leq r$ .

1.2F5algebragiven

Positivity and harmonicity. Define $g (z) : = \log \frac{∣ z - a ∣}{r}$ for $z \in Ω$ . Then $g (z) > 0$ on $Ω$ , since $∣ z - a ∣ > r$ ; and $g$ is harmonic on $Ω$ , because $\log ∣ z - a ∣$ is harmonic on the open set $C ∖ {a} \supseteq Ω$ by [F5] and subtracting the constant $\log r$ leaves its Laplacian zero.

1.3givenalgebra

Boundary values, local boundedness and the infinity normalization. If $ξ \in \partial Ω$ , that is $∣ ξ - a ∣ = r$ , then for $z \in Ω$ , $∣ g (z) - 0 ∣ = \log \frac{∣ z - a ∣}{r} \to \log \frac{r}{r} = 0$ as $z \to ξ$ , and for every $δ > 0$ one has $\sup {g (z) : z \in Ω, ∣ z - ξ ∣ < δ} \leq \log \frac{r + δ}{r} < + \infty$ ; moreover $g (z) - \log ∣ z ∣ = \log \frac{∣ z - a ∣}{∣ z ∣} - \log r \to - \log r$ as $∣ z ∣ \to \infty$ , because $\frac{∣ z - a ∣}{∣ z ∣} \to 1$ and $\log$ is continuous at $1$ .

2.1step 1.1step 1.2F1

Identification with the equilibrium potential. On $Ω$ one has $∣ z - a ∣ > r$ , so by [F1] $U^{μ_{K}} (z) = \log \frac{1}{∣ z - a ∣}$ there and $V_{K} - U^{μ_{K}} (z) = \log \frac{1}{r} - \log \frac{1}{∣ z - a ∣} = \log \frac{∣ z - a ∣}{r} = g (z) .$

2.2step 1.2step 1.3F2F7

The explicit function is a Green function. Steps 1.2 and 1.3 give properties 1, 2 and 3 of [F2] for $g$ with Robin constant $V_{K} = \log \frac{1}{r}$ ; property 4 holds with the exceptional set $E : = \emptyset$ , because step 1.3 gives the boundary limit $0$ at every point of $\partial Ω$ , and $\emptyset$ is Borel and capacity-polar by [F7]. Hence $g$ is a Green function of $Ω$ with pole at infinity and Robin constant $V_{K}$ in the sense of [F2].

3.1step 1.1step 2.1step 2.2F4

Uniqueness and the notation. By step 1.1, $K$ is compact with $cap (K) = r > 0$ , so [F4] applies and gives: the Green function of $Ω$ with pole at infinity exists, every function satisfying properties 1-4 of [F2] equals $V_{K} - U^{μ_{K}}$ , and the notation $g_{Ω} (\cdot, \infty)$ is licensed with $g_{Ω} (z, \infty) = V_{K} - U^{μ_{K}} (z)$ on $Ω$ . By step 2.2 the function $g$ satisfies properties 1-4, so $g = V_{K} - U^{μ_{K}} = g_{Ω} (\cdot, \infty)$ on $Ω$ by step 2.1, and the boundary and normalization assertions are step 1.3.

4.1step 2.1step 3.1F1F4F8∎

Assembly and choice. Assertions of the Statement are exactly steps 2.1 and 3.1 (identification, notation, boundary limit on the entire circle and the infinity normalization); the boundary set is empty, so no exceptional set is needed. The Axiom of Choice is used only through [F1] and [F4], which by [F8] also supply Dependent and Countable Choice to their equilibrium-measure and Frostman inputs; the radial computations of steps 1.2, 1.3 and 2.1 are choice-free.

Remarks

Direct radial computation, not the general quasi-everywhere machinery. The properties of $g$ are verified here by direct radial computation: positivity and harmonicity came from $\log ∣ z - a ∣$ being harmonic off its centre, the boundary limit $0$ holds at every boundary point because $g = \log (∣ z - a ∣ / r)$ extends continuously to the closed exterior with value $0$ on the circle, and the normalization at infinity is the elementary limit $\log (∣ z - a ∣ / ∣ z ∣) \to 0$ . In particular the exceptional set in property 4 of Green function with a pole at infinity may be taken empty here; the general quasi-everywhere uniqueness theorem (Green function at infinity from the equilibrium potential) is invoked only for the uniqueness clause, where the ordinary maximum principle on the exterior alone would not suffice for candidates whose boundary limit is assumed only quasi-everywhere.

Logarithmic Potential, Capacity, and Riesz Decomposition: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Capacity of a disc and its circular equilibrium measure

Statement

Facts & Assumptions

Verification

Remarks

Arcsine equilibrium measure and capacity of a segment

Statement

Facts & Assumptions

Verification

Remarks

Chebyshev extremals and the exact disk Fekete polynomial

Statement

Facts & Assumptions

Verification

Remarks

Chebyshev extremal nodes converge to the arcsine equilibrium measure

Statement

Facts & Assumptions

Verification

Remarks

Finite and countable planar sets have zero logarithmic capacity

Statement

Facts & Assumptions

Verification

Remarks

Two Cantor sets with different logarithmic capacities

Statement

Facts & Assumptions

Verification

Remarks

Riesz measure of a log modulus records the holomorphic zeros

Statement

Facts & Assumptions

Verification

Remarks

Infinity-pole Green function recovered from a circular conductor

Statement

Facts & Assumptions

Verification

Remarks

Sources