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Frostman inequalities and quasi-everywhere equilibrium equality

Statement

Assume the Axiom of Choice. Let $K \subseteq C$ be compact with $cap (K) > 0$ and let $μ_{K}$ be its equilibrium measure (Existence and uniqueness of the equilibrium measure). Then

$U^{μ_{K}} (z) \leq V_{K} for every z \in C,$

and $U^{μ_{K}} (z) = V_{K}$ outside a Borel capacity-polar subset of $K$ ; the exceptional set may be taken to be a countable union of compact sets of capacity zero. Here $U^{μ_{K}}$ , $V_{K}$ and cap are those of Logarithmic potential and energy of a positive compactly supported measure, Robin constant and logarithmic capacity of a compact set and the polarity convention of Capacity-polar sets, quasi-everywhere, and subharmonic polar sets.

The Axiom of Choice is spent through the equilibrium-measure existence theorem Existence and uniqueness of the equilibrium measure, applied to $K$ and to the nonpolar compact subsets of $K$ that occur in the argument, and through the Evans-potential lemma Compact capacity-zero sets and subharmonic minus-infinity loci used for the countable-union closure of the polar class; that lemma assumes only Dependent Choice, which the Axiom of Choice supplies. The potential and energy estimates themselves are choice-free.

Facts & Assumptions

Given: a nonempty compact set $K \subseteq C$ with $cap (K) > 0$ , its equilibrium measure $μ_{K}$ , the Axiom of Choice, and the potential, energy, capacity and polarity 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 Support of a finite Borel measure on the plane.

[F1]

Let $ρ, σ$ be finite positive Borel measures carried by a common compact set $L$ , and take $R > \max {1, diam L}$ . Then $k_{R} = k + \log R \geq 0$ on $L \times L$ . If $0 \leq σ \leq ρ$ , product measure monotonicity gives $σ \otimes σ \leq ρ \otimes ρ$ , hence $I (σ) \leq I (ρ) + (ρ (C)^{2} - σ (C)^{2}) \log R,$ which is finite when $I (ρ) < + \infty$ . If $ρ$ and $σ$ have equal total mass and finite energies, Countable Choice and Strict positivity of logarithmic energy for a zero-mass signed charge give a finite mixed energy and $I (ρ, σ) = \frac{1}{2} (I (ρ) + I (σ) - I (ρ - σ)) \leq \frac{1}{2} (I (ρ) + I (σ)) .$ No pointwise monotonicity $U^{σ} \leq U^{ρ}$ is asserted: the unshifted kernel changes sign (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

For a finite positive Borel measure $ν$ of compact support, $U^{ν} (z) = \int k (z, w) d ν (w)$ with $k (z, w) = \log \frac{1}{∣ z - w ∣} \in (- \infty, + \infty]$ and diagonal value $+ \infty$ , $p_{ν} = - U^{ν}$ is the subharmonic normalisation, $I (ν) = \iint k d ν d ν \in (- \infty, + \infty]$ computed from the shifted nonnegative kernel $k_{R} = k + \log R$ for $R > diam supp ν$ , and the mixed energy $I (ν, ρ) = \iint k d ν d ρ$ is symmetric (Logarithmic potential and energy of a positive compactly supported measure). If $diam supp ν > 0$ then $k \geq - \log diam supp ν$ on $supp ν \times supp ν$ , so $I (ν) = \iint k d ν d ν > - \infty$ .

[F2]

$V_{K} = \inf_{ν \in P (K)} I (ν) \in (- \infty, + \infty]$ and $cap (K) = e^{- V_{K}}$ when $V_{K} < + \infty$ and $0$ otherwise; so $cap (K) > 0$ is equivalent to $V_{K} < + \infty$ (Robin constant and logarithmic capacity of a compact set).

[F3]

The support $S = supp μ$ of a finite positive Borel measure $μ$ is closed, carries $μ$ , is contained in every closed carrier, and $μ \neq 0$ if and only if $S \neq \emptyset$ ; in particular every ball about a point of $S$ has positive $μ$ -measure (Support of a finite Borel measure on the plane).

[F4]

Capacity-polar means that every compact subset has capacity zero; quasi-everywhere means outside a Borel capacity-polar set (Capacity-polar sets, quasi-everywhere, and subharmonic polar sets, Robin constant and logarithmic capacity of a compact set).

[F5]

Assume the Axiom of Choice: every nonempty compact $K$ with $cap (K) > 0$ has exactly one equilibrium measure $μ_{K}$ , and $I (μ_{K}) = V_{K} = \inf_{P (K)} I < + \infty$ ; if $cap (K) = 0$ then $V_{K} = + \infty$ and no equilibrium measure is asserted (Existence and uniqueness of the equilibrium measure).

[F6]

$p_{ν}$ is subharmonic on $C$ for every finite positive compactly supported $ν$ , hence upper semicontinuous with values in $[- \infty, \infty)$ , and $U^{ν} = - p_{ν}$ is lower semicontinuous with values in $(- \infty, + \infty]$ ; consequently every sublevel set ${U^{ν} \leq c}$ is closed (Distributional Laplacian of a compact logarithmic potential, Subharmonic functions on plane domains).

[F7]

If $ν \neq 0$ is finite positive, compactly supported and $U^{ν} \leq M$ on $supp ν$ for some real $M$ , then $U^{ν} \leq M$ on all of $C$ (Maximum principle for a compact logarithmic potential).

[F8]

Assume Dependent Choice. A compact set $E \subseteq C$ has $cap (E) = 0$ if and only if there are a complex domain $Ω \supseteq E$ and a function subharmonic on $Ω$ with $E \subseteq {u = - \infty}$ ; and if $(F_{j})_{j \geq 1}$ is a specified sequence of compact sets with $cap (F_{j}) = 0$ for every $j$ , then there is a function $u$ subharmonic on all of $C$ , not identically $- \infty$ , with $u = - \infty$ on $⋃_{j} F_{j}$ (Compact capacity-zero sets and subharmonic minus-infinity loci).

[F9]

The Axiom of Choice implies Dependent Choice, which implies Countable Choice (AC implies DC implies countable choice), the choice principle used by [F6].

Proof

technique · direct

1.1F2F3F5given

By [F2] and [F5] the hypothesis $cap (K) > 0$ gives $V_{K} < + \infty$ and the equilibrium measure $μ : = μ_{K} \in P (K)$ with $I (μ) = V_{K}$ ; $μ$ is carried by $K$ , so by [F3] its support $S = supp μ$ is a nonempty compact subset of $K$ with $μ (C ∖ S) = 0$ and $μ (B) > 0$ for every ball $B$ about a point of $S$ .

2.1step 1.1F1F9

Since $diam K > 0$ the kernel is bounded below on $K \times K$ , so [F1] gives $I (μ) = \iint k d μ d μ = \int_{C} U^{μ} d μ$ by Tonelli and $\int_{S} U^{μ} d μ = I (μ) = V_{K}$ ; moreover, by [F1] with the common carrier $K$ and $R > \max {1, diam K}$ , every finite positive measure $σ \leq μ$ carried by $K$ has $I (σ) \leq I (μ) + (μ (C)^{2} - σ (C)^{2}) \log R < + \infty$ , and the mixed energy of two finite positive compactly supported measures with finite energy is finite; in particular $I (μ, ν) < + \infty$ for every $ν \in P (K)$ with $I (ν) < + \infty$ .

3.1step 2.1F1F2F5

Minimality inequality: for every $ν \in P (K)$ with $I (ν) < + \infty$ and $I (μ, ν) < + \infty$ one has $I (μ, ν) \geq V_{K}$ . Indeed, for $t \in [0, 1]$ the convex combination $μ_{t} : = (1 - t) μ + t ν$ lies in $P (K)$ , so $I (μ_{t}) \geq V_{K} = I (μ)$ by [F2] and [F5]; expanding the double integral of $μ_{t} \otimes μ_{t}$ with [F1] gives $I (μ_{t}) = (1 - t)^{2} I (μ) + 2 t (1 - t) I (μ, ν) + t^{2} I (ν)$ , so subtracting $I (μ)$ , dividing by $2 t > 0$ and letting $t ↓ 0$ yields $I (μ, ν) - I (μ) \geq 0$ , that is $I (μ, ν) \geq V_{K}$ .

4.1step 1.1step 2.1step 3.1F1F3F6assume-hypcontradiction

Claim: $U^{μ} (x_{0}) \leq V_{K}$ for every $x_{0} \in S$ . Suppose not, and choose $η > 0$ with $U^{μ} (x_{0}) > V_{K} + η$ (if $U^{μ} (x_{0}) = + \infty$ any $η > 0$ will do); by [F6] the set ${U^{μ} > V_{K} + η}$ is open, so there is $r > 0$ with $U^{μ} > V_{K} + η$ on the disc $B : = B (x_{0}, r)$ ; put $m : = μ (K \cap B)$ , so $0 < m \leq 1$ by step 1.1. If $m = 1$ then $μ$ is carried by $K \cap B$ , so by step 2.1, where $\int_{C} U^{μ} d μ = I (μ) = V_{K}$ is a finite real number, $V_{K} = \int_{K \cap B} U^{μ} d μ \geq m (V_{K} + η) = V_{K} + η > V_{K}$ , a contradiction; hence $0 < m < 1$ . The restriction $σ : = μ ⁣ ↾_{K ∖ B}$ satisfies $0 \leq σ \leq μ$ and $σ (C) = 1 - m > 0$ , so $ν : = σ / (1 - m) \in P (K)$ has $I (ν) = I (σ) / (1 - m)^{2} < + \infty$ and $I (μ, ν) < + \infty$ by step 2.1, while the pointwise bound on $B$ and $\int_{C} U^{μ} d μ = V_{K}$ give $I (μ, ν) = \frac{1}{1 - m} \int_{K ∖ B} U^{μ} d μ = \frac{1}{1 - m} (V_{K} - \int_{K \cap B} U^{μ} d μ) \leq \frac{V_{K} - m (V_{K} + η)}{1 - m} = V_{K} - \frac{m η}{1 - m} < V_{K},$ contradicting step 3.1; so no such $x_{0}$ exists.

5.1step 1.1step 4.1F7

Since $μ \neq 0$ and $U^{μ} \leq V_{K}$ on $S = supp μ$ by step 4.1, the maximum principle [F7] with $M : = V_{K}$ gives $U^{μ_{K}} (z) \leq V_{K}$ for every $z \in C$ , which proves the first assertion and in particular gives the finiteness $I (μ, ν_{F}) = \int U^{μ} d ν_{F} \leq V_{K}$ for every $ν_{F} \in P (K)$ below.

6.1step 3.1step 5.1F4F5F6contradiction

For $n \geq 1$ the set $E_{n} : = {z \in K : U^{μ} (z) \leq V_{K} - 1 / n}$ is compact, because $K$ is compact and ${U^{μ} \leq c}$ is closed by [F6]; if $F \subseteq E_{n}$ were compact with $cap (F) > 0$ , then [F5] applied to $F$ would give an equilibrium measure $ν_{F}$ with $supp ν_{F} \subseteq F$ , and steps 3.1 and 5.1 would give the contradictory chain $V_{K} \leq I (μ, ν_{F}) = \int U^{μ} d ν_{F} \leq V_{K} - 1 / n < V_{K}$ ; hence every compact subset of $E_{n}$ has capacity zero.

7.1step 5.1step 6.1F4F8F9∎

By step 6.1 each $E_{n}$ has the property that every compact subset has capacity zero, in the sense of [F4]. Hence $E : = ⋃_{n \geq 1} E_{n}$ is capacity-polar: if $F \subseteq E$ is compact, then $F = ⋃_{n \geq 1} (F \cap E_{n})$ is a specified countable union of compact sets $F \cap E_{n} \subseteq E_{n}$ of capacity zero, so the specified- $F_{σ}$ clause of [F8] supplies a function subharmonic on $C$ that equals $- \infty$ on all of $F$ , and the compact clause of [F8], applied with $Ω = C$ , gives $cap (F) = 0$ . Thus $E$ is a Borel capacity-polar subset of $K$ (it is a countable union of compact sets), and for $z \in K ∖ E$ one has $U^{μ} (z) > V_{K} - 1 / n$ for every $n \geq 1$ , hence $U^{μ} (z) \geq V_{K}$ , while step 5.1 gives $U^{μ} (z) \leq V_{K}$ ; therefore $U^{μ} = V_{K}$ on $K ∖ E$ , which is the second assertion.

Remarks

Why the two halves are different. The inequality $U^{μ} \leq V_{K}$ everywhere is the part that uses the minimality of $μ$ through the competitor $ν$ obtained by deleting a small disc of large potential; the reverse inequality $U^{μ} \geq V_{K}$ needs no minimality beyond the perturbed-copy inequality of step 3.1 and in fact only holds quasi-everywhere, as the isolated-point example of Capacity-polar sets, quasi-everywhere, and subharmonic polar sets shows.

Choice. The Axiom of Choice enters through [F5] for $K$ and again for the compact sets $F \subseteq E_{n}$ of step 6.1, and through [F8] in step 7.1, whose Evans-potential lemma assumes only Dependent Choice; the running argument is otherwise choice-free, and [F6] needs only Countable Choice, which [F9] supplies.

Sharpness of the exceptional set. The exceptional set is a countable union of compact zero-capacity sets, and step 7.1 shows through [F8] that such a union is again capacity-polar (the definition alone tests only compact subsets and does not supply the countable-union closure); so the statement is exactly the classical "equals $V_{K}$ quasi-everywhere on $K$ ".

Depends on

Used by

Reciprocity inequality for logarithmic potentials Proposition
Green function at infinity from the equilibrium potential Theorem

Dependency tree · two levels

66 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. B. Saff, Logarithmic Potential Theory with Applications to Approximation Theory, §1 (standard reference, not scraped)
B. Khoruzhenko, LTCC Potential Theory notes, §3 (standard reference, not scraped)
C. Kuehn, Introduction to Potential Theory via Applications, §2.3 (standard reference, not scraped)