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Reciprocity inequality for logarithmic potentials

Statement

Assume the Axiom of Choice. Let $K \subseteq C$ be compact with $cap (K) > 0$ and let $μ_{K}$ be its equilibrium measure. Then for every compactly supported Borel probability measure $σ$ on $C$ ,

$\inf_{z \in K} U^{σ} (z) \leq V_{K} = \log \frac{1}{cap (K)} .$

The Axiom of Choice enters only through the existence of the equilibrium measure and through Frostman's theorem; the reciprocity identity itself and the infimum estimate are choice-free.

Facts & Assumptions

Given: a compact set $K \subseteq C$ with $cap (K) > 0$ , its equilibrium measure $μ_{K}$ , a compactly supported Borel probability measure $σ$ on $C$ , the logarithmic kernel $k (z, w) = \log \frac{1}{∣ z - w ∣}$ with diagonal value $+ \infty$ , the potential $U^{ν} (z) = \int k (z, w) d ν (w)$ and the mixed energy $I (ν, ρ) = \iint k d ν d ρ$ of Logarithmic potential and energy of a positive compactly supported measure, and the Axiom of Choice (The Axiom of Choice).

[F1]

For a finite positive Borel measure $ν$ of compact support, $U^{ν} (z) \in (- \infty, + \infty]$ , and for $R > diam supp ν$ one has $k_{R} = k + \log R \geq 0$ on the product of the support with itself and $I (ν) = \iint k_{R} d ν d ν - ν (C)^{2} \log R$ , independently of $R$ ; for a pair $ν, ρ$ of such measures and $R > diam (supp ν \cup supp ρ)$ the same shifted kernel is nonnegative on the product of the two supports and $I (ν, ρ) = \iint k_{R} d ν d ρ - ν (C) ρ (C) \log R$ , independently of $R$ (Logarithmic potential and energy of a positive compactly supported measure).

[F2]

$V_{K} = \inf_{ν \in P (K)} I (ν)$ and $cap (K) = e^{- V_{K}}$ when $V_{K} < + \infty$ and $0$ otherwise, so $cap (K) > 0$ is equivalent to $V_{K} < + \infty$ , and then $V_{K} = \log \frac{1}{cap (K)} \in R$ (Robin constant and logarithmic capacity of a compact set).

[F3]

Assume the Axiom of Choice: a compact nonempty $K$ with $cap (K) > 0$ has exactly one equilibrium measure $μ_{K}$ , and $I (μ_{K}) = V_{K}$ ; the measure $μ_{K}$ is a Borel probability measure carried by $K$ , so its support is a nonempty compact subset of $K$ (Existence and uniqueness of the equilibrium measure, Probability measures and probability spaces).

[F4]

Assume the Axiom of Choice: with $μ_{K}$ as above, $U^{μ_{K}} (z) \leq V_{K}$ for every $z \in C$ (Frostman inequalities and quasi-everywhere equilibrium equality).

[F5]

Tonelli's theorem computes the integral of a nonnegative product-measurable function on a $σ$ -finite product as either iterated integral (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product); the support $supp ν$ of a finite positive Borel measure $ν$ is closed, carries $ν$ , and every closed set carrying $ν$ contains it (Support of a finite Borel measure on the plane).

Proof

technique · direct

1.1F2F3F5givenalgebra

By [F2] and [F3] the hypothesis $cap (K) > 0$ gives $V_{K} \in R$ , the equilibrium measure $μ : = μ_{K}$ , and $supp μ \subseteq K$ ; since $K$ and $supp σ$ are compact and nonempty, $D : = diam (K \cup supp σ) < + \infty$ , and as $cap (K) > 0$ the set $K$ is not a singleton, so $D > 0$ ; fix $R > D$ .

1.2F4givenalgebra

Frostman bound. By [F4] one has $U^{μ} \leq V_{K}$ pointwise on $C$ , and $σ$ is a probability, so $\int U^{μ} d σ \leq \int V_{K} d σ = V_{K}$ .

2.1step 1.1F1F5

On $supp μ \times supp σ$ one has $∣ z - w ∣ \leq D < R$ , so the shifted kernel $k_{R} = k + \log R$ is a nonnegative Borel function there; Tonelli [F5] applied to the product measure $μ \otimes σ$ therefore gives $\iint k_{R} d μ d σ = \iint k_{R} d σ d μ$ , both sides being elements of $[0, + \infty]$ .

2.2step 1.1F1F3F5givenalgebra

The infimum bound. For $z \in K$ and $w \in supp σ$ , one has $k (z, w) \geq \log \frac{1}{D}$ , including $z = w$ , where the kernel has value $+ \infty$ . Integrating this pointwise bound against the probability $σ$ , carried by its support, gives $U^{σ} (z) \geq \log \frac{1}{D} > - \infty$ on $K$ . Thus $m : = \inf_{z \in K} U^{σ} (z) \in [\log \frac{1}{D}, + \infty]$ and $m \leq \int_{K} U^{σ} d μ$ : for finite $m$ , integrate $U^{σ} \geq m$ against the probability $μ$ carried by $K$ ; for $m = + \infty$ , the potential is $+ \infty$ everywhere on $K$ and its integral is $+ \infty = m$ .

3.1step 1.1step 2.1F1givenalgebra

For each $z$ one has $\int_{C} k_{R} (z, w) d σ (w) = \int_{C} k (z, w) d σ (w) + σ (C) \log R = U^{σ} (z) + \log R$ as extended reals, since $k_{R} = k + \log R$ pointwise and $σ (C) = 1$ ; integrating against the probability $μ$ gives $\iint k_{R} d σ d μ = \int U^{σ} d μ + \log R$ , and symmetrically $\iint k_{R} d μ d σ = \int U^{μ} d σ + \log R$ , the identities being understood in $(- \infty, + \infty]$ with $+ \infty + \log R = + \infty$ .

4.1step 2.1step 3.1algebra

Reciprocity. Comparing the two expressions for the common value in step 2.1 gives $\int U^{σ} d μ + \log R = \int U^{μ} d σ + \log R$ ; subtracting the finite real number $\log R$ yields the reciprocity identity $\int_{C} U^{σ} d μ = \int_{C} U^{μ} d σ \in (- \infty, + \infty]$ .

5.1step 4.1step 2.2step 1.2F2F3F4given∎

Combining steps 4.1, 2.2 and 1.2, $\inf_{z \in K} U^{σ} (z) \leq \int U^{σ} d μ = \int U^{μ} d σ \leq V_{K}$ , and $V_{K} = \log \frac{1}{cap (K)}$ by [F2]; this is the assertion. The Axiom of Choice was used only through [F3] and [F4].

Remarks

What reciprocity does and does not require. The identity $\int U^{σ} d μ = \int U^{μ} d σ$ is a Fubini statement for the kernel on the product of the two supports; no finiteness of $I (σ)$ is assumed, and the value $+ \infty$ is allowed on both sides. The shifted kernel $k_{R} = k + \log R$ is nonnegative on that product, which is what makes Tonelli applicable without any integrability hypothesis.

Sharpness of the inequality. The inequality $\inf_{K} U^{σ} \leq V_{K}$ is the classical reciprocity inequality of Saff, Proposition 1.13; equality holds for $σ = μ_{K}$ whenever $U^{μ_{K}} = V_{K}$ at every point of $K$ . Quasi-everywhere equality alone (Frostman inequalities and quasi-everywhere equilibrium equality) does not suffice for the infimum over all of $K$ : exceptional polar points may have smaller potential.

Depends on

Used by

Monic polynomial lower bounds for the Chebyshev constant and capacity Lemma

Dependency tree · two levels

37 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)