A Fourier integral formula for logarithmic energy

A formula which expresses logarithmic energy of Borel measures on R^n in terms of the Fourier transforms of the measures is established and some applications are given. In addition, using similar techniques a (known) formula for Riesz energy is reinvented.


Introduction and main result
The notions of logarithmic potential and logarithmic energy play a central role in potential theory in particular in dimension two as well as in free probability.Recommended introductions are [14] and [21] for the classical theory and [18] for the more recent theory of free probability.We consider arbitrary dimension n ∈ N and write M(R n ) for the set of all complex Borel measures on R n and M + (R n ) for the subset of nonnegative µ.With |µ| denoting the total variation of µ ∈ M(R n ) and |x| denoting the euclidean norm of x ∈ R n we write denote the (n − 1)-sphere, σ n−1 the surface measure of S n−1 and ω n−1 the corresponding area, given by 2π n/2 /Γ(n/2).In particular, S := S 1 is the unit circle in C = R 2 .Writing a k := S ζ k dµ(ζ) for Fourier coefficients of a complex measure µ supported on S, a basic fact is that the logarithmic energy can be expressed by This is an identity if µ ∈ M 0 (R 2 ) or µ ≥ 0 and serves as definition of I(µ) for arbitrary complex µ (cf.[13], p. 35, [11]).In particular, in all cases The identity can be seen by expanding for 0 < r < 1 and taking the limit r → 1 (cf.[4], Section 2.4).It might come as a surprise that a 0 = µ(S) does not enter the scene.This can be explained by the fact that the logarithmic potential of the arclength measure σ 1 vanishes on S and, as a consequence, I(µ) equals I(µ + λσ 1 ) for arbitrary µ and scalars λ (cf.[17], p. 119).
Our aim is to establish a continuous n-dimensional version of a formula for the logarithmic energy of appropriate measures µ in terms of the Fourier transform and the generalized hypergeometric function K n , defined by with the Pochhammer symbol (α, k) := Γ(α + k)/Γ(α).In particular, one obtains K 1 = cos and in the case n = 2, in which the logarithmic potential coincides with the Newtonian, K 2 = J 0 , where J α denotes the Bessel function of first kind and order α.More generally, K n can be expressed in terms of J n/2−1 as With these notations, the main result on mutual logarithmic energy reads as follows: In particular, for ) is locally integrable at the origin with respect to dξ/|ξ| n , then the integrand function on the right hand side in (2) has the same property and the double-sided limit reduces to a one-sided lim N →∞ .This holds if µ ν is Dini continuous at 0, which is in particular the case if µ and ν have compact support.
We will frequently use the following substitution rule for the n-dimensional Lebesgue measure, a proof of which can be found e.g. in [6], p. 78: Due to the fact that J α (t) = O(t −1/2 ) as 0 < t → ∞ for α ≥ 0 (see [19], 10.17.3), the function t → K n (t)/t is absolutely integrable at +∞ for n ≥ 2 and improperly integrable for n = 1.Hence, lim exists and so, by monotonicity, for arbitrary complex Borel measures the limit N → ∞ in (3) also exists as value in (−∞, ∞].This implies that the function | µ| 2 is integrable at ∞ with respect to dξ/|ξ| n if I(µ) is finite and that extends the definition of logarithmic energy to arbitrary complex Borel measures having compact support.
Choosing dµ(x) = ϕ(x) dx, where ϕ belongs to the Schwartz space S, we obtain from (2) with the Dirac measure δ at 0 and δ = 1 where the integral is improper in the case n = 1.Since ( ϕ) ∧ = (2π) n ϕ(− •), replacement of ϕ by ϕ implies that we have as temperate distribution, where (cf. [23], p. 44, 258).In this way, (log for s > 0 and c n (s) a suitable constant (see e.g.[3], p. 161).By comparing the two representations it turns out that In Section 3 the integrals are calculated inductively from their values for n = 1, 2.
Remark 1.2.From the main theorem in [16] it follows that for measures µ, ν with the integral on the left hand side > −∞ if I(µ), I(ν) ∈ R. In particular, this implies that for arbitrary µ, ν ∈ M 0 (R n ) we have as the example in [16], p. 3340, shows.
In Section 2 we give the proof of Theorem 1.1 (which uses only classical methods) and Section 3 contains some applications.Applying similar techniques, in Section 4 we present an alternative proof of a known Fourier integral formula for Riesz energy.

Proof of Theorem 1.1
A main goal is that the formulas for nonnegative measures hold both in the cases that the energy is finite and is not.We start with three quite simple tools, where the first one is an "improper" version of the Cauchy-Frullani theorem (cf.[20]).Recall that L loc (0, ∞) is the set of all Lebesgue measurable functions g : (0, ∞) → C such that g1 K is integrable for all compact K ⊂ (0, ∞).Moreover, we write g ∞ for the essential supremum of |g|.
If, in addition, g is essentially bounded, then Proof.We may assume that 0 < a < 1.For fixed 0 < ε < N < +∞ we have The first term is u ln(1/a).According to Cauchy's criterion, the second term vanishes as ε tends to 0 and the third as N tends to ∞.Moreover, if g is essentially bounded we obtain that . Lemma 2.2.Let g ∈ L loc (0, ∞) be real-valued with u := ess sup g < ∞ and such that t → (u − g(t))/t is integrable at 0. Then, for all N > 0 we have Proof.The assertion follows from the (essential) nonnegativity of u − g and Lemma 2.3.Let g satisfy the assumptions of Lemma 2.1 and Lemma 2.2 with the same u.Then, there are R, c > 0 such that for each a > R and each Proof.We choose R > 0 such that where the constant c > 0 is chosen such that sup Remark 2.4.With the aid of the functions K n one can express the Fourier transform of the surface measure σ n−1 .More precisely, since for n ∈ N, we have (see e.g.[24], p. 154, or [9], p. 428) (6) In particular, Due to the fact that J α (t) = O(t −1/2 ) as t → ∞ for α ≥ 0 (see [19], 10.17.3) one sees that K n | (0,∞) satisfies the assumptions of Lemma 2.1 with u = 1 and, since K n (t) = 1+O(t 2 ) as t → 0, also the assumptions of Lemma 2.2 are fulfilled.Finally, because K ′ n (0) = 0 and Now, we are in a position to give the Proof of Theorem 1.1: Fix 0 < ε < N < +∞.Then, by Fubini's theorem and ( 4) Another application of Fubini's theorem and (6) give us First, suppose that Then we necessarily have (|µ| ⊗ |ν|)({(x, y) ∈ R n × R n : x = y}) = 0. Therefore, if we take g = K n and u = g(0) in Lemma 2.1 (see Remark 2.4) the dominated convergence theorem yields that In this case, The same argument implies together with Lemma 2.1, Remark 2.4 and the dominated convergence theorem that lim exists and is finite.Therefore, we only have to show that lim inf By Remark 2.4, we know that for |x − y| < 1 and sufficiently large N and small ε.Therefore, we can apply Fatou's lemma (notice that µ ⊗ ν is finite) and get with Then, The same argument implies together with Lemma 2.1 and the dominated convergence theorem that lim exists and is finite for all T ≥ 1.If we pick g = K n in Lemma 2.3, then there is some R > 1 such that for all |x − y| ≥ R and 0 for some constant c ∈ R independent of x, y, ε and N. Therefore, we can apply Fatou's lemma since µ ⊗ ν is finite and get lim sup

Consequences of Theorem 1.1
Let us start with two illustrating examples, which also show that some interesting integrals involving J 0 emerge form Theorem 1.1: According to (6), Theorem 1.1 and (4) imply that Since K 2 = J 0 and since, as already mentioned in the introduction, I(σ 1 ) = 0, we obtain Applying Lemma 2.1 with g = J 2 0 we get, more generally, ∞ 0 (J 2 0 (ar) − J 0 (r)) for a > 0. The standard one-dimensional example in which µ is known is the arcsine distribution Here is (see [5], p. 11) J 0 = µ, but then with the Dirac measure δ 0 also Since I(µ) = I(µ ⊗ δ 0 ), by choosing n = 2 in Theorem 1.1 and using (8) we have This is of course folklore but usually proved in a quite different way by exploiting some amount of potential theory in the plane.More precisely, one shows that µ is the equilibrium measure of [−1, 1], that is, µ minimizes logarithmic energy among all Borel probability measures on [−1, 1], and that the minimum is ln 2 or, in other words, that the logarithmic capacity of [−1, 1] is 1/2 = e − ln 2 .Also, taking n = 1 in Theorem 1.1 and using (7) leads to (see e.g.[1], Example 5.5, [12], p. 278).According to Lemma 2.1, we end at for arbitrary a > 0. In particular, this implies that for n = 1 the function K 1 = cos in Theorem 1.1 may be replaced by ξ → J 0 (2|ξ|).
A next consequence of Theorem 1.1 is a characterization for the finiteness of the logarithmic energy for nonnegative measures.
According to the assumption and the smoothness of K n , the limit in the equation above exists if and only if lim is finite, we conclude that I(µ) ∈ R if and only if lim As already mentioned in the introduction, the limit ε → 0 in the preceding corollary exists for arbitrary Borel measures having compact support.As a consequence, the equivalence statement of the corollary also holds for compactly supported complex measures.
Since, according to continuity, | µ| 2 is not locally integrable with respect to dξ/|ξ| n at the origin if µ(R n ) = 0, we have

Riesz Energy
Logarithmic energy may be viewed as the limit case α → 0 of Riesz energies.For positive α and x ∈ R n we write Moreover, we write M α (R n ) for the set of all complex Borel measures with 1 |x − y| α d|µ|(x) d|µ|(y) < +∞ .
For µ ∈ M α (R n ) ∪ M + (R) the Riesz potential p µ,α is defined µ almost everywhere and is called the Riesz energy of order α of µ.It is a well-known fact (see [15], p. 162, [14] Here the situation is more convenient compared to the logarithmic case due to the fact that ξ → |ξ| α | µ| 2 (ξ) is locally integrable with respect to dξ/|ξ| n at the origin.Arbitrary complex measures are considered in [10], where Riesz energy is defined by the formula (10).Finally, according to the pre-Hilbert space structure of the space of signed measures in M α (R n ) endowed with the inner product (µ, ν) → p µ,α dν (see e.g.[14], p. 82), the mixed integral p µ,α dν exists in C whenever µ, ν ∈ M α (R n ) (and of course in [0, ∞] for nonnegative µ, ν).
Applying similar techniques as in the proof of Theorem 1.1 one can show the following result, in which α is restricted to be less than (n + 1)/2 due to the fact that K n has to be (at least improperly) integrable at ∞ with respect to t α−1 dt.
is finite if and only if | µ| 2 is locally integrable at ∞ with respect to dξ/|ξ| n .