On Asymptotic Growth of the Support of Free Multiplicative Convolutions

Let µ be a compactly supported probability measure on R + with expectation 1 and variance V. Let µ n denote the n-time free multiplicative convolution of measure µ with itself. Then, for large n the length of the support of µ n is asymptotically equivalent to eV n, where e is the base of natural logarithms, e = 2.71. .. 1 Preliminaries and the main result First, let us recall the definition of the free multiplicative convolution. Let a k denote the moments of a compactly-supported probability measure µ, a k = t k dµ, and let the ψ-transform of µ be ψ µ (z) = ∞ k=1 a k z k. The inverse ψ-transform is defined as the functional inverse of ψ µ (z) and denoted as ψ (−1) µ (z). It is a well-defined analytic function in a neighborhood of z = 0, provided that a 1 = 0. Suppose that µ and ν are two probability measures supported on R + = {x|x ≥ 0} and let ψ (−1) µ (z) and ψ (−1) ν (z) be their inverse ψ-transforms. Then, as it was first shown by Voiculescu in [5], the function f (z) := 1 + z −1 ψ (−1) µ (z) ψ (−1) ν (z) is the inverse ψ-transform of a probability measure supported on R +. (Voiculescu used a variant of the inverse ψ-transform, the S-transform.) This new probability measure is called the free multiplicative convolution of measures µ and ν, and denoted as µ ⊠ ν. The significance of this convolution operation can be seen from the fact that if µ and ν are the distributions of singular values of two free operators X and Y, then µ ⊠ ν is the distribution of singular values of the product operator XY (assuming that the algebra containing X and Y is tracial). For more details about free convolutions and free probability theory, the reader can consult [2], [4], or [6]. We are interested in the support of the n-time free multiplicative convolution of the measure 415

(z) .It is a well-defined analytic function in a neighborhood of z = 0, provided that a 1 = 0. Suppose that µ and ν are two probability measures supported on R + = {x|x ≥ 0} and let ψ (−1) µ (z) and ψ (−1) ν (z) be their inverse ψ-transforms.Then, as it was first shown by Voiculescu in [5], the function is the inverse ψ-transform of a probability measure supported on R + .(Voiculescu used a variant of the inverse ψ-transform, the S-transform.)This new probability measure is called the free multiplicative convolution of measures µ and ν, and denoted as µ ⊠ ν.The significance of this convolution operation can be seen from the fact that if µ and ν are the distributions of singular values of two free operators X and Y, then µ ⊠ ν is the distribution of singular values of the product operator XY (assuming that the algebra containing X and Y is tracial).For more details about free convolutions and free probability theory, the reader can consult [2], [4], or [6].We are interested in the support of the n-time free multiplicative convolution of the measure µ with itself, which we denote as µ n : Let L n denote the upper boundary of the support of µ n .
Theorem 1. Suppose that µ is a compactly-supported probability measure on R + , with the expectation 1 and variance V. Then where e denotes the base of natural logarithms, e = 2.71 . . .
Remarks: 1) Let X i be operators in a von Neumann algebra A with trace E. Assume that X i are free in the sense of Voiculescu and identically distributed, and let for all sufficiently large n.This result also holds if we relax the assumption E (X * i X i ) = 1 and define 2) Theorem 1 improves the author's result in [3], where it was shown that L n /n ≤ cL where c is a certain absolute constant and L is the upper bound of the support of µ.Theorem 1 shows that the asymptotic growth in the support of free multiplicative convolutions µ n is controlled by the variance of µ and not by the length of its support.The idea of proof of Theorem 1 is based on the fact that the radius of convergence of Taylor series for ψ n (z) is 1/L n .Therefore the function ψ n (z) must have a singularity at the boundary of the disc |z| = 1/L n .Since all the coefficients in this Taylor series are real and positive, the singularity is z n = 1/L n .Therefore, the study of L n is equivalent to the study of the singularity of ψ n (z) which is located on R + and which is closest to 0. By Proposition 5.2 in [1], we know that for all sufficiently large n, the measure µ n is absolutely continuous on R + \ {0} , and its density is analytic at all points where it is different from zero.For these n, the singularity of ψ n (z) is neither an essential singularity nor a pole.Hence, the problem is reduced to finding a branching point of ψ n (z) which is on R + and closest to zero.The branching point of ψ n (z) equals a critical value of ψ Since by Voiculescu's theorem, therefore we can find critical points of ψ Thus, our task is to estimate the root u n of this equation which is real, positive and closest to 0, and then study the asymptotic behavior of z n = ψ (−1) n (u n ) as n → ∞.This study will be undertaken in the next section.

Proof of Theorem 1
Notation: L and L n are the least upper bounds of the support of measures µ and µ n , respectively; V and V n are variances of these measures; ψ (z) and ψ n (z) are ψ-transforms for measures µ and µ n , and ψ (−1) (u) and ψ (−1) n (u) are functional inverses of these ψ-transforms.When we work with ψ-transforms, we use letters t, x, y, z to denote variables in the domain of ψ-transforms, and b, u, v, w to denote the variables in their range.
Lemma 2. For all positive z such that z < 1/ (2L) , it is true that where c 1 and c 2 depend only on L.
Proof: Clearly, E X k ≤ L k .Using the Taylor series for ψ (z) and ψ ′ (z), we find that for all positive z such that z < 1/ (2L) , which implies the statement of this lemma.QED.
Lemma 3.For all positive u such that u < 1/ (12L) , it is true that where c 3 depends only on L.
Proof: Let the Taylor series for ψ (−1) (u) be Using the Lagrange inversion formula, it is possible to prove that see, e.g., proof of Lemmas 3 and 4 in [3].This implies that the Taylor series for ψ (−1) (u) are convergent in the disc |u| < (6L) −1 .Hence, in this disc, which implies the statement of this lemma.QED.The proof of Theorem 1 uses the following proposition.Its purpose is to estimate the critical point of ψ (−1) n (u) from below.Later, we will see that this estimate gives the asymptotically correct order of magnitude of the critical point.(u) which belongs to R + and which is closest to 0. Then for all ε > 0, there exists such n 0 (L, V, ε) , that for all n > n 0 , Proof of Proposition 4: Claim: Let ε be an arbitrary positive constant.Let x n = (n (1 + 2V + 2ε)) −1 and b n = ψ (x n ) .Then for all n ≥ n 0 (ε, L, V ) and all u ∈ [0, b n ], the following inequality is valid: If this claim is valid, then since u n is the smallest positive root of equation ( 1), therefore we can conclude that u n > b n = ψ (x n ).By Lemma 2, it follows that for all sufficiently large n (Indeed, note that the last inequality has 2ε and ε on the left-hand and right-hand side, respectively.Since Lemma 2 implies that ψ (z) ∼ z for small z, therefore this inequality is valid for all sufficiently large n.) Hence, Proposition 4 follows from the claim, and it remains to prove the claim.Proof of Claim: Let us re-write inequality (2) as where z = ψ (−1) (u) .
Using Lemma 2, we infer that inequality (3) is implied by the following inequality: ,

Proposition 4 .
Let u n be the critical point of ψ (−1) n