Continuity and growth of free multiplicative convolution semigroups

Let µ be a compactly supported probability measure on the positive half-line and let µ (cid:2) t be the free multiplicative convolution semigroup. We show that the support of µ (cid:2) t varies continuously as t changes. We also obtain the asymptotic length of the support of these measures.


Introduction
Let µ and ν be probability measures on [0, ∞). The free convolution µ ν represents the distribution of the product of two positive operators in a tracial W * -probability space whose distributions are µ and ν respectively. We refer to [15] for an introduction to free probability theory.
Given a probability measure µ on [0, ∞) not being a Dirac measure at 0, it is known [3] that, for any t > 1, the fractional free convolution power µ t is defined appropriately, such that it interpolates the discrete convolution semigroup {µ n } n∈Z + , where µ n = µ · · · µ is the n-fold free convolution. This is the multiplicative analogue of Nica-Speicher semigroup [12] defined firstly for the free additive convolution. The free convolution semigroups obey many regularity properties and has been studied extensively. See [16] for a survey on free convolutions and other topics in free probability theory.
Denote by supp(µ) the topological support of the measure µ. We will prove the following result about continuity of the support of µ t , which is the analogue of the work [17] for free multiplicative convolution on the positive half line. We then study the asymptotic size of the support. Theorem 1.2. Let µ be a compactly supported probability measure on [0, ∞) with mean m 1 (µ) = ∞ 0 s dµ(s) = 1. We denote by ||µ t || = max{m : m ∈ supp(µ t )}. Then lim t→∞ ||µ t ||/t = eV, where V is the variance of µ. Theorem 1.2 generalizes Kargin's work [11] (see also [1]) to continuous semigroups. Our proof is different from Kargin's proof, but uses the density formula for free convolution semigroups [10,19]. In addition, we obtain formulas for the limits of the left and right edges of the support of (µ t ) 1/t in Proposition 4.2, where the measure (µ t ) 1/t is the push-forward of µ t under the map x → x t . This was also studied by Tucci [13] and Haagerup-Möller [9] for discrete semigroups.
The free multiplicative convolution on the unit circle is usually studied together with the positive half line case. It was shown in [2] that many results can be deduced from results on free additive convolutions. Since analogue results were known in additive case, a separate work on free multiplicative convolution on the unit circle become unnecessary. Hence we focus on measures on the positive half line in our article.
An estimation for the size of support of free additive convolution semigroups was obtained in [5,8]. In light of the proof of Theorem 1.2, it is very likely higher order asymptotic expansion can be obtained for free additive convolution semigroups. We plan to investigate it in a forthcoming work as it may have some applications in quantum information theory.
The paper is organized as follows. In Section 2, we collect some known regularity results about free multiplicative convolution semigroups. In Section 3, we give the proof for Theorem 1.1. We study the asymptotic behaviour of free multiplicative convolution semigroups and give the proof for Theorem 1.2 in Section 4.

Free convolution on the positive half line
Let µ be a probability measure on [0, ∞). The ψ-transform of µ is the moment generating function of µ defined as which is analytic on Ω = C\[0, ∞). The η-transform of µ is defined as η µ = ψ µ /(1 + ψ µ ) on the same domain as the ψ-transform. It is know [3] that the map η is a map from C + to itself when it is restricted to C + . Any probability measure µ on [0, ∞) can be recovered from its η-transform by Stieltjes inversion formula. Indeed, we have the identity where G µ is the Cauchy transform of µ. If µ is not a Dirac measure at 0, then η µ (z) > 0 for z < 0, and therefore η µ |(−∞, 0) is invertible. Let η −1 µ be the inverse of η µ and set Σ µ (z) = η −1 µ (z)/z, where z < 0 is sufficiently small. The free convolution of two such probability measures µ and ν is determined by Σ µ ν (z) = Σ µ (z)Σ ν (z). In particular, the n-th order free multiplicative convolution power µ n of µ satisfies the identity Σ µ n (z) = Σ n µ (z), (2.2) where z < 0 is sufficiently small.
We now briefly recall the construction of µ t which interpolates the relation (2.2) as follows. Let κ µ (z) = z/η µ (z) for z ∈ Ω and one can write κ µ (z) = exp(u(z)), where u is an analytic function on Ω and can be expressed as where a = − log |η µ (i)| and ρ is a finite positive Borel measure on [0, ∞) following [10,Proposition 4.1]. To eliminate the trivial case, we assume that ρ = 0 in this article. We define Φ t (z) := z exp[(t − 1)u(z)]. It turned out that the function Φ t is the right inverse of Voiculescu subordination function ω t [3]. More precisely, we have Φ t (ω t (z)) = z, and η µ t (z) = η µ (ω t (z)) for all z ∈ Ω and t > 1. It turns out that the function ω t can be regarded as the ηtransform of a -infinitely divisible measure on [0, ∞) and the function η µ t can be retrieved from ω t . We refer to [3,10] for more details.
The following result was proved in [3].
1. The functions η µ t and ω t can be extended as continuous functions defined on C + .

A point
3. The nonatomic part of µ t is absolutely continuous and its density is continuous except at the finitely many points x such that η µ t (x) = 1.
4. The density of µ t is analytic at all points where it is different from zero.
The study of regularity property of free convolutions relies on Voiculescu's subordination result [3,4,7,14]. By a careful study of boundary behaviour of subordination functions, we were able to give a formula for the density function of absolutely continuous part of µ t in [10]. To describe our result, we need some auxiliary functions studied in [10].

It is clear that
The functions defined above can be used to describe the image set Ω t = ω t (C + ). The set Ω t is in fact the connected component of Φ −1 t (C + ∪ (−∞, 0)) having the negative half line as part of its boundary. Moreover, we proved that The following result is one of main results in [10].
Suppose that µ is a probability measure on [0, ∞) not being a Dirac measure at 0 and t > 1.
Then the following statements hold.
(1) The measure (µ t ) ac is equal to the closure of S t .
(2) The density of (µ t ) ac is analytic on the set S t and is given by (3) The number of components in supp(µ t ) ac is a decreasing function of t ∈ (1, ∞).

Continuity of free convolution semigroups
In this section, we assume that the probability measure µ on [0, ∞) is compactly supported and µ is not a Dirac measure.
Moreover, when m 1 (µ) = 1, we have We then deduce the first equation.
(2) Given any b > 0, the sets V + t ∩ [0, b] are Hausdorff continuous with respect to t.
Proof. By the definition of η-transform, we have η µ (0) = 0 and η µ (0) = m 1 (µ) < ∞. The implies that η µ is real on some interval [0, a) under the assumption that µ is compactly supported. It follows that u is also real on [0, a) and hence ρ([0, a)) = 0 by Stieltjes inversion formula. Hence, the function g is finite in a neighborhood of zero and g(r) = ∞ 0 This proves the first assertion. We prove the second claim by contradiction. Assume that we have t n → t ∈ (α, β) We then have a series r n ∈ (V + tn ∩ [0, b])\B (V + t ). We may assume that r n → r by passing to a subsequence if necessary. Lemma 3.2 implies that supp(ρ) ⊂ B (V + t ) and hence we can take the limit g(r) = lim n→∞ g(r n ) = lim where we used the choice of r n ∈ V + tn . On the other hand, g ≤ 1 t−1 and g is strictly convex on any open interval contained in (V + t ) c . This contradiction finishes the proof. Proof. For 0 < c < π, we define V + t,c = {r ∈ V + t : A t (r) ≥ c}. Given > 0, we will start to prove that there exists δ > 0 such that The first inequality follows from the fact that the function g(r, θ) is a decreasing function of θ. To prove the second inequality by contradiction, we assume that there exists a series t n > t, t n → t and r n , r ∈ V + t,c ∩ (0, b) such that r n → r and A tn (r n ) > A t (r n ) + . We then have g(r n , A t (r n ) + ) ≥ g(r n , A tn (r n )) = 1 t n − 1 .
due to the fact that the integrand in (2.4) is bounded away from zero and A t is continuous. On the other hand, we have g(r, A t (r)) = 1 t−1 and g(r, θ) is a strictly decreasing function of θ. This contradiction yields the second inequality in (3.1).
Given > 0, using the similar argument as above, we can prove that there exists We claim that sup{A s (r) : r ∈ (0, b)\V + t,c } ≤ 2c if s − t is small enough. Assume that is not the case, then there exists a series t n → t and r n → r, where r n ∈ (0, b)\V + t,c , such that A tn (r n ) > 2c. We have g(r n , 2c) ≥ g(r n , A tn (r n )) = 1 t n − 1 .
Taking the limit, we have g(r, 2c) ≥ 1/(t − 1), which implies that A t (r) ≥ 2c. As the cluster set of r n / ∈ V + t,c , r is either not in V + t,c or an end point of V + t,c , it must satisfy A t (r) ≤ c. This contradiction proves our claim. The desired assertion follows by above results and applying Proposition 3.3. Assuming that 0 ∈ supp((µ s ) ac ) for some s > 1, we claim that ρ is not compactly supported. Indeed, if ρ is compactly supported, then lim r→∞ g(r) = 0 and hence the set V + s is compact and the set supp((µ s ) ac ) is bounded away from zero by the part (1) of Theorem 2.2, which contradicts the assumption. The fact that the measure ρ is not compactly supported in turn implies that V t is not compact by the fact that supp(ρ) ⊂ V + t (see Lemma 3.2). According to Theorem 2.2, supp((µ t ) ac ) is the closure of the set {1/h t (r) : r ∈ V + t }. We see that 0 ∈ supp((µ t ) ac ) for all t > 1 in this case.  Recall that h t (r) = Φ t (re iAt(r) ). Fix r ∈ (b, c) and denote z 1 = re iAt(r) and z 2 = re iAs(r) .
We have Let δ < t − 1 and s ∈ (t − δ, t + δ). To estimate the first factor, we note that for any z = re iθ ∈ Ω s ∩ Ω t and We now choose an arc γ, disjoint from (−∞, 0), of the circle centered at 0 of radius r that connects z 1 and z 2 , and obtain The proof of Proposition 3.4 implies, in particular, that lim s→t A s (r) = A t (r) uniformly on (b, c). Hence, the arc γ can be chosen uniformly small for all r ∈ (b, c) and the right hand side of (3.3) tends to zero uniformly on (b, c) as s → t. We then estimate the second factor exp (t − s)u(z 1 ) . We note that h t ([b, c]) is a compact set which does not contain 0. Recall that Φ t (z) = z exp[(t − 1)u(z)] and h t (r) = Φ t (re iAt(r) ). Hence, lim s→t exp (t − s)u(z 1 ) = 1 uniformly for r ∈ (b, c). The desired result then follows.
We are now in a position to prove the main result of this section.

Estimation of norm of free multiplicative convolution semigroups
We give an estimation of the size of the support of µ t for a compactly supported probability measure on [0, ∞). .
which yields that lim t→∞ (tα t ) = 1/V thanks to Lemma 3.1. We then obtain where we used (4.1). This proves the desired result. Proof. The set V + t is bounded away from zero and ∞. Let α t = min{r : r ∈ V + t } > 0 and β t = max{r : r ∈ V + t } < ∞, we have lim t→∞ α t = 0 and lim t→∞ β t = 0 by Theorem 2.2.
By rescaling the measure and applying Theorem 1.2, we have lim t→∞ t −1 m −t 1 b t = eV /m 2 1 .
Taking the logarithm of this and dividing by t, one easily shows that lim t→∞ (b t ) 1/t = m 1 . For a 1/t t , consider the push-forward µ * of the measure µ by the map x → 1/x and use the property that (µ * ) t = (µ t ) * . We deduce that lim t→∞ This finishes the proof.  [9] (see also [13]), where weak limits of rescaled discrete semigroups were studied. Our method is not suitable to study weak limits, but works to estimate the asymptotic bound of free multiplicative convolution semigroups. The interested reader can easily generalize results in [9] to continuous free multiplicative convolution semigroups using their method.

Remark 4.4.
Free multiplicative convolution semigroups can also be defined for measures on the unit circle [3]. Let µ be a probability measure on the unit circle not being a Dirac measure, it is known in [18,Proposition 3.26] that µ t converges to the uniform measure on the unit circle as t → ∞.