FEE POBABILITY AND ANDOM MATICES Lecture 4: Free Harmonic Analysis, October 4, 007 The Cauchy Transform. Let C + = {z C Im(z > 0} denote the complex upper half plane, and C = {z Im(z < 0} denote the lower half plane. Let ν be a probability measure on and for z let G(z = z t dν(t G is the Cauchy transform of the measure ν. Let us briefly check that the integral converges to an analytic function on C +. Lemma. G is an analytic function on C + with range contained in C. Since z t Im(z and ν is a probability measure the integral is always convergent. If Im(w 0 and z w < Im(w / then for t we have z w t w < Im(w Im(w = so the series n=0 ( z w t w n converges uniformly to t w on z w < t z Im(w /. Thus (z t = n=0 (t w (n+ (z w n on z w < Im(w /. Hence [ ] G(z = (t w (n+ dν(t (z w n n=0 is analytic on z w < Im(w /. Finally note that for Im(z > 0, we have for t, Im((z t < 0, and hence Im(G(z < 0. Thus G maps C + into C. Lemma. (i lim yg(iy = i and (ii sup y G(x + iy = y y 0,x Proof. (i y Im(G(iy = = ( y Im dν(t = iy t dν(t +(t/y y y + t dν(t dν(t where y and since ( + (t/y we may apply Lebesgue s Dominated Convergence Theorem.
LECTUE 4: FEE HAMONIC ANALYSIS We have y e(g(iy = yt dν(t. But for all y>0and for all t y + t yt y + t and yt/(y + t converges to 0 as y. Therefore y e(g(iy 0 as y, again by the Dominated Convergence Theorem. (ii Fory>0 and z = x + iy, y G(z y z t dν(t = y (x t + y dν(t Thus sup y G(x + iy. By (i however, the supremum is. y 0,x Theorem. Suppose a<b. Then lim y 0 + π b a Im(G(x + iy dx = ν((a, b + ν({a, b} If ν and ν are probability measures with G ν = G ν, then ν = ν. Proof. We have Im(G(x + iy = ( Im x t + iy Thus b Im(G(x + iy dx = a = = b a (b t/y dν(t = y dx dν(t (x t + y (a t/y tan ( b t y + x d xdν(t y (x t + y dν(t tan ( a t y where we have let x =(x t/y. So let f(y, t = tan ((b t/y tan ((a t/y and 0 t/ [a, b] f(t = π/ t {a, b} π t (a, b dν(t
LECTUE 4: FEE HAMONIC ANALYSIS 3 Then lim y 0 + f(y, t =f(t, and, for all y>0and for all t, we have f(y, t π. So by Lebesgue s Dominated Convergence Theorem b lim Im(G(x + iy dx = lim f(y, t dν(t y 0 + a y 0 + = f(t dν(t = π (ν((a, b + ν({a, b} This proves the first claim. The first claim shows that if G ν = G ν then ν and nu agree on all intervals and thus are equal. emark. The proof of the next theorem depends on a fundamental result of. Nevalinna which provides an integral representation for an analytic function from C + to C +. Suppose that φ : C + C + is analytic, then there is a unique Borel measure σ on and real numbers α and β, with β 0 such that for z C + φ(z =α + βz + t z dσ(t This integral representation is achieved by letting ξ = iz+, and thus iz z = i +ξ. Then one defines ψ by ψ(ξ =iφ(z and obtains an analytic function ψ mapping the open unit disc, D, into the complex right ξ half plane. Since ψ the real and imaginary parts of ψ are harmonic conjugates, it is enough to work with the real part. Now e(ψ isa positive harmonic function on D; when e(ψ is harmonic on an open set containing D one gets the measure from Cauchy s Integral Theorem; in general one dilates e(ψ to get a harmonic function on D, and then takes a limit as the dilation shrinks to e(ψ. This integral representation is usually attributed to G. Herglotz (9. The details can be found in the book of Akhiezer and Glazman. Theorem (. Nevanlinna. Suppose G : C + C is analytic and sup y G(x + iy =. Then there is a unique probability measure on such that G(z = dν(t. z t Proof. By the remark above there is a unique finite measure on such that G(z =α + βz + +tz dσ(t with α and β 0. z t Next we will use the condition ( sup y G(x + iy = y>0,x Considering first the imaginary part we get that β = 0 and for all N, N y ( + t dσ(t. Thus ( + N t +y t dσ(t. Let N. I. Akhiezer and I. M. Glazman, Theory of Linear Operators in Hilbert Space (96, vol., Ch. VI, 58
4 LECTUE 4: FEE HAMONIC ANALYSIS ν(e = E +t dσ(t. From the real part of ( we get that for all y>0 y α + t(y t + y dν(t Therefore t( y α = lim y dσ(t t + y = lim t y y dσ(t = tdσ(t +(t/y Hence G(z = t + +tz tz + t z t dσ(t = ++tz dσ(t z t = z t ( + t dσ(t = z t dν(t emark. If {ν n } n is a sequence of finite Borel measures on we say that {ν n } n converges in distribution to the measure ν if for every f C b ( (the continuous bounded functions on we have lim n f(t dνn (t f(t dν(t. We say that {ν n } n converges vaguely to ν if for every f C 0 ( (the continuous functions on vanishing at infinity we have lim n f(t dνn (t f(t dν(t. Convergence in distribution implies vague convergence but not conversely. However if ν is a probability measure then {ν n } n converges vaguely to ν does imply that {ν n } n converges to ν in distribution. Theorem. Suppose that (ν n n is a sequence of probability measures on with G n the Cauchy transform of ν n. Suppose {G n } n converges pointwise to G on C +. If lim y yg(iy = i then there is a unique probability measure ν on such that ν n ν in distribution, and G(z = dν(t. z t Proof. {G n } n is uniformly bounded on compact subsets of C + (as G(z Im(z for the Cauchy transform of any probability measure, so by Montel s theorem {G n } n is relatively compact in the topology of uniform convergence on compact subsets of C +, thus, in particular, {G n } n has a subsequence which converges uniformly on compact subsets of C + to an analytic function, which must be G. ThusG is analytic, as it is the uniform limit of analytic functions. Now for z C +, G(z C. Also for each n N, x and y 0 y G n (x + iy. see Kai Lai Chung Probability, nd ed. (984 3.3 and 3.4.
LECTUE 4: FEE HAMONIC ANALYSIS 5 Thus x, y 0, y G(x + iy. So in particular, G is nonconstant. If for some z C +, Im(G(z = 0 then by the minimum modulus principle G would be constant. Thus G maps C + into C. Hence by Nevanlinna s theorem there is a unique probability measure ν such that G(z = dν(t. z t Now by the Helley selection theorem 3 there is a subsequence {ν nk } k converging vaguely to some measure ν. For fixed z the function t /(z t isinc 0 (. Thus for Im(z > 0, G nk (z = dν z t n k (t d ν(t. Therefore G(z = dν(t i.e. ν = ν. Thus{ν z t z t n k } k converges vaguely to ν. Since ν is a probability measure {ν nk } k converges vaguely to ν. Thus{ν n } converges in distribution to ν. Corollary. If {ν n } n is a sequence of probability measures on with G n G pointwise on C +. Then {ν n } converges vaguely to a finite measure ν. The analytic relation between moments and free cumulants involves finding a functional inverse for the Cauchy transform of a probability measure, ν. One cannot do this in general without making some assumptions about ν. A fairly general condition was found by Hans Maassen in 99 4 Theorem (H. Maassen. Let ν be a probability measure on with σ = t dν(t < and tdν(t =0. Then G is a bijection between {z Im(z >σ} and { z Im(z > σ} i σ Compactly Supported Measures. When the probability measure ν has compact support then one can find a region in C + upon which the Cauchy transform is univalent. Suppose ν is a probability measure on and supp(ν [ r, r]. Then ν has moments m n = t n dν(t of all orders and m n r n. Let M(z =+m z + m z + be the moment generating function. If 3 E. Lukacs, Characteristic Functions, nd ed., Griffin, 97 4 Addition of Freely Independent andom Variables, J. Funct. Anal., 06 (99, 409-438.
6 LECTUE 4: FEE HAMONIC ANALYSIS z <r the series converges. If z >rthen z <r and G(z = z t dν(t = z n t n dν(t z n 0 = z n t n dν(t = z n m n = ( z z z M z n 0 Let f(z =zm(z =G( =z + m z z + m z 3 +. Then f(0=0 and f (0 =. Suppose z, z (4r then f(z f(z ( f(z f(z z z e z z [ ] d f(z + t(z z = e dt dt z z = 0 0 n 0 e(f (z + t(z z dt Now e(f (z = e( + zm +3x m + z r 3 z r = (+( z r+3( z r + = So for z < (4r we have ( z r e(f (z ( /4 = 9 Hence for z, z < (4r we have f(z f(z z 9 z. In particular f is one-to-one on {z z < (4r }. Also f(z = z +m z + m z + ( z ( ( + r z + r z + = z ( z = /4 3 z r z Thus for z =(4r we have f(z (6r. Moreover, since f is one-to-one on {z z (4r }, the curve γ = {f(e it /(4r 0 t π} only wraps once around the origin. Thus f maps {z z < (4r } to the interior of the curve γ. Hence {z z < (6r } {f(z z < (4r }
LECTUE 4: FEE HAMONIC ANALYSIS 7 We will denote the inverse of f under composition by f.thusf is analytic on {z z < (6r } and has a simple zero at z =0. Thus K(z = f (z is meromorphic on {z z < (6r } and has a simple pole at z =0. Now K(G(z=/(f (G(z=/(f (f(z = z and G(K(z = f(f (z = z. Theorem. Suppose supp(ν [ r, r], then G ν is univalent on the open disc with centre 0 and radius (6r Let (z =K(z. Then is analytic on {z z < z (6r }, and ( G z + (z = z = (G(z + G(z which is the same relation as was found combinatorially. Bercovici and Voiculescu found a theorem that relates convergence in distribution of a sequence of probability measures with support in [ r, r] to the uniform convergence of the corresponding -transforms on a disc. Theorem (Bercovici & Voiculescu. Let {ν n } n be a sequence of probability measures on with compact support, let n be the corresponding sequence of -transforms. Then i r s.t. supp(ν n [ r, r] and ν with supp(ν [ r, r] such that ν D n ν if and only if ii s>0 such that { n } n converges uniformly on {z z <s} to an analytic function on {z z <s}. Example: The Semi-circle Law. As an example of Stieltjes inversion let us take a familiar example and calculate its Cauchy transform and then using only the Cauchy transform find the density by using Stieltjes inversion. The density of the semi-circle law is given by dν dt = the moments are given by 4 t sdt on [, ] π m n = { 0 n odd t n dν(t = n even c n/
8 LECTUE 4: FEE HAMONIC ANALYSIS where the c n s are the Catalan numbers: c n = ( n n + n Now let M(z be the moment generating function then M(z = M(z =+c z + c z 4 + c m c n z (m+n = m,n 0 k 0 Now we saw in lecture three that c m c n = c k+ so therefore m+n=k ( m+n=k c m c n z k M(z = c k+ z k = c z k+ z (k+ k 0 k 0 z M(z = M(z orm(z =+z M(z Therefore z M(z M(z + = 0. Solving the quadratic equation we have M(z = ± 4z and M(0= z Now + 4t as t 0, t whereas 4t = t + 4t ast 0 Therefore we choose the minus sign M(z = 4z z Now G(z = ( z M = z z 4z = z z 4 z Also + z M(z = M(z implies ( ( ( + z M = z z z M(z
LECTUE 4: FEE HAMONIC ANALYSIS 9 i.e. +G(z = zg(z orz = G(z+ G(z Thus K(z =z + and (z =z i.e. all so all cumulants of the z semi-circle law are 0 except κ, which equals. Now let us apply Stieltjes inversion to G(z. To do this we have to choose a branch of z 4. To achieve this we choose a branch for each of z and z +. Now z = z / e iθ/ where θ is the argument of z. Choosing a branch means connecting to with a curve along which θ will have a jump discontinuity.. z θ (z 0.8 0.6 0.4 0. θ (z - - -0. We choose θ, the argument of z, to be such that 0 θ < π. Similarly we choose θ, the argument of z +, such that 0 θ < π. Thus θ + θ is continuous on C \ [,. However e i(θ +θ / is continuous on C \ [, ] because e i(0+0/ ==e i(π+π/, so there is no jump as the half line [, is crossed. Now z 4= z z += z 4 / e iθ/ e iθ/ = z 4 / e i(θ +θ / Im( (x + iy 4 = (x + iy 4 / sin((θ + θ / lim Im( (x + iy 4 = x 4 / y 0 = { 0 x > x < { 0 x > 0 4 x x < lim y 0 ( x + iy (x + Im(G(x + iy = lim Im iy 4 + y 0 + { 0 x > = 4 x x <
0 LECTUE 4: FEE HAMONIC ANALYSIS Therefore { 0 x > lim Im(G(x + iy = y 0 + π 4 x π x < Hence we recover our original density. Stieltjes Inversion at an Atom. When the Cauchy transform has a pole at a the limit calculation using the dominated convergence theorem used above will not apply. A situation that frequently occurs is that G(z has a simple pole at a, i.e. G(z =f(a/(z a with f analytic on a neighbourhood of a, then ν has an atom at a with weight f(a. Theorem. Suppose G(z =f(z/(z a with a a real number and f analytic on a neighbourhood, B(a, r, ofa. Then a+r lim Im(G(x + iy dx = f(a y 0 + π a r Proof. Let C be the boundary of the rectangle with vertices at a+y+ir, a y + ir, a y ir, and a + y ir. We will parameterize the four sides as follows. C a + r + iy C a C 4 a r iy C 3 C : the top side; γ (t =t+iy where t decreases from a+r to a r; C : the left side; γ = a r + it where t decreases from y to y; C 3 : the bottom side; γ 3 (t =t iy where t increases from a r to a + r C 4 : the right side; γ 4 (t =a + r + it where t increases from y to y Then by Cauchy s integral theorem πi C f(z dz = f(a z a
LECTUE 4: FEE HAMONIC ANALYSIS So let us calculate separately each integral over C i.now a r a+r G(z dz = G(x + iy and G(z dz = G(x iy C a+r C 3 a r Thus a+r G(z dz = G(x iδ G(x + iy dx C +C 3 a r = a+r a r i Im(G(x + iy dx Thus a+r Im(G(x + iy dx = G(z dz a r π πi C +C 3 So we only need to prove that for i = or 4 lim G(z dz =0 y 0 + C i we shall only do the case i =. Since f is analytic on B(0,r there is M>0such that f(z M for z C.Thus y G(z dz M My dt C y r it r Thus for i = and 4 lim G(z dz =0 y 0 + C i