X
k=0
σ(n)z(X(n)z)k .
If we let Λk be the multilinear form
Λk(z1, . . . , zk) = σ[z1Xz2. . . Xzk], then for every n and every z ∈ B(n)(0(n), 1/M ), we have
G˜σ(z) =
∞
X
k=0
Λ(n)k (z, . . . , z).
Therefore, by Lemma 3.5.10, we have Λk = ∆kG˜σ(0, . . . , 0). The general formula for
∆kG˜σ(0(n0), . . . , 0(nk)) follows from Lemma 3.5.2.
Lemma 4.3.5. If kzk < 1/M , then we have
G˜σ(z)
≤ kσ(1)kkzk 1 − M kzk In particular,
kzk < 1/(M + ) =⇒
G˜σ(z)
≤ kσ(1)k
.
Proof. This follows by applying the triangle inequality to the geometric series expansion.
4.4 Analytic characterization
The following theorem is due to Williams [Wil17, Theorem 3.1] and Anshelevich-Williams [AW16, Theorem A.1]. While we have nothing to add in terms of the proof, we aim to improve the organization and explanation of details and thus to present the proof in a more
“textbook-like” manner.
Theorem 4.4.1. Let G(n) : H(n)+ (B) → Mn(B). The following are equivalent:
(1) G = Gσ for some generalized law σ with rad(σ) ≤ M if and only if the following conditions hold.
(2) The following conditions hold:
(a) G is fully matricial.
(b) G maps H(n)+ (B) into (n)−,0(B).
(c) ˜G(z) = G(z−1) has a fully matricial extension to B(0, 1/M ).
(d) This extension satisfies ˜G(0) = 0 and ˜G(z∗) = ˜G(z)∗.
(e) For every > 0, there exists C > 0 such that kzk ≤ 1/(M + ) implies k ˜G(z)k ≤ C. Proof of (1) =⇒ (2). Assume that (1) holds. We have already shown that (a), (b), (c), and (e) hold in Lemmas 4.2.7, 4.3.3, 4.3.5. Moreover, (d) follows from power series expansion in Lemma 4.3.3.
The proof of (2) =⇒ (1) is more involved, so we will prove several lemmas before concluding the proof of the Theorem. First, we define the map σ : BhXi → B. The correct choice of σ is clear in light of Lemma 4.3.4.
Lemma 4.4.2. Let G satisfy (2) of Theorem 4.4.1. Define σ : BhXi → B by σ(z0Xz1. . . Xzk) = ∆k+1G(0, . . . , 0)[z˜ 0, . . . , zk].
Then any R > M is an exponential bound for σ(n).
Proof. Because kzk ≤ 1/R implies k ˜G(z)k ≤ CR−M, we have by Lemma 3.4.2 that k∆kG(0˜ (n), . . . , 0(n))k ≤ CR−MRk.
Next, we show that σ extends to the analytic completion of BhXi. Fix R > M . As in the proof of Theorem 2.6.5, we define a norm on Mn(BhXi) = Mn(B)hX(n)i by
kF (X(n))kR= inf ( n
X
j=1
p(Fj) : Fj monomials and f =
n
X
j=1
Fj
) ,
where p(z0X(n)z1. . . X(n)zk) = Rkkz0k . . . kzkk for z0, . . . , zk ∈ Mn(B). We denote the completion by BhXi(n)R . Recall that this is a Banach ∗-algebra.
Lemma 4.4.3. Fix R > M . Then the map σ(n) defined above extends to a bounded map BhXi(n)R → Mn(B). Moreover, if kzkR < 1/R, then 1 − X(n)z is invertible in BhXi(n)R and we have
G(z) = σ˜ (n)[z(1 − X(n)z)−1].
Proof. The first claim follows because kσ(n)(F (X))k ≤ CR−MkF (X)kR since R is an expo-nential bound for σ. Next, suppose that kzkR ≤ 1/R. Then because the geometric series (1 − X(n)z)−1 converges in BhXi(n)R , we see that 1 − X(n)z is invertible. Moreover, a direct power series computation shows that ˜G(z) = σ(n)[z(1 − X(n)z)−1] after we invoke Lemma 3.5.7.
With these preparations in order, we can begin to prove complete positivity of σ. We start out by proving that certain symmetric moments are positive.
Lemma 4.4.4. Suppose that G satisfies (2) of Theorem 4.4.1 and define σ as in Lemma 4.4.2. Let B0 and B1 be self-adjoint elements of Mn(B) with B1 ≥ > 0. Then
σ(n)h
B1(X(n)+ B0)2k
B1i
≥ 0.
Proof. Fix B0 and B1 and let φ be a state on Mn(B). Consider the scalar-valued function g : H+→ H− given by
g(ζ) = φ ◦ G(B1−1ζ − B0).
Now we analyze the behavior of g at ∞. Note that ζ−1B1−1− B0 is invertible in BhXi(n)R if ζ is small enough. In fact, for sufficiently small ζ, we have k(ζ−1B1−1− B0)−1k < 1/R. Thus, we have
g(1/ζ) = φ ◦ ˜G((B1−1ζ−1− B0)−1)
= φ ◦ σ[(B−11 ζ−1− B0)−1(1 − X(n)(B1−1ζ−1− B0)−1)−1]
= φ ◦ σ[(B−11 ζ−1− B0− X(n))−1]
= φ ◦ σ[B1ζ(1 − (X(n)+ B0)B1ζ)−1]
=
∞
X
k=0
ζk+1φ ◦ σ[(B1(X(n)+ B0))kB1],
where the intermediate steps are performed in BhXi(n)R . In particular, ˜g(ζ) = g(1/ζ) extends to be analytic in a neighborhood of 0. Because ˜G preserves adjoints, we have g(ζ) = g(ζ).
Therefore, g is the Cauchy-Stieltjes transform of some compactly supported measure ρ on R. Moreover, by examining the power series coefficients of ˜g at 0, we have
φ ◦ σh
B1(X(n)+ B0)2k
B1i
= Z
R
t2kdρ(t) ≥ 0.
Because this holds for every state φ, we have σ[(B1(X(n) + B0))2kB1] ≥ 0 by Proposition 2.1.8 (5).
Lemma 4.4.5. Let G satisfy (2) and let σ be as above. Let F (Y ) = C0Y C1. . . Y Ck be a monomial in Mn(B)hY i and let B0 ∈ Mn(B) be self-adjoint. Then
σ(n)
F X(n)+ B0∗
F X(n)+ B0
≥ 0.
Proof. Let us write Y = X(n)+ B0. Consider the matrix terms δ will be much larger than the off-diagonal terms, while the extra diagonal term δ−4kCk∗Ck is already positive. Therefore, by the previous lemma,
σ(n)h matrix multipication, treating each n × n block as a unit. The top left block of the product will be the sum of terms of the form
(Cδ)1,i1Y (Cδ)i1,i2Y . . . (Cδ)ik−2,ik−1Y (Cδ)ik−1,1
since Y(k+1) is a block diagonal matrix. Because Cδ is tridiagonal, the sequence of indices must have |ij−1 − ij| ≤ 1. We can picture such a sequence as a path in the graph with vertices {1, . . . , k + 1} and edges between j and j + 1 and a self-loop at each vertex j.
All the entries in Cδ are O(δ) except the bottom right entry with the term δ−4kCk∗Ck. Thus, any path which yields a term larger than O(δ) must reach the last vertex k + 1 and use the self-loop at the vertex k + 1. But if we travel along the path at a speed ≤ 1, the
To finish the proof that σ(n)(P (X)∗P (X)) ≥ 0 for every P , we will use the following matrix amplification trick to reduce to the case of a monomial.
Lemma 4.4.6. Let P (X) ∈ Mn(BhXi) be a polynomial of degree d. Denote expressed as in the conclusion of the lemma.
First, we claim that Γ(1) contains the monomials in BhXi. Let p(X) = a0Xa1. . . Xak be
Finally, we claim that Γ(n) is closed under addition. Suppose that P (X) and Q(X) are in Γ(n). Then there exist integers r and s and matrices B1, . . . , Bd ∈ M2r(B) and C1, . . . ,
and Q(X) 0
0 0
= C0Xb(s)C1Xb(s). . . Bd−1Xb(s)Bd. Then observe that
P (X) + Q(X) 0
0 0
= SB0 0 0 C0
Xb(r+s)B1 0 0 C1
. . . bX(r+s)Bd 0 0 Cd
S∗ Where
S =
1n×n 0n×(n−r) 1n×n 0n×(n−s)
0(r+s−n)×n 0(r+s−n)×(n−r) 0(r+s−n)×n 0(r+s−n)×(s−n)
Altogether, we have shown that Γ = S∞
n=1Γ(n) contains the 1 × 1 monomials of degree
≤ d, is closed under P 7→ P ⊗ Ei,j, and is closed under addition. This implies that Γ contains all matrix polynomials of degree ≤ d as desired.
Conclusion to the proof of Theorem 4.4.1. Suppose that G satisfies (2) of the theorem and let σ : BhXi → B be given as in Lemma 4.4.2. To show that σ is completely positive, choose a polynomial P (X) ∈ Mn(BhXi). Let
B0 =0 1 1 0
∈ M2(B).
Then by Lemma 4.4.6, we can write P (X) in the form
P (X) 0
0 0
= C0(X(2m)+ B0(m))C1(X(2m)+ B0(m)) . . . Cd−1(X(2m)+ B0(m))Cd, where Cj ∈ M2m(B). Thus, by Lemma 4.4.5, we have
σ(2m)Cd∗(X(2m)+ B0(m))Cd−1∗ . . . (X(2m)+ B0(m))C0∗
C0(X(2m)+ B0(m)) . . . Cd−1(X(2m) + B(m)0 )Cd ≥ 0, which implies that σ(n)(P (X)∗P (X)) ≥ 0.
Next, we have shown in Lemma 4.4.2 that σ exponentially bounded by R whenever R > M . Therefore, σ is a generalized law with rad(σ) ≤ M .
It remains to show that the Cauchy transform of σ is the original function G. It follows from Lemma 4.4.3 that ˜G(z) = ˜Gσ(z) when kzk < 1/R. If we let z0 = 2iR, then we have z ∈ B(n)(z0, R) implies that Im z ≥ R + for some > 0 which implies that z−1 ∈ B(n)(0, 1/R).
Hence, we have G = Gσ on B(z0, R). So by the identity theorem (Theorem 3.9.7), we have G = ˜Gσ on the whole matricial upper half-plane.
We now give an analytic characterization of when the generalized law σ is a law, and hence complete the analytic characterization of the Cauchy-Stieltjes transforms of B-valued laws.
Lemma 4.4.7. Let σ be a B-valued generalized law. Then the following are equivalent.
(1) σ is a law.
(2) ∆ ˜Gσ(0, 0)[z] = z for all z ∈ B.
(3) For each n, limkzk→0z−1G˜σ(z) = 1n, where the limit occurs in norm and is taken over all invertible z in Mn(B).
Proof. We have ∆ ˜Gσ(0, 0)[z] = σ(z) for z ∈ B. We also know by Corollary 2.6.7 that σ is a law if and only if σ|B = id. This implies that (1) ⇔ (2).
(1) =⇒ (3). If σ is a law, then
z−1G˜σ(z) = z−1σ(n)[z(1 − X(n)z)−1] = σ(n)[(1 − X(n)z)−1], which is fully matricial in a neighborhood of zero, and hence (3) holds.
(3) =⇒ (1). Fix an invertible operator z ∈ B. Then we have for ζ ∈ C that
ζ→0lim 1 ζ
G˜σ(ζz) = z lim
ζ→0(ζz)−1G˜σ(ζz) = z.
On the other hand,
limζ→0
1 ζ
G˜σ(ζz) = lim
ζ→0
∞
X
k=0
ζkσ[z(Xz)k] = σ[z].
Therefore, σ[z] = z. Any element of B can be written as a linear combination of invertible operators and hence σ|B = id, which means that σ is a law.
We also have the following corollary of Theorem 4.4.1 which is helpful for estimating the radius of generalized laws.
Corollary 4.4.8. Suppose that σ and τ are B-valued generalized laws and Im Gσ(z) ≥ Im Gτ(z). Then
(1) Gτ(z) − Gσ(z) is the Cauchy-Stieltjes transform of some generalized law ρ.
(2) rad(σ) ≤ rad(τ ).
(3) For Im z ≥ , we have kGσ(z) − Gτ(z)k ≤ kσ(1) − τ (1)k/.
Proof. (1) Observe that Gτ − Gσ maps H+(B) into H−(B). Moreover, ˜Gτ − ˜Gσ extends to be fully matricial in a neighborhood of 0 in a way which preserves adjoints. Therefore, there is a generalized law ρ such that Gτ − Gσ = Gρ. Now ρ = τ − σ as maps BhXi → B.
(2) This follows from Lemma 2.6.8 since τ = σ + ρ.
(3) This follows by applying the estimate for Lemma 4.2.7 to Gρ(z).