Hilbert space methods for the construction ofanalytic matrix functions

Zinaida Lykova

Newcastle University, UK

Jointly with J. Agler (UCSD), D. Brown (Newcastle) and N. J.Young (Leeds, Newcastle)

Winnipeg, July 2019

Some analytic interpolation problems

Let Ω be a domain in Cd . We seek a method to resolve thefollowing question.Given distinct points λ1, . . . , λn in the unit disc D and targetpoints w1, . . . ,wn in Ω, does there exist an analytic maph : D→ Ω such that

h(λj) = wj for j = 1, . . . , n?

When Ω is a matrix ball this is the Nevanlinna-Pick problem, whichis solved by Pick’s Interpolation Theorem.Other domains Ω arise in the robust stabilization problem incontrol engineering. Of especial interest is the case that Ω is theset of N × N matrices of spectral radius < 1.

Synopsis

• The spectral Nevanlinna-Pick problem

• The symmetrized bidisc

• The rich saltire – some function spaces and mappings

• A solvability criterion for the spectral Nevanlinna-Pick problem

• The tetrablock

The spectral Nevanlinna-Pick problem

Let r(A) denote the spectral radius of a square matrix A.Given distinct points λ1, . . . , λn in the unit disc D and matricesW1, . . . ,Wn in CN×N , determine whether there exists an analyticmap F : D→ CN×N such that

F (λj) = Wj for j = 1, . . . , n

and r(F (λ)) < 1 for all λ ∈ D.

This is a long-standing problem. No satisfactory solution iscurrently known.We study the case that N = 2. Denote by Ω the domain

W ∈ C2×2 : r(W ) < 1.

The symmetrized bidisc

is the domain

Gdef= (z + w , zw) : z ,w ∈ D.

It is a nonconvex, polynomially convex domain.A 2× 2 matrix A belongs to Ω if and only if (tr A, detA) ∈ G .The closed symmetrised bidisc is

Γ = (z1 + z2, z1z2) : z1, z2 ∈ D.

Interpolation into Ω and G

Theorem(Agler,Young 2004) Let λ1, . . . , λn be distinct points in D andlet W1, . . . ,Wn be 2× 2 complex matrices, none of them a scalarmultiple of the identity. Let sj = tr Wj , pj = detWj for each j .The following conditions are equivalent.(i) There exists an analytic 2× 2 matrix function F in D such that

F (λj) = Wj for j = 1, . . . , n (1)

and the spectral radius r of the matrix F (λ)

r(F (λ)) ≤ 1 for all λ ∈ D; (2)

(ii) there exists an analytic function h : D→ Γ such that

h(λj) = (tr Wj , detWj), j = 1, 2, ..., n. (3)

Thus the interpolation problems for Ω and G are equivalent.

A strategy for the 2× 2 spectral N-P problem is as follows

• Reduce to an interpolation problem in the space Hol(D, Γ) ofanalytic functions from the disc D to the closed symmetrised bidiscΓ.

• Present a duality between the space Hol(D, Γ) and a subset ofthe Schur class S2 of the bidisc. The functions Φ(z , ·) induce aduality between Hol(D, Γ) and a subset of S2.Here Φ : C3 → C is defined by

Φ(z , s, p) =2zp − s

2− zsfor zs 6= 2.

• Use Hilbert space models for S2 to obtain a necessary andsufficient condition for solvability the 2× 2 spectralNevanlinna-Pick problem.

A rich structure for ΓThe rich structure related to the construction of analytic matrixfunctions can be summarised diagrammatically as

S2×2 ←→ R

l l

Hol(D, Γ) ←→ S2,

(4)

Where Hol (D, Γ) := the set of holomorphic functions from D to Γ,

S2×2 := the set of holomorphic 2× 2 matrix functions F

on D such that ||F (λ)|| ≤ 1 for every λ ∈ D,S2 := the set of holomorphic functions from D2 to D,R := the set of pairs of positive kernels on the bidisc

subject to a certain boundedness condition.

Analytic lifting of h ∈ Hol(D, Γ) to F =[Fij]∈ S2×2

In the case that s2 6= 4p, the H∞ function 14s

2 − p is nonzero andso it has an inner-outer factorization, expressible in the form

14s

2 − p = F12F21

where F12 is outer, F12(0) > 0 and

|F12| = |F21| a.e. on T.

Let

F =[Fij]

=

[12s F12F21

12s

]. (5)

Then tr F = s and detF = p. One can show that ‖F‖∞ ≤ 1 on D.

Duality between Hol(D, Γ) and S2 – I

Since Φ(D× Γ) ⊂ D−, if h = (s, p) ∈ Hol(D, Γ) then the function

(z , λ) 7→ Φ(z , h(λ)) =2p(λ)z − s(λ)

−zs(λ) + 2for z , λ ∈ D

belongs to S2. For any fixed λ ∈ D, the map z 7→ Φ(z , h(λ)) is alinear fractional self-map f (z) = az+b

cz+d of D with the property“b = c”. We say that a linear fractional map f of the complexplane has the property “b = c” if f (0) 6=∞ and either f is aconstant map or, for some a, b and d in C,

f (z) =az + b

bz + dfor all z ∈ C ∪ ∞.

Duality between Hol(D, Γ) and S2 – II

Proposition

Let G be an analytic function on D2. There exists a functionh ∈ Hol(D, Γ) such that

G (z , λ) = Φ(z , h(λ)) for all z , λ ∈ D (6)

if and only if G ∈ S2 and, for every λ ∈ D, G (·, λ) is a linearfractional transformation with the property “b = c”.

Thus one can define Lower E : Hol (D, Γ)→ S2 by

Lower E (h)(z , λ) = Φ(z , h(λ)) =2zp(λ)− s(λ)

2− zs(λ)

for h = (s, p) ∈ Hol (D, Γ).

The Schur class of the bidisc

Every function in S2 has a Hilbert space model (Agler, 1990).That is to say, if ϕ ∈ S2 then there exist a separable Hilbert spaceM, a Hermitian projection P on M and an analytic mapu : D2 →M such that

1− ϕ(µ)ϕ(λ) = 〈(I − µ∗PλP)u(λ), u(µ)〉 for all λ, µ ∈ D2, (7)

where, for λ = (λ1, λ2) ∈ D2, λP denotes λ1P + λ2(I − P). Thetriple (M,P, u) is called a model of ϕ. The function Φ(z , h(λ))has the property that it is linear fractional in z for every λ ∈ D. Inconsequence, the projection P has rank one for some model of thefunction.

Hilbert space model for Φ(z , h(λ))

TheoremLet h ∈ Hol(D, Γ). There exists a unique function

F =

[F11 F12F21 F22

]∈ S2×2 such that

(tr F , detF ) = h (8)

and F11 = F22, |F12| = |F21| a.e. on T, F12 is either 0 or outerand F12(0) ≥ 0.Moreover, for all µ, λ ∈ D and all w , z ∈ C such that1− F22(µ)w 6= 0 and 1− F22(λ)z 6= 0, F satisfies the identity

1− Φ(w , h(µ))Φ(z , h(λ)) = (1− wz)γ(µ,w)γ(λ, z)

+ η(µ,w)∗ (I − F (µ)∗F (λ)) η(λ, z), (9)

where

γ(λ, z) = (1− F22(λ)z)−1F12(λ) and η(λ, z) =

[zγ(λ, z)

1

]. (10)

Maps between Hol(D, Γ) and S2×2

One can define Left S : S2×2 → Hol (D, Γ) by

F =

[F11 F12F21 F22

]7→ (tr F , detF ).

The map Left N : Hol (D, Γ)→ S2×2 is defined by

Left N (h) = F ∈ S2×2 for h ∈ Hol (D, Γ),

where F as in the previous Theorem.

Kernels associated with Φ(z , h(λ)) on D2

Definition

R :=(N,M) : N and M are positive kernels on D2,

N has the rank at most 1 and

1− (1− wz)N(z , λ,w , µ)− (1− µλ)M(z , λ,w , µ) ≥ 0

for all z , λ,w , µ ∈ D.

Note that, for the kernels associated with Φ(z , h(λ)) fromTheorem 2

N(z , λ,w , µ) = γ(µ,w)γ(λ, z)

and

M(z , λ,w , µ) = η(µ,w)∗I − F (µ)∗F (λ)

1− µλη(λ, z),

we have (N,M) ∈ R.

Γ-inner functions

DefinitionA Γ-inner function is an analytic function h : D→ Γ such that theradial limit

limr→1−

h(rλ) ∈ bΓ (11)

for almost all λ ∈ T.

Here bΓ is the distinguished boundary of G (or Γ). It is thesymmetrisation of the 2-torus:

bΓ = (z + w , zw) : |z | = |w | = 1.

By Fatou’s Theorem, the radial limit (11) exists for almost allλ ∈ T with respect to Lebesgue measure.

Criteria for the solvability of spectral N-P problems

TheoremLet n ≥ 1, let λ1, . . . , λn be distinct points in D and let (sj , pj) ∈ Γfor j = 1, . . . , n. The following three conditions are equivalent.

1. There exists an analytic function h : D→ Γ satisfying

h(λj) = (sj , pj) for j = 1, . . . , n; (12)

2. there exists a rational Γ-inner function h satisfying equations(12);

3. for every triple of distinct points z1, z2, z3 in D, there existpositive 3n-square matrices N = [Ni`,jk ]n,3i ,j=1, `,k=1 of rank at

most 1 and M = [Mi`,jk ]n,3i ,j=1, `,k=1 such that, for 1 ≤ i , j ≤ nand 1 ≤ `, k ≤ 3,

1−(

2z`pi − si2− z`si

)2zkpj − sj2− zksj

= (1−z`zk)Ni`,jk+(1−λiλj)Mi`,jk .

(13)

Interpolation into the tetrablock

The tetrablock is a domain E in C3 that plays a similar role to Gfor another interpolation problem arising in robust control theory.E ∩ R3 is a regular tetrahedron.

There is also a rich saltire for E: simply replace Hol(D, Γ) byHol(D, E) and modify the mappings in the lower triangle.

There is a solvability criterion for interpolation problems inHol(D, E). All constructions and proofs are similar to thosedescribed above.

A µDiag-synthesis problem

The µDiag-synthesis problem is an optimisation problem over aclass of holomorphic matrix functions on the open unit disc D.

Let Diag =

(z 00 w

): z ,w ∈ C

and M ∈M2(C), define

µDiag(M) = (inf ||E || : E ∈ Diag and I −ME is singular)−1.

If I −ME is non-singular for all E ∈ Diag then we setµDiag(M) = 0.

QuestionFor j = 1, . . . , n, let λj ∈ D and Wj ∈M2(C) be such thatµDiag(Wj) ≤ 1. When is there a holomorphic F : D→M2(C)such that F (λj) = Wj for j = 1, . . . , n, and µDiag(F (λ)) ≤ 1 forλ ∈ D?

The tetrablock

Abouhajar, White and Young showed that this case ofµDiag-synthesis is related to the tetrablock. The open tetrablock isthe set

E := x ∈ C3 : there is an M ∈M2(C) such that ||M|| < 1

and x = (m11,m22, detM).

The closed tetrablock is the closure of E, that is, the set

E = x ∈ C3 : there is an M ∈M2(C) such that ||M|| ≤ 1

and x = (m11,m22, detM).

We denote the set of holomorphic functions from D to E byHol (D,E).

Interpolation problem in Hol (D,E)

The case of µDiag-synthesis is equivalent to an interpolationproblem in Hol (D,E).

Theorem (Abouhajar, White, and Young, 2007)

Let λ1, . . . , λn ∈ D and Wj =

[w j11 w j

12

w j21 w j

22

]∈M2(C) be such that

µDiag(Wj) ≤ 1 and w j11w

j22 6= detWj for j = 1, . . . , n. The

following are equivalent.

1. There is a holomorphic 2× 2 matrix function F on D suchthat F (λj) = Wj for j = 1, . . . , n, and µDiag(F (λ)) ≤ 1 for allλ ∈ D.

2. There is a holomorphic function x : D→ E such thatx(λj) = (w j

11,wj22, detWj) for j = 1, . . . , n.

Criterion for solvability

Let λ1, . . . , λn be distinct points in D and let

Wj =

[w j

11 w j12

w j21 w j

22

]∈M2(C)

be such that µDiag(Wj) ≤ 1 and w j11w

j22 6= detWj for j = 1, . . . , n. Set

(x1j , x2j , x3j) = (w j11,w

j22, detWj) ∈ E for each j . The following are equivalent.

(i) There is a holomorphic 2× 2 matrix function F on D such that

F (λj) = Wj for each j , and µDiag(F (λ)) ≤ 1 for all λ ∈ D;

(ii) there is an x ∈ Hol (D,E) such that x(λj) = (x1j , x2j , x3j) for each j ;

(iii) for some triples of distinct points z1, z2, z3 in D, there are positive 3n-square matrices N = [Nil,jk ]n,3i,j=1,l,k=1 of rank 1

and M = [Mil,jk ]n,3i,j=1,l,k=1 such that[

1− zlx3i − x1ix2izl − 1

zkx3j − x1jx2jzk − 1

]≥ [(1− zlzk)Nil,jk ] +

[(1− λiλj)Mil,jk

].

Proof of Main Theorem(iii) =⇒ (ii): Since N is positive and has rank 1 there are γjk ∈ Csuch that, for 1 ≤ j ≤ n and 1 ≤ k ≤ 3,

Nil ,jk = γilγjk .

Since M is positive there is a Hilbert space H of dimension at most3n and vjk ∈ H such that, for 1 ≤ j ≤ n and 1 ≤ k ≤ 3,

Mil ,jk = 〈vjk , vil〉H .

By assumption

1− zlx3i − x1ix2izl − 1

zkx3j − x1jx2jzk − 1

= (1− zlzk)Nil ,jk + (1− λiλj)Mil ,jk .

Let Ψ(zk , x1j , x2j , x3j) =x3jzk−x1jx2jzk−1 for 1 ≤ j ≤ n and 1 ≤ k ≤ 3. We

can show that the Grammians of the following vectors are equal.Ψ(zk , x1j , x2j , x3j)γjkvjk

∈ C2 ⊕ H and

1zkγjkλjvjk

∈ C2 ⊕ H.

It can be shown there is a unitary operator L on C2 ⊕ H with

L :

Ψ(zk , x1j , x2j , x3j)γjk

vjk

7→ 1

zkγjk

λjvjk

for 1 ≤ j ≤ n and 1 ≤ k ≤ 3. Write L =

[A BC D

]. Then

(Ψ(zk , x1j , x2j , x3j)

γjk

)= A

(1

zkγjk

)+ Bλjvjk

and

vjk = C

(1

zkγjk

)+ Dλjvjk .

Thus, for 1 ≤ j ≤ n and 1 ≤ k ≤ 3,(Ψ(zk , x1j , x2j , x3j)

γjk

)= (A + Bλj(I − Dλj)

−1C )

(1

zkγjk

).

Set Ξ(λ) = A + Bλ(I − Dλ)−1C =

[a(λ) b(λ)c(λ) d(λ)

]for all λ ∈ D.

Then, since L is unitary and H is finite dimensional, Ξ is a rational2× 2 inner function. It follows that x = (a, d , det Ξ) is a rationalE-inner function. Moreover, for 1 ≤ j ≤ n and 1 ≤ k ≤ 3,(

Ψ(zk , x1j , x2j , x3j)γjk

)= Ξ(λj)

(1

zkγjk

)=

(a(λj) + b(λj)zkγjkc(λj) + d(λj)zkγjk

),

and so, for 1 ≤ j ≤ n and 1 ≤ k ≤ 3,

Ψ(zk , x1j , x2j , x3j) = a(λj) + b(λj)zk(1− d(λj)zk)−1c(λj).

Hence, for 1 ≤ j ≤ n and z ∈ D,

x1j − x3jz

1− x2jz=

a(λj)− (a(λj)d(λj)− b(λj)c(λj))z

1− d(λj)z.

It follows that x(λj) = (x1j , x2j , x3j) for each 1 ≤ j ≤ n.

(i) =⇒ (iii): Suppose there is an x = (x1, x2, x3) ∈ Hol (D,E) suchthat x(λj) = (x1j , x2j , x3j) for each 1 ≤ j ≤ n. Then it can be

shown there is a function F =

[x1 f1f2 x2

]∈ S2×2 such that

1−Ψ(w , x(µ))Ψ(z , x(λ)) = (1− wz)γ(µ,w)γ(λ, z)

+(1− µλ)η(µ,w)∗I − F (µ)∗F (λ)

1− µλη(λ, z)

for all z , λ,w , µ ∈ D, where Ψ(z , x(λ)) = x3(λ)z−x1(λ)x2(λ)z−1 ,

γ(λ, z) =f2(λ)

1− x2(λ)zand η(λ, z) =

[1

γ(λ, z)z

].

In particular, for 1 ≤ i , j ≤ n and 1 ≤ l , k ≤ 3,

1−Ψ(zl , x1i , x2i , x3i )Ψ(zk , x1j , x2j , x3j) = (1− zlzk)γ(λi , zl)γ(λj , zk)

+(1− λiλj)η(λi , zl)∗ I − F (λi )

∗F (λj)

1− λiλjη(λj , zk).

The 3n-square matrices

N = [Nil ,jk ]n,3i ,j=1,l ,k=1 :=[γ(λi , zl)γ(λj , zk)

]n,3i ,j=1,l ,k=1

and

M = [Mil ,jk ]n,3i ,j=1,l ,k=1 :=

[η(λi , zl)

∗ I − F (λi )∗F (λj)

1− λiλjη(λj , zk)

]n,3i ,j=1,l ,k=1

are positive. Moreover, N has rank 1 and

1− x3izl − x1ix2izl − 1

x3jzk − x1jx2jzk − 1

= (1− zlzk)Nil ,jk + (1− λiλj)Mil ,jk

for 1 ≤ i , j ≤ n and 1 ≤ l , k ≤ 3.

Boundedness of N and M

We have ||F (λj)|| =

∣∣∣∣∣∣∣∣[x1(λj) f1(λj)f2(λj) x2(λj)

]∣∣∣∣∣∣∣∣ ≤ 1,

γ(λj , zk) =f2(λj)

1− x2jzkand η(λj , zk) =

[1

γ(λj , zk)zk

]for all 1 ≤ j ≤ n and 1 ≤ k ≤ 3. Hence |f2(λj)| ≤ 1 and so

|γ(λj , zk)| ≤ 1

|1− x2jzk |≤ 1

1− |x2j |for each j and k .

Moreover, for each j and k,

||η(λj , zk)||2C2 = 1 + |γ(λj , zk)zk |2 ≤ 1 +1

(1− |x2j |)2.

It follows that

|Nil ,jk | = |γ(λi , zl)| |γ(λj , zk)| ≤ 1

(1− |x2i |)(1− |x2j |)

and

|Mil ,jk | ≤||I − F (λi )

∗F (λj)|||1− λiλj |

||η(λi , zl)||C2 ||η(λj , zk)||C2

≤ 2

|1− λiλj |

√1 +

1

(1− |x2i |)2

√1 +

1

(1− |x2j |)2

for all 1 ≤ i , j ≤ n and 1 ≤ l , k ≤ 3.There are software packages that can determine whether a pair(N,M) satisfies the desired matrix inequality. These bounds show,to find such a pair, it is sufficient to search over a compact set.

References

1. A. A. Abouhajar, M. C. White and N. J. Young, A Schwarzlemma for a domain related to µ-synthesis, J. Geom. Anal. 17 (4)(2007), 717-750.2. J. Agler, Zinaida A. Lykova and N. J. Young, A case ofµ-synthesis as a quadratic semidefinite program, Siam J. ControlOptim., 51, (3) (2013), 2472–2508.3. D. C. Brown, Z. A. Lykova and N. J. Young, A rich structurerelated to the construction of holomorphic matrix functions,Journal of Functional Analysis, 272(4) (2017), 1704–1754.

Thank you