Solving Continuous Time Leech Problems for Rational Operator Functions

The main continuous time Leech problems considered in this paper are based on stable rational finite dimensional operator-valued functions G and K. Here stable means that G and K do not have poles in the closed right half plane including infinity, and the Leech problem is to find a stable rational operator solution X such that G(s)X(s)=K(s)(s∈C+)andsup{‖X(s)‖:ℜs≥0}<1.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} G(s)X(s) = K(s) \quad (s\in \mathbb {C}_+) \quad \hbox {and}\quad \sup \{ \Vert X(s) \Vert :\Re s \ge 0 \} < 1 . \end{aligned}$$\end{document}In the paper the solution of the Leech problem is given in the form of a state space realization. In this realization the finite dimensional operators involved are expressed in the operators of state space realizations of the functions G and K. The formulas are inspired by and based on ideas originating from commutant lifting techniques. However, the proof mainly uses the state space representations of the rational finite dimensional operator-valued functions involved. The solutions to the discrete time Leech problem on the unit circle are easier to develop and have been solved earlier; see, for example, Frazho et al. (Indagationes Math 25:250–274 2014).


Introduction
Throughout this paper U, E and Y are all finite dimensional Hilbert spaces. (1.1) Moreover, L 2 + (U) is the Hilbert space formed by the set of all Lebesgue measurable functions g(t) with values in U, t ≥ 0 and such that Let G in H ∞ (U, Y) and K in H ∞ (E, Y) be rational functions. Let T G and T K denote the corresponding Wiener-Hopf operators, . To be precise, G(s) = D 1 + Lg • (s) and K(s) = D 2 + Lk • (s) (for s ∈ C + ). (1.2) Here Lg • and Lk • denote the Laplace transform of the functions g • and k • , respectively, that is, It is emphasized that in our problems g • and k • are integrable operator valued functions over the interval [0, ∞). Following engineering notation, the time function g(t) is denoted with a lower case, while its Laplace transform G(s) is denoted by a capital. A function X in H ∞ (E, U) is called a solution to the Leech problem associated with data G and K whenever G(s)X(s) = K(s) (s ∈ C + ) and X ∞ ≤ 1. (1.4) The Leech problem is an example of a metric constrained interpolation problem, the first part of (1.4) is the interpolation condition, and the second part is the metric constraint.
In an unpublished note from the early 1970's (and then eventually published) [20] Leech proved that the problem is solvable if and only if the operator T G T * G − T K T * K is nonnegative. Later the Leech theorem was derived as a corollary of more general results; see, e.g., [21, page 107], [9, Sect. VIII.6], and [1,Sect. 4.7]. But in Leech's work and in the other results just mentioned the problem was solved for H ∞ functions in the open unit disc. In that case, T G and T K are Toeplitz operators. Here we are working with H ∞ functions in the open right half plane.
One can use a Cayley transform to obtain a right half plane solution from an open unit disc solution. Moreover, one can directly use the commutant lifting theorem with the unilateral shift on the appropriate H 2 space determined by the multiplication operator s −1 s+1 , to arrive at state space solutions to the In other words, Λ is strictly contractive if and only if is strictly positive. Multiplying both sides by T G T * G 1/2 , we see that Λ is strictly contractive if and only if T G T * G − T K T * K is strictly positive. Throughout this paper we assume that Λ defined in (1.5) is strictly contractive, or equivalently, T G T * G − T K T * K is assumed to be strictly positive. In fact, we will develop a state space method involving an algebraic Riccati equation to determine when Λ is strictly contractive. Then motivated by the central solution for the commutant lifting theorem, we will compute a solution X for our Leech problem with data G and K. In fact what we shall present is an analog of Theorem IV.4.1 in [10], which uses a central solution based on a discrete time setting. For the non-discrete time setting the null space for the backward shift in Theorem IV.4.1 in [10] is replaced by the Dirac delta function. An explanation of the role of the Dirac delta is given in Sect. 14 "(Appendix 2)". We will directly show that our solution is indeed a solution to our Leech problem. The next theorem is our first main result for the non-discrete case. Here δ(t) is the Dirac delta function and L is the Laplace transform.
(i) Then X is a solution to our Leech problem. To be precise, X is a function in H ∞ (E, U) satisfying G(s)X(s) = K(s) and X ∞ < 1. is the outer spectral factor for the function I − X * (−s)X(s), that is I − X(−s) * X(s) = Θ(−s) * Θ(s). (1.9) In this paper we shall derive state space formulas for X(s) and Θ(s). Furthermore, we will directly show that X(s) is indeed a strictly contractive solution to our Leech problem, that is, G(s)X(s) = K(s) and X ∞ < 1 and Θ is the outer spectral factor for I − X(−s) * X(s).
The formula for X(s) = U (s)V (s) −1 in Eq. (1.7) was motivated by [10,Theorem IV.4.1], which is a consequent of the central solution to the Sz.-Nagy-Foias commutant lifting theorem. It is emphasized that the setting in [10,Theorem IV.4.1] was developed for the discrete time case, that is, when the Hardy spaces are in the open unit disc, and the corresponding operators are Toeplitz operators. Similar type formulas for U and V are also presented in the band method for the discrete time Nehari problem (see [18]). To arrive at X(s) = U (s)V (s) −1 in (1.7), we replaced the Toeplitz operators by Wiener-Hopf operators and the kernel of the backward shift by the Dirac delta function. This leads to the formula in (1.7) for our solution to the Leech problem. Of course, there are several issues that arise when one makes these adjustments to arrive at X(s) in (1.7). First and foremost is that the Dirac delta function δ(t) is not a well defined function. Moreover, we did not present any justification on why replacing a Toeplitz operators with Wiener-Hopf operators and using the Laplace transform instead of a Fourier transform should lead to a solution X(s) = U (s)V (s) −1 to our Leech problem with data G(s) and K(s). However, we will put all of these issues to rest. First we will use the formula X(s) = U (s)V (s) −1 in (1.7) to derive the state space realization for X(s) in Theorem 3.2. Then, once we have these state space formulas, we will directly verify that X(s) = U (s)V (s) −1 in Theorem 3.2 is indeed a solution to our Leech problem. Moreover, we will also directly verify that all the other results in Theorem 3.2 hold. This direct verification of the solution should eliminate any doubt one may have concerning our solution X(s) to the continuous time Leech problem. Finally, it is noted that we can adapt Theorem IV.4.1 in [10] with the corresponding unilateral shifts on H 2 of the right half plane to obtain Theorem 1.1. However, we decided to simply give the result and verify directly that it holds on the Leech problem.
The paper consists of 14 sections including the introduction, which contains the first main theorem (Theorem 1.1). Throughout, beginning in Sect. 2, the emphasis will be on Leech problems that are based on finite dimensional state space realizations. In the third section the two main theorems are presented, and the first one is proved in the fourth section. The Proof of Theorem 1.1 and the proof of the second theorem in Sect. 3 are based on Lemmas 5.1, 6.1 and 7.1, using formulas (5.3), (6.2), and (7.4). Furthermore, the solution X is given by (8.2), and then Sects. 10 and 11 prove Theorem 3.2. The Proof of Theorem 1.1 is presented in Sect. 12. The Appendices 1 and 2 in Sects. 13 and 14 respectively, present classical results that are used in the paper. "Appendix 1" treats a Riccati equation. In "Appendix 2" the definition of the Dirac delta function δ for the specific class of operators that we 32 Page 6 of 37 A.E. Frazho et al. IEOT need is developed. In fact, the results in Sect. 14 are a special case of the general theory of Dirac delta functions. For completeness assume that T G T * G − T K T * K is a positive operator not necessarily strictly positive, or equivalently, T * G y ≥ T * K y for all y in L 2 + (Y). Then there exists a contraction Λ mapping L 2 + (E) into H = ker(T G ) ⊥ such that Λ * T * G = T K , or equivalently, T G Λ = T K . By choosing the appropriate unilateral shifts one can use the Sz.-Nagy-Foias commutant lifting theorem to show that there exists a function X in H ∞ (E, U) such that Λ = P H T X and X ∞ = T X ≤ 1. (1.10) In particular, X is a solution to our Leech problem. So there exists a solution to our Leech problem if and only if T G T * G − T K T * K is positive. Finally, we compute a solution to our Leech problem when T G T * G − T K T * K is strictly positive. The strictly positive hypothesis is due to standard numerical issues with solving Riccati equations.

The State Space Setup
Throughout this paper, we assume that G and K are stable rational matrix functions. Given this we will establish a state space method involving a special Riccati equation to determine when the operator To develop a solution to our rational Leech problem with data G and K, we use some classical state space realization theory from mathematical systems theory (see, e.g., Chapter 1 of [8] or Chapter 4 in [4]). For our G and K this means that the matrix function G K admits a state space representation of the following form: (2.1) As expected, I denotes the identity operator. Throughout A is a stable operator on a finite dimensional space X . By stable we mean that all the eigenvalues for A are in the open left half plane {s ∈ C : (s) < 0}. Moreover, B 1 , B 2 , C, D 1 and D 2 operate between the appropriate spaces. Since G and K are stable rational operator valued functions, G and K have no poles in the closed right half plane {s ∈ C : (s) ≥ 0}. The realization (2.1) is called minimal if there exists no realization of G K as in (2.1) with the dimension of the state space X smaller than the one in the given realization. In that case, the dimension of the state space X of A is called the McMillan degree of G K . If the realization (2.1) is minimal, then the matrix A is automatically stable. We will use the realization (2.1) of G(s) K(s) to obtain alternative formulas for the functions U (s) and V (s) in (1.7). These alternative state space formulas will be given in Eqs. (6.2) and (5.3) below.
The observability operator W obs mapping X into L 2 minimal. All we need is that A is stable and the pair {C, A} is observable, or equivalently, W obs is one to one. However, from a practical perspective, one would almost always work with a minimal realization. This guarantees that A is stable, makes the state space computations more efficient. As a first step towards our main result we obtain Theorem 3.1 below, which presents a necessary and sufficient condition for T G T * G − T K T * K to be strictly positive in terms of the operators in (2.1) and related matrices. To accomplish this we need the rational matrix function with values in Y given by Here G(s) = G(−s) * and K(s) = K(−s) * . Note that R has no pole on the imaginary axis {iω : −∞ < ω < ∞}. In Sect. 13 "(Appendix 1)" we will show that R admits the following state space representation: Here R 0 on Y and Γ mapping Y into X are defined by where Δ is the unique solution of the Lyapunov equation Since A is stable, this Lyapunov equation is indeed solvable. The solution is unique and given by With our realization for R(s) we associate the following algebraic Riccati equation: For the moment, let us assume that R 0 is strictly positive. Then we say that Q is a stabilizing solution to the algebraic Riccati (2.9) if Q is a strictly positive operator solving (2.9) and the operator A 0 on X defined by is stable. If a stabilizing solution exists, then it is unique. If Q is a stabilizing solution, then W 0 is the observability mapping X into L 2 + (Y) corresponding to the pair {C 0 , A 0 } defined by Finally, T R denotes the non-causal Wiener-Hopf operator determined by R, that is,

Main Theorems
The first main theorem presented in this paper is Theorem 1.
is strictly positive if and only if the following two conditions hold.
(i) The operator T R is strictly positive, or equivalently, R 0 given by (2.5) is strictly positive and there exists a stabilizing solution Q to the algebraic Riccati equation (2.9), that is, Q 0 and the operator A 0 defined by (2.10), i.e., In this case, the Wiener-Hopf operator T R is strictly positive and the unique stabilizing solution to the algebraic Riccati equation is given by The inverse of the operator Finally, R(s) = Φ(s)Φ(s) where Φ is the invertible outer function given by  Theorem 3.2. Let G and K be stable rational matrix functions, and let G K be given by the minimal (stable) realization (2.1). Furthermore, assume that T G T * G − T K T * K is strictly positive, or equivalently, assume that items (i) and (ii) of Theorem 3.1 hold. Then the following holds.
(i) A solution X to the Leech problem with data G and K is given by the following stable state space realization: The operator A is stable, and X ∞ < 1. Finally, the McMillan degree of X is less than or equal to the McMillan degree of G K .
Then Θ is an invertible outer function, that is, both Θ(s) and Θ(s) −1 are functions in H ∞ (E, E). Furthermore, Θ is the outer spectral factor for I − XX. To be precise, Theorem 3.1 will be proved in the next section. The Proof of Theorem 3.2 will be finished at the end of Sect. 11.

Proof of Theorem 3.1
Throughout the section G in H ∞ (U, Y) and K in H ∞ (E, Y) are rational functions, and we assume that G K is given by the minimal stable realization in (2.1). We first prove three lemmas. The first deals with the rational matrix function R(s) defined by (2.3).
In other words, for (α) > 0: Using this with the corresponding result for K, we have Notice that . This with (4.2) readily implies that In other words, we have Multiplying with α + α and taking limits α → iω on the imaginary axis, shows  In particular, the operator T G T * G − T K T * K is strictly positive if and only if the operator T R − W obs ΔW * obs is strictly positive. Proof. We first recall some elementary facts concerning Hankel operators. To this end, let In a similar way, let H K be the corresponding Hankel operator mapping L 2 + (E) into L 2 + (Y) determined by K. Let W con,1 mapping L 2 + (U) into X and W con,2 mapping L 2 + (E) into X be the controllability operators defined by Let P j be the controllability Gramian defined by P j = W con,j W * con,j for j = 1, 2. Notice that P j is the unique solution to the Lyapunov equation The Wiener-Hopf operators T GG * and T KK * are given by the following identities: see (4.4). This with Δ = P 1 − P 2 , yields (4.3).

Lemma 4.3. Let M and T be self-adjoint operators on a Hilbert space H, and let T be strictly positive. Assume that
Proof. Multiplying M = T − W NW * on both sides by T −1/2 yields Let P be the orthogonal projection onto the range of T −1/2 W . Then Notice that This readily implies that The previous equation decomposes T −1/2 MT −1/2 into the orthogonal sum of two self-adjoint operators. Therefore M is strictly positive if and only if Assume that M is strictly positive, or equivalently, Q −1 − N is strictly positive. Then In other words, M −1 is given by the formula in (4.7).
, we see that the matrix Q −1 − Δ is strictly positive, and hence item (ii) in Theorem 3.1 is fulfilled. Furthermore, in this case the inversion formula (4.7) yields the formula to compute the inverse of . Conversely, assume items (i) and (ii) are satisfied. Item (i) gives T R 0. Then again using Lemma 4.3, we see that item (ii) implies that The rest of this paper is devoted to establishing the state space formulas (3.5) for our solution X of the Leech problem and proving both Theorems 1.1 and 3.2.

A State Space Realization for V (s)
Our solution to the Leech problem is motivated by Recall that Λ is a strict contraction if and only if T G T * G − T K T * K is strictly positive. The following result provides a state space realization for V (s).
Recall that A 0 and C 0 are given by (2.10) and that Q solves (2.9). Formula (5.3) for V (s) will play a fundamental role in computing our solution X(s) for the Leech problem.
Proof of Lemma 5.1. By assumption, the operator T G T * G − T K T * K is strictly positive, or equivalently, conditions (i) and (ii) of Theorem 3.1 hold. Hence T R is strictly positive, the unique stabilizing solution to the algebraic Riccati Eq. (2.9) is given by Q = W * obs T −1 R W obs and (see Remark 2.1) Q −1 − Δ is also strictly positive. Moreover, Theorem 3.1 shows that Furthermore, the inverse is given by In other words, This readily implies that x and x ∈ X ; see Part (iii) of Theorem 13.4 in Sect. 13 "(Appendix 1)" for a discussion of W 0 in the general Riccati setting. In particular, W * obs W 0 = Q. Hence The calculations involving the Dirac delta function δ(t) are explained in Sect. 14 "(Appendix 2)". Using this we have In other words, Let us simplify the expression for B. To this end, Using this and (5.5) gives And therefore Now observe that This readily implies that Now that we have a formula for B, notice that

Further notice
For the last equality consult the formulas (13.16) and (13.17) in Sect. 13. Thus we have This with (5.5) and (5.4) yields By taking the Laplace transform of the previous result, we arrive at the state space realization for V (s) = L(I − Λ * Λ) −1 δ (s) in Eq. (5.3) of Lemma 5.1 that we have been looking for, that is,

A State Space Realization for U (s)
The following result provides a state space formula for the function U (s).
Lemma 6.1. Assume that T G T * G − T K T * K is strictly positive, or equivalently, that items (i) and (ii) of Theorem 3.1 hold. Consider the function U (s) defined by Then a state space realization for U (s) is given by Once we have established our state space realizations for U (s) and V (s), we will directly verify that U (s)V (s) −1 is a stable rational solution to our Leech problem, that is, Proof of Lemma 6.1. By assumption T G T * G − T K T * K is strictly positive, or equivalently, conditions (i) and (ii) of Theorem 3.1 hold, or equivalently, Λ is a strict contraction. By employing (5.4), we have In other words, By consulting (5.5), we have The last equality follows by using the properties of the Dirac delta function in Sect. 14 "(Appendix 2)". Replacing K with G in (5.9), we obtain By taking the Laplace transform with (6.1), we have Here U (s) = LΛ(I − Λ * Λ) −1 δ (s). This yields the state space realization for U (s) in Eq. (6.2) above, and completes the Proof of Lemma 6.1.

A State Space Realization for V (s) −1
To compute our solution X(s) = U (s)V (s) −1 to the Leech problem, we need to take the inverse of V (s). Recall that Using standard state space techniques, The "feedback operator" A is defined by 3) The following result expresses V (s) −1 is a slightly different form.
Moreover, the operator A is stable. In particular, V (s) is an invertible outer function, that is, both V (s) and its inverse V (s) −1 are functions in H ∞ (E, E).
Proof of (7.4). Let us derive the formula for A in (7.4). Later in Sect. 9, we will show that A is stable. Recall that Δ is the solution to the Lyapunov equation and that Q is the stabilizing solution of the Riccati Eq. (2.9); see also (13.14) in Sect. 13. To simplify (7.4), let us compute Using this we have Furthermore, using (2.10) (see also (13.9) in Sect. 13 "(Appendix 1)") and (13.4) yields the following By applying the definitions of C u and C v in (6.7) and (7.1), we have Using this we obtain from (7.3) that In other words, To simplify the formula involving A in (7.8), we need to work on C u − D † 1 D 2 C v . To this end, notice that To simplify the last expression, observe that Using this in our previous formula, we have In other words, Using Ξ = I − ΔQ −1 in (7.8), we obtain the result that we have been looking for, that is, Multiplying by Ξ on the right, yields the following formula for A in (7.4): To complete the Proof of Lemma 7.1, it remains to show that A is stable. This will be proved in Sect. 9.
Because A is similar to A, it follows that the operator A is also stable. In particular, V (s) −1 is a function in H ∞ (E, E). Using the state space formula for V (s) −1 in (7.2), with A defined in (7.11), we arrive at the state space formula for V (s) −1 in (7.4).

The State Space Realization for X(s) = U (s)V (s) −1
We are now ready to compute X(s) = U (s)V (s) −1 , which will turn out to be our solution to the Leech problem. For convenience recall that V and U are given by

Finally, the operator A is stable. In particular, X(s) is a rational function in H ∞ (E, U).
This establishes the formula for X(s) in (3.5) in Theorem 3.2. In Sect. 11 we will show that X ∞ < 1. Derivation X(s) in Proposition 8.1. To compute X(s) = U (s)V (s) −1 , first notice that By employing standard state space techniques, we obtain

This with the definition of
Recall that Ξ = (I − ΔQ) −1 . Using this we have see (7.9) and (7.11). This yields the formula for X(s) in (8.2). In Sect. 9 we will prove that A is stable.

Proof of the Stability of A using a Lyapunov Equation
In this section, we will directly show that V −1 is analytic on the closed right half plane. To accomplish this we will prove that A is stable. Since A is similar to A, the operator A is also stable. This guarantees that V (s) −1 is a function in H ∞ (E, E); see (8.6). Since A 0 is stable, V (s) is also a function in H ∞ (E, E); see (8.1). In particular, V (s) is an invertible outer function. U). In Sect. 11 we will show that X ∞ < 1. Recall Throughout this section Q is the stabilizing solution for the algebraic Riccati Eq. (2.9).
In order to show that A is stable, we will first establish the following lemma. Then A satisfies the Lyapunov equation where the operator F , the strict contraction Z and the isometry E are given by Proof. Because Q −1 − Δ is strictly positive (see item (ii) in Theorem 3.1), the operator is also strictly positive. The first step is to prove that Using the formulas for C u and C v in (8.1) and the definition of Γ in (2.6) we have that Recall that Q = W * obs W 0 . This identity is a consequence of the algebraic Riccati Eq. (13.14); see the defination of W 0 in (2.11), the Lyapunov Eqs. (13.20) and (13.23) with Theorem 13.4 in Sect. 13 "(Appendix 1)". Because Q is self-adjoint Q = W * 0 W obs . This yields the Lyapunov equation QA + Therefore (9.7) holds.
Recall [see (9.3)] that P = Q − QΔQ = QΞ −1 , and observe that For the last equality in this calculation, we used the formula for C v in (8.1). So we have By taking the adjoint we see that Note that (9.7) gives Adding the equalities (9.8), (9.9) and (9.10) yields Now observe that, using the formula for C u in (8.1) with D 1 D † 1 = I, This yields the Lyapunov equation that we have been looking for, that is, Next we will transform (9.11) into (9.4). Recall that In other words, The operator E given by So we can rewrite the Lyapunov Eq. (9.11) as which can be rewritten as (9.4).

Proposition 9.2. The operator A in (9.2) is stable.
Proof. Recall that A satisfies the Lyapunov equation where P and F are given by (9.3) and (9.5), respectively. Assume that λ is an eigenvalue for A with eigenvector x, that is, Ax = λx. Using this in the Lyapunov Eq. (9.13), we have Notice that P = Q Q −1 − Δ Q is strictly positive. Hence (Px, x) > 0 and . We will prove that (λ) is nonzero. Assume that (λ) = 0, then we have that F x = 0. So in particular (I − Z * Z) 1/2 C v x = 0. Because Z is strictly contractive, (I − Z * Z) 1/2 is invertible, and thus C v x = 0. Using this with the definition of A, we see that Thus λ is an eigenvalue for the stable operator A 0 . But this means that (λ) < 0, which contradicts the assumption that (λ) = 0. We conclude that (λ) < 0 and therefore λ is a stable eigenvalue for A. This proves that the operator A is stable.
Finally, since A is similar to A, the operator A is also stable.

A Direct Proof of G(s)X(s) = K(s)
So far we have derived a state space formula for X(s); see (3.5) in Theorem 3.2. Because A is stable and A is similar to A, the operator A is stable and X(s) is a rational function in H ∞ (E, U). In this section, we will directly prove that G(s)X(s) = K(s). For convenience recall that In other words, Using Therefore we have

G(s)X(s) = K(s).
So to show that X(s) is indeed a solution to our Leech problem, it remains to show that X ∞ ≤ 1. In fact, we will show in the next section that X ∞ < 1.

The Outer Spectral Factor for I − X(s)X(s)
In this section, we will show that Θ(s) = D Because V (s) is an invertible outer function, Θ(s) is also an invertible outer function, that is, both Θ(s) and Θ(s) −1 are functions in H ∞ (E, E). This sets the stage for the following result.  To prove that (11.4) holds, set Using this notation, the state space formula for U (s) and V (s) in (11.1) and We claim that To verify this, notice that the formulas for V (s) and U (s) in (11.6) yield (11.8) The last equality follows from To prove this equality, observe that This yields (11.9).
To complete the proof, it remains to show that the right hand side of (11.8) is equal to D −1 v , or equivalently, By consulting the Lyapunov Eq. in (9.7), we obtain Since Q − QΔQ = QΞ −1 , we have Moreover, we also have Now observe that Likewise its adjoint This with (11.8) shows that Recall that X(s) = U (s)V (s) −1 . By applying D v to both sides, we see that By taking the appropriate inverse, we arrive at Therefore Θ is the outer spectral factor for I − X(−s) * X(s). Finally, because Θ is an invertible outer function, we also have X ∞ < 1. This completes the Proof of Proposition 11.1.
Since Proposition 11.1 is proved, we have also finished the Proof of Theorem 3.2.

Proof of Theorem 1.1
The Proof of Theorem 1.1 concerns rational functions G in H ∞ (U, Y) and where U (s) and V (s) are the functions defined by (1.7). Moreover, the rational functions G and K admit a minimal stable realization (2.1). Given these data we have to prove items (i) and (ii).
According to Lemma 6.1 the function U (s) has a realization (6.2) and according to Lemma 5.1 the function V (s) has a realization (5.3). Thus it follows from Proposition 8.1 that X(s) = U (s)V (s) −1 has a realization (8.2) and X(s) is analytic on C + . Moreover, by Sect. 10, we have GX = K. Proposition 11.1 gives that X ∞ < 1 and completes the proof that X is indeed a solution of the Leech problem for G and K. Thus item (i) is proved. Finally, Proposition 11.1 also shows us that formula (1.9) holds true with Θ Φ(s).
Finally, recall that an operator T on X is strictly positive, denoted by T 0, if T is positive and its inverse exists and is also a positive operator.
) has a stabilizing solution. In that case the unique spectral factor Φ(s) is given by Here stabilizing solution means that Q 0 and A 0 is stable, where Clearly, R −1/2 0 on both sides of (13.7) can be eliminated. We inserted R −1/2 0 in (13.7) to match the hypothesis in [5,Theorem 14.8]. Notice that the Riccati Eq. (13.8) can be rewritten as (13.10) or, using (13.9), as We do not prove the above theorem, but we will verify that Φ(s)Φ(s) = R(s).
To this end first notice that (13.11) is equivalent to . (13.12) Next observe that In other words, R(s) = Φ(s)Φ(s). Notice the similarity of this computation with the Proof of Lemma 13.1. We mention that Theorem 13.2 also is a special case of [19,Theorem 19 The following theorem provides the main result of this section and is used on several places in this paper. It shows when T R is strictly positive for any rational function R(s) given by a state space realization (13.3).

Theorem 13.4. Let R(s) be any rational function given by the state space realization
where {C, A} is observable, A is stable and Γ is an operator from Y into X . Then the following statements are equivalent (a) T R is strictly positive operator on L 2 + (Y). (b) R 0 0 and there exists a stabilizing solution to the algebraic Riccati equation Here stabilizing solution means that Q 0 and A 0 , given by (13.9), is stable. (i) The spectral factor Φ and its inverse are given by The stabilizing solution Q to the algebraic Riccati Eq. (13.14) is unique, and given by . (13.17) In particular, the pair {C 0 , A 0 } is observable. (iv) The following holds Proof. The proof of this theorem is split into 6 parts. Part 1. We prove that statement (c) implies statement (a). We have R = ΦΦ and thus We conclude that T R 0. Part 2. Let us show that statement (a) implies statement (c). From Theorem 13.3 we conclude that since T R is invertible, the function R has a canonical Wiener-Hopf factorization, R = R − R + . Without loss of generality we may assume that R ± (∞) = R 1/2 0 . Then R + and R − are uniquely determined. Since T R is strictly positive, we have that T R = T * R = T * R+ T * R− and thus R(s) = R + (s) R − (s). The uniqueness of the factorization yields that R − = R + . Put Φ = R + and we conclude that the factorization is a spectral factorization of R. Part 3. In this part we show that the statements (b) and (c) are equivalent. We apply Theorem 13.2. Note that if we have the spectral factorization R = ΦΦ, then R 0 0. But then we also have (13.7) and conclude that (13.8) has a stabilizing solution. Therefore (13.14) has a stabilizing solution. Conversely, according to Theorem 13.2, statement (b) implies (c). Part 4. We have established the equivalence of the statements (a), (b) and (c) and note that statement (i) immediately follows from Theorem 13.2. Part 5. Let us establish the identity in (13.18) in statement (iv). To accomplish this we employ the realization for R(s) in (13.3). Using ΓC 0 = A − A 0 , a standard calculation shows that As before, let Q be the stabilizing solution of the algebraic Riccati Eq. (13.14).
Recall that the operator A 0 is stable and T R is strictly positive. Using C 0 and A 0 in (13.9), the Riccati Eq. (13.14) can be rewritten as Using the Lyapunov Eq. (13.20), we have The last equality follows from Γ * Q = C − R 0 C 0 ; see (13.9). In other words, This with the preceding identity and (13.19) yields the result that we have been looking for, that is, This proves (13.18) and statement (iv). Part 6. The observability operator for the pair {C 0 , A 0 } is the operator W 0 mapping X into L 2 + (Y) defined by (for x ∈ X ); (13.21) (see also the identity (3.19) in [13]).
Recall that (13.18) has been proved. Note that C(sI −A) −1 is a stable rational operator valued function, while Γ * (sI + A * ) −1 Q is a rational function, which is analytic on the open left half plane {s ∈ C : (s) < 0}. and has the value zero at infinity. Moreover, This with (13.18) implies that T R W 0 x is equal to W obs x for each x ∈ X .
Since T R is invertible, we also have W 0 = T −1 R W obs . Throughout we assumed that {C, A} is observable, or equivalently, W obs is one to one. Therefore W 0 = T −1 R W obs is also one to one. The Lyapunov equation in (13.20) In other words, By employing W 0 = T −1 R W obs , we see that Q = W * obs T −1 R W obs . This proves statements (ii) and (iii).

Appendix 2: The Dirac Delta Function δ
In this section we introduce a simple form of the Dirac delta for a specific class of operators. The continuity properties of these operators allow for an elementary definition of the Dirac delta. Let F(U, Y) be the Hilbert space formed by the set of all linear operators mapping U into Y under the Frobenius or trace norm, that is, if M is in F(U, Y), then M 2 F = trace(M * M ). Moreover, L 2 + (F(U, Y)) is the Hilbert space formed by the set of all square integrable Lebesgue measurable functions over the interval [0, ∞) with values in F(U, Y). Consider the state space systems {A 1 , B 1 , C 1 , D} and {A 2 , B 2 , C 2 , D}, where for j = 1, 2 the operator A j is a stable operator on X j , and B j maps U into X j , while C j maps X j into Y and D maps U into Y. The state spaces X j , input space U and output space Y are all finite dimensional complex vector spaces of possibly different dimension of the form C . Let We define the corresponding kernel function k(t) by k(t) = C 1 e A1t B 1 , t ≥ 0, C 2 e −A2t B 2 , t < 0.
The corresponding Wiener-Hopf operator T F mapping L 2 + (F(U, U)) into the space L 2 + (F(U, Y)) is given by (14.2) where h ∈ L 2 + (F(U, U)). Let M U ,Y be the linear space consisting of all Wiener-Hopf operators of the form (14.2). Now fix an operator valued function h in the space L 2 + (F(U, U)). Then h defines a linear transformation L h from M U ,Y into the space L 2 + (F(U, Y)) by L h (T F ) = T F h. If in (14.1) we have D = 0, then the Dirac delta function L δ is formally defined by (for 0 ≤ t < ∞).
Thus we defined the transformation L δ mapping M U ,Y into L 2 + (F(U, Y)). The next step is to extend the definition of δ to the case when D is not equal to 0. We extend L 2 + (F(U, Y)) to a direct sum If h is in L 2 + (F(U, U)), then T F h is in the second component L 2 + (F(U, Y)) of the previous space. For F (s) given by (14.1), we define the linear map L δ T F by L δ T F (t) = D C 1 e A1t B 1 ∈ F(U, Y) L 2 + (F(U, Y)).
We denote this as T F δ (t) = Dδ(t) + C 1 e A1t B 1 (t ≥ 0). (14.4) Note that the sum is formal here and that the δ(t) in Dδ(t) emphasizes this. Finally, it is noted that formally, the Laplace transform of δ(t) equals one, that is, Lδ (s) = 1. Let F (s) = D + C 1 (sI − A 1 ) −1 B 1 , with the operator A 1 a stable operator on X 1 , and B 1 maps U into X 1 , while C 1 maps X 1 into Y and D maps U into Y. Then and the kernel function k * for this operator T * F therefore is given by We conclude from that (T * F δ)(t) = D * δ(t) because k * (t) = 0 for t ≥ 0. The Dirac delta function viewed as a limit For the classical approach to the Dirac delta function, for any a > 0, let ξ a be the function defined by Finally, it is well know and easy to establish that the Laplace transform of δ(t) equals one, that is, Lδ (s) = 1. (This follows from the fact that Lξ a (s) converges to 1 for fixed s.) We will also check the expression for T * F δ (t). In fact, we already defined that To formally verify this, observe that This yields T * F δ (t) = D * δ(t) in (14.7). Finally let F j (s) = D j + C(sI − A j ) −1 B j for j = 1, 2. Then it is easy to verify that the following holds assuming all the spaces are compatible: • T F2 T F1 h = T F2 (T F1 h) when h is in L 2 + (F(U 1 , U 1 )). • For convenience we will denote D 1 δ also as δD 1 . So D 2 D 1 δ, D 2 δD 1 and δD 2 D 1 all are notations for the same element in the first component of a direct sum as (14.3).