PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 63 (77), 1998, 55–65

PERTURBED INTERPOLATION PROBLEMS

Dušan Georgijević

Communicated by Miroljub Jevtić

Abstract. A sort of perturbations of interpolation problems is used, which enables to transfer the known results on interior interpolation problems of Pick– Nevanlinna type to boundary value interpolation problems. It is shown that the nonnegativity and the so called absence-of-residues property of the Pick kernel are necessary and sufficient condition for solvability of a matricial directional boundary value interpolation problem of Pick–Nevanlinna type.

1. Introduction. The classical Pick–Nevanlinna interpolation problem [1],

[2] involves finding a condition ensuring the existence of a complex function ψ, analytic and with a nonnegative imaginary part in the open upper half-plane C^{+} of the complex plane, taking prescribed values at points of a given finite set E ⊂ C^{+}. The problem has been solved independently by Pick [1] and Nevanlinna [2]. Pick established the condition in the form of nonnegativity of a certain matrix (the so called Pick matrix), while Nevanlinna expressed his condition as a sequence of inequalities.

The Pick–Nevanlinna and similar interpolation problems have been studied very much. A variety of generalizations of the mentioned problem have been considered. Nevanlinna already considered multiple interpolation on an infinite set E [2]. Next generalization allowed the demanded function to be meromorphic [3]. In a latter paper [4] Nevanlinna “moved” the set E to the boundary of the domain C^{+}. The so arisen problem is called the Pick–Nevanlinna boundary value interpolation problem. It turned out to be natural in such a problem to prescribe also some estimate of the radial derivative of the unknown function at each point in E. Loewner [5] solved the boundary problem with interpolation data given on a full interval of the real axis. Rosenblum and Rovnyak [6] studied a different type of boundary interpolation problems, in which E is a Borel subset of the boundary.

AMS Subject Classification (1991): Primary 30E05; Secondary 30E25, 47A56.

Key words and phrases: matricial directional boundary value interpolation problem of Pick–Nevanlinna type, perturbation of interpolation problems.

Fedčina [7] introduced the next considerable generalization of the problem, where the unknown function is a matrix function and one prescribes not its full values but rather only values in given directions at points in E (directional or tangential matricial interpolation problem). Such problems were thereafter very intensively studied, in view of manifold applications to problems of technics and physics.

Among other numerous generalizations of the problem, we mention here only the problems on multilply connected domains and the interpolation problems for functions of several independent variables. As well, a great number of various approaches and methods were applied.

The interior interpolation problems (i.e., the problems in which E ⊂ C^{+}) are better studied than the boundary problems (where E ⊂ R). In this paper we consider a sort of perturbations of interpolation problems which enables us to transfer known results on the interior problems to the boundary ones. In this way we will establish our result which asserts that nonnegativity of an analogue to the Pick matrix (called Pick kernel), together with the so called absence-of-residues property, is necessary and sufficient for solvability of a directional matricial boundary value interpolation problem. Ball and Helton [8] proved a similar result, in a more general setting, by a different method.

2. Interpolation. Let us describe the problem which we will consider. Let H be a finite dimensional Hilbert space and let L(H) denote the set of

all linear mappings from H into H. Let E be a finite set of points on the real axis R, let for each x ∈ E a subspace H_{x} of H be given, and two linear mappings: φ : H_{x} → H and φ^{1}(x) : H_{x} → H_{x}, such that P_{x}φ(x)P_{x} and φ^{1}(x) are self-adjoint, where P_{x} denotes the orthogonal projection in H onto H_{x}.

The problem is to find an operator function ψ : C^{+} → L(H) having the properties: (i) ψ is analytic on C^{+}; (ii) Im ψ(z) ≥ 0, z ∈ C^{+}; (iii) for each x ∈ E, there exists lim_{y↓0} ψ(x + iy) =: ψ(x), Im ψ(x) = 0 and ψ(x)/H_{x} = φ(x); (iv) for each

x ∈ E, there exists lim_{y↓0} \frac{1}{y} Im ψ(x + iy) =: ψ′(x) and ⟨ψ′(x)a, a⟩_{H} ≤ ⟨φ^{1}(x)a, a⟩_{H},

a ∈ H_{x}.

(The above radial limits ψ(x) and ψ′(x) exist at least in the sense of weak operator convergence. Since H is finite dimensional, it follows that these radial limits exist also in the sense of uniform operator convergence.)

Denote the above boundary value interpolation problem by I(φ, φ^{1}).

In connection with the problem I(φ, φ^{1}) we introduce an operator valued kernel P which will be called Pick kernel. Let

P(φ, φ^{1}; x, t) = P(x, t) := \frac{1}{t − x}P_{t}[φ(t)^{∗} − φ(x)]P_{x}

for t ≠ x, x, t ∈ E, and

P(φ, φ^{1}; x, x) = P(x, x) := P_{x}φ^{1}P_{x}, x ∈ E.

If P ≥ 0, i.e., if P is positive semidefinite, then there exists a unique functional Hilbert space h(φ, φ^{1}) = h, whose reproducing kernel is P [9]. (The proof given in [9] concerns the scalar case, but it works also in the general.) The elements of h are functions defined on E and valued in H. In fact, they are linear combinations of functions of the form P(x, ·)a, x ∈ E, a ∈ H_{x}.

The kernel P will be said to have the absence-of-residues property if the following holds: whenever

(1)  ∑_{x∈D} P(x, ·)a_{x} = 0

for some set D ⊂ E and some nonzero vectors a_{x}, a_{x} ∈ H_{x}, x ∈ D, and

(2)  a_{v} := ∑_{x∈D} \frac{1}{v − x}a_{x} ∈ H

for some v ∈ E \ D, then is

(3)  φ(v)a_{v} = ∑_{x∈D} \frac{1}{v − x}φ(x)a_{x}.

If an operator function ψ : C^{+} → B(H) satisfies (i) and (ii) (without assuming that H is finite dimensional), then there exists a unique functional Hilbert space H(ψ) with the reproducing kernel

K(w, z) := \frac{1}{z − w̄} [ψ(z) − ψ(w)^{∗}], z, w ∈ C^{+}.

(B(H) stands for the set of all bounded linear operators on H.) In the scalar case: dim H = 1, this fact is established in [10, Th. 5]) and in the general case it can be established by a procedure quite analogous to that in [10].

For convenience of the reader, we shall present this procedure here. The main tool will be the following representation of ψ, called the Herglotz representation: there exists a unique Hermitian nondecreasing, left continuous operator function μ : [−∞, +∞] → B(H), such that μ(−∞) = 0 and

(4)  ψ(z) = ∫_{−∞}^{+∞} \frac{1 + tz}{t − z} dμ(t) + Re ψ(i), z ∈ C^{+},

where the integral exists in the sense of the uniform operator norm. Such a representation for the scalar case can be found, for example, in [11, pp. 40 and 67]. The general case follows easily from the scalar one. See also [12].

From (4) it follows that

K(w, z) = ∫_{−∞}^{+∞} \frac{1 + t^{2}}{(t − z)(t − w̄)} dμ(t), z, w ∈ C^{+}.

According to this, whenever w_{1}, w_{2},…, w_{n} are points in C^{+} and a_{1}, a_{2},…, a_{n} some vectors in H, then we have

∑_{j,k=1}^{n}⟨K(w_{j}, w_{k})a_{j}, a_{k}⟩_{H} = ∑_{j,k=1}^{n} ∫_{−∞}^{+∞} \frac{1 + t^{2}}{(t − w_{k})(t − w̄_{j})}⟨dμ(t)a_{j}, a_{k}⟩_{H}

= ∫_{−∞}^{+∞}(1 + t^{2})⟨dμ(t) ∑_{j=1}^{n} \frac{1}{t − w̄_{j}} a_{j}, ∑_{k=1}^{n} \frac{1}{t − w̄_{k}} a_{k}⟩_{H} ≥ 0.

This shows that K ≥ 0, which implies, by [9], that there exists a unique functional Hilbert space whose reproducing kernel is K.

Theorem 1. A necessary and sufficient condition for the problem I(φ, φ^{1}) to have a solution is that the kernel P is nonnegative and has the absence-of-residues property.

Note that in the case H_{x} ≡ H, x ∈ E, the nonnegativity of P alone is necessary and sufficient for solvability of the problem I(φ, φ^{1}), as in the scalar case. Namely, in that case the nonnegativity of P implies the absence-of-residues property. This can be proved in the same way as in the necessity part of the below proof, for the kernel K/E^{2}. (See the derivation of (6) from (5).)

Proof . Necessity. Assume that ψ is a solution to the problem. Then ψ satisfies (i) and (ii) and there exists the space H(ψ). According to [13, Remark 1], (iii) and (iv) imply that the functions in z

K(x, z)a := \frac{1}{z − x}[ψ(z) − ψ(x)]a, x ∈ E, a ∈ H,

belong to H(ψ) and that ⟨K(x, ·)a, K(t, ·)b⟩_{H(ψ)} = ⟨K(x, t)a, b⟩_{H}, x, t ∈ E, a, b ∈ H, with K(x, x) := ψ′(x), x ∈ E. Then we have P(x, t) = P_{t}K(x, t)P_{x} for x ≠ t, x, t ∈ E, and P(x, x) ≥ P_{x}K(x, x)P_{x}, x ∈ E, which shows that P ≥ K, where K(x, t) := P_{t}K(x, t)P_{x}, x, t ∈ E. Thus, the nonnegativity of P follows from the nonnegativity of (the extended) K.

In order to show that P has the absence-of-residues property, let (1) and (2) hold. Since P ≥ K (as we have seen above), it follows that

∑_{x∈D} K(x, ·)a_{x} = 0,

which implies

∑_{x∈D} K(x, ·)a_{x} = 0.

In particular, we have

(5)  ∑_{x∈D}⟨K(x, ·)a_{x}, K(v, ·)a⟩_{H(ψ)} = 0

for any a ∈ H. From (5), we obtain further

∑_{x∈D} \frac{1}{v − x}[⟨ψ(v)a_{x}, a⟩_{H} − ⟨ψ(x)a_{x}, a⟩_{H}] = 0,

and, according to (iii),

(6)  ⟨φ(v)a_{v}, a⟩_{H} − ⟨∑_{x∈D} \frac{1}{v − x}φ(x)a_{x}, a⟩_{H} = 0.

Since (6) holds for any vector a in H, it follows that (3) holds. The absence-of-residues property of P is established.

Sufficiency. First case: H_{x} ≡ H, x ∈ E; P > 0. For any y > 0 set φ(x+iy) := φ(x) + iφ^{1}(x), x ∈ E, and

P_{y}(x, t) := \frac{1}{t − x + 2iy} [φ(t + iy) − φ(x + iy)^{∗}], x, t ∈ E.

It is clear that

lim_{y↓0} P_{y} = P.

Since P > 0, i.e., P is positive definite, we can make P_{y} positive by choosing a sufficiently small y. (Namely, we may replace the positivity of P_{y}, resp. P, by the positivity of the Gram matrix G_{y}, resp. G, of the set of vectors P_{y}(x, ·)e_{j}, x ∈ E, 1 ≤ j ≤ dim H, resp. P(x, ·)e_{j}, x ∈ E, 1 ≤ j ≤ dim H, using an orthonormal basis (e_{j}) in H, and then make the Sylvester determinants of G_{y} positive, as those of G are positive.) For any y > 0 we pose the problem of finding an operator function ψ : C^{+} → L(H) having the properties (i), (ii) and

(iii_{y})  ψ(x + iy) = φ(x + iy), x ∈ E.

If the Pick kernel P_{y} (i.e., the Pick matrix) of the interior problem (i)–(iii_{y}) is positive, then this problem has a solution (for ex. [14]). Let ψ_{y} denote a solution to this problem. Since for any a ∈ H, a ≠ 0, the scalar functions

[⟨ψ_{y}(z)a, a⟩_{H} − i⟨a, a⟩_{H}]/[⟨ψ_{y}(z)a, a⟩ + i⟨a, a⟩_{H}]

form a normal family, we can choose a sequence (y_{n}) of positive numbers tending to 0 and such that the sequence of functions ⟨ψ_{n}a, a⟩_{H}, n ∈ N, (ψ_{n} := ψ_{y_{n}}) converges uniformly on compact subsets of C^{+}, where the limit can be ∞ in the whole C^{+} or nowhere in C^{+}. Starting with an orthonormal basis (e_{j}) in H, putting above e_{j}, e_{j} ± e_{k}, e_{j} ± ie_{k} (1 ≤ j, k ≤ dim H) instead of a, and passing several times to a subsequence, we can get a sequence (y_{n}) such that y_{n} → 0, n → ∞, and that ⟨ψ_{n}e_{j}, e_{k}⟩_{H} converges uniformly on compact subsets of C^{+} as n → ∞, for any j, k, 1 ≤ j, k ≤ dim H. This implies that for any two vectors a, b ∈ H, ⟨ψ_{n}a, b⟩_{H} converges uniformly on compact subsets of C^{+}.

Denote the reproducing kernel of the space H(ψ_{n}) by K_{n}, n ∈ N. Thus

K_{n}(w, z) = \frac{1}{z − w̄} [ψ_{n}(z) − ψ_{n}(w)^{∗}], z, w ∈ C^{+}.

If y > 0 and y_{n} ≤ y, then we have

(7)  K_{n}(x + iy, x + iy) ≤ K_{n}(x + iy_{n}, x + iy_{n})

(see [13, Remark 2] ) and therefore

|⟨K_{n}(x+iy_{n}, x+iy)a,b⟩_{H}|^{2} ≤ ⟨K_{n}(x+iy_{n}, x+iy_{n})a, a⟩_{H} ·⟨K_{n}(x+iy, x+iy)b, b⟩_{H}
≤ ⟨K_{n}(x + iy_{n}, x + iy_{n})a, a⟩_{H} · ⟨K_{n}(x + iy_{n}, x + iy_{n})b, b⟩_{H}
= ⟨φ^{1}(x)a, a⟩_{H} · ⟨φ^{1}(x)b, b⟩_{H}, x ∈ E, a, b ∈ H.

Since

K_{n}(x + iy_{n}, x + iy) = \frac{1}{i(y + y_{n})}[ψ_{n}(x + iy) − φ(x + iy_{n})^{∗}],

it follows

(8) |⟨[ψ_{n}(x + iy) − φ(x + iy_{n})^{∗}]a, b⟩_{H}|^{2} ≤ (y + y_{n})^{2}⟨φ^{1}(x)a, a⟩_{H} · ⟨φ^{1}(x)b, b⟩_{H},

x ∈ E, a, b ∈ H. This shows that ⟨ψ_{n}(x + iy)a, b⟩_{H} cannot tend to ∞, and, consequently, that ψ_{n} converges to a function ψ : C^{+} → L(H) satisfying (i) and (ii).

Let n → ∞ in (8). Then we obtain

|⟨[ψ(x + iy) − φ(x)]a, b⟩_{H}|^{2} ≤ y^{2}⟨φ^{1}(x)a, a⟩_{H}⟨φ^{1}(x)b, b⟩_{H},

which shows that

lim_{y↓0} ψ(x + iy) = φ(x), x ∈ E,

i.e., that ψ satisfies (iii).

Rewrite (7) in the form K_{n}(x+iy, x+iy) ≤ φ^{1}(x), and let n → ∞: K(x+iy, x + iy) ≤ φ^{1}(x), x ∈ E. Since K(x + iy, x + iy) is Hermitian nonincreasing in y [13, Remark 2], it follows that there exist the limits

(9)  lim_{y↓0} K(x + iy, x + iy) =: K(x, x), x ∈ E,

and that

(10)  K(x, x) ≤ φ^{1}(x), x ∈ E.

Since K(x + iy, x + iy) = \frac{1}{y} Im ψ(x + iy), the relations (9) and (10) show that (iv) is also satisfied.

Thus ψ is a solution to the problem I(φ, φ^{1}).

Second case: H_{x} ≡ H, x ∈ E. Take an ε > 0 and add εI to φ^{1}(x), to obtain φ_{ε}^{1}(x) := φ^{1}(x) + εI, x ∈ E. Denote the corresponding Pick kernel by P_{ε}. It is clear that P_{ε} > 0.

According to the preceding case, the problem I(φ_{ε}, φ_{ε}^{1}) has a solution ψ_{ε}, for each ε > 0. Just as above, we can choose a sequence (ε_{n}) tending to 0 and such that ψ_{n} := ψ_{ε_{n}} converges uniformly on compact subsets of C^{+} to an operator function ψ : C^{+} → L(H). Now ⟨ψ_{n}(z)a, a⟩_{H} cannot tend to ∞, because of ψ_{n}(x) = φ(x). Then ψ satisfies (i) and (ii).

Reasoning quite analogously as in the preceding case, we show that

lim_{y↓0} ψ(x + iy) = φ(x)

and

lim_{y↓0} \frac{1}{y} Im ψ(x + iy) ≤ φ^{1}(x), x ∈ E.

Thus ψ satisfies also (iii) and (iv) and therefore it is a solution to the problem I(φ, φ^{1}).

Third case: H_{x} ≢ H for some x ∈ E; for each equality of the form (1) in h it must be a_{v} ∈ H_{v} for any v ∈ E \ D. Then we can extend the kernel P in the following way.

Let v ∈ E be such that H_{v} ≢ H and let a_{0} be a nonzero vector in H ⊖ H_{v}. Denote by H_{v}^{e} the linear span of H_{v} ∪ {a_{0}}, set φ_{e}(v)a_{0} := φ(v)^{∗}a_{0} and φ_{e}(v)a := φ(v)a for a ∈ H_{v}, and then extend φ_{e}(v) to H_{v}^{e} by linearity. It is easily seen that then the operator P_{v}^{e}φ_{e}(v)P_{v}^{e} is self-adjoint, where P_{v}^{e} denotes the orthogonal projection in H onto H_{v}^{e}.

Denote by h_{v} the subspace of h generated by functions of the form P(x, ·)a, x ∈ E \ {v}, a ∈ H_{x}. Let F_{0} be the h_{v} function for which

⟨F_{0}(x), a⟩_{H} = \frac{1}{x − v} [⟨a_{0}, φ(x)a⟩_{H} − ⟨φ_{e}(v)a_{0}, a⟩_{H}], x ∈ E \ \{v\}, a ∈ H_{x}.

To show that F_{0} is well-defined, it is enough (and necessary) to check that the functional

f_{0} : P(x, ·)a → ⟨F_{0}(x), a⟩_{H}, x ∈ E \ {v}, a ∈ H_{x},

can be extended to a linear functional on h_{v} (i.e., that F_{0} gives the Riesz representation of a linear functional on h_{v}). The functional f_{0} really can be linearly extended to h_{v}, since for any equality of the form (1) in h_{v}, i.e., such that v ∉ D, it holds

∑_{x∈D}⟨F_{0}(x), a_{x}⟩_{H} = 0.

Namely, the last equality is equivalent to the following:

⟨a_{0}, ∑_{x∈D} \frac{1}{v − x}φ(x)a_{x}⟩_{H} − ⟨φ_{e}(v)a_{0}, a_{v}⟩_{H} = 0,

which is true since it must be a_{v} ∈ H_{v} and

φ(v)a_{v} = ∑_{x∈D} \frac{1}{v − x}φ(x)a_{x}.

Now, set ⟨φ_{e}^{1}(v)a_{0}, a⟩_{H}:=⟨a_{0}, φ_{e}^{1}(v)a⟩_{H}:=⟨F_{0}(v), a⟩_{H} for any vector a ∈ H_{v}, ⟨φ_{e}^{1}(v)a_{0}, a_{0}⟩_{H} := ∥F_{0}∥_{h}^{2} +1, ⟨φ_{e}^{1}(v)a, b⟩_{H} := ⟨φ^{1}(v)a, b⟩_{H} for a, b ∈ H_{v}, and extend ⟨φ_{e}^{1}(v)a, b⟩_{H} to H_{v}^{e} × H_{v}^{e} by linearity and antilinearity. It is easy to see that then ⟨φ_{e}^{1}(v)a, b⟩_{H} is an Hermitian bilinear form on H_{v}^{e} × H_{v}^{e} and that its restriction to H_{v} × H_{v} coincides with ⟨φ^{1}(v)a, b⟩_{H}.

Let further φ_{e}(x) := φ(x) and φ_{e}^{1}(x) := φ^{1}(x) for x ∈ E \ {v}, and let P_{e} be the kernel P(φ_{e}, φ_{e}^{1}; ·, ·).

The kernel P_{e} is nonnegative and has the absence-of-residues property. Really, nonnegativity follows from the fact that P_{t}P_{e}(x, t)P_{x} = P(x, t), x, t ∈ E, that P is nonnegative, and that

(11)  P_{t}P_{e}(v, t)a_{0} = F_{0}(t), t ∈ E,

and

(12)  ⟨P_{e}(v, v)a_{0}, a_{0}⟩_{H} > ∥F_{0}∥_{h}^{2}.

Denote by h_{e} the corresponding space h(φ_{e}, φ_{e}^{1}). Then h is isomorphic to the functional Hilbert subspace h̃ of h_{e} whose reproducing kernel is P_{e}(x, t)P_{x}, x, t ∈ E, and an isomorphism from h̃ onto h is the mapping Φ : F → F_{P}, F ∈ h̃, where F_{P} (t) := P_{t}F (t), t ∈ E.

It follows from (11) and (12) that the function P_{e}(v, ·)a_{0} is not in h̃, since otherwise ΦP_{e}(v, v)a_{0} = F_{0} and ⟨P_{e}(v, v)a_{0}, a_{0}⟩_{H} = ∥F_{0}∥_{h}^{2}. This means that P_{e}(v, ·)a_{0} cannot be a linear combination of functions of the form P_{e}(x, ·)a, x ∈ E, a ∈ H, so that there is no new equality of the form (1) in h_{e}, in comparison with those in h (more precisely, in h̃). It follows that P_{e} has the absence-of-residues property, since P has.

Evidently, every solution of the problem I(φ_{e}, φ_{e}^{1}) is also a solution of the initial problem I(φ, φ^{1}).

If H_{v}^{e} ≢ H or H_{u} ≢ H for some u ∈ E \ {v}, then P_{e} can be extended in the same way as P above. Continuing in this way, we come, in finitely many steps, to a problem of the type I(φ, φ^{1}) for which H_{x} ≡ H, x ∈ E, and whose every solution is also a solution of the initial problem I(φ, φ^{1}). Thus the case under consideration is reduced to the preceding one.

Fourth case: there is an equality of the form (1) in h such that a_{v} ∉ H_{v} for some v ∈ E \ D. Then we can extend the kernel P in the following way.

Starting with an equality (1) and a v ∈ E \ D such that a_{v} ∉ H_{v}, we set

φ_{e}(v)a_{v} := ∑_{x∈D} \frac{1}{v − x}φ(x)a_{x}.

The absence-of-residues property of P implies independence of the vector φ_{e}(v)a_{v} on the relation (1), for fixed v and a_{v}.

To establish this, it is enough to show that for any equality of the form (1) such that a_{v} = 0 for some v ∈ E \ D it must be

(13)  ∑_{x∈D} \frac{1}{v − x}φ(x)a_{x} = 0.

This is true whenever P has the absence-of-residues property, since (13) is the condition (3) in the case a_{v} = 0.

Consider the following h function:

F_{0} := ∑_{x∈D} \frac{1}{v − x}P(x, ·)a_{x}.

Compute its inner product with any function of the form P(u, ·)a, u ∈ E \ {v}, a ∈ H_{u}:

⟨F_{0}, P(u, ·)a⟩_{h} = ∑_{x∈D} \frac{1}{v − x}⟨P(x, u)a_{x}, a⟩_{H} + \frac{1}{u − v} ∑_{x∈D}⟨P(x, u)a_{x}, a⟩_{H}

= ∑_{x∈D} \frac{u − x}{(u − v)(v − x)}⟨P(x, u)a_{x}, a⟩_{H}

= \frac{1}{u − v} ∑_{x∈D\\{u\}} \frac{1}{v − x}[⟨a_{x}, φ(u)a⟩_{H} − ⟨φ(x)a_{x}, a⟩_{H}]

= \frac{1}{u − v} [⟨a_{v}, φ(u)a⟩_{H} − ⟨φ_{e}(v)a_{v}, a⟩_{H}],

because of ⟨a_{u}, φ(u)a⟩_{H} = ⟨φ(u)a_{u}, a⟩_{H}.

Thus,

⟨F_{0}(u), a⟩_{H} = \frac{1}{u − v} [⟨a_{v}, φ(u)a⟩_{H} − ⟨φ_{e}(v)a_{v}, a⟩_{H}, u ∈ E \ \{v\}, a ∈ H_{u}.

Verify that ⟨φ_{e}(v)a_{v}, a⟩_{H} = ⟨a_{v}, φ(v)a⟩_{H}, a ∈ H_{v}. It follows from (1) that

∑_{x∈D}⟨P(x, v)a_{x}, a⟩_{H} = 0,

i.e.,

∑_{x∈D} \frac{1}{v − x}[⟨a_{x}, φ(v)a⟩_{H} − ⟨φ(x)a_{x}, a⟩_{H}] = 0,

which gives ⟨a_{v}, φ(v)a⟩_{H} − ⟨φ_{e}(v)a_{v}, a⟩_{H} = 0, a ∈ H_{v}, and we are done.

As well, multiplying (1) by the function F_{0} (innerly), we obtain ⟨φ_{e}(v)a_{v}, a_{v}⟩_{H} = ⟨a_{v}, φ_{e}(v)a_{v}⟩_{H}.

Denote by H_{v}^{e} the linear span of H ∪ {a_{v}}, set φ_{e}(v)a := φ(v)a for a ∈ H_{v}, and extend φ_{e}(v) to H_{v}^{e} by linearity. Evidently, then the operator P_{v}^{e}φ_{e}(v)P_{v}^{e} is self-adjoint, where P_{v}^{e} denotes the orthogonal projection in H onto H_{v}^{e}.

Now, set ⟨φ_{e}^{1}(v)a_{v}, a⟩_{H}:=⟨a_{v}, φ_{e}^{1}(v)a⟩_{H}:=⟨F_{0}(v), a⟩_{H} for any vector a ∈ H_{v}, ⟨φ_{e}^{1}(v)a_{v}, a_{v}⟩_{H} := ∥F_{0}∥_{h}^{2} +1, ⟨φ_{e}^{1}(v)a, b⟩_{H} := ⟨φ^{1}(v)a, b⟩_{H} for a, b ∈ H_{v}, and extend ⟨φ_{e}^{1}(v)a, b⟩_{H} to H_{v}^{e} × H_{v}^{e} by linearity and antilinearity. It is easy to see that then ⟨φ_{e}^{1}(v)a, b⟩_{H} is an Hermitian bilinear form on H_{v}^{e} × H_{v}^{e} and that its restriction to H_{v} × H_{v} coincides with ⟨φ_{e}^{1}(v)a, b⟩_{H}.

For x ∈ E \ {v}, let φ_{e}(x) := φ(x) and φ_{e}^{1}(x) := φ^{1}(x). Let P_{e} be the kernel P(φ_{e}, φ_{e}^{1}; ·, ·).

The kernel P_{e} is nonnegative and has the absence-of-residues property. This is proved in the same way as in the previous case.

Evidently, every solution of the problem I(φ_{e}, φ_{e}^{1}) is also a solution of the initial problem I(φ, φ^{1}).

If there exists some equality of the form (1) in h_{e}, such that a_{v} ∉ H_{v} for some v ∈ E \ D, then we can continue to extend the kernel in the same way as above. So we reduce, in finitely many steps, the problem I(φ, φ^{1}) to a problem of the type I(φ, φ^{1}) such that there exists no equality of the form (1) with a_{v} ∉ H_{v} for some v ∈ E \ D.

The proof is completed.

The interpolation problem (i)–(iii) considered in the sufficiency part of the above proof is obtained by a perturbation of the problem I(φ, φ^{1}). That is the earlier mentioned perturbed interpolation problem.

We can easily extend our result to the case when the set E is infinite, in a quite natural way, using an adequate limit process. This can lead to a generalization of the classical Loewner theorem [5], where we prescribe the full matrix values of ψ and ψ′ at points of an open subset of R. That generalization will be published elsewhere.

References

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[8] J. A. Ball and J. W. Helton, Interpolation problems of Pick–Nevanlinna and Loewner types for meromorphic matrix functions: parametrization of the set of all solutions, Integral Equations Operator Theory 9 (1986), 155–203.

[9] N. Aronszajn, Theory of reproducing kernels, Trans. Amer. Math. Soc. 68 (1950), 337–404.

[10] L. de Branges, Hilbert Spaces of Entire Functions, Prentice-Hall, Englewood Cliffs, N. J., 1968.

[11] K. Hoffman, Banach Spaces of Analytic Functions, Prentice-Hall, Englewood Cliffs, N. J., 1962.

[12] M. S. Brodskii, Triangular and Jordan Representations of Linear Operators, Trans. Math. Monographs, Vol. 32, Amer. Math. Soc., Providence, R.I., 1971.

[13] D. R. Georgijević, Radial limits in some analytic spaces, (submitted)

[14] H. Dym, J contractive matrix functions, reproducing kernel Hilbert spaces and interpolation, CBMS Regional Conference Series in Mathematics, 71, AMS, Providence, R. I., 1989.

Katedra za matematiku (Received 27 12 1996) Mašinski fakultet (Revised 22 02 1998) 11000 Beograd Yugoslavia