Continuous Dependence of Szegő Kernel on a Weight of Integration

The weighted Szegő kernel was investigated in a few papers (see Nehari in J d’Analyse Mathématique 2:126–149, 1952; Alenitsin in Zapiski Nauchnykh Seminarov LOMI 24:16–28, 1972; Uehara and Saitoh in Mathematica Japonica 29:887–891, 1984; Uehara in Mathematica Japonica 42:459–469, 1995). In all of these, however, only continuous weights were considered. The aim of this paper is to show that the Szegő kernel depends in a continuous way on a weight of integration in the case when the weights are not necessarily continuous. A topology on the set of admissible weights will be constructed and Pasternak’s theorem (see Pasternak-Winiarski in Studia Mathematica 128:1, 1998) on the dependence of the orthogonal projector on a deformation of an inner product will be used in the proof of the main theorem.


Introduction
This paper is a continuation of [10].For the reader's convenience we will recall preliminaries from that article.The reader is also referred to publications devoted to various applications of reproducing kernel Hilbert spaces ( [9]).The main result of this paper is Theorem 2.2 which shows that Szegő kernel depends continuously on a weight of integration.
Let be a bounded domain with the boundary of class C 2 .Suppose that the function μ : ∂ → R is measurable and almost everywhere greater than 0 with respect to the surface measure dS on ∂ .Such a function will be called a weight.By L 2 (∂ , μ) we will denote a set of classes of functions f : ∂ → C, square-integrable in the sense that where the integral is understood as an integral of a scalar field.The set L 2 (∂ , μ) with an inner product given by f |g μ := ∂ f (w)g(w)μ(w)dS is a Hilbert space.Now let us consider the space A( ) of continuous functions f : Of course L 2 H (∂ , μ) can change as a set with a change of μ.However Proposition 1.1 If μ 1 , μ 2 are weights and there exist m, M > 0, such that a.e. on ∂ , then for any f as vector spaces, we will write μ 1 ≈ μ 2 .It is easy to show that this is an equivalence relation.
Each element of B( , 1) has a unique holomorphic prolongation to by Poisson integration (see [3] for more details), so it is also true for any element from B( , μ), because B( , μ) ⊂ B( , 1) for any μ.
is a bounded domain of class C 2 , therefore ∂ has finite Lebesgue measure).We will denote the set of all such prolongations by B( , μ) (where B( , μ) ⊂ A( )).
A good question to ask is how to find a holomorphic prolongation of functions from L 2 H (∂ , μ)\B( , μ) for an arbitrary μ?We will answer this question in a moment.
We will use the same symbol for a function and its prolongation, which should not be misleading.
Let μ be a weight with the following property: (CB) for any compact set X ⊂ there exists C X > 0, such that for any f ∈ B( , μ) and z ∈ X Then for functions from L 2 H (∂ , μ)\B( , μ) we can define their prolongation to in the following way: Let ( f n ) be a sequence of functions from B( , μ).Let f be the limit of this sequence in L 2 H (∂ , μ).Since by (CB) the sequence of functions { f n | } is a locally uniformly Cauchy sequence on , the function is well defined and holomorphic on .
From now on, if μ fullfills (CB), we will interpret L 2 H (∂ , μ) as a set of functions on .

Definition 1.1 Let μ be a weight satisfying (CB). A function (if it exists) S
It is true (as for any reproducing kernel Hilbert space) that if S μ and S μ are Szegő kernels of the same space, then S μ = S μ and if the Szegő kernel exists, then it is given uniquelly by the formula where {ϕ i } i∈I is an arbitrary complete orthonormal system of L 2 H (∂ , μ).Hence for any z, w ∈ we have S μ (w, z) = S μ (z, w) and by Hartogs's theorem on separate analyticity the function × (z, w) → S 0 (z, w) := S μ (z, w) is holomorphic, where = {w ∈ C N : w ∈ }.So S μ is real analytic on × , holomorphic with respect to the first N variables and antiholomorphic with respect to the last N variables.Moreover for any z ∈ we have It is a natural question to ask which conditions μ must satisfy in order for L 2 H (∂ , μ) to be a reproducing kernel Hilbert space.Definition 1. 2 We will say that a weight μ is Szegő admissible (S-admissible for short) if there exists Szegő kernel of L 2 H (∂ , μ) space.Indeed, by the reproducing property, the Cauchy-Schwarz inequality, and again the reproducing property: On the other hand, for f = S μ (z, •), the above inequality becomes equality.
⇐ (CB) means that the point evaluation functionals and the function Then μ is S-admissible.
For more details and the proof see [10].

Main Results
At the beginning, we will introduce the appropriate topology in the set SAW(∂ ) of S-admissible weights on ∂ .Let us recall that we denote by ≈ the equivalence relation on SAW(∂ ) defined as follows: for any ) is equal as a vector space to L 2 H (∂ , μ 2 ).Just as in the work [4] it can be proved that in this case the norms in L 2 H (∂ , μ 1 ) and L 2 H (∂ , μ 2 ) are equivalent, i.e. there are positive constants c and C, such that for any f ∈ L 2 H (∂ , μ 1 ) we have See [11] for details about weighted Szegő space.
For any μ ∈ SAW(∂ ), denote by SAW(∂ , μ) the equivalence class of μ with respect to the relation ≈.Note that SAW(∂ , μ) contains infinitely many elements, because for any function g ∈ L ∞ (∂ ) such that essinf z∈∂ g(z) > 0 the ordinary product gμ is an element of SAW(∂ , μ) (see Proposition (1.1)) and if On SAW(∂ , μ) we consider the map: where Her(H) denotes the real Banach space of all continuous hermitian forms on a Hilbert space H with the standard Banach space norm: (It is possible that for some weights zero is the only element of our space.In that case the norm of the operator is equal to zero).We denote by τ μ the weakest topology on SAW(∂ , μ) with respect to which the map B μ is continuous.By the Lax-Milgram Theorem each inner (hermitian) product −, • ν equivalent to −, • μ on L 2 H (∂ , μ) uniquely determines an invertible positive definite continuous operator Moreover, if Her + (L 2 H (∂ , μ)) denotes the cone in Her(L 2 H (∂ , μ)) of all positive definite hermitian forms (the set of hermitian products) on L 2 H (∂ , μ) equivalent to −, • μ , then the map is an isometry (onto its image) with respect to the standard norms in Her(L 2 H (∂ , μ)) and in the space L(L 2 H (∂ , μ)) of all bounded endomorphisms of L 2 H (∂ , μ).Therefore the map μ is a homeomorphism.Hence τ μ is the weakest topology with respect to which the map μ • B μ is continuous.
On the other hand, if μ 1 ≈ μ and ν ∈ SAW(∂ , μ) = SAW(∂ , μ 1 ), then We can write is the map of composition with constant invertible operator and therefore τ μ = τ μ 1 .This means that the topology τ μ does not depend on the choice of an equivalence class representative from SAW(∂ , μ).
Let us consider the family μ∈SAW(∂ ) τ μ of subsets of SAW(∂ ).It is, of course, the base of the same topology τ on SAW(∂ ).From now on we will consider SAW(∂ ) as a topological space endowed with this topology.
Note that for any μ ∈ SAW(∂

SAW(∂ , ν).
Moreover, from the definition of SAW(∂ ) it follows almost immediately that for any ν 1 , ν 2 ∈ SAW(∂ , μ) and any t ∈ [0, 1] we have In addition, the map is evidently continuous and therefore the map is also continuous.Hence SAW(∂ , μ) is connected and consequently it is a connected component of SAW(∂ ) with respect to τ .
It may happen for some μ ∈ SAW(∂ ) that B μ is not a 1-1 map.In this case τ μ is not a Hausdorff topology.In extreme cases it may happen that L 2 H (∂ , μ) = {0} and B μ ≡ 0. For example let f : [0, 1] → R be a measurable and almost everywhere positive function which is not integrable on any interval.(It is well-known that such functions exist; see e.g.[2], Example 26(c), page 327).Let γ : [0, 1] → ∂ be a parametrization of ∂ .Then, for the weight μ := f • γ −1 , the space L 2 H (∂ , μ) is equal to {0} as a vector space and its reproducing kernel is a zero function of two variables.Indeed, it is clear that any non-zero function which is continuous on and holomorphic on is not an element of B( , μ).
On the other hand, in cases that are important for applications (for example, when μ is bounded from above and below by non-zero constants), B μ is a 1-1 mapping and τ μ is Hausdorff.Indeed, any weight bounded from above and below by non-zero constants is an element of SAW(∂ , 1), as a consequence of Proposition 1.1.For such μ all polynomials are elements of L 2 H (∂ , μ) and it is easy to see that B 1 is an injection.
Our results are true in all cases.Now let us recall this theorem (see Theorem 5.1.in [6] for more details): Theorem 2.1 Let H be a Hilbert space and V be a closed vector subspace of H. Let P(•) denote the mapping that assigns to each positive defined and invertible operator A ∈ L(H) the projection of H onto V orthogonal with respect to the hermitian product f , g A := f , Ag , f , g ∈ H.

Then P(•) is analytic with respect to the natural analytic structure on an open set of all positively defined and invertible operators in H.
We are ready to prove main result of this section.Let g ν := (I − P ν ) f , where I denotes the identity operator of L 2 H (∂ , μ).Then g ν = 0 for any ν ∈ SAW(∂ , μ).Using the fact that the subspace (ker E z ) ⊥ν

Theorem 1 . 1
The weight μ is S-admissible if and only if Condition (CB) is satisfied.Proof ⇒ comes directly from Definition 1.1.In fact we can show that the smallest possible constant C X in Condition (CB) is max z∈X S μ (z, z) .