By Yuri Arlinskii, Sergey Belyi, Eduard Tsekanovskii

This e-book is dedicated to conservative realizations of varied sessions of Stieltjes, inverse Stieltjes, and common Herglotz-Nevanlinna capabilities as impedance features of linear structures. the most function of the monograph is a brand new method of the conclusion concept profoundly regarding built extension concept in triplets of rigged Hilbert areas and unbounded operators as state-space operators of linear structures. The connections of the conclusion idea to structures with accretive, sectorial, and contractive state-space operators in addition to to the Phillips-Kato sectorial extension challenge, the Krein-von Neumann and Friedrichs extremal extensions are supplied. between different effects the ebook includes functions to the inverse difficulties for linear platforms with non-self-adjoint Schrödinger operators, Jacobi matrices, and to the Nevanlinna-Pick process interpolation.

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**Example text**

Therefore, φ = 0 and hence Dom(A) ∩ (U + I)N 0 It is easy to see that (U + I)Ni ⊂ Dom(A). 10) Dom(A˙ ∗ ) = Dom(A) + Ni + N−i . Let f = g + ϕ + ψ, ˙ ϕ ∈ Ni , and ψ ∈ N−i . Then where g ∈ Dom(A), 1 1 f = g + (U − I)(U −1 ψ − ϕ) + (U + I)(U −1 ϕ + ϕ) = x + (U + I)y, 2 2 where x = g + 12 (U − I)(U −1 ψ − ϕ) and y = 12 (U + I)(U −1 ϕ + ϕ). This implies that Dom(A˙ ∗ ) ⊆ Dom(A) + (U + I)Ni . Since the inverse inclusion is obvious, we conclude that Dom(A˙ ∗ ) = Dom(A) + (U + I)Ni . 42), Dom(A) ∩ (U + I)N i Dom(A˙ ∗ ) = Dom(A) This completes the proof of statement 2.

But A˙ ∗ N ⊂ Dom(A) PL N = L. Taking into account that N ∩ H0 = {0} and H = H0 ⊕ L, we get the equality ˙ H = H0 +N. A closed symmetric operator A˙ is said to be an O-operator if both its semideﬁciency indices equal zero. For such an operator M = N. 3. 17). The proof easily follows from the deﬁnition of a regular O-operator. 5 Closed symmetric extensions ˙ Then Let A be a closed symmetric extension of a symmetric operator A. (Af, g) = (f, Ag), (∀f, g ∈ Dom(A)), and, in particular ˙ g) = (f, Ag) = (f, P Ag), (Af, ˙ g ∈ Dom(A)).

Therefore (h, fn ) → (h, f ) for any h ∈ H+ . In particular, if h ∈ L⊥ ∩H+ , then (h, fn ) = 0, and consequently, (h, f ) = 0. Thus, L⊥ ∩ H+ ⊂ (L (−) ⊥ (−) ⊥ ) . (−) Conversely, let h ∈ (L ) . Then h ∈ H+ and (h, f ) = 0 for any f ∈ L , and, in particular, for any f ∈ L. Hence h ∈ L⊥ and h ∈ L⊥ ∩ H+ . 2). The second part is being proved similarly by substituting L⊥ for L. 4) can be proved in a similar way. 3) by the respective substitution of L(⊂ H+ ) for L⊥ (⊂ H− ). 1 can be interpreted as asserting the commutativity of the following diagram: ←− R −−−−→ R ⏐ ⏐ ⏐↑ ↓⏐ −→ R+ ←−−−− R− Here R+ , R, and R− are the classes of all (+)-closed, (·)-closed, and (−)-closed linear manifolds in H+ , H, and H− , respectively.