By Mats Gyllenberg, Lars-Erik Persson

Providing the lawsuits of the twenty-first Nordic Congress of Mathematicians at Luleå collage of know-how, Sweden, this remarkable reference discusses fresh advances in research, algebra, stochastic methods, and using pcs in mathematical study.

**Read Online or Download Analysis, algebra, and computers in mathematical research: proceedings of the Twenty-first Nordic Congress of Mathematicians PDF**

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**Additional info for Analysis, algebra, and computers in mathematical research: proceedings of the Twenty-first Nordic Congress of Mathematicians**

**Example text**

This property is equivalent to S ⊂ S ⊂ T ∗ for densely deﬁned closed operators S and T . An example of a dual pair is provided by the minimal operators associated with a diﬀerential expression and its formal adjoint. The existence of m-accretive extensions of a dual pair of densely deﬁned accretive operators was proved in [Phillips, 1969]. Fractional-linear transformations of a dual pair of accretive operators lead to aparametrization of all contractive extensions of a dual pair of non-densely deﬁned contractions.

30) ∗ ∗ + D+ = Sm + K(D+ ) ⊆ S ∗ ∩ dom U . Moreover, satisﬁes T ∗ = Sm n+ (T ) ≥ dim K(D+ ) = ∞ and n− (T ) ≥ dim D+ = ∞. Next the type of a reduction operator is characterized by means of maximal symmetric relations which are contained in its domain. 4]. 16 Let U be a reduction operator for S ∗ and assume that n+ (S), n− (S) ≤ ℵ0 . Then U is of type II if and only if there exist maximal symmetric extensions S1 ⊆ dom U and S2 ⊆ dom U of S such that n+ (S1 ) = n+ (S2 ) or n− (S1 ) = n− (S2 ). 15, since T ⊆ T ∗ ⊆ dom U has defect numbers n+ (T ) = n− (T ) = ∞ and all maximal symmetric extensions of T are automatically contained in dom U .

12]. 26 Let U be a reduction operator for S ∗ , let N be a neutral subspace of the Kre˘ın space (K, (jW ·, −)) and let K be the angular operator of N with respect to K+ W: + N = {f + + Kf + : f + ∈ PW N}. Then the symmetric extension Sm := U −1 (N ∩ ran U ) of S satisﬁes n+ (Sm ) = 0 ⇔ n− (Sm ) = 0 ⇔ ran U ∩ K− W ⊆ ran (K+ − K); −1 ran U ∩ K+ ). W ⊆ ran (K− − K Proof It is shown that n+ (Sm ) = 0 if and only if ran U ∩ K− W ⊆ ran (K+ − K); the other case is analogous. Hence, assume that ran U ∩ − K− ∈ ran U ∩ K− W ⊆ ran (K+ − K).