By Kondratiev A. S., Mazurov V. D.

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2) it follows that Im a(ν)b(ν) = −ν, (ν ∈ R \ {0}). 3) In particular, a and b do not vanish anywhere on R \ {0}. We can now determine the spectral matrix for this problem. 4). From this we conclude that the spectral matrix P (λ) has zero entries except for the one in the upper left corner, which we denote by ρ(λ). The second component of Ff now plays no role in the Plancherel formula. Indeed, deﬁne ∞ F1 f (λ) = f (t) s1 (λ, t) dt, 0 then we have the following. 1 F1 extends to an isometry from the space L2 ( ] 0, ∞ [ ) onto L2 (R, dρ).

His formulation follows from the one above by the observation that ρ admits a unique decomposition ρ = ρd + ρc with ρc a continuous monotonically increasing function with ρc (0) = 0, and with ρd a right-continuous monotonically increasing function which is constant on each interval where it is continuous. 4, so that ρc = 0, so that the above gives rise to a discrete decomposition. In case L is of the limit point type at ∞, the decomposition is of mixed discrete and continuous type. 3) 0 with the interpretation that the integral converges as an integral with values in L2 (R, dρ).

Weyl, eigenfunction expansions, symmetric spaces 31 −1 (d) In the space Hom(E¯λ∗ , Eλ ) the inverse h−1 λ,x converges to hλ,b as x → b. 2) it follows by taking the limit for x → b that ¯ f,g [f, g]λ,b = lim [f, g]x = [f, g]c + (λ − λ) x→b [c,b] for all f, g ∈ Eλ . This establishes the existence of the limit hλ,b . As hλ,x is a Hermitian form for every x ∈ ] a, b [ , the limit is Hermitian as well. (b, d) Fix a basis f1 , f2 of Eλ and write f (z) := z1 f1 +z2 f2 , for z ∈ C2 . For x ∈ [c, b] we deﬁne the Hermitian matrix Hx by i[f (z), f (w)]x = Hx z , w .

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