By Flanigan F.
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N=0 Then φ is clearly holomorphic for λ > x , but we know it extends holomorphically to λ > r(x) and tends to 0 as λ tends to infinity. Let ψ(λ) = φ(1/λ). Then ψ extends 34 analytically to zero with value zero and defines an analytic function on the open ball of radius 1/r(x) about zero, as does, therefore, ∞ f (λn xn ). ψ(λ)/λ = n=0 This shows that for each |λ| < 1/r(x) and each f ∈ X ∗ , f (λn xn ) is bounded. By the uniform boundedness principle, the set of elements λn xn are bounded in X, say by K.
Our first results don’t require the full structure of in the algebra of operators on a Banach space, but just an arbitrary Banach algebra structure, and so we start there. The spectrum and resolvent in a Banach algebra. Let X be a Banach algebra with an identity element denoted 1. We assume that the norm in X has been normalized so that 1 = 1. The two main examples to bear in mind are (1) B(X), where X is some Banach space; and (2) C(G) endowed with the sup norm, where G is some compact topological space, the multiplication is just pointwise multiplication of functions, and 1 is the constant function 1.
35 Spectral Theorem for bounded self-adjoint operators in Hilbert space. We now restrict to self-adjoint operators on Hilbert space and close with a version of the Spectral Theorem for this class of operator. ” (American Mathematical Monthly 70, 1963) both in the relatively elementary statement of the theorem and the outline of the proof. First we note that self-adjoint operators have real spectra (not just real eigenvalues). Proposition. If H is a Hilbert space and T ∈ B(H) is self-adjoint, then σ(T ) ⊂ R.