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functional analysis - Spectrum of a positive operator in $B (H . . . @TryingToLearn : If you are interested in an excellent introduction to Spectral Theory that is aimed toward Operator Algebras, then you can't go wrong with William B Arveson's "A Short Course in Spectral Theory" Arveson was an amazing talent in Operator Algebras, and he had a way of making things look simple in a field that is not simple
operator theory - Why is the ultraweak topology on $B (H)$ called . . . To motivate my thoughts, there is also the strong operator topology on B(H) B (H) which is induced by the seminorms u ↦ ∥u(x)∥ u ↦ ‖ u (x) ‖ for x ∈ H x ∈ H And indeed, the strong topology is stronger than the weak topology So this terminology is actually consistent with the behaviour of the topologies
functional analysis - Closed unit ball of $B (H)$ is not compact in . . . In operator theory we prove that closed unit ball of B(H) B (H) is compact in weak operator topology and is closed in strong operator topology But a book of operator theory states that closed unit ball of B(H) B (H) is not compact in strong operator topology Is there any straightforward proof for this statement? thanks for your guidance
Is the strong topology on $B(H)$ first countable? In fact the latter deals with U(H) U (H) while the OP here deals with B(H) B (H) Their difference is so relevant that one question has an affirmative answer and the other has a negative one I was about to answer by proving that B(H) B (H) is not first countable but was prevented from doing so by the closure
functional analysis - Wot topology on $B (H)$ is not metrizable . . . Let H H be a infinite dimensional Hilbert space and B(H) B (H) be the space of bounded and linear operators on H H I know that weak operator topology (wot) and strong operator topology (sot) are metrisable on bounded subsets of B(H) B (H)
Dual and bidual spaces of $B(H)$ in norm topology Since B(H) B (H) is not separable in the norm-topology if H H is infinite-dimensional these spaces will be very large Specifically B(H) B (H) contains a subspace isomorphic to ℓ∞(N) ℓ ∞ (N) (The subspace is the diagonal operators wrt some Hilbert-basis ) The dual space will thus contain a very large space