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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
Proving that $\lim_ {h\to 0 } \frac {b^ {h}-1} {h} = \ln {b}$ I think you need only the derivative of $\exp$ (but, even if we need derivative of $\ln$, is it forbidden ?) Anyway, I removed my answer since it involves L'Hopital's rule and Rustyn explicitly wrote without L'Hopital's rule
Matrix operator rewriting - Mathematics Stack Exchange If I understand correctly, each operator acts in its own Hilbert space, that is S = (S, 0) S = (S, 0) and T = (0, T) T = (0, T) therefore + + in the defintion of A A is a direct sum