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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
$H$ is normal whenever $Ha\\not = Hb \\implies aH\\not =bH$ If H H be a subgroup of a group G G such that Ha ≠ Hb H a ≠ H b implies that aH ≠ bH a H ≠ b H Then how can I show that gHg−1 ⊂ H g H g 1 ⊂ H ∀ ∀ g ∈ G g ∈ G? This is what I have done: For any h ∈ H h ∈ H, (ghg−1)(gh−1g−1) = e (g h g 1) (g h 1 g 1) = e where e e is the identity element of the group G G
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properly infinite $C^*$-algebra $B(H)$ - Mathematics Stack Exchange There is a conclusion:If H H is an infinite dimensional Hibert space,then B(H) B (H) is a properly infinite C∗ C ∗ -algebra According to the definition,we need to find mutually orthogonal projections e, f ∈ B(H) e, f ∈ B (H) such that e ≤ idH, f ≤ idH e ≤ i d H, f ≤ i d H and e, f e, f are equivalent to idH i d H