678 reputation
314
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location Denmark
age 25
visits member for 2 years, 5 months
seen 2 days ago

student


Sep
7
accepted The supremum norm is submultiplicative
Sep
7
comment The supremum norm is submultiplicative
That shortens it :) thanks.
Sep
7
awarded  Custodian
Sep
7
revised The supremum norm is submultiplicative
added 20 characters in body
Sep
7
reviewed Approve suggested edit on The supremum norm is submultiplicative
Sep
7
asked The supremum norm is submultiplicative
Sep
5
comment Is a bounded Borel function of a normal operator normal
I still cannot seem to grasp this. Why are points E-null? Do you maybe have a proof that the transformed spectrum is the essential range?
Sep
4
accepted Is a bounded Borel function of a normal operator normal
Sep
2
comment Is a bounded Borel function of a normal operator normal
Just to be sure, I expect you to be right on this, but let me present my argument for why you are not so you can tell me what I get wrong. $f(M)=\int_{\sigma(M)} f(\lambda) dE(\lambda)$ so if I change $f$ on a $E$-nullset $f(M)$ would not change, but $f(\sigma(T))$ would. If there are no $E$-null sets this though can not happen. I.e. the essential range and ordinary range coincide.
Sep
2
comment Is a bounded Borel function of a normal operator normal
I just want to make sure that I appreciate your comment no matter - I would not have thought about the issue otherwise.
Sep
2
comment Is a bounded Borel function of a normal operator normal
First of all thanks for the answer, interesting that the result is indeed true. Regarding the second part actually, we have proven that the relevant spectral measure only has the empty set as null set, would that not remove the problem? Else I guess the result should be that $\sigma(S)=\text{EssRan}_g(\sigma(T))$.
Sep
2
accepted Intuition $C^*$-identity
Sep
2
asked Is a bounded Borel function of a normal operator normal
Aug
31
comment $E[(ae^x-b)^+]$, $X$ is $\mathcal{N}(0,1)$ distributed
I agree. Actually I asked out of own interest - though I misunderstood the question slightly so it is not so relevant anyways.
Aug
30
comment $E[(ae^x-b)^+]$, $X$ is $\mathcal{N}(0,1)$ distributed
What is the simplest way to realize your formula? :) If this is indeed a one-shot it would be quite neat.
Aug
30
asked Intuition $C^*$-identity
Aug
19
accepted Characterizing C* algebra generated by elements.
Aug
19
comment Characterizing C* algebra generated by elements.
Its an $C^*$-algebra and contained in all $C^*$ algebras $A'$ such that $A_0 \subset A'$? I was stumbled about the requirements on the norm, but they will just be inherited from $A$ and maintained in the limit. Right? If so thanks for making me realize it instead of just telling me :)
Aug
19
comment Characterizing C* algebra generated by elements.
Yea, that was what i meant. The result is as good as one could hope. Do you by any chance have a proof or reference? :)
Aug
19
asked Characterizing C* algebra generated by elements.