5h
revised Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
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14h
comment $X$ complete normed space $\implies\mathrm B(X,Y)$ complete normed space?
I don't know. But the implication you are talking about is valid for all nonzer Banach space X. I've seen this proof in several functional analysis books, but I don't remember the unit xac reference.
16h
comment $X$ complete normed space $\implies\mathrm B(X,Y)$ complete normed space?
Any opertator in $\mathcal{B}(X,Y)$ is of the form $T(z)=z y$ for some $y\in Y$. Consider Cauchy sequence in $\mathcal{B}(X,Y)$ with no limit. It induces a Cauchy sequence in $Y$ with no limit.
16h
comment $X$ complete normed space $\implies\mathrm B(X,Y)$ complete normed space?
Well, that was erroneous argument. Now I suggest even more simpler example.
16h
revised $X$ complete normed space $\implies\mathrm B(X,Y)$ complete normed space?
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20h
comment Is $\mathcal{B}(H)$ complemented in $\ell_\infty(I, H)$
If you're interested there is a simpler proof for absence of lust for non-type I C^*-algebras: Any *-subalgebra $M_n$ in a $C^*$-algebra $A$ in contractively complemented. So lust constant of A is bounded below by lust constant of $M_n$. By results of Gordon and Lewis the latter tends to infinity for large $n$. So $A$ have lust iff it contains no big copies of $M_n$. Such $C^*$-algebras are called subhomogeneous.
21h
comment Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
@TomekKania, I've added the definition.
21h
revised Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
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1d
revised Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
edited tags
2d
comment When $C(X)$ is an injective $C(X)$-module?
Thank you prof. Ozawa and excuse me for such late reply.
2d
comment Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
@DavidHandelman, I'm interested in functional analytic version of injectivity. Anyway thank you for mentioning purely algebraic case also.
2d
revised Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
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2d
comment Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
Imean classical relative injectivity. The answer is, clearly, no for strict injectivity.
2d
asked Is $\mathcal{K}(H)$ injective $\mathcal{B}(H)$-module?
Jan
23
accepted When $C(X)$ is an injective $C(X)$-module?
Jan
22
comment n- dimensional normed linear space isomorphic to n-dimensional Euclidean space
this paper could be interesting to you
Jan
20
revised Checking that $C_{0}(X)$ is a vector space
edited title
Jan
20
comment Boundedness of linear operators in $L^p$
Yes$\phantom{}\phantom{}$
Jan
20
comment Boundedness of linear operators in $L^p$
$T$ is not necessarily bounded because any infinite dimensional Banach space admits a discontinuous linear operator. Even bijectivity doesn't help: choose Hamel basis in $L_p$ of vectors with norm 1, take countable subfamily. Multiply $n$-th vector of subfamily by $n$. The rest left unchanged. Extend this mapping by linearity. It is bijective but not bounded.
Jan
20
comment Trace class operator
you should ask Martin Argerami. His a main specialist on operator algebras here.
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