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3h
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.
4h
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.
5h
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.
5h
revised $X$ complete normed space $\implies\mathrm B(X,Y)$ complete normed space?
deleted 346 characters in body
8h
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.
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.
Jan
19
comment Proving that the triangle inequality holds for a metric on $\mathbb{C}$
@IvoTerek then why don't you post an answer here in English?
Jan
19
comment When does a homeomorphism preserve Cauchy Sequence?
If you want to speak of Cauchy-stuff you need to fix a uniformity, not the other way round (choose uniformity to fit into the topology).
Jan
19
revised A fan, a horn, and a snowflake - unusual math terms
added 3 characters in body
Jan
19
comment When does a homeomorphism preserve Cauchy Sequence?
Therefore this question should be understood in terms of uniform spaces
Jan
19
comment Trace class operator
It was answered here a long time ago. You search along the site.
Jan
18
revised A fan, a horn, and a snowflake - unusual math terms
added 68 characters in body
Jan
18
comment Compact operator and a sot convergent sequence of operators
May be this can help
Jan
18
comment Takhtajan's “Quantum Mechanics for Mathematicians”
Don't read it, my advisor said that it sometimes contain errors that could be catched only by experts. I vaguely remember an example where author implicitly supposed that all characters of $\ell_\infty$ are point evaluations.
Jan
17
answered A fan, a horn, and a snowflake - unusual math terms
Jan
16
comment isometrically isomorphism
@Fundamental, for example? you can't destroy the property of vector space being zero-dimensional (I think it is a nice property) without destroying that space :)