Feb
1
revised Can this equation be solved without numeric calculation?
edited tags
Feb
1
comment Double dual of the space of bounded operators on Hilbert space
Possible duplicate of Complementability of von Neumann algebras
Jan
28
awarded Revival
Jan
28
comment There is no bounded linear surjection between $\ell_p$ spaces
It IS closed. There is a theorem that a adjoint of surjective operator between Banach spaces is closed.
Jan
28
comment There is no bounded linear surjection between $\ell_p$ spaces
@Svetoslav , since $T^*$ is an embedding, then $\ell_\infty\cong \operatorname{Im}(T^*)\subset\ell_{p'}$. Thus we have a non-separable subspace (that is $\operatorname{Im}(T^*)$) inside a separable space $\ell_{p'}$. Contradiction.
Jan
26
answered There is no bounded linear surjection between $\ell_p$ spaces
Jan
26
comment Is weakly continuous implies strong continuous?
@anonymous, as far as I know all weak-to-norm continuous operators are finite rank
Jan
13
accepted Antiproximanal subspace of $L_1[0,1]$
Jan
13
comment Antiproximanal subspace of $L_1[0,1]$
@BillJohnson, why don't you post this as answer?
Jan
13
revised An interesting integral to determine the sign
This question is not about functional analysis, so I deleted the fa tag.
Jan
13
comment Antiproximanal subspace of $L_1[0,1]$
I was always sure that this trick works only for some $x\notin Y$, not for all of them. Thank you.
Jan
12
asked Antiproximanal subspace of $L_1[0,1]$
Jan
10
revised an isomorphism from $L^\infty(\mathbb{T})$ to $L^\infty([-1,1],\displaystyle\frac{2}{\pi } \sqrt{1-t^2}\mathrm{d}t)$
edited body
Jan
10
comment the isomorphism of $L^\infty$ spaces
This is another question. Please ask it via standard form. Not in comments.
Jan
10
revised Prove a functional series is pointwise and/or uniformly convergent
edited tags
Jan
9
comment the isomorphism of $L^\infty$ spaces
No it would take too long
Jan
9
answered the isomorphism of $L^\infty$ spaces
Jan
3
revised If X* is separable X is also separable.X is normed vector space
added 13 characters in body; edited tags
Dec
31
comment If ${\left\| A \right\|_1} \le 1$ Can we prove that all ${\sigma _i}(A)\le 1$?
it is not, that's why i deleted it
Dec
17
comment Isometry of the adjoint operator
no take T to be a non zero functional on X (as you can see Y must be $\mathbb{C}$)
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