Reputation
1,106
Next privilege 2,000 Rep.
Edit questions and answers
Badges
3 15
Newest
 Yearling
Impact
~15k people reached

13h
awarded  Yearling
Mar
16
comment Proof that $S$ is a partial order when it is the reflexive closure of a strict partial order
You are welcome. Indeed, it is not mentioned in the statement, but in the first line of the proof :).
Mar
16
comment Proof that $S$ is a partial order when it is the reflexive closure of a strict partial order
Yes, it is, and that's exactly the content of Theorem 4.5.2.
Mar
16
answered Proof that $S$ is a partial order when it is the reflexive closure of a strict partial order
Mar
16
comment Branch cut choice for $(z+\frac{1}{z})^{\frac{1}{3}}$
Hm.... Finding branch like that means finding a branch $\log_\tau$ of $\log$ such that for any $|z|>2$ one has arg$(z+1/z)\neq\tau$. I don't think that's possible.
Mar
16
comment The Banach–Mazur distance for finite-dimensional $\ell_p$
@BenWallis If I am not mistaken, the reason is that, regardless how large $k_n$ is, an $n$-dimensional subspace is going to be isometric with $l_2^n$, and the Banach-Mazur distance between $l_2^n$ and $l_p^n$ goes to $\infty$ when $p\neq 2$. Thus, you cannot find a bound that depends on $p$ alone.
Mar
15
answered How do I change the order of integration of this integral?
Mar
15
comment The Banach–Mazur distance for finite-dimensional $\ell_p$
Yes, the conjecture is false for $p\neq 2$. By the way, by Dvoretzky, any infinite dimensional $X$ contains, for every $n$, almost isometric copies of $l_2^n$.
Feb
7
answered A Property about orthogonal projections
Feb
7
comment Algebraic and orthogonal complements
$Z$ is an algebraic complement of $Y\subset X$ if any $x\in X$ can be written uniquely as $x=y+z$, with $y\in Y$ and $z\in Z$. A subspace $Y$ has many algebraic complements, but they are all isomorphic. Being isomorphic means that they have the "same dimension", which is to say bases in any algebraic complement having the same cardinality. When we speak about orthogonal complement, it is assumed additional structure exists (an inner product space), and the orthogonal complement is defined in terms of the inner product.
Feb
6
answered Algebraic and orthogonal complements
Feb
1
asked Isomorphisms with invariant linearly independent dense subset.
Feb
1
comment Every isomorphism on a separable Banach space has a completely invariant dense subset
Thank you. Can you pick $D$ such that is linearly independent? Don't tell me is still trivial :)
Feb
1
accepted Every isomorphism on a separable Banach space has a completely invariant dense subset
Feb
1
comment Every isomorphism on a separable Banach space has a completely invariant dense subset
Of course, I see now. Thanks!
Feb
1
comment Every isomorphism on a separable Banach space has a completely invariant dense subset
I can see that $T(D)\subset D$. But why equality? If you exclude $n=0$ (so no $E$ in $D$), why is still dense? Probably I am missing something easy :)
Feb
1
comment Every isomorphism on a separable Banach space has a completely invariant dense subset
@Fundamental You are right, it was trivial. Modified it to what actually I am trying to prove. Still trivial?
Feb
1
revised Every isomorphism on a separable Banach space has a completely invariant dense subset
deleted 2 characters in body
Feb
1
asked Every isomorphism on a separable Banach space has a completely invariant dense subset
Jan
28
comment Theorem about equivalent norms.
The only way this is true is if the norms are identical. Either one of the two conditions implies that.