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Distinguished Professor of Mathematics at Texas A&M University


17h
revised A lower bound on $\|\cdot\|_{p_{\ast}}$ image of $\ell^{q_{\ast}}$ vectors
added 5 characters in body
1d
answered A lower bound on $\|\cdot\|_{p_{\ast}}$ image of $\ell^{q_{\ast}}$ vectors
1d
comment A lower bound on $\|\cdot\|_{p_{\ast}}$ image of $\ell^{q_{\ast}}$ vectors
Funny, from the obvious argument of using $\ell_{q_*}$ normalized characteristic functions of disjoint sets of size of order $n/T$ I get a lower bound of $C n^{1/{p_*}-1/{q_*}}/T^{1/q}$, which is better than the bound I mentioned that uses something non trivial (the Kashin decomposition). I guess one of us (probably I) made a computational mistake.
2d
awarded  Good Answer
2d
comment A lower bound on $\|\cdot\|_{p_{\ast}}$ image of $\ell^{q_{\ast}}$ vectors
I have been traveling internationally to a conference the last couple of days and haven't had time to think about this or write anything down. Manana (which I think translates into English as "some indeterminate time in the future"). :)
Oct
18
comment A lower bound on $\|\cdot\|_{p_{\ast}}$ image of $\ell^{q_{\ast}}$ vectors
OK. The problem looks interesting. What relationship between T and n is interesting for you? For fixed T, when $p\not= q$, it is clear that your parameter tends to infinity as $n\to \infty$ (a lower bound is $C n^{1/p_* - 1/q_*}/T^{1/2}$). Here I assume that $n \ge 2T$ when $q\not =2$.
Oct
18
answered A lower bound on $\|\cdot\|_{p_{\ast}}$ image of $\ell^{q_{\ast}}$ vectors
Oct
17
awarded  Nice Answer
Oct
17
revised Extreme unit linear functional not norming a vector
Corrected grammar and spelling.
Oct
16
comment Extreme unit linear functional not norming a vector
Yes, if the dual of the space is separable or, more generally, if every separable subspace has separable dual ("Asplund space"). Then the dual ball is the norm closed convex hull of its extreme points. No, in general (consider $C[0,1]).
Oct
16
answered Extreme unit linear functional not norming a vector
Oct
15
revised Approximating operators on Banach spaces by bounded operators on a proper dense subspace
Made correct the statement about what is "widely believed to exist".
Oct
15
answered Extending compact operators
Oct
15
comment A Banach space with all Hilbertian subspaces complemeneted
Yes, or any random proportional dimensional subspace.
Oct
14
comment A Banach space with all Hilbertian subspaces complemeneted
No; if $k(n)^{|1/2 - 1/p(n)|}$ stays bounded the space is isomorphic to $\ell_2$. But I was wrong about this being an example. The space does contain an uncomplemented Hilbertian subspace. Sorry about that.
Oct
13
comment A Banach space with all Hilbertian subspaces complemeneted
Such a space $X$ must be of type $2-\epsilon$ for all $\epsilon >0$ (since by Krivine's theorem, otherwise $\ell_p$ is finitely representable in $X$ and $L_p$ contains an uncomplemented Hilbert space). It need not be of type 2 (consider $(\sum_n \ell_{p(n)}^{k(n)})_2$ with $p(n) \uparrow 2$ and $k(n) \to \infty $ quickly).
Oct
10
comment Infinite dimensional subspaces of $L^1$
Are you trying to understand which subspaces of $L_p$ embed into $\ell_p$? They are characterized for $1<p<\infty$ and much is known for $p=1$.
Oct
7
answered Reverse Hausdorff Young for nonnegative functions
Oct
3
comment What is between super-reflexivity and reflexivity?
When you do not state a specific question, how do you expect to get an answer? MO is not a good site for a fishing expedition.
Oct
2
revised What is between super-reflexivity and reflexivity?
edited tags