bio  website  math.tamu.edu/~johnson 

location  Texas A&M  
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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
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answered  A lower bound on $\\cdot\_{p_{\ast}}$ image of $\ell^{q_{\ast}}$ vectors 
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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. 
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awarded  Good Answer 
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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 
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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 
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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 
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A Banach space with all Hilbertian subspaces complemeneted
Yes, or any random proportional dimensional subspace. 
Oct 14 
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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 
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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 
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What is between superreflexivity 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 superreflexivity and reflexivity?
edited tags 