bio | website | math.tamu.edu/~johnson |
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location | Texas A&M | |
age | ||
visits | member for | 5 years, 5 months |
seen | 19 hours ago | |
stats | profile views | 10,001 |
Distinguished Professor of Mathematics at Texas A&M University
May 15 |
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Sard's Theorem For Banach Spaces
Look at MathSciNet reviews of papers by Sean Bates to see which ones are relevant for you. |
May 12 |
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Rearrangments of Fourier series
Maybe, but it is trivial that if $X$ has a basis, then for any $f$ in $X$ there is another basis s.t. $f$ is the first vector of the new basis. |
May 12 |
answered | Rearrangments of Fourier series |
May 10 |
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A question on p-summing operators
My answer says "no". |
May 9 |
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On $p$-summable sequences with respect to operator ideals
added 11 characters in body |
May 9 |
answered | A question on p-summing operators |
May 4 |
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Uncountability of the set of subsets of $\mathbb N$
This should be on Math.StackExchange instead of MO, but... For each real $x$, consider the set of rationals less than $x$. |
Apr 27 |
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SubGROUPs of Banach spaces, when are they dense in a vector subspace?
Yes, their Acta paper. |
Apr 27 |
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SubGROUPs of Banach spaces, when are they dense in a vector subspace?
They can. In fact, in separable $L_1$ predual (in particular, the universal space $C[0,1]$) can be isometrically embedded as a norm one complemented subspace of Aff $K$ for some $K$. This is an old result of Lazar and Lindenstrauss. |
Apr 22 |
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A question involving Mazur's Lemma
The Banach-Saks theorem is not true for every Banach space. There exist normalized unconditional bases that converge weakly to zero and yet the norm of the average of any n terms has norm at least 1/2. |
Apr 21 |
awarded | Nice Answer |
Apr 13 |
answered | Fixed point theorems |
Apr 9 |
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Are “most” operators on an infinite-dimensional complex Banach space “diagonalizable”?
The Argyros-Haydon spaces I mentioned have Schauder bases, so the finite rank operators are dense in the compact operators--I neglected to mention that. I don't know whether on every Banach space the closure of the diagonal operators contains the compact operators. |
Apr 8 |
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Are “most” operators on an infinite-dimensional complex Banach space “diagonalizable”?
There are infinite dimensional Banach spaces on which every bounded linear operator is of the form $\lambda I + K$ with $K$ a compact operator. On such a space the diagonal operators are dense. |
Apr 5 |
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On the modulus of convexity of mixed-norm $\ell_{p_1,p_2}$ spaces
When $p=\log n$, $\ell_p^n$ is $\ell_\infty^n$ up to a constant $C$ (the base of the logarithm). So you are looking at $\ell_\infty^n(\ell_1^m)$ in a $C$-equivalent norm. |
Apr 2 |
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On the modulus of convexity of mixed-norm $\ell_{p_1,p_2}$ spaces
A fairly easy way of seeing that you get the expected is to take advantage of the fact that in Banach lattices you can calculate the moduli of convexity and smoothness in terms of properties of the lattice ($p$ convexity; $p$ concavity). Look at volume 2 of Lindenstrauss-Tzafriri. I think it contains what you need; at least it will refer to the relevant papers. |
Apr 2 |
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Between Tietze's and Dugundji's Extension Theorems
Similarly, Q1 has a positive answer for separable Frechet spaces, since they are all homeomorphic to the countable product of lines. |
Apr 2 |
answered | Between Tietze's and Dugundji's Extension Theorems |
Apr 1 |
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How large can a set of nearly equidistant points be?
JL says any $m$ points in a Hilbert space $1+\epsilon$-embeds into a Hilbert space of dimension at most $c\epsilon^{-2} \log m$ for some constant $c$ independent of $m$ and $\epsilon$. That is sharp up to $\log (1/\epsilon)$. |
Apr 1 |
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How large can a set of nearly equidistant points be?
Apply JL to an orthonormal set in a high dimensional space; all the pairwise distance are the same, so you get almost the same in the low dimensional space. |