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# Bill Johnson

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 Name Bill Johnson Member for 3 years Seen 30 mins ago Website Location Texas A&M Age
 19h comment Continuity of lattice operations in Banach latticesIn $L_2(0,1)$ take '$X_n = \{1_{A_i} : 1 \le i \le 2^n \}$', where $A_i = [(i-1)/2^n, i/2^n]$'. This gives an order bounded counterexample to Q2 which can be generalized to any order continuous non discrete Banach lattice. But the lattice generated by the $X_n$ is not totally bounded. 19h accepted Continuity of lattice operations in Banach lattices 20h answered Continuity of lattice operations in Banach lattices 2d comment Explicit form of the homeomorphism between $C[0,1]$ and $C[0,1]\setminus 0$I would try to adapt the construction for $\ell_2$ to $c_0$ and use the fact that $c_0$ is isometric to a norm one complemented subspace of $C[0,1])$ in a natural way. Jun13 revised Range of the Fourier transform on L^1Edited so as to bump to the first page. Jun13 comment Range of the Fourier transform on L^1Easier, Yemon: $C_0$ contains a subspace isomorphic to $c_0$ while no $L_1$ space does (e.g. by cotype or weak sequential completeness or...). Jun13 answered An almost orthogonality principle for L^p Jun11 answered An almost orthogonality principle for L^p Jun4 comment almost projective Banach space, complex scalarsWell, for example from $E^{*}$ being one injective you get that $E^{**}$ is contractively complemented in an $L_1$ space, which gives what I said. Jun4 comment $(1+\epsilon)$-injective Banach spaces, complex scalarsWhy a down vote for this good question? Jun4 comment $(1+\epsilon)$-injective Banach spaces, complex scalarsSure, because for dual spaces you can use weak$^*$ compactness. A similar proof shows that a dual space that is $\pi_{\lambda}^\infty$ for every $\lambda > 1$ is $1$-injective. Jun3 comment What is the largest possible operator norm of a sparse (0,1)-matrix?Sure, Survit. That is the Hilbert-Schmidt, which dominates the operator norm, and the norms are the same for rank one operators. More generally, $k^{1/2}$ is the right answer for all $k \le n$ for the same reason--put all the ones in one column. I guess when $n^2 \ge k > n$ you get the max when you fill up as many columns as possible with ones and put the left over ones into one column. Jun3 comment almost projective Banach space, complex scalarsI don't recall seeing anything for the complex case for this question or your one on injective spaces. This one looks easier to me. You get that the inclusion from $E$ into its second dual isometrically factors through some $L_1$ space and that $E$ is almost isometric to $\ell_1(\Gamma)$. Is that enough to get $E$ contractively complemented in some some $L_1$ space? That would do it, yes? Aren't the contractively complemented subspaces of $L_1$ spaces treated in Lacey's book? I am travelling and cannot easily check. May31 comment Canonical embedding of a closed subspace of a reflexive spaceLooks like a homework problem, Peter. May25 accepted Extension of equivalent norms May25 comment Unambiguous “weak” vector valued $L^{+\infty}$ spaces?More generally, if $E'$ is separable, then weak$^*$ measurability into $E'$ gives strong measurability. This is in books; Diestel-Uhl comes to mind. It follows from the fact that the unit ball is weak$^*$ measurable when $E'$ is separable. May25 answered Extension of equivalent norms May24 comment Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$For $x$ in $D$ let $N(x)$ be the set of $n$ s.t. $f_n(x) < f(x) + 1$. Since $D$ is infinite, $\cap_{x \in D} N(x)$ can be empty. May24 comment Dual space of Bochner space: is there an easier proof to show they’re isometric?No, this is all there is to it. I wonder why this question has hung around for four days. I just voted to close. May24 comment Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$Well, one mistake in your proof is that the final inequality in Step 3 only holds for $n$ sufficiently large, where "sufficiently large" depends on $x$. May24 comment Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$The intervals do intersect, but the functions are a small perturbation of a disjoint sequence of functions (multiply $f_n$ by the characteristic function of the complement of the union of the supports of $f_m$ for $m>n$). May24 revised Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$added tag. May24 answered Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$ May23 comment Almost orthogonal vectors in subsets of euclidean spaceWell, for a random subset (where random can mean various things) the Johnson-Lindenstrauss lemma gives $\exp ( b \epsilon^2 m)$ for some positive constant $b$ independent of $n$ and $\epsilon$, Dustin. Presumably the OP is interested in more precise results and low dimensions. May14 accepted Absolutely 2-summable operator on a Hilbert space May14 revised Absolutely 2-summable operator on a Hilbert spaceadded 1 characters in body May14 answered Absolutely 2-summable operator on a Hilbert space May14 comment Absolutely 2-summable operator on a Hilbert spaceConsider the identity from $\ell_2^{2^n}$ to $\ell_\infty^{2^n}$ and compute what you get with the unit vector basis and the Walsh basis. May14 revised Absolutely 2-summable operator on a Hilbert spaceadded 2 characters in body May13 comment A characterization of Hilbert spaces?Nice, Sergei. If some duality map is biLipschitz, then must the Banach space be isomorphic to a Hilbert space? I have a (possibly false) recollection that this is true but don't see a proof. May10 comment Duality map in strictly convex Banach spacesThe OP was basically told this already in a deleted answer. It is far past the time that this thread should have been closed. May10 comment A Johnson-Lindenstrauss lemma for finite fields?into $\ell_1^n$ with $n \le \delta m$ with distortion depending only on $\delta$. May10 comment A Johnson-Lindenstrauss lemma for finite fields? How can scaling help? $\{1,2,\dots, N\}$ embeds isometrically into $F_2^N$, while in $F_2^n$ there are only $n$ distances. Every finite subset of $L_1$ embeds into $F_1^N$ with arbitrarily small distortion (allowing scaling, of course) if $N$ is sufficiently large, so you are asking more than having $m$ element subsets of $L_1$ embed into $\ell_1^n$ with $n$ small relative to $m$. Here there are negative results due to Brinkman-Charikar (simplified by Lee-Naor and further simplified by Schechtman and me). Only recently was it proved that $m$ element subsets of $L_1$ embed... May8 accepted weak* continuity of linear maps May6 comment Could we interpolate the compactness of compact operators?Backticks often produce forgiveness of formatting transgressions even when you are not aware of having committing them. May6 revised Could we interpolate the compactness of compact operators?Formatting; punctuation. May5 comment Inductive limit of C*-algebras with injective connecting mapsNo, for trivial reasons. Take a strictly increasing sequence $K_n$ of cardinal numbers and consider $A_n=c_0(K_{2n-1})$, $B_n=c_0(K_{2n})$. May5 revised spectral radius monotonicityImproved grammar and formatting. May4 comment Sz.-Nagy dilation for uniformly convex Banach spacesThe point is that, modulo abstract nonsense, for spaces that possess the approximation property, all the difficulty is in the finite dimensional version of the problem. BTW, I doubt this has a positive answer even in the setting of $L_p$ spaces. May4 comment Sz.-Nagy dilation for uniformly convex Banach spacesIf you can prove it for all finite dimensional spaces with a given common modulus of uniform convexity with the dilation spaces all having a (possibly different) common modulus of uniform convexity, then you get the same result for all spaces that have the approximation property and have that given common modulus of uniform convexity. This follows from a standard Banach space ultraproduct argument (basically the same one I used to extend the finite dimensional Akcoglu-Sucheston theorem, which they already knew, to the infinite dimensional setting).... May2 comment Is every submartingale a convex function of a martingale?No, but there are results in that direction: projecteuclid.org/… May2 accepted Complemented subspaces of $\ell_p(I)$ for uncountable $I$ May2 comment Extension of power bounded operators over a finite subspaceIn general you cannot do better than the projection constant of $\ell_p^n$, which is of order $n^{1/p}$ when $p \ge 2$ and $n^{1/2}$ otherwise. To see that, look at $Y= \ell_p^n \oplus_p C(K)$ and take an isometric copy $E$ of $\ell_p^n$ in $C(K)$. On $\ell_p^n \oplus_p E$ let $T$ be an isometry that maps $\{0\} \oplus E$ to $\ell_p^n \oplus \{0\}$ and $\ell_p^n \oplus \{0\}$ to $\{0\} \oplus E$. Apr10 comment On hyperplanes of $L\infty$Accepting an answer, especially an incomplete one, is not necessary; even undesirable, because a better answer might be posted. Just show that you are watching for information on the question you asked and respond to questions. Apr8 comment On hyperplanes of $L\infty$Is the exact value known for the analogous problem for $C[0,1]$? BTW: It is considered impolite by many users, including myself, to ask a question on MO and then vanish for days. Apr8 accepted On hyperplanes of $L\infty$ Apr6 comment Finite codimensional subspaces of L(X,Y)So if $X$ does not have a subspace isomorphic to $\ell_1$ and $Y = \Bbb{R}$, then it is true (because then each functional in $X^{**}$ is a pointwise limit of elements in $X$ by the Odell-Rosenthal theorem). What else do you know about the question, Kevin? Apr6 comment Finite codimensional subspaces of L(X,Y)There is something wrong with your final paragraph, Kevin. A normed closed subspace of $X^*$ need not be weak$^*$ closed--consider the kernel of a functional in $X^{**}\sim X$`. Apr3 answered On hyperplanes of $L\infty$ Mar29 awarded ● Pundit