User bill johnson - MathOverflow

Answer by Bill Johnson for Extension of equivalent norms

2013-05-25T04:29:14Z

Answer by Bill Johnson for Counterexample of non-negative sequence weakly converging in $\mathscr{M}^1$ but not $L^1$

2013-05-24T04:18:17Z

For $n>100$ let $F_n = (k/n)_{k=1}^{n -1}$ and for $x= k/n $ in $F_n$ let $I_{k,n}$ be a symmetric interval around $x$ having length $n^{-n}/(n-1)$. Set $f_n = n^n \sum_{k=1}^{n-1} 1_{I_{k,n}}$ . It is clear that $f_n$ converges weak$^*$ to $1_{[0,1]}$. But the $(f_n)$ are essentially disjointly supported and hence are equivalent to the unit vector basis for $\ell_1$.

Answer by Bill Johnson for Absolutely 2-summable operator on a Hilbert space

2013-05-14T18:58:28Z

I answered the OP's question in a comment, but the right question is whether you can estimate the $2$-summing norm of an operator $T$ from a Hilbert space $H$ by taking $\sup (\sum \|Te_n\|^2)^{1/2}$ , where the sup is over all ON bases $(e_n)$ for $H$. In fact, if $H$ is infinite dimensional, this sup is equal to the $2$-summing norm (combine a dilation argument with the fact that $T$ is essentially zero on an infinite dimensional subspace). More interesting is that the the sup is at least $2^{-1/2}$ times the $2$-summing norm of $T$. This is the gist of Tomczak's lemma (see Theorem 18.4 in her book "Banach-Mazur distances and finite-dimensional operator ideals").

Answer by Bill Johnson for On hyperplanes of $L\infty$

2013-04-03T15:25:24Z

IIRC, if $K$ is a compact Hausdorff space that has no isolated points, then every projection from $C(K)$ onto a hyperplane has norm at least two. Maybe Dan Amir proved this? Anyway, in your situation the proof goes like this. A projection $P$ from $L^\infty$ onto $H$ has the form $Pf = f -(\int f) g$, where $\int g = 1$. Let $E=[g\ge 1]$ and observe that $g$ has positive measure. Take a subset $F$ of $E$ that has measure $\epsilon > 0$ and let $h=1_F - 1_{F^c}$. Then $H$ has norm one in $L^\infty$ and $\int h = -1 + 2\epsilon$. Thus $Ph$ is at least $2 - 2\epsilon$ on $F$.

Answer by Bill Johnson for B(H) as a direct sum of a closed two sided ideal and a subalgebra

2013-03-11T22:57:08Z

$K(H)$, the compact operators on $H$, is the only proper closed ideal in $B(H)$ when $H$ is a separable infinite dimensional Hilbert space, and $K(H)$ is not complemented in $B(H)$ (because if it were the diagonal compact operators would be complemented in the diagonal bounded operators, which is to say that $c_0$ would be complemented in $\ell_\infty$).

Answer by Bill Johnson for weak*-compactness of unit ball in equivalent norm

2013-02-24T22:26:00Z

No. In fact, any non reflexive space can be equivalently renormed so that it is not isometric to a dual space. See

Davis, William J.; Johnson, William B. A renorming of nonreflexive Banach spaces. Proc. Amer. Math. Soc. 37 (1973), 486–488.

Answer by Bill Johnson for Quotients with unconditional bases

2013-02-20T23:41:39Z

The original Gowers-Maurey HI space GM is reflexive. Ferenczi proved that the dual (and also every quotient) of GM is HI. Then GM $\oplus$ GM is not HI but cannot have a quotient with unconditional basis, for then its dual would have a subspace with an unconditional basis. But if $X\oplus X$ has a subspace with an unconditional basis, then also $X$ has a subspace with an unconditional basis (the two complementary projections onto the copies of $X$ in the direct sum $X \oplus X$ cannot both be strictly singular on the same subspace).

I would guess that the experts even know that there is an unconditionally saturated reflexive space $X$ whose dual is HI and thus every quotient of $X$ cannot have an unconditional basis.

Ferenczi, V. Quotient hereditarily indecomposable Banach spaces. Canad. J. Math. 51 (1999), no. 3, 566–584.

Answer by Bill Johnson for Complexifying a real Banach space and its dual

2013-02-12T01:26:17Z

Not an answer, but this is a bit long for a comment.

A Banach space is a dual space iff there is a total family of continuous linear functionals so that the unit ball of the space is compact in the weak topology on the space generated by the family of functionals. From this it is easy to see that if $E_\Bbb{C}$ is a dual space, then $E_\Bbb{C}$ is a dual space when considered as a real Banach space, which implies that there is a norm on $E\oplus E$ which is a dual norm and the projections onto the copies of $E$ have norm one.

This suggests the following questions, which as far as I know are open problems.

If $E\oplus E$ is isomorphic to a dual space, is $E$ isomorphic to a dual space? This question is equivalent to: if $E_\Bbb{C}$ is isomorphic to a dual space, is $E$ isomorphic to a dual space?
Same as (1), but with the additional condition that $E$ be separable.
Is every complemented subspace of a separable dual space isomorphic to a dual space?

Answer by Bill Johnson for How well do random projections preserve the distance between a point and a linear subspace?

2013-02-02T22:55:04Z

Gideon Schechtman and I thought through this. First, you need to multiply the random projection $P$ by a factor that is of order $(d/\ell)^{1/2}$ to have the distances preserved; i.e., where you wrote $P$ you should have written this constant times $P$. Secondly, you need $\ell \ge k$ else the image of the $k$ dimensional subspace will be onto the range of $P$ with probability one. Thirdly, to almost preserve the distance of $y$ to the $k$ dimensional subspace you only need to almost preserve the distance to an $\epsilon$ net of the unit sphere of the subspace, which has cardinality like $\epsilon^{-k}$, which JL says you can do by taking $\ell$ larger than some constant times $k$.

Answer by Bill Johnson for Subspaces of $l_p$ and Banach-Mazur distance

2013-01-13T18:05:18Z

(1) is correct. It follows from the fact that there is a sequence $(E_n)$ of finite dimensional subspaces of $\ell_p$ s.t. $\gamma_p(E_n) \to \infty$. Here $\gamma_p(X)$ is the factorization constant of the identity on $X$ through an $L_p$ space; that is, $\gamma_p(X)= \inf \|T\|\cdot \|S\|$, where the infimum is over all $T:X\to Y$, $S:Y:\to X$, and $Y$ an $L_p$ space.

For fixed $n$, set $X_n=\ell_p(E_n)$. Since $E_n$ is finite dimensional, the space $X_n$ is isomorphic to $\ell_p$ (and embeds isometrically into $\ell_p$) but the Banach-Mazur distance of $X_n$ to $\ell_p$ is at least $\gamma_p(E_n)$.

There remains the sticky point of producing a sequence $(E_n)$ as above. One way is to take a subspace $X$ of $\ell_p$ that fails the approximation property and let $E_n$ be a sequence of finite dimensional subspaces of $X$ with $E_1\subset E_2 \subset \dots$ and $\cup E_n$ dense in $X$. Or, for $p<2$, $E_n$ can be (Banach-Mazur arbitrarily close to) $\ell_r^n$ with $p<r<2$ , because $\ell_r$ embeds isometrically into $L_p(0,1)$ and there are various ways of checking that $\gamma_p(\ell_r^n)\to \infty$. Since I do not know your background (you have, unfortunately, chosen to remain anonymous), I do not know which approach of producing $(E_n)$ is best for you.

Answer by Bill Johnson for Absolute norms and 1-unconditional sums

2013-01-06T20:34:35Z

What Yemon says is correct. The "right" space to use is the space $Y$ of all elements in $Z$ such that only finitely many terms are non zero--you can always complete at the end.

Unconditional sums can be much more complicated than absolute sums. In an absolute sum, if you have linear operators $T_n$ on $X_n$ s.t. $\sup_n \|T_n\| < \infty$ and define $T$ on $Y$ by $Tx = (T_n x(n))$ for $x=(x(n))$ in $Y$, then $T$ is a bounded linear operator on $Y$. This is not true for unconditional sums. In fact, the Kalton-Peck space is (the completion of) an unconditional sum of a sequence of 2-dimensional spaces (which can all be taken to be two dimensional Hilbert spaces) and yet does not have an unconditional basis!

Answer by Bill Johnson for Non-super reflexive space

2013-01-04T20:49:29Z

The first question is easy: Every non reflexive space has a separable non reflexive subspace (e.g. by the Eberlein-Smulian theorem or by R. C. James' characterization of non reflexivity).

The second question was a longstanding open problem that was solved by James in the 1970s. Pisier and Xu gave another proof--you can find their paper by using MathSciNet. Their approach is more conceptual and uses interpolation theory but is not easy.

Answer by Bill Johnson for closed complement

2012-12-13T04:02:52Z

For the first question, see Theorem 2.5.5 in the book of Albiac and Kalton. The second question is immediate from the first and the easy fact that $C[0,1]$ has a complemented subspace isometric to $c_0$.

Answer by Bill Johnson for Embedding of $\ell_p$ into infinite direct sums

2012-11-30T23:24:15Z

The answer is no. Let $P_n$ be the natural projection from $Z_{1p} :=\ell_1(\ell_p)$ onto the sum of the first $n$ copies of $\ell_p$. Let $Z$ be any subspace of $Z_{1p}$ that contains no isomorphic copy of $\ell_p$. Then the restriction of $P_n$ to $Z $ is strictly singular, so there is a norm one vector $x_n$ in $Z $ with $\|P_n x_n\|<1/n$. Do a standard gliding hump argument to deduce that $x_n$ has a subsequence equivalent to the unit vector basis of $\ell_1$.

Product of Borel sigma algebras

2010-09-24T18:11:10Z

If $X$ and $Y$ are separable metric spaces, then the Borel $\sigma$-algebra $B(X \times Y)$ of the product is the the $\sigma$-algebra generated by $B(X)\times B(Y)$. I am embarrassed to admit that I don't know the answers to:

Question 1. What is a counterexample when $X$ and $Y$ are non separable?

Question 2. If $X$ is an uncountable discrete metric space, does $B(X)\times B(X)$ generated the Borel $\sigma$-algebra on $X \times X$?

Question 3. If $X$ and $Y$ are metric spaces with $X$ separable, does $B(X)\times B(Y)$ generated the Borel $\sigma$-algebra on $X \times Y$?

Answer by Bill Johnson for Complemented subspaces of $\ell_p(I)$ for uncountable $I$

2012-11-09T13:53:07Z

This is just an exercise in the reflexive range once you know the proof in the separable case. Suppose $X$ is a subspace of $\ell_p(I)$ with density character $\aleph$. Since each vector in $X$ has countable support, WLOG $I=\aleph$. If $X$ is complemented, you just need to check that $X$ has a complemented subspace isomorphic to $\ell_p(\aleph)$, for then you can apply the decomposition method. The separable case is known, so assume that $\aleph$ is uncountable. Take a maximal set of disjointly supported unit vectors in $X$. If the cardinality of the set is less than $\aleph$, then the union $A$ of their supports has cardinality less than $\aleph$. Let $P$ be the band projection onto the functions supported on $A$. Since $X$ has density character greater than $|A|$, this projection is not one to one (the adjoint cannot have dense range; this is where reflexivity helps).

Answer by Bill Johnson for Isometric embeddings of $\ell_q^m$ into $\ell_p$ and $L_p$ for $p,q\in[1,+\infty]$

2012-10-29T23:33:15Z

See $$ $$ MR0417756 (54 5804) 46B05 Dor, L. E. Potentials and isometric embeddings in $L_1$. Israel J. Math. 24 (1976), no. 3-4, 260–268 $$ $$ for a complete answer to your question.

Answer by Bill Johnson for Does the Fourier series of an $L^1$ function converge to the function weakly in $L^1$?

2012-10-29T14:16:19Z

No. If the partial sum projections $S_n$ converged in the weak operator topology, they would be pointwise weakly bounded hence pointwise norm bounded whence uniformly bounded. That would give convergence pointwise strongly.

Answer by Bill Johnson for Direct proof of injectivity of $L_\infty$

2012-10-23T21:00:17Z

Write $L_\infty$ as the closure of a net (directed by inclusion) of finite dimensional $\ell_\infty$ spaces. Compose the operator into $L_\infty$ with norm one projections onto these subspaces and extend. Use weak$^*$ compactness of the unit ball of $L_\infty$ to pass to a limit of a subnet of these operators. $$ $$ Basically the same argument works for any dual space that is $\mathcal{L}_{\infty,\lambda}$ for all $\lambda > 1$.

Answer by Bill Johnson for C*-algebras with no nontrivial endomorphisms

2012-10-16T14:05:47Z

Look at $$ $$ Cook, H. Continua which admit only the identity mapping onto non-degenerate subcontinua. Fund. Math. 60 1967 241–249. $$ $$ The title and review give the main result, which does not give the example you seek. However, IIRC, Howard also constructed a non-degenerate compact metric space on which the only continuous self maps are the identity and constant functions. I can't access the mentioned paper to check if it is there.

Answer by Bill Johnson for Projections onto $n$-codimensional subspaces of a Banach space: norms.

2012-10-15T15:37:32Z

In many books $^*$ you can find the result that there is a projection of norm at most $\sqrt{n}$ onto any $n$ dimensional subspace of a Banach space. For reflexive spaces, this gives immediately that every $n$ codimensional subspace is the range of a projection that has norm at most $\sqrt{n} +1$. For non reflexive spaces, by using the principle of local reflexivity (which also is in many books), you get for any $\epsilon > 0$ the estimate $\sqrt{n} +1 + \epsilon $. $$ $$ $*$ See, for example, Albiac and Kalton, ``Topics in Banach space theory", Theorem 12.1.6. In this book you can also find the principle of local reflexivity.

Answer by Bill Johnson for BM-distances between $B(\ell_p^n)$ and $\ell^{n^2}_p$

2012-10-07T16:47:58Z

For the first question, $n^{1/2}$ is the right order for $p=1,2,\infty$. The case $p=\infty$ is arguably the easiest, because $B(\ell_\infty^n)=\ell_\infty^n(\ell_1^n)$ isometrically and the Banach-Mazur distance between $\ell_\infty^n$ and $\ell_1^n$ is of order $n^{1/2}$. That gives the upper bound, and the lower bound is also true because $\gamma_\infty(\ell_1^n)$ (the factorization constant of the identity on $\ell_1^n$ through an $\ell_\infty$-space) is of order $n^{1/2}$.
$$ $$ Also $B(\ell_1^n)=\ell_\infty^n(\ell_1^n)$ isometrically, and you get the lower bound from the fact that $\gamma_1(\ell_\infty^n)$ is of order $n^{1/2}$. For the upper bound assume that $n$ is a power of two (I think standard arguments from this case give the general case but did not try to think it through). In $L_1^N(\ell_1^n)$ consider the basis $w_k\otimes e_j$, $1\le j,k \le n$, where $(w_k)$ is the Walsh basis for $L_1^n$. Take the basis to basis mapping from $L_1^N(\ell_1^n)$ onto $\ell_\infty^n(\ell_1^n)$ that maps $w_k\otimes e_j$, $1\le j \le n$, onto the $k$-th copy of $\ell_1^n$. This mapping has norm at most $n^{1/2}$ because the Walsh basis has an upper $\ell_2$ estimate, and the inverse has norm one. $$ $$ For $B(\ell_2^n)$, Yemon's comment gives a lower estimate of $n^{1/2}$ (in fact, for all $1\le p \le 2$ because $\gamma_p(\ell_\infty^n)$ is of order $n^{1/2}$ in this range). The upper estimate is just the fact that the norm of an operator in $\ell_2^n$ is at most $n^{1/2}$ times the Hilbert-Schmidt norm of the operator. $$ $$ For all values of $p$, if you write a vector in $\mathbb{R}^{n^2}$ as an $n$ by $n$ matrix $A$, then the $\ell_p^{n^2}$ norm of $A$ is the $p'$-summing norm of $A$ considered as an operator from $\ell_p^n$ into $\ell_{p'}^{n}$. That is what is used above in the $p=2$ case, but I don't see that this helps for other values of $p$. For $2<p<\infty$ you get a lower bound from the fact that $B(\ell_p^n)$ contains $\ell_\infty^n$ and $\ell_{p'}^n$ isometrically, but for $p=4$ that gives a lower bound of only $n^{1/4}$.

Answer by Bill Johnson for Do random projections (approximately) preserve convexity?

2012-09-10T17:35:24Z

Negative result:

See p. 377 in Chapter 15 of Matousek's book, which can be found here. In short, if you want the image of the $k$ points to be between the surface of a convex body $K$ and the surface of $DK$ for some $D>1$, you need the operator to have rank at least $k^{f(D)}$ for some function $f$.

Positive result on a related problem:

Johnson, William B.; Lindenstrauss, Joram; Schechtman, Gideon On Lipschitz embedding of finite metric spaces in low-dimensional normed spaces. Geometrical aspects of functional analysis (1985/86), 177–184, Lecture Notes in Math., 1267, Springer, Berlin, 1987,

it is proved that for some constant $C$, if you have $k$ points on the surface of a symmetric convex body, then you can put the points isometrically into a suitable $\ell_\infty^m$ in such a way that a random projection of order rank $k^{1/D}$ will place the points between the surface of a symmetric convex body $K$ and the surface of $CDK$; see the paper for a precise statement. I don't think symmetry places much of a role here. We were interested in the embedding of points into a Banach space and so did not think about general convex bodies. The embedding theorem we proved was later made obsolete by Matousek when he proved that and metric space with size $k$ embeds into $\ell_\infty^{n}$ with distortion $D$ with $n$ about $Dk^{1/(2D)} \log k$ (see p. 404 at the above given link).

Answer by Bill Johnson for Non-probabilistic proof of the Johnson–Lindenstrauss lemma

2012-09-11T12:46:15Z

You might be interested in something Jelani Nelson wrote me in an email on Oct. 13, 2011:

"Another notion of derandomizing JL is the following: come up with a distribution over embeddings that can be sampled using as few random bits as possible so that, for any vector $x$ in $R^d$, a random vector has its $\ell_2$ norm preserved up to $1+\epsilon$ with probability $1-\delta$ by a random embedding from that distribution (embedding into $O(\epsilon^{-2}\log(1/\delta))$ dimensions). Existentially, it can be shown that there exists such a distribution which can be sampled from using $O(log(d/\delta))$ random bits. An actual explicit such distribution would imply the two works above, since our deterministic algorithm could just try all embeddings in the support of the distribution (there would be poly$(d/\delta)$ of them) and take the best one, with $\delta = 1/n^2$.

Obtaining a distribution that can be sampled using $O(\log(d/\delta))$ random bits is open. The best we have to date are:

-- Daniel M. Kane, Raghu Meka, Jelani Nelson: Almost Optimal Explicit Johnson-Lindenstrauss Families. APPROX-RANDOM 2011: 628-639. (requires $O(\log d + \log(1/\delta)*\log\log(1/\delta) + \log(1/\delta)\log(1/\epsilon))$ random bits)

-- Zohar Shay Karnin, Yuval Rabani, Amir Shpilka: Explicit Dimension Reduction and Its Applications. IEEE Conference on Computational Complexity 2011: 262-272. (requires $(1+o(1))\log d + O(\log(1/(\epsilon\delta))$ random bits).

In fact, combining the approaches of both works can get $(1+o(1))\log d + O(\log(1/\delta)\log\log(1/\delta) + \log(1/\delta)\log(1/\epsilon))$."

I suggest contacting Jelani directly if you want more information.

Answer by Bill Johnson for Extension of lipschitz functions along a curve

2012-09-05T19:50:49Z

It is not always possible to extend when $X$ is a Banach space. Take a Banach space $Y_n$ which contains an $n$ dimensional subspace $E_n$ such that every projection from $Y_n$ onto $E_n$ has norm at least $C_n$ with $C_n\to \infty$. ($Y_n$ can e.g. be $L_1$ and $E_n$ the span of $n$ IID gaussian random variables; then $C_n$ is of order $n^{1/2}$.) Let $X_n = Y_n \oplus_2 E_n$. For the curve in $Y_n$ take any curve in the unit sphere of $E_n \oplus {0}$ that contains an $\epsilon_n$ net $A_n$ of the unit sphere of $E_n \oplus {0}$. For $f_n$ take the natural isometry from $E_n \oplus {0}$ onto $ {0} \oplus E_n $ restricted to the curve. Let $F_n$ be an extension of $f_n$ to a Lipschitz mapping on $X_n$; WLOG $F_n$ maps into $ {0} \oplus E_n $ since this is a norm one complemented subspace of $X_n$. Let $G_n$ be the positively homogeneous extension of the restriction of $F_n$ to the unit sphere of $X_n$. Then the Lipschitz constant of $G_n$ is at most three times the Lipschitz constant of $F_n$. Compose $G_n$ with the obvious isometry from $ {0} \oplus E_n $ onto $E_n \oplus {0}$. The restriction of this map to $Y_n$ gives a positively homogenous mapping from $Y_n$ into $E_n$ that is the identity on $A_N$. By the arguments in $$ $$ Johnson, William B.(1-OHSN); Lindenstrauss, Joram(IL-HEBR) Extensions of Lipschitz mappings into a Hilbert space. Conference in modern analysis and probability (New Haven, Conn., 1982), 189–206, Contemp. Math., 26, Amer. Math. Soc., Providence, RI, 1984 $$ $$

we conclude that if $\epsilon_n$ is sufficiently small, there is a projection from $Y_n$ onto $E_n$ whose norm is no worse than something like ten times the Lipschitz constant of $G_n$.

All of this shows that you cannot get Lipschitz extensions with controlled norms. Take an infinite direct sum to get an example where you cannot get any Lipschitz extension.

Answer by Bill Johnson for Complemented Subspaces and Riesz-Thorin interpolation

2012-08-14T20:52:11Z

You do get complementation in the case you mention. Two key facts that you did not state explicitly but which follow easily from your hypotheses are that every unit vector $e_n$ is in the linear span of at most $K$ of the $a_i$ and $b_i$, and the span of $K$ of the $a_i$ and $b_i$ is contained in the span of at most $N=2K^2$ unit vectors. Using these facts, it is not hard to verify that the orthogonal projection $P$ onto $A$ is bounded in the $\ell_1$, which by interpolation and duality gives you what you want. To check boundedness in $\ell_1$, you just have to give a bound on $\|Pe_n\|_1$ that is independent of $n$. But on the span of $N$ unit vectors, the $\ell_1$ norm is dominate by $N^{1/2}$ times the $\ell_2$ norm.

EDIT: As Antoine points outs, my answer is wrong.

Answer by Bill Johnson for Around Banach isomorphism theorem

2012-07-08T17:44:59Z

No. If $E$ has codimension less than the continuum in a Banach space then such an $f$ must be open. This was proved by Saxon and Levin; see the proposition on page 95 of

this paper

which is

Saxon, Stephen; Levin, Mark, Every countable-codimensional subspace of a barrelled space is barrelled. Proc. Amer. Math. Soc. 29 1971 91–96.

So for an example, take any discontinuous linear functional on a Banach space and let $E$ be its kernel.

Answer by Bill Johnson for Basic sequences in $\ell_p$

2012-03-30T20:14:59Z

The answer is yes also for $L_p$, but I don't know a good book reference. For $2<p<\infty$ , this is contained the paper of Kadec and Pelczynski--it is their second dichotomy theorem. Actually, they get that a normalized weakly null sequence has a subsequence that is either equivalent to an orthonormal sequence (in which case its closed span is automatically complemented) or has, for every $\epsilon > 0$, a subsequence that is $1+\epsilon$-equivalent to the unit vector basis for $\ell_p$ and spans a subspace that is $1+\epsilon$-complemented.

For $1\le p <2$, I think the result was pointed out by Pelczynski but I don't know a reference. It follows from arguments like those in Wojtaszczyk's book characterizing weak compactness in $L_1$. You can find an outline of the argument in a paper I wrote with G. Schechtman:

Multiplication operators on L(Lp) and lp-strictly singular operators, J. European Math. Society 10 1105-1119 (2008), which you can download from my home page.

EDIT July 7, 2012: The result above that I attributed to Pelczynski is actually due to Enflo and Rosenthal:

Enflo, Per; Rosenthal, Haskell P. Some results concerning Lp(μ)-spaces. J. Functional Analysis 14 (1973), 325–348.

Answer by Bill Johnson for surjectivity of operators on l^infty

2012-07-06T15:16:26Z

Here is an answer to an easier but related question.

Proposition. There is a one to one operator $T$ from $\ell_1(2^{\aleph_0})$ into $\ell_\infty$ that has dense range.

Of course, such an operator cannot be surjective because $\ell_1(2^{\aleph_0})$ is not isomorphic to $\ell_\infty$.

My proof of the Proposition uses an old result of Bill Davis and mine (Remark 4 in

Davis, W. J.; Johnson, W. B. On the existence of fundamental and total bounded biorthogonal systems in Banach spaces. Studia Math. 45 (1973), 173–179):

$\ell_\infty$ has a biorthogonal system $(x_\alpha,x_\alpha^*)_{\alpha<2^{\aleph_0}}$ with $\|x_\alpha\|=1$ and $\sup_\alpha \|x_\alpha^*\|<\infty$ such that the linear span of $(x_\alpha)$ is dense in $\ell_\infty$.

To prove the Proposition, define $T$ to be the norm one linear extension of the map $e_\alpha \mapsto x_\alpha$, where $(e_\alpha)$ is the unit vector basis for $\ell_1(2^{\aleph_0})$. This mapping obviously has dense range and is one to one because every biorthogonal system is countably linearly independent.

Here is a variation on the OP's question:

Is there a one to one bounded linear operator from $\ell_\infty$ into itself that has dense range but is not surjective?

The interest in the variation is that this question is easily seen to be equivalent to:

Are there quasi-complementary copies of $\ell_\infty$ in $\ell_\infty$ that are not complementary?

(Recall that two closed subspaces of a Banach space are said to be quasi-complementary if their sum is dense and their intersection is ${0}$.)

Answer by Bill Johnson for Fundamental problems whose solution seems completely out of reach

2012-07-03T00:05:25Z

Is every complemented subspace of $C[0.1]$ isomorphic to $C(K)$ for some compact metric space $K$?

Is every infinite dimensional complemented subspace of $L_1[0.1]$ isomorphic either to $L_1[0.1]$ or to $\ell_1$?

Comment by Bill Johnson

2013-05-24T16:20:52Z

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.

Comment by Bill Johnson

2013-05-24T16:12:48Z

No, this is all there is to it. I wonder why this question has hung around for four days. I just voted to close.

Comment by Bill Johnson

2013-05-24T12:24:09Z

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 $x$.

Comment by Bill Johnson

2013-05-24T12:11:55Z

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 $m>n$).

Comment by Bill Johnson

2013-05-23T18:41:03Z

Well, 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.

Comment by Bill Johnson

2013-05-14T03:29:56Z

Consider 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.

Comment by Bill Johnson

2013-05-13T07:43:57Z

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.

Comment by Bill Johnson

2013-05-10T05:39:21Z

The OP was basically told this already in a deleted answer. It is far past the time that this thread should have been closed.

Comment by Bill Johnson

2013-05-10T05:35:30Z

into $\ell_1^n$ with $n \le \delta m$ with distortion depending only on $\delta$.

Comment by Bill Johnson

2013-05-10T05:34:17Z

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...

Comment by Bill Johnson

2013-05-06T18:46:00Z

Backticks often produce forgiveness of formatting transgressions even when you are not aware of having committing them.

Comment by Bill Johnson

2013-05-05T19:36:54Z

No, for trivial reasons. Take a strictly increasing sequence $K_n$ of cardinal numbers and consider $A_n=c_0(K_{2n-1})$, $B_n=c_0(K_{2n})$.

Comment by Bill Johnson

2013-05-04T20:16:13Z

The 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.

Comment by Bill Johnson

2013-05-04T20:10:41Z

If 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)....

Comment by Bill Johnson

2013-05-03T18:15:15Z

You should not encourage such questions by giving answers to them.

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