28
$\begingroup$

Given Banach spaces $X$ and $Y$, the Banach-Mazur distance between $X$ and $Y$ is defined as $$ d(X,Y) = \inf\{ \|\varphi\|\|\varphi^{-1}\| : \varphi\colon X\to Y \text{ isomorphism} \}. $$ We consider the Banach spaces $\ell_\infty=\ell_\infty(\mathbb{N})$ and $L_\infty=L_\infty([0,1],\lambda)$, where $\lambda$ is the Lebesgue measure on $[0,1]$.

What is $d(\ell_\infty,L_\infty)$?

Improve this question
$\endgroup$
16
  • 2
    $\begingroup$ Is there any reason to think that it is finite ? $\endgroup$
    – Denis Serre
    Commented Jun 8, 2016 at 21:00
  • 2
    $\begingroup$ @DenisSerre: $l^\infty$ and $L^\infty$ are isomorphic as Banach spaces. I think this is due to Pelczynski. $\endgroup$
    – Nik Weaver
    Commented Jun 8, 2016 at 21:12
  • 2
    $\begingroup$ Yes, it is well known that they are isomorphic. This follows from the injectivity of these spaces and Pelczynski decomposition method. The details can be found, for instance, in Albiac-Kalton. I am not sure if the answer to the question is known though. $\endgroup$
    – Bunyamin Sari
    Commented Jun 8, 2016 at 21:12
  • 2
    $\begingroup$ Reading the "construction" given by Theo Buehler (see @RobertIsrael's comment) my guess would be that one can get a BM-distance of $\leq 4$ by careful book-keeping of the isomorphisms, since both $Y=\ell_\infty$ and $X=L_\infty$ are $1$-injective, and since $X\cong_1 X\oplus_\infty X$ and $Y\cong_1 Y\oplus_\infty Y$ $\endgroup$
    – Yemon Choi
    Commented Jun 9, 2016 at 15:23
  • 2
    $\begingroup$ Is there an argument that ZFC actually uniquely specifies the value of $d(\ell_\infty,L_\infty)$? $\endgroup$
    – James E Hanson
    Commented Sep 10, 2021 at 21:15

2 Answers 2

Reset to default
10
$\begingroup$

Not a complete answer, feel free to edit. (Likely, the answer is not known anyway.) The distance is at least 2. Look at both spaces as $C(K)$ spaces. Corresponding $K$'s are not homeomorphic, see the discussion here https://math.stackexchange.com/questions/207435/isometry-between-l-infty-and-ell-infty

So by Amir-Cambern theorem (near isometry property of $C(K)$ spaces) the distance is at least 2. The theorem says if there is an isomorphism between $C(K_1)$ and $C(K_2)$ with distortion strictly less than 2, then $K_1$ and $K_2$ are homeomorphic.

Improve this answer
$\endgroup$
10
  • 2
    $\begingroup$ I don't think it is right that the unit ball of $L_\infty$ has more extreme points than the unit ball of $\ell_\infty$. Here's the argument in the real case. A point in the unit ball of $L_\infty$ is $2\mathbf 1_B-1$, where $B$ is a measurable set. Two such functions agree if the $B$'s differ by a set of measure 0. Any Lebesgue measurable $B$ agrees with a Borel $B$ up to a set of measure 0. The cardinality of the Borel $\sigma$-algebra on $[0,1]$ is $\aleph_1$, the same as the cardinality of the collection of $\pm 1$ sequences. $\endgroup$
    – Anthony Quas
    Commented Jun 9, 2016 at 0:05
  • 1
    $\begingroup$ Choose an extreme point of the unit ball, and call it $e$. We can define a partial ordering on our space by $x \ge 0$ if $\|\|x\| e - x \| \le \|x\|$. This makes $L_\infty$ or $\ell_\infty$ into a Banach lattice (equivalent by an isometry to the usual lattice structure). Now $\ell^\infty$ has the property that it has minimal nonnegative elements of norm $1$, i.e. $x$ such that the only element $0 \le y \le x$ with $\|y\| = 1$ is $x$. But $L_\infty$ does not have such elements. So $\ell_\infty$ and $L_\infty$ are not isometric. $\endgroup$
    – Robert Israel
    Commented Jun 9, 2016 at 0:40
  • 2
    $\begingroup$ It is not true, Fedor. Take a sequence $p_n$ that strictly decreases to $1$ and consider $X= (\sum_{n=1}^\infty \ell_{p_n}^2)_2$. Then the BM distance of $X$ to $X\oplus_2 \ell_1^2$ is one but the spaces are not isometrically isomorphic because the latter one is not strictly convex. $\endgroup$
    – Bill Johnson
    Commented Jun 9, 2016 at 6:37
  • 2
    $\begingroup$ Another way to express Robert Israel's comment is to say that the Stone space of $\ell_\infty$ has isolated points while the Stone space of $L_\infty$ does not. $\endgroup$
    – Bill Johnson
    Commented Jun 9, 2016 at 6:41
  • 4
    $\begingroup$ @YemonChoi Arbitrarily big, e.g., the distance between $C(\omega)$ and $C(\omega^n)$ tends to infinity as $n$ does. $\endgroup$
    – Bunyamin Sari
    Commented Jun 9, 2016 at 15:34
2
$\begingroup$

The distance is strictly greater than $2$ --- as is true for any compact, nonhomeomorphic $K_1$ and $K_2$ when $K_1$ is totally disconnected (Cohen-Chu, Stud. Math. 1995, p.6). The question of an upper bound is more puzzling. @HannesThiel comment 6/13/2016 suggests distance $\le 16$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .