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

  • 2
    Is there any reason to think that it is finite ? – Denis Serre Jun 8 '16 at 21:00
  • 2
    @DenisSerre: $l^\infty$ and $L^\infty$ are isomorphic as Banach spaces. I think this is due to Pelczynski. – Nik Weaver Jun 8 '16 at 21:12
  • 1
    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. – Bunyamin Sari Jun 8 '16 at 21:12
  • 1
    Yes, they are isomorphic and therefore the distance is finite. – Hannes Thiel Jun 8 '16 at 21:16
  • 1
    Banach spaces $l^\infty$ and $L^\infty$ are isomorphic, but isomorphisms are not constructive. Something like the Hahn-Banach theorem is required to prove the existence of an isomorphism. – Gerald Edgar Jun 9 '16 at 0:46

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.

  • 2
    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. – Anthony Quas Jun 9 '16 at 0:05
  • 1
    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. – Robert Israel Jun 9 '16 at 0:40
  • 1
    @AnthonyQuas this lattice is defined using norm and linear structures only, thus Robert's argument looks fine. – Fedor Petrov Jun 9 '16 at 6:12
  • 1
    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. – Bill Johnson Jun 9 '16 at 6:37
  • 1
    @YemonChoi Arbitrarily big, e.g., the distance between $C(\omega)$ and $C(\omega^n)$ tends to infinity as $n$ does. – Bunyamin Sari Jun 9 '16 at 15:34

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.