# What is the Banach-Mazur distance between $\ell_\infty$ and $L_\infty$?

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

• Is there any reason to think that it is finite ? – Denis Serre Jun 8 '16 at 21:00
• @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
• 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
• Yes, they are isomorphic and therefore the distance is finite. – Hannes Thiel Jun 8 '16 at 21:16
• 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.
• 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
• 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
• 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
• 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. – Bill Johnson Jun 9 '16 at 6:41