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Consider the hyperplane $H=\{f\in L^\infty: \int f = 0\}$ of $L^\infty = L^\infty[0,1]$. My question is:

1. What is the Banach-Mazur distance between $H$ and $L^\infty$? Are there "natural" isomorphisms between these two spaces?

Since $L^\infty$ is 1-injective, we have the lower bound $$ inf\{\|P\|: \mbox{$P$ is a projection from $L^\infty$ onto $H$}\}\leq d_{BM}(H,L^\infty). $$ Recalling that the canonical projection $P:L^\infty\rightarrow H$ given by $Pf(x)= f(x) - \int f$ has norm $2$, we can pose the following related question:

2. Given a (linear, continuous) projection $P:L^\infty\rightarrow H$, do we always have $\|P\|\geq 2$?

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Do you know whether the B-M distance is strictly larger than 2? My guess is that the exact value is not known. –  Bill Johnson Apr 3 '13 at 15:37

2 Answers 2

up vote 4 down vote accepted

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 $F$.

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For a pointed metric space $(M,d,0)$, we denote by $Lip_0(M)$ the Banach space of all real-valued Lipschitz functions $f$ defined on $M$ and such that $f(0)=0$. Recall that $Lip_0 [0,1] = L^\infty$ (see, for example, N. Weaver's book Lipschitz Algebras). It's not hard to see that the hyperplane $H$ defined above is isometrically isomorphic to $Lip_0([0,1]/\{0,1\})$, where $[0,1]/\{0,1\}$ the metric space obtained by collapsing $\{0,1\}$ to one point (see again Weaver for quotient metric spaces), that is, $[0,1]/\{0,1\}$ is the unit circle with the length distance. $Lip_0 ([0,1]/\{0,1\})$ can be seen as the subspace of $Lip_0 [0,1]$ formed by the functions $f$ such that $f(1)=0$. Take the obvious linear Lipschitz extension operator $T$ from $Lip_0 \{0,1\}$ to $Lip_0 [0,1]$ and define $\Phi: Lip_0 \{0,1\} \oplus_\infty Lip_0 ([0,1]/\{0,1\}) \rightarrow Lip_0 [0,1]$ by $\Phi(f,g)\doteq T(f)+g$. It is straightforward that $\Phi$ is an onto isomorphism with $\|\Phi\|=2$ and $\|\Phi^{-1}\|=2$. Since $Lip_0\{0,1\} \oplus_\infty Lip_0 ([0,1]/\{0,1\})$ is isometrically isomorphic to $Lip_0 ([0,1]/\{0,1\})$, it follows that $d_{BM}(H,L^\infty)\leq 4$.

Combining with Bill Johnson's answer we then have $2\leq d_{BM}(H,L^\infty) \leq 4$, and the search for the exact value continues.

From the above construction we get an explicit isomorphism between $H$ and $L^\infty$; not a very natural one though.

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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. –  Bill Johnson Apr 8 '13 at 17:15
    
I don't know about the C[0,1] problem either... if I find something about it, I'll post. Thank you for the etiquette advice. I'm a new user and I should learn this. After reading your answer and checking it, I felt like thinking about the upper bound before giving a feedback. I understand that that could be interpreted as impolite; and definitely I should clicked on the 'accept answer' button as soon as I checked it, so I apologize! –  Pedro Kaufmann Apr 9 '13 at 8:19
    
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. –  Bill Johnson Apr 10 '13 at 2:09

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