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Let $X,Y$ be isomorphic Banach spaces.

The Banach-Mazur distance:

$$ d(X,Y)=\inf\{\|T\| \cdot \|T^{-1}\|: T:X\longrightarrow Y \ \text{is an isomorphism} \}$$

can be rewritten as:

$$ d(X,Y)=\inf\{\|T^{-1}\|: T:X\longrightarrow Y \ \text{is an isomorphism}, \|T\|=1\} $$

If $X,Y$ are finite dimensional spaces the infimum is reached.

But if $X,Y$ are infinite dimensional spaces the infimum is reached ?

Any hints would be appreciated.

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Wojtaszczyk offers a hint for exercise II.E.3 of Banach Spaces for Analysts, which asks to find non-isometric spaces $X$ and $Y$ whose Banach-Mazur distance is $1$: "Take $(p_n)_{n=1}^\infty$ and $(q_n)_{n=1}^\infty$ two disjoint sequences, dense in $[1,1.5]$ such that $p_1=1$. Take $X=\bigl(\sum_{n=1}^\infty \ell^5_{p_n}\bigr)_2$ and $Y=\bigl(\sum_{n=1}^\infty \ell^5_{q_n}\bigr)_2$. Show that $Y$ does not contain $\ell_1^5$ isometrically." –  David Mitra Jul 17 '13 at 18:54
@DavidMitra is it necessary to have $5$ in the construction? Do you know the solution? –  Norbert Jul 17 '13 at 20:59

1 Answer 1

up vote 5 down vote accepted

Let's follow the hint given by Wojtaszczyk.

Take $(p_n)$ and $(q_n)$ two disjoint sequences, dense in $[1,1.5]$ such that $p_1=1$. Take $X=\left(\sum \ell_{p_n}^5\right)_2$ and $Y =\left(\sum \ell_{q_n}^5\right)_2$.

The Banach-Mazur distance is $1$ because for every $\epsilon>0$ the interval $[1,1.5]$ can be partitioned into subintervals of size $\epsilon$, each of which meets both sequences countably many times.

Now, it seems to me that the reason $Y$ does not contain an isometric copy of $\ell_1^5$ is that $Y$ is strictly convex. (If I'm right then neither $1.5$ nor $5$ are of importance; $1.5$ could be any number in $(1,\infty)$ and $5$ could be an integer $\ge 2$.)

Indeed, suppose that $(y_n)$ and $(z_n)$ are two nonzero elements of $Y$ such that
$$\sum \left\|\frac{y_n+z_n}{2}\right\|_{q_n}^2 = \frac14 \sum \left\|y_n\right\|_{q_n}^2 + \frac14 \sum \left\|z_n\right\|_{q_n}^2 \tag1$$ From $$\sum \left\|\frac{y_n+z_n}{2}\right\|_{q_n}^2 \le \sum \left( \frac{\|y_n\|_{q_n}+\|z_n\|_{q_n}}{2}\right)^2 \le \frac14 \sum \left\|y_n\right\|_{q_n}^2 + \frac14 \sum \left\|z_n\right\|_{q_n}^2 \tag2$$ we see that equality holds throughout. Since $\ell_2$ is strictly convex, equality in the second half of (2) implies that there exists $\lambda> 0$ such that $ \|z_n\|_{q_n}=\lambda \|y_n\|_{q_n}$ for all $n$. Since each $\ell_{q_n}$ is also strictly convex, equality in the first half of (2) implies that $z_n=\lambda y_n$. Thus, $Y$ is strictly convex.

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Ex 5.44 in Banach space theory. The basis for linear and non-linear analysis. M. Fabian, P. Habala, P. Hajek, V. Montesinos, V. Zizler also saves the day. –  Norbert Oct 26 '13 at 20:41

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