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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

2 Answers 2

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 $\|(y_n+z_n)\|_Y = \|(y_n)\|_Y+\|(z_n)\|_Y$, meaning that $$\sqrt{\sum \left\| y_n+z_n \right\|_{q_n}^2} = \sqrt{\sum \left\|y_n\right\|_{q_n}^2} + \sqrt{\sum \left\|z_n\right\|_{q_n}^2} \tag1$$ By the triangle inequality and Minkowski inequality for $\ell^2$, $$\sqrt{\sum \left\| y_n+z_n \right\|_{q_n}^2} \le \sqrt{\sum (\left\| y_n\|_{q^n}+ \|z_n \right\|_{q_n})^2}\le \sqrt{\sum \left\|y_n\right\|_{q_n}^2} + \sqrt{\sum \left\|z_n\right\|_{q_n}^2} \tag{2}$$ where equality must hold throughout by (1). 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
Why is $\sum \left( \frac{\|y_n\|_{q_n}+\|z_n\|_{q_n}}{2}\right)^2 \le \frac14 \sum \left\|y_n\right\|_{q_n}^2 + \frac14 \sum \left\|z_n\right\|_{q_n}^2$ true for the second half of (2)? Should that be $\sum \left( \frac{\|y_n\|_{q_n}+\|z_n\|_{q_n}}{2}\right)^2 \le \frac12 \sum \left\|y_n\right\|_{q_n}^2 + \frac12 \sum \left\|z_n\right\|_{q_n}^2$? – Ti Wen Oct 11 at 23:26

For completeness, I add the details of the proof of $d_{BM}(X,Y)=1$ for the example suggested by Wojtaszczyk. The key is the following

Lemma. The identity map $\ell_p^n\to \ell_q^n $ has norm $1$ when $p\le q$, and norm $n^{1/p-1/q}$ when $p\ge q$.

Proof. (a) Suppose $p\le q$. Let $x$ be a vector of unit $p$-norm. Since each coordinate of $x$ has magnitude at most $1$, it follows that $|x_i|^q\le |x_i|^p$. Hence $\|x\|_q\le 1$.

(b) Suppose $p\ge q$. By Jensen's inequality, applied to the convex function $\phi(t)=t^{p/q}$, implies $$ \frac{1}{n} \sum (|x_i|^q)^{p/q} \ge \left(\frac{1}{n} \sum |x_i|^q\right)^{p/q} $$ hence $\|x\|_q \le n^{1/q-1/p}\|x\|_p$.

In the construction considered here, $n=5$ and $p,q\in [1,1.5]$. Divide $[2/3,1]$ into subintervals of length $<\epsilon$, for small $\epsilon$. If $1/p$ and $1/q$ fall into the same subinterval, the Banach-Mazur distance between $\ell_p^5$ and $\ell_q^5$ is at most $5^\epsilon=1+O(\epsilon)$.

Since the sequences $p_k$ and $q_k$ are dense in $[1,1.5]$, each of subintervals mentioned above contains countably many exponents of either kind. Use a bijection between them to define an isomorphism. The norm of this isomorphism is at most $5^\epsilon$ in either direction.

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