The Banach-Mazur distance is not reached

Question

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.

Answer 1

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.