Two infinite dimensional spaces whose Banach-Mazur distance equal to 1(details in the proof)

85 Views Asked by At

Sorry for this "duplicated" question. Two years ago, there was a post concerning about Banach-Mazur distance between infinite dimensional spaces and the answer was accepted. But I am really a beginner for Banach space theory and don't quite understand the answer given there.

The question is

Let $(p_n)$ and $(q_n)$ be two disjoint sequences, dense in $[1,1.5]$ such that $p_1=1$. Take $X=\left(\sum \ell_{p_n}^5\right)_2$ and $Y =\left(\sum \ell_{q_n}^5\right)_2$. Show that $X$ and $Y$ are isomophic but not isometric to each other and $d_{BM}(X,Y)=1$.

He gives a hint to the proof of $d_{BM}(X,Y)=1$:

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.

I know we are supposed to find an isomophism $T:X\rightarrow Y$ such that $\|T\| \cdot \|T^{-1}\|\le 1+\epsilon$. But there are two maps $T \text{ and } T^{-1}$ here, which makes the proof very difficult . Maybe I should fix $\|T^{-1}\|=1$ by scaling, but this doesn't reduce any workload for me.

For the proof of the non-isometric part, he was trying to show that $Y$ is not strictly convex. However, I don't believe that the second half of the following inequality is true:

$$\sum \left\|\frac{y_n+z_n}{2}\right\|_{q_n}^2 \le \sum \left( \frac{\|y_n\|_{q_n}+\|z_n\|_{q_n}}{2}\right)^2 \le \frac14 \sum \left\|y_n\right\|_{q_n}^2 + \frac14 \sum \left\|z_n\right\|_{q_n}^2 \tag2$$

I think it should be

$$\sum \left\|\frac{y_n+z_n}{2}\right\|_{q_n}^2 \le \sum \left( \frac{\|y_n\|_{q_n}+\|z_n\|_{q_n}}{2}\right)^2 \le \frac12 \sum \left\|y_n\right\|_{q_n}^2 + \frac12 \sum \left\|z_n\right\|_{q_n}^2 \tag2$$

But then his proof doesn't seem to work.

Every answer and helpful hint will be appreciated! And if you know any other examples for the same purpose, please feel free to post them.