# Separable Banach spaces isometric to quotient of a Banach space

We know that every separable Banach space is isometrically isomorphic to a quotient space of $$(\ell^1,\|.\|_1)$$. We also know that the norm defined by $$\|x\|=(\|x\|_1^2+\|x\|_2^2)^{1/2}$$ for all $$x\in \ell^1$$ is equivalent to $$\|.\|_1$$. My question is that is every separable Banach space isometrically isomorphic to a quotient of $$(\ell^1,\|.\|)$$?

The standard way of proving the result that I stated above is as follows:

Let $$\{x_n:n\in \mathbb{N}\}$$ be a dense subset of $$S_X$$, where $$X$$ is a separable Banach space. Then $$T((\lambda_n))=\sum\limits_{n=1}^{\infty}\lambda_n x_n \text{ for all }(\lambda_n)\in \ell^1,$$ is a continuous linear map from $$\ell^1$$ onto $$X$$. Consequently, $$\ell^{1}/\ker T$$ is linearly homeomorphic to $$X$$. It can also be shown that $$\ell^1/\ker T$$ and $$X$$ are actually isometric. I tried to mimic the same proof for my question too, but couldn't succed. Any help is appreciated.

Following

Dowling, P. N.(1-MMOH); Lennard, C. J.(1-PITT-MS) Every nonreflexive subspace of L1[0,1] fails the fixed point property.
Proc. Amer. Math. Soc. 125 (1997), no. 2, 443--446,

say that a norm $$\|\cdot \|$$ on $$\ell^1$$ is asymptotically isometrically equivalent to the $$\ell^1$$ norm provided that there exists $$\lambda_n \uparrow 1$$ with $$\lambda_1>0$$ so that for all sequences $$(a_n)$$ of scalars, $$\sum_n \lambda_n |a_n| \le \| \sum_n a_n e_n \| \le \sum_n |a_n|,$$ where $$(e_n)$$ is the usual unit vector basis. Suppose $$\| \cdot\|$$ satisfies this condition for such a sequence $$(\lambda_n)$$. Let $$(x_n)$$ be a dense sequence in the unit ball of an arbitrary separable Banach space $$X$$ and define an operator $$Q$$ from $$(\ell^1, \|\cdot \|)$$ to $$X$$ by mapping $$e_n$$ to $$\lambda_n x_n$$ and extending by linearity and continuity. Then $$Q$$ is a norm one linear operator from $$(\ell^1, \|\cdot \|)$$ to $$X$$ such that the image of the unit ball is a dense subset of the unit ball of $$X$$, and hence $$Q$$ is a quotient mapping.

Your norm on $$\ell^1$$ is not is asymptotically isometrically equivalent to the $$\ell_1$$ norm. However, look at the closed span $$Y$$ of $$(\sum_{k\in F_n} e_k)_n$$, where $$F_n$$ are disjoint finite sets of natural numbers and the cardinalities of $$F_n$$ increase to $$\infty$$. Then $$Y$$ under your norm is isometric to an asymptotically isometric $$\ell^1$$ space. Moreover, $$Y$$ is norm one complemented in your space because the unit vector basis is a symmetric basis in your space, so every subspace spanned by a constant coefficient block basis is contractively complemented.

• This is your homework problem, Anupam. It is not a difficult problem, but here is a hint: The block basis, call it $y_n$, for $Y$ I gave you satisfies $\|y_n\|_1/\|y_n\|_2 \to \infty$. That is all that is needed. Apr 23 '20 at 14:46
• What I have understood is as follows : Since Y is norm one complimented in $(\ell^1,\|.\|)$, therefore $Y$ is isometric to $(\ell^1,\|.\|)$ modulo $\ker P$, where $P:X->Y$ is the projection. Also $Y$ is isometric to the space $\ell^1$ with a norm which is asymptotically isometrically equivalent to $\ell^1$-norm. Again $Q$ from $\ell^1$ to X is a quotient map. Thus $\ell^1$ with the mentioned norm modulo kernel of $Q$ is isometric to $X$. Now how to conclude? @Bill Johnson. Apr 24 '20 at 13:31
• I do not understand your problem. $X$ is isometrically a quotient of $Y$ and $Y$ is isometrically a quotient of $(\ell^1, \|\cdot \|)$, so $X$ is isometrically a quotient of $(\ell^1, \|\cdot \|)$. Apr 24 '20 at 17:54
• Yes, I have understood now. One last query. As mentioned by you, $Y$ is isometric to an asymptotically isometric $\ell^1$ space-does this mean that there exists a sequence $(\lambda_n)$ with $\lambda_1>0$ and $\lambda_n\to 1$ such that $$\sum\limits_{n=1}^{\infty}\lambda_n|t_n|\leq \|\sum\limits_{n=1}^{\infty}t_ny_n\|\leq \sum\limits_{n=1}^{\infty}|t_n|$$ for all $(t_n)\in \ell^1$?, where $Y$ is the closed linear span of $(y_n)$ and $y_n=\sum\limits_{k\in F_n}e_k$, $F_n$ are disjoint finite sets of natural numbers and the cardinalities of $F_n$ increase to $\infty$. @Bill Johnson. Apr 25 '20 at 12:39
• Almost. In what you wrote $y_n$ should be replaced by $y_n/\|y_n\|$. Note that $\|y_n\| = (|F_n|^2 + |F_n|)^{1/2}$. Apr 25 '20 at 15:16