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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.

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The answer is yes.

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.

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  • $\begingroup$ 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. $\endgroup$ – Bill Johnson Apr 23 at 14:46
  • $\begingroup$ 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. $\endgroup$ – Anupam Apr 24 at 13:31
  • $\begingroup$ 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 $(\ell^1, \|\cdot \|)$. $\endgroup$ – Bill Johnson Apr 24 at 17:54
  • $\begingroup$ 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. $\endgroup$ – Anupam Apr 25 at 12:39
  • $\begingroup$ Almost. In what you wrote $y_n$ should be replaced by $y_n/\|y_n\|$. Note that $\|y_n\| = (|F_n|^2 + |F_n|)^{1/2}$. $\endgroup$ – Bill Johnson Apr 25 at 15:16

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