I want to find an example of a non-separable normed space $X$ and a closed subspace $M$ of $X$ such that $X/M$ is separable.
The first think came to my mind is $\ell^{\infty}$. Can I find a closed subspace $M$ of $\ell^{\infty}$ so that $\ell^{\infty}/M$ is isometrically isomorphic to $\ell^1$? Can I think of $M$ to be kernal of any continuous linear operator from $\ell^{\infty}$ onto $\ell^{1}$?
Please help me out.
$\ell_1$ is not isomorphic to a quotient of $\ell_\infty$. Indeed, the latter one is a Grothendieck space, hence so is every its quotient. By the Eberlein-Smulyan theorem, a separable space is Grothendieck if and only if it is reflexive. Certainly, $\ell_1$ is not reflexive.
As a matter of fact, $\ell_p$ is a quotient if and only if $p\geqslant 2$. To see this, take a sequence of independent $p$-random variables in $L_1$ ($p\leqslant 2$). They span an isometric copy of $\ell_p$ in $L_1$ so $\ell_{q}$, where $q$ is the conjugate exponent to $p$, is a quotient of $L_\infty$, the dual of $L_1$. You may now embed $L_\infty$ into $\ell_\infty$ using the fact that $L_1$ is a quotient of $\ell_1$. As $L_\infty$ is injective, there is a projection from $\ell_\infty$ onto every copy of $L_\infty$, so you may compose any such projection with a quotient from $L_\infty$ onto $\ell_q$
To see that $\ell_p$ for $p<2$ is not a quotient of $\ell_\infty$, note that $\ell_p$ is a quotient of $\ell_\infty$, $(\ell_p)^*$ embeds into $(\ell_\infty)^*$, which has cotype 2. Thus, the conjugate exponent of $p$ must be at most 2.
