If $X$ is a uniformly rotund space , then for any closed subspace $M$ of $X$, $X/M$ is uniformly rotund. Does this hold for a locally uniformly rotund space? That is if $X$ is locally uniformly rotund, is it true that $X/M$ is locally uniformly rotund? I couldn't prove it nor could I found a counter example to disprove this. Can anyone please help me out? 



Quotient space of a locally uniformly rotund space
Consider any set $\Gamma$ and the Banach space $X = \ell\_1(\Gamma)$. Then the norm $$\|x\|^2 = \|x\|\_{\ell\_1(\Gamma)}^2 + \|x\|\_{\ell\_2(\Gamma)}^2$$ is LUR (and equivalent to the original norm on $\ell\_1(\Gamma)$). Now pick a space $Y$ without a LUR renorming (for example, $\ell\_\infty$). If $\Gamma$ has cardinality continuum, then there exists a linear surjection $T\colon \ell\_1(\Gamma)\to \ell\_\infty$, so that we can take $M = \ker T$. If you work with the norm $\|\cdot\|$ on $X$, you may easily deduce that $X / M$ is not LUR.


With a little bit more care you may get a separable counterexample too (with $\Gamma = \mathbb N$).


