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?
1 Answer
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 $\Gamma = \mathbb N$).
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$\begingroup$ Suppose $\Gamma=\mathbb{N}$. Let $X$ be the space $\ell^{1}(\mathbb{N})$ with the LUR renorming as suggested by you. Is it possible to show that $\ell^{1}(\mathbb{N})$ with the canonical norm is linearly isometric to some quotient of $X$? I am not getting the proper function. @ Tomek Kania $\endgroup$– AnupamCommented Apr 20, 2020 at 5:54