We know that if X is a uniformly rotund space then for every closed subspace M of X, X/M is uniformly convex. Does the similar assertion true for a locally uniformly rotund space ? If not is there any known sufficient conditio for that?
The answer is negative. Take as $X$ the space $\ell_1$ in the LUR renorming $\|x\| = \sqrt{\|x\|_1^2 + \|x\|_2^2}$ where $ \|x\|_p$ denotes the canonical $\ell_p$ norm. This $X$ possesses quotient spaces that are not strictly convex. With some additional effort one can demonstrate that $X$ has a quotient isometric to $\ell_1$ in canonical norm, which implies that $X$ has as its quotients all separable Banach spaces.
I've answered the same question here.
https://mathoverflow.net/questions/344626/quotient-space-of-a-locally-uniformly-rotund-space
It would be good to acknowledge that you asked the same question elsewhere.
Dear Tomasz Kania , the norm $\|x\|' = \|x\|_1 + \|x\|_2$ from your answer on MathOverflow is strictly convex, but is not LUR. To get a LUR norm one needs,
for example, $\|x\| = \sqrt{\|x\|_1^2 + \|x\|_2^2}$ like in my answer.
The reference why it is LUR:
Lemma 13.26 from the book "Banach Space Theory. The Basis for Linear and Nonlinear Analysis" by Fabian, Habala, Hájek, Montesinos and Zizler.
Of course, yes you are right. I had in mind exactly this excerpt from this very book. Corrected, thank you.
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