Does every Banach space admit a continuous (not necessarily equivalent) strictly convex norm?

Question

Trying to find and answer to this question, I have encountered two more-studied problems.

The first is to find when a Banach space admits an equivalent uniformly convex norm. The answer is that for example separable spaces always do, but nonseparable spaces might not. This does not satisfy me because I only want the existence of a continuous strictly convex norm. I am happy for it to generate a strictly weaker topology.

I have also found this paper that says that is we are interested in uniform and not strict convexity, then even admitting a single uniformly convex function is enough to make the space equivalent to a uniformly convex one. However this is about uniform convexity which is stronger than strict convexity.

The second problem is about compact convex subsets of a Banach space admitting a continuous strictly convex function. Hervé proved this is the case iff the set is metrisable. The proof is Theorem I.4.3 of "Erik M.Alfsen Compact Convex Sets and Boundary Integrals". This does not sound useful to me either, because all the sets I'm interested in are both noncompact and metrisable.

Does anyone know whether every Banach space admits a continuous strictly convex norm? Or do I need to put on my learning goggles and give the paper a detailed read?

Answer 1

If $\|\cdot\|_1$ is a continuous strictly convex norm on $(X,\|\cdot\|_0)$, then $\|x\|_2=\|x\|_0 + \|x\|_1$ defines an strictly convex norm on $X$ equivalent to $\|\cdot\|_0$. Therefore, a space like $\ell_\infty/c_0$ that does not admit an equivalent strictly convex norm, it does not admit a continuous strictly convex norm.

J. Bourgain proved that $\ell_\infty/c_0$ does not admit an equivalent strictly convex norm here.

Answer 2

I think that any continuous norm on $l^{\infty} (\Gamma)$ with $\Gamma$ uncountable is not strictly convexifiable. This is a concequence of the following two facts.

Fact 1: The space $l^{\infty} (\Gamma)$ with $\Gamma$ uncountable is not strictly convexifiable.

Fact 2: If $X$ is a Banach space and $T:l^{\infty} (\Gamma) \rightarrow X $ (with $\Gamma $ uncountable ) is a bounded one to one operator then $X$ contains isomorphically a copy of $l^{\infty} (\Gamma)$ with $\Gamma$ uncountable.Hence $X$ is not strictly convexifiable.

Remark: There is an unclear point in this answer. This is the requirement that $X$ is a Banach space which could not be valued for continuous and non equivalent norms on $l^{\infty} (\Gamma)$.However I believe that the result is correct.

Edit : The non completeness of $l^{\infty} (\Gamma)$ with the new norm is not a problem. The exact conclusion in Fact 2 is that there exists an uncountable $ \Delta \subset \Gamma $ such that $T| l^{\infty} (\Delta)$ is an isomorphism.