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Banach space $X$ (or its norm) is said to be strictly convex if its unit sphere $S_X$ does not contain any nontrivial line segment. There is also stronger notion of uniform convexivity. We say that space $X$ is uniformly convex if for any $\varepsilon > 0$ there exists $\delta > 0$ such that for any $x,y \in S_X$ $$ \|x - y\| \geq \varepsilon \implies \Bigl \|\frac{x+y}{2} \Bigr \| \leq 1 - \delta. $$ It follows from parallelogram identity that every inner product space is uniformly convex. Moreover, it is known that $L_p$ spaces are uniformly convex for $1 < p < \infty$. On the other side there are classical spaces, such as $C[0,1], L_1, c_0, c$, which are not strictly convex (when equiped with their standard norms). Some of these spaces admit equivalent and strictly convex renorming. For example in $c_0$ space there is norm $$ \|x\|_{sc} = \sup_{n \in \mathbb{N}}|x_n| + \Bigl( \sum_{n=1}^{\infty} \frac{1}{2^n} |x_n|^2 \Bigr)^{\frac{1}{2}}$$ which is strictly convex and equivalent to the classical norm $\|x\| = \sup_{n \in \mathbb{N}} |x_n|$.

$\textbf{My questions}:$ Does every Banach space admit strictly convex (not necessarily equivalent) renorming? If not, is there some class of B spaces, for which such renorming exists? What about uniformly convex renorming?

1 Answers1

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Question 1. Does every Banach space admit a strictly convex equivalent norm?

Not in general. In fact, in [Chapter II.7, 1] it is shown that $$ \text{for $\Gamma$ an uncountable set, $\ell_{\infty}(\Gamma)$ admits no equivalent strictly convex norm.}$$ As noted by the OP, this fails for $\Gamma=\mathbb N$ since $\ell^\infty(\mathbb N)$ does admit an equivalent strictly convex norm. Day [3] showed that

Theorem. If there exists an injective bounded linear operator $$T \colon X \to c_0(\Gamma)$$ for some set $\Gamma$, then $X$ admits an equivalent strictly convex norm.

Here, $c_0(\Gamma)$ is the set of all bounded real functions $f$ such that for every $ε>0$ the set $(|f|>ε)$ is finite. $\ell_\infty(\mathbb N)$ falls into that category with $\Gamma = \mathbb N,$ $c_0(\mathbb N) = \{ (x_n) \colon x_n \to 0\}$ and $$T((x_n)_{n=1}^\infty)=(\tfrac 1n x_n)_{n=1}^\infty.$$

However, for separable spaces this is indeed the case:

Fact. Any separable Banach space $X$ has an equivalent strictly convex norm.

This does not extend to the non separable case, as shown by the example of $\ell_\infty(\Gamma)$ with $\Gamma$ uncountable.

The proof is quite simple (compare with your example in $c_0$): Let $(x_n)$ be a dense set in $X$. Use Hahn-Banach to find $f_n \in X^*$ with norm one such that $f_n(x) =\|x_n\|$. Let $T \colon X \to \ell_2$ be given by $$T(x) = \left( \tfrac {1}{2^n} f_n(x) \right)_{n=1}^\infty \in \ell_2, \quad x \in X.$$ Then $T$ is bounded and injective. Then $|x|:= \|x\|+\|Tx\|$ is an equivalent strictly convex norm. Hint: To see this, one can use the fact that $(X,|\cdot|)$ is strict convexity iff $|x+y|=|x|+|y|$ implies that $x=ay$ for some $a>0$.

In fact, much more can be said in the separable case:

Theorem (7.1 in [1]). Any separable Banach space admits an equivalent norm that (among other things) is Locally Uniformly Convex (LUC) i.e.,
$$ \| x_n\| = \|x\|=1 \text { and } \|x_n +x\| \to 2 ~ \text{ imply } ~ \|x_n-x\| \to 0.$$ In the (stronger) case that $X^*$ is separable then $X$ admits an equivalent norm such that both $X$ and $X^*$ are LUC.

and in the reflexive case:

Theorem (Troyanski [4]). If $X$ is reflexive, then both $X$ and $X^*$ admit an equivalent LUC (and Frechet differentiable) norm.


Question 2. Does every Banach space admit an equivalent uniformly convex (UC) norm?

Again, the answer is no. In fact, there is a characterization of such spaces proved by Enflo [2]: they are precisely the super-reflexive spaces. Super-reflexivity is a stronger notion of reflexivity that, essentially, means that every space $E$ that is "locally represented" in $X$ must be reflexive (one can take $E=X^{**}$ to infer that super-reflexivity implies reflexivity). See this for an explicit example of a reflexive space which is LUC but not UC.

[1] Smoothness and Renormings in Banach Spaces

[2] Banach spaces which can be given an equivalent uniformly convex norm

[3] Day, M. M. (1955). Strict Convexity and Smoothness of Normed Spaces. Transactions of the American Mathematical Society, 78(2), 516–528.

[4] S . L . T ROYANSKI , ‘On locally uniformly convex and dif ferentiable norms in certain non-separable Banach spaces’, Studia Math . 37 (1971) 173 – 180 .

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    Thank you very much. This is really great stuff. It is interesting that even though $\ell_{\infty}(\mathbb{N})$ is not separable, it admits strictly convex equivalent norm (given by the same formula as the one from my example in space $c_0$). I wonder if there is some generalization of the first statement that you quoted. –  Oct 31 '22 at 13:15
  • That is interesting indeed! Unfortunately, I am not aware of any generalizations at the moment – Evangelopoulos Foivos Oct 31 '22 at 13:23
  • @triplebackflip See my updated answer – Evangelopoulos Foivos Nov 10 '22 at 12:06