Functional Analysis – Strictly Convex Renorming of Banach Space

Question

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?

Answer 1

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

Let $(x_n)$ be a dense set in $X$. Use the Hahn-Banach theorem 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 $$X \ni x \mapsto ( \tfrac {1}{2^n} f_n(x))_{n=1}^\infty \in \ell_2.$$ Then $T$ is bounded and injective. Show that $|x|:= \|x\|+\|Tx\|$ is an equivalent strictly convex norm. Hint: Use the fact that $(X,|\cdot|)$ is strict convexity iff $|x+y|=|x|+|y|$ implies that $x=ay$ for some $a>0$.

For $\Gamma$ uncountable, $\ell_{\infty}(\Gamma)$ admits no strictly convex norm.

See Chapter II,7 in [1]

There exist reflexive, non separable Banach spaces that admit no uniformly convex norm

See [2] and [3]

$\textbf{Edit:}$ As noted by OP, although $\ell_\infty(\mathbb N)$ is not separable, it admits a strictly convex renorm. Day [4] showed that

If there exists an injective bounded linear operator $$T \colon X \to c_0(\Gamma)$$ for some set $\Gamma$, then $X$ admits a strictly convex renorm. 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 $$T((x_n)_{n=1}^\infty)=(\tfrac 1n x_n)_{n=1}^\infty$$

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If you are willing to drop uniform convexity for locally uniform convexity, i.e.,
$$ \| x_n\| = \|x\|=1 \text { and } \|x_n +x\| \to 2 ~ \text{ imply } ~ \|x_n-x\| \to 0$$ then things are much nicer:

Every separable Banach space admits an equivalent norm that is locally uniformly convex.

See Theorem 2.6 (i) in [1]. This does not extend to the non separable case, as shown by the example of $\ell_\infty(\Gamma)$ with $\Gamma$ uncountable.

In fact, Troyonski [5] showed that

If $X$ is reflexive, then both $X$ and $X^*$ admit an equivalent locally uniformly convex (and Frechet differentiable) norm.

[1] Smoothness and Renormings in Banach Spaces

[2] Reflexive Banach spaces not isomorphic to uniformly convex spaces

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

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

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