I first want to recall the moduli of uniform smoothness (US), uniform convexity (UC), asymptotic uniform smoothness (AUS), and asymptotic uniform convexity (AUC). Throughout, let $X$ be an infinite dimensional Banach space and let $\text{codim}(X)$ denote the set of (closed) subspaces $Y$ of $X$ such that $\dim X/Y<\infty$. Fix $\sigma,\tau>0$.
$$\rho(\sigma)=\sup\Bigl\{\frac{\|x+\sigma y\|+\|x-\sigma y\|}{2}-1:\|x\|,\|y\|=1\Bigr\}$$
$$\delta(\tau)=\inf\Bigl\{1-\Bigl\|\frac{x+y}{2}\Bigr\|:x,y\in B_X, \|x-y\|\geqslant \tau\Bigr\}$$
$$\overline{\rho}(\sigma)=\underset{x\in S_X}{\sup}\underset{Y\in \text{codim}(X)}{\inf}\underset{y\in B_Y}{\sup} \|x+\sigma y\|-1$$
$$\overline{\delta}(\tau)=\underset{x\in S_X}{\inf}\underset{Y\in \text{codim}(X)}{\sup}\underset{y\in S_Y}{\inf}\|x+\tau y\|-1.$$
The moduli are related through the usual Fenchel duality (that is, each is equivalent but in general not equal to the Fenchel dual of the other). Being US is equivalent to $\inf_{\sigma>0}\rho(\sigma)/\sigma=0$ and being UC is equivalent to $\delta(\tau)>0$ for all $\tau>0$. Similarly, being AUS is the condition that $\inf_{\sigma>0}\overline{\rho}(\sigma)/\sigma=0$, and being AUC is equivalent to $\overline{\delta}(\tau)>0$ for all $\tau>0$.
Of course, it is known that a Banach space $X$ admits an equivalent UC (or US) norm iff it is superreflexive (and in this case it actually admits a single norm which has both properties). This was shown first by Enflo, and then by Pisier. This, of course, implies reflexivity, so there is no point distinguishing between weak and weak$^*$ topologies in this case. Generalizing and blurring the proofs between US and UC as well as the works of Enflo and Pisier, we first imagine the situation in which we define an equivalent ecart on $X$. That is, a positive-homogeneous functional $|\cdot|:X\to [0,\infty)$ which satisfies $a|x|\leqslant \|x\|\leqslant b|x|$ for all $x\in X$ and some $a,b>0$ (independent of $x$), but not necessarily satisfying the triangle inequality, which also satisfies inequalities related to US/UC. For example, it may be the case that, somehow, for some $c>0$ and $q\in [2,\infty)$, for every $0<\tau<1$, the ecart satisfies $$\Bigl|\frac{x+y}{2}\Bigr|^q \leqslant 1-c\tau^q$$ for any $x,y$ with $|x|,|y|\leqslant 1$ and $|x-y|\geqslant \tau$. Then we can define $K=\{x\in X:|x||\leqslant 1\}$ and take as our equivalent norm the Minkowski functional $\mu$ of the closed, convex hull $\overline{\text{co}}(K)$ of $K$.
If one is trying to prove the existence of an equivalent US/UC norm, one must ask whether the good US/UC property of the ecart is retained by $\mu$ (with possibly slightly worse constants/quantifications, ie replacing the constant $c$ above for the ecart with $c/2$ for $\mu$).
For smoothness properties, it is the case that the smoothness inequality, if anything, gets better and not worse, when passing from $|\cdot|$ to $\mu$. And this seems quite clear to me geometrically that convexifying smooths things out. However, for convexity, the situation seems to be quite different. Indeed, a huge computational difficulty of Enflo's work (in my opinion, the hardest part) was proving that the UC inequality is retained (with much worse constants) by $\mu$.
How did Pisier get around this? By not having to. Pisier's inequality on the ecart actually implies that it is already a norm.
I want to understand the geometry underlying this phenomenon. To me, Pisier's proof does the best job of getting the underlying geometry across, but not in a way that carries over. The underlying phenomenon from Enflo's argument is, to me, impenetrable. It also is quantitatively suboptimal.
The reason I would like to understand the geometry is because the question of the existence of equivalent AUC norms is still open. The existence of equivalent AUS norms is fully solved, even up to the optimal power type, and the proof does involve first constructing an ecart such that $$\underset{\lambda}{\lim\sup}|x+\sigma y_\lambda|\leqslant 1+C\sigma^p$$ for some $C>0$ some $1<p<\infty$, every $\sigma>0$, every $x$ with $|x|\leqslant 1$, and every weakly null net $(y_\lambda)$ with $\sup_\lambda \|y_\lambda\|\leqslant 1$. This property is clearly retained after convexifying. Something even better can be done for AUF spaces like $c_0$.
But neither Pisier's magic, nor Enflo's computations, nor the niceness of the AUS case carry over to the AUC situation. Our starting point is an ecart equivalent to the original norm having the property that for some $1\leqslant q<\infty$, some $c>0$, it holds that $$\underset{\lambda}{\lim\sup}|x+\tau y_\lambda|^q \geqslant 1+c\tau^q$$ for every $\tau>0$, every $x\in K:=\{y\in X:|y|=1\}$, and every weakly null net $(y_\lambda)$ in $X$ with $\inf_\lambda \|y_\lambda\|\geqslant 1$. In this case, we would like to know that the inequality holds for all $x$ in the closed, convex hull of $K$ rather than just $K$, possibly with a smaller value of $c$.
The ideal response to this question would be an explanation for how we can actually fill in this gap. An acceptable response would be some explanation of the underlying geometry of why convexifying does not kill uniform convexity.
