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Definition. [LUR point and LUR space] Let $X$ be a normed space. A point $x_0$ on the unit sphere $S_X$ of $X$ (that is, the unit sphere centered at $0$ in $X$) is said to be \emph{locally uniformly convex} or \emph{LUR} (Locally Uniformly Rotund) if, for every $\varepsilon > 0$, there exists $\delta > 0$ such that for every $x \in S_X$, $$ \|x - x_0\|_X \geq \varepsilon \quad \Rightarrow \quad \left\| \frac{x + x_0}{2} \right\|_X \leq 1 - \delta. $$

If every point of $S_X$ is LUR, we simply say that $X$ is LUR.

I would like to know if the statement is true, as well as if its demonstration is correct.

Afirmation. Let $X$ be a Banach space and let $(y_n)_{n \in \mathbb{N}} \subseteq S_X$. Under these conditions, if $X$ is locally uniformly convex, then for every sequence $(x_n)_{n \in \mathbb{N}} \subseteq S_X$ such that $$ \lim_{n \to \infty} \left\| \frac{x_n + y_n}{2} \right\| = 1 \quad \Longrightarrow \quad \lim_{n \to \infty} \|x_n - y_n\| = 0. $$ Proof. Assume that $X$ is locally uniformly convex and let $(y_n)_{n \in \mathbb{N}} \subseteq S_X$. Suppose, by contradiction, that there exist $\varepsilon > 0$ and subsequences $(x_{n_k})_{k \in \mathbb{N}}$ of $(x_n)_{n \in \mathbb{N}}$ and $(y_{n_k})_{k \in \mathbb{N}}$ of $(y_n)_{n \in \mathbb{N}}$ in $X$ such that $$ \|x_{n_k}\| = \|y_{n_k}\| = 1 \quad \text{and} \quad 1 - \left\| \frac{x_{n_k} + y_{n_k}}{2} \right\| \to 0, \quad \text{but} \quad \|x_{n_k} - y_{n_k}\| \geq 2\varepsilon. $$ This cannot happen, because the fact that $\|x_{n_k} - y_{n_k}\| \geq 2\varepsilon$ and $1 - \left\| \frac{x_{n_k} + y_{n_k}}{2} \right\| \to 0$ contradicts the assumption that $X$ is locally uniformly convex. Therefore, $$ \lim_{n \to \infty} \|x_n - y_n\| = 0. $$

Thanks for the help!If anyone knows if another statement similar to this is worth it, I would be grateful!

vinipenalty27
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The characterization you are looking for is the following: $X$ is Locally Uniformly Convex if and only if for every $x \in S_X$ and any sequence $y_n \in S_X$, $$ \left\| \frac 12 (x+y_n) \right\| \to 1 \text{ implies that } \|x-y_n\|\to 0 .$$ The property you are trying to prove, namely, for any two sequences $x_n,y_n \in S_X$, $$ \left\| \frac 12 (x_n+y_n) \right\| \to 1 \text{ implies that } \|x_n-y_n\|\to 0 $$ holds if and only if $X$ is Uniformly Convex, which is a stronger notion of convexity than locally uniformly convex. I cant think of an easy example of a locally uniformly convex space that is not uniformly convex but such spaces most certainly exist.

In fact, any reflexive (or separable) space $X$ can be given an equivalent locally uniformly convex norm but not every reflexive space can be given an equivalent uniformly convex norm (in fact, such spaces that admit uniformly convex renormings are called super-reflexive).

Edit: I think the following is a classic example; see page 113 in Nonlinear Functional Analysis by Deimling. Take any sequence $p_n \in (1,\infty)$ and consider the $\ell^2$-direct sum of the sequence spaces $\ell^{p_n}$ $$X=\left( \bigoplus_{n=1}^\infty \ell^{p_n} \right)_{\ell^2} = \left \{(x_n) \in \prod_{n\ge1} \ell^{p_n} \colon \, \|(x_n)\| = \left(\sum_{n\ge1} \|x_n\|_{\ell^{p_n}} \right)^{1/2}<\infty \right\}.$$ Then $X$ is reflexive and locally uniformly convex. But it is unifomly convex if and only if $p_n \in [a,b]$ for some $1<a\le b<\infty$.

Evangelopoulos Foivos
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