Does there exists an example of a Banach space that is compactly LUR; but not LUR

Question

We know that a Banach space $X$ is locally uniformly rotund(LUR) if for $x, x_n \in S_X$ with $\Vert x+x_n \Vert \to 2$, we have $x_n \to x$. In the same case, if $(x_n)$ has a convergent subsequence, we say that $X$ is compactly LUR. Here $S_X$ is the unit sphere of $X$. It may be noted that every LUR Banach space is compactly LUR, but not conversely. Also, if $X$ is rotund and compactly LUR, then it is LUR. However, I am having trouble finding an example of a space that is CLUR but not LUR. Please guide me. Thank you.

Answer 1

Definition A Banach space $X$ is strictly convex if for all $x,y\in S_X$ with $\|x+y\|=2$ we have that $x=y$.

Proposition A Banach space $X$ is LUR iff it is CLUR and strictly convex.

Proof: Let $X$ be LUR. Then it is CLUR. Also, take any $x,y\in S_X$ with $\|x+y\|=2$. Then $\lim_{n\to\infty} \|x+y\|=2$. By LUR we have $\lim_{n\to\infty} y=x$. Therefor $X$ is strictly convex.

Conversely, let $X$ be CLUR and strictly convex. Take any sequence $x_n$ so that $\|x_n+x\|\to 2.$ Then every subsequence $x_{n_k}$ satisfies $\|x_{n_k}+x\|\to 2.$ By CLUR there is a further subsequence $x_{n_{k_j}}$ so that $x_{n_{k_j}}\to \ell$ for some $\ell\in X$. By continuity of norm, we have that $\|\ell+x\|=2$ and so $\ell=x$. Since every subsequence has a further subsequence that converges to $x$, $x_n\to x$.

So we just need to find an example of a CLUR space that is not strictly convex. One such example is the two dimensional space $\mathbb R^2$ equipped with the $L^\infty$ norm. It is CLUR because given any $x_n\in S_X$ at all there is a convergent subsequence by Bolzano-Weierstrass. However let $x_n:=((-1)^n,1)$ and $x=(0,1)$. In this case, $\|x_n+x\|=\|((-1)^n,2)\|=2$ but $x_n$ is not convergent.