Is there a non-reflexive Banach space which is strictly convex?

Question

I just come up with the fact that a space being strictly convex, does not implies it is reflexive (at least I never saw a proof of it).

How can one construct a example of a non-reflexive Banach space which is strictly convex?

Thank you

Answer 1

I think it should work:

One way is to start with these results

Lemma.Let $(K, d)$ be a compact metric space and $\{ t_n \} \subseteq K$ dense and countable. For all $x\in X$ set $\lVert x \rVert = \sqrt{ \lVert x \rVert_{C^0}^2 + \sum_n \frac{1}{2^n}\lvert x(t_n) \rvert^2}$. Then $\lVert \cdot \rVert_{C^0}$ and $\lVert \cdot \rVert$ are equivalent and $\lVert \cdot \rVert$ is a strictly convex norm.

Proposition.Let $X$ be a separable Banach space, $(B_{X'}, \sigma(X',X)) = ( K, d)$ a compact metrizable set. $T : X \to C(K)$ defined by $Tx(f) = \langle f, x \rangle_{X'X}$ is a linear isometry. Hence $X$ has an equivalent strictly convex norm.

Now choose $X$ such that there exists a bounded sequence $\{ x_n \} \subseteq B_X$ that does not admit a subsequence weakly convergent in $B_X$. Hence, by Kakutani's theorem, $X$ can't be reflexive. Such an example could be $X = C([0,1])$. In fact, it is a separable Banach space; $B_{X'}$ is compact in the $\sigma(X',X)$ topology by Banach-Alaouglu theorem. Moreover, $B_{X'}$ is metrizable in $\sigma(X', X)$ because $X$ is separable, but the sequence defined by $x_n(t) := t^n$ is not weakly convergent.

Notation. $B_{X'}$ is the unit ball in $X'$; $X'$ is the dual of $X$; $\sigma( X', X )$ is the weak-$\star$ topology of $X'$; $C(K)$ is the space of continuous functions defined on $K$.