A non-separable strictly convex space with a separable pre-dual I am asked to find an example of a non-separable strictly convex space with a separable pre-dual.
So please give me some hints or some references about such problems.
Thanks in advance.
 A: Clarkson (1936) was the first to prove that if a normed space $X$ is separable, then $X$ is isomorphic to a strictly convex space. We will use the following, which can be found in the paper of Day mentioned below and which follows immediately from Clarkson's work.

Fact: if $X$ is separable, then $X^*$ is isomorphic to a strictly convex space.

Sketch: the unit sphere of $X$ is separable. So take a countable dense set $\{x_n\,;\,n\geq 1\}$ in the latter. Now consider for every $x^*\in X^*$ the sequence $Tx^*:=\left(\frac{x^*(x_n)}{2^n} \right)_{n\geq 1}$. This yields a linear injection into the strictly convex space $\ell^2$. The new norm
$$
\|x^*\|_{sc}:=\|x^*\|+\|Tx^*\|_2
$$
is equivalent to the dual norm $\|x^*\|=\sup_{\|x\|\leq 1}|x^*(x)|$ and it is strictly convex. $\Box$
Remark: then it suffices to renorm $X$ by $\|x\|:=\sup_{\|x^*\|_{sc}\leq 1}|x^*(x)|$ to get an equivalent norm on $X$ for which the dual is $X^*$ with this strictly convex norm. As pointed out by Martin, Klee expanded on Clarkson's idea to show that all this could be done in such a way that $X$ be smooth.

Application: $\ell^\infty$ is neither separable nor strictly convex. But it is the dual of the separable space $\ell^1$. So by the fact above, $\ell^\infty$ can be renormed into a strictly convex space. If you want a noncommutative example, take $B(H)$ and renorm it as the dual of $S_1(H)$ the separable space of trace-class operators.

References: see the free access article by Day for details and more on strict convexity and smoothness in normed spaces. See the appendix of this  other free access paper of Klee (thanks Martin again) for more details.
