# Is the dual of a strictly convex Banach space strictly convex?

I'm doing Ex 3.27.4 in Brezis's book of Functional Analysis. The question leads to below statement for which I don't know if it's true or not.

Let $$(E, | \cdot |)$$ be a strictly convex Banach space. Then its dual $$(E', \| \cdot \|)$$ is also strictly convex.

Could it become true if we impose further that $$E$$ is separable?

This is not true in general. If $$E'$$ is strictly convex, then $$E$$ is smooth (i.e., for any nonzero $$x \in E$$, there exists unique $$x^* \in E'$$ with norm one, such that $$\langle x^*, x \rangle=||x||$$). So, if $$E$$ is not smooth then $$E'$$ is never strictly convex. To see that this is true, suppose that $$E$$ is not smooth. Hence, there exists $$x \in X$$ with norm one, and two distinct functionals $$x_1^*,x_2^* \in E'$$ with norm one such that $$\langle x^*_1,x\rangle=\langle x_2^*,x \rangle =1$$. This then implies that $$\tfrac 12 \langle x_1^*+x_2^* ,x \rangle =1,$$ and so $$|| \tfrac 12 (x_1^*+x_2^* )|| \ge 1.$$ But, it is also true that $$|| \tfrac 12 (x_1^* +x_2^*)|| \le 1$$, and so $$|| \tfrac 12 (x_1^* +x_2^*)|| =1$$. This contradicts the assumption that $$E'$$ is strictly convex.