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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?

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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.

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