Let $E$ be a separable Banach space with norm $\|\cdot\|$. The dual norm on $E^*$ is also denoted $\|\cdot\|$.
Let $(a_n) \subset B_E$ be a dense subset of $B_E$ with respect to the strong topology.
Let $(b_n) \subset B_{E^*}$ be a countable subset of $B_{E^*}$that is dense in $B_{E^*}$ for the $\text{weak}^*$ topology $\sigma(E^*, E)$.
Given $f \in E^*$, set$$\|f\|_1 = \left\{\|f\|^2 + \sum_{n = 1}^\infty |\langle f, a_n\rangle|^2\right\}^{1\over2}.$$Given $x \in E$, set$$\|x\|_2 = \left\{\|x\|_1^2 + \sum_{n = 1}^\infty {1\over{2^n}}|\langle b_n, x\rangle|^2\right\}^{1\over2},$$where $\|x\|_1 = \sup_{\|f\|_1 \le 1} \langle f, x\rangle$.
Why does the fact that the dual norm of $\|\cdot\|_2$ is strictly convex follow from results in Problem 4 in the textbook?
