I'm doing Ex 3.27 in Brezis's book of Functional Analysis.
Let $(E, | \cdot |)$ be a separable Banach space and $(E', \| \cdot \|)$ its dual space. Let $(a_n)$ be a countable dense subset of the closed unit ball $B_E$ in norm topology. Let $(b_n)$ be a countable dense subset of the closed unit ball $B_{E'}$ in weak$^\star$ topology. Construct an equivalent norm $|\cdot|_2$ on $E$ that is strictly convex and whose dual norm $\| \cdot \|_2$ is also strictly convex.
Below is my attempt:
Consider $\varphi:E \to \ell^2, x \mapsto (2^{-n/2} \langle b_n, x \rangle)_n$. Then $\varphi$ is linear injective. We define a graph norm $| \cdot|_1$ by $|x|_1 := \sqrt{|x|^2 + |\varphi(x)|^2_{\ell^2}}$. We can verify that $|\cdot|_1$ is strictly convex and that $|x| \le |x|_1 \le \sqrt{2} |x|$. Let $\| \cdot\|_1$ be the dual norm of $|\cdot|_1$. It follows that $\frac{1}{\sqrt 2} \|f\| \le\| f \|_1 \le \|f\|$.
Consider $\psi:E' \to \ell^2, f \mapsto (2^{-n/2} \langle f, a_n \rangle)_n$. Then $\psi$ is linear injective. We define a graph norm $\| \cdot \|_1$ by $\| f \|_2 := \sqrt{\|f\|_1^2 + |\psi(f)|^2_{\ell^2}}$. Similar to above, we get $\|\cdot\|_2$ is strictly convex and $\frac{1}{\sqrt 2} \|f\| \le \|f\|_2 \le \sqrt 2 \|f\|$. Let $| \cdot|_2$ be the dual norm of $\|\cdot\|_2$, i.e., $$ | x |_2 := \sup_{\|f\|_2 = 1} \langle f, x \rangle, \quad \forall x \in E. $$
It follows that $\frac{1}{\sqrt 2} |x| \le |x|_2 \le \sqrt 2 |x|$.
Because the construction of $| \cdot |_2$ involves $| \cdot|_1$ which is strictly convex, I guess that $| \cdot |_2$ inherits this property. However, I'm unable to prove it. Could you please leave me some hints to finish this exercise?