Corresponding norm from a dual norm? Let $(X,N_1)$ be a Banach space (separable if necessary) and let $(X^*,N_1^*)$ be its dual space. Here $N_1^*$ denotes the classical dual norm associated to $N_1$.
Let $N_2^*$ be an equivalent norm to $N_1^*$ on $X^*$. Can we construct a norm $N_2$ on $X$ such that $N_2^*$ is the corresponding dual norm associated to $N_2$?
My idea is the following :
$$ N_2 (x) := \sup_{\substack{f \in X^* \\ N_2^* (f) \leq 1}} \vert f(x) \vert. $$
However, with such a definition of $N_2$, I do not manage to prove that $N_2^*$ is the dual norm associated to $N_2$ ...
Thank you for any help.
 A: The answer can be found in "Calculus without derivatives" of Prof. Jean-Paul Penot. I refer to Lemma 3.94 p.251.
The answer is the following : an equivalent norm on the dual $X^*$ is the dual norm of an equivalent norm on $X$ if and only if it is weak-* lower semicontinuous.
The major idea is the following. Let us consider the notations introduced in the above question. 
--> If $N_2^*$ is a norm on $X^*$ equivalent to $N_1^*$, then one can easily prove that $N_2$ is a norm equivalent to $N_1$. 
--> Now, let us denote by $\tilde{N}_2$ the dual norm associated to $N_2$. Our aim is to prove that $N_2^* = \tilde{N}_2$. As detailed by @anonymous, one can easily prove that $\tilde{N}_2 \leq N^*_2$. 
--> Now, let us assume by contradiction that there exists $f \in X^*$ such that $\tilde{N}_2 (f) < N^*_2 (f)$. Then, there exists $g \in X^*$ such that $0 < \tilde{N}_2 (g) \leq 1 < N^*_2 (g)$. Since $N^*_2$ is weak-* lower semicontinuous, its unit ball B is convex and closed in $X^*$ with the weak-* topology. As a consequence, from the Hahn-Banach separation theorem, we can separate $\{ g \}$ and $B$. Since the dual of $X^*$ (endowed with the weak-* topology) coincides with $X$, we obtain that there exists $x_0 \in X$ such that 
$$ f(x_0) \leq \alpha - \varepsilon < \alpha + \varepsilon \leq g(x_0)$$ for all $f \in B$. In particular, $x_0 \neq 0$. Then, from definition of $N_2$, we get that $$N_2 (x_0) < g(x_0) \leq \tilde{N}_2 (g_0) N_2 (x_0)$$ and we obtain that $\tilde{N}_2 (g_0) > 1$. This is a contradiction.
The conclusion follows.
