This question is related to an intermediate step in the proof of Exercise 3.27 Subproblem 4 of [1,Brezis].
Let $(E, \|\cdot\|_E)$ be a separable Banach space that is not necessarily reflexive. Its dual space is denoted as $(E^\star, \|\cdot\|_{E^\star})$. Consider the function $\phi: E \to \mathbb{R}$ defined as $\phi(x) = \frac{1}{2}\|x\|^2_E$. Its conjugate $\phi^*: E^\star \to \mathbb{R}$ is then given by $$\phi^*(f) = \sup_{x \in E}\Big\{\langle f, x\rangle - \phi(x) \Big\}.$$
If E is reflexive, by Holder's inequality and Corollary 1.3 of [1,Brezis], we can conclude that $\frac{1}{2}\|f\|^2_{E^\star} = \phi^*(f)$. [2]
Question 1. If E is not reflexive (as assumed above), will the equality $\frac{1}{2}\|f\|^2_{E^\star} = \phi^*(f)$ still hold? (My personal guess to this question is No. )
Question 2. Consider $\|\cdot\|_E$ takes the following specific form.
Let $(b_n) \subseteq \mathbb{B}_{E^\star}$ be a countable subset of $\mathbb{B}_{E^\star}$ that is dense in $\mathbb{B}_{E^\star}$ for the weak$^\star$ topology $\sigma(E^\star, E)$. The norm $\|x\|_E$ is defined as $$\|x\|_E^2 = \sum_{n=1}^{\infty}\frac{1}{2^n}\big| \langle b_n, x\rangle \big|^2 = 2\phi(x). $$
Under such setup, will the equality $\frac{1}{2}\|f\|^2_{E^\star} = \phi^*(f)$ still hold? (This answer should be affirmative, as it is a result used in the proof of Exercise 3.27 Subproblem 4 of [1,Brezis]).
My attempt to prove this equality is:
For any $x \in E$, let $\iota_x: E^\star \to \mathbb{R}$ denote the evaluation map, i.e., $\iota_x(f) = f(x), \forall f \in E^\star$. Additionally, let $J: E \to E^{\star\star}$ denote the canonical embedding from $E$ to its bidual $E^{\star\star}$.
\begin{align*} \phi^*(f) &= \sup_{x \in E}\Big\{\langle f, x\rangle - \phi(x)\Big\} = \sup_{x \in E} \Big\{ \langle \iota_{x}, f\rangle - \phi(\iota_x)\Big\} \overset{(b)}{\leq} \sup_{\xi \in E^{\star\star}} \Big\{ \langle \xi, f\rangle - \frac{1}{2}\|\xi\|^2_{E^{\star\star}}\Big\} \\ & \overset{(a)}{\leq} \sup_{\xi \in E^{\star\star}} \Big\{ \|\xi\|_{E^{\star\star}} \cdot \|f\|_{E^\star} - \frac{1}{2}\|\xi\|^2_{E^{\star\star}}\Big\} = \frac{1}{2} \|f\|^2_{E^\star}, \end{align*} where the equality in $(a)$ can be achieved using the similar arguments for the reflexive case. However, I have some difficult time in proving the equality in $(b)$ can be achieved. I understand that for a fixed $\xi \in E^{\star\star}$, we can approximate $\langle \xi, f\rangle - \phi(\xi)$ by a sequence of evaluation maps $(\iota_{x_n})_{n \in \mathbb{N}}$, where with some abuse of notation, $\phi(\xi) := \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2^n}\big| \langle \xi, b_n\rangle \big|^2$. In other words, $\langle \xi, f\rangle - \phi(\xi)$ can be approximated by $\big( \langle \iota_{x_n}, f\rangle - \phi(\iota_{x_n}) \big)_{n \in \mathbb{N}}$, since $J(E)$ is dense in $E^{\star\star}$ in the weak$^\star$ topology $\sigma(E^{\star\star}, E^\star)$ [3]. Two gaps remain:
(1) Do we expect that same argument holds when the R.H.S. of $(b)$ is $\sup_{\xi \in E^{\star\star}}$, i.e., the statement that "fix $\xi \in E^{\star\star}$" may not be feasible in the first place.
(2) Regarding the approximation above, a gap to be filled is $\phi(\xi)$ and $\frac{1}{2}\|\xi\|^2_{E^{\star\star}}$. When $\xi \in J(E)$, then we have $\phi(\xi) = \frac{1}{2}\|\xi\|^2_{E^{\star\star}}$. Yet, I have some difficulty in figuring out what will happen when $\xi \in E^{\star\star}\backslash J(E)$ based on the definition of bidual norm $\|\xi\|_{E^{\star\star}} = \sup_{\|f\|_{E^\star} = 1} \big| \langle \xi, f\rangle \big|$.
[1] H. Brézis, Functional analysis, Sobolev spaces and partial differential equations. Springer,2011, vol. 2, no. 3.
[2] Fenchel Conjugate of a norm squared
[3] Relationship between a Banach Space $X$ and its bidual $X^{**}$
