I'm doing Ex 1.1.3 in Brezis's book of Functional Analysis. Could you have a check on my attempt?
Let $E$ be a n.v.s. and $E'$ its dual. The duality map $F$ is defined for every $x \in E$ by
$$
F(x) = \left\{f \in E' \mid \|f\| = |x| \text { and } \langle f, x\rangle = |x|^{2} \right\}.
$$
Then $F(x) = \left\{f \in E' \mid \|f\| \le |x| \text { and } \langle f, x\rangle = |x|^{2} \right\}$. Prove that
$$
F(x) = \{f \in E' \mid \frac{1}{2} |y|^{2} – \frac{1}{2} |x|^{2} \ge \langle f, y-x\rangle \quad \forall y \in E\}.
$$
I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.
Best Answer
Let $f \in E'$ such that $\|f\| = |x|$ and $\langle f, x\rangle=\|x\|^{2}$.
Then $\langle f, y-x\rangle = \langle f, y\rangle - |x|^2 \le \|f\| |y| - |x|^2 = |x||y| - |x|^2$.
Also, $|x||y| - |x|^2 \le \frac{1}{2} |y|^{2} - \frac{1}{2} |x|^{2}$. This implies $\langle f, y-x\rangle \le \frac{1}{2} |y|^{2} - \frac{1}{2} |x|^{2}$.
Let $f \in E'$ such that $\frac{1}{2} |y|^{2} - \frac{1}{2} |x|^{2} \ge \langle f, y-x\rangle$ for all $y \in E$.
It follows that $\frac{1}{2} |kx|^{2} - \frac{1}{2} |x|^{2} \ge \langle f, kx-x\rangle$ and thus $\frac{(k-1)(k+1)}{2} |x|^2 \ge (k-1) \langle f, x \rangle$ for all $k$. With $k>1$, we get $\frac{(k+1)}{2} |x|^2 \ge \langle f, x \rangle$ and thus $|x|^2 \ge \langle f, x \rangle$. With $k<1$, we get $\frac{(k+1)}{2} |x|^2 \le \langle f, x \rangle$ and thus $|x|^2 \le \langle f, x \rangle$. Hence $|x|^2 = \langle f, x \rangle$.
Now we have $\frac{1}{2} |y|^{2}+ \frac{1}{2} |x|^{2} \ge \langle f, y \rangle$ for all $y \in E$. This implies $$ |x|^2 \ge \sup_{y \in E} [2 \langle f, y \rangle - |y|^2] = \|f\|^2. $$
I use the characterization proved in this thread.