Is it true that $\|f\|^2 = \sup_{x\in E} \left [ 2 \langle f, x \rangle - |x|^2 \right ]$? Let $(E, |\cdot|)$ be a normed linear space and $(E', \| \cdot \|)$ its dual. Then
$$
\|f\| := \sup_{x\in E} \frac{\langle f, x \rangle}{ |x|}, \quad \forall f \in E'.
$$
We have
$$
\left [ \frac{\langle f, x \rangle}{ |x|} \right ]^2 \ge 2 \langle f, x \rangle - |x|^2, \quad \forall f \in E', x\in E.
$$
It follows that
$$
\|f\|^2 \ge \sup_{x\in E} \left [ 2 \langle f, x \rangle - |x|^2 \right ].
$$

Is it true that
$$
\|f\|^2 = \sup_{x\in E} \left [ 2 \langle f, x \rangle - |x|^2 \right ]?
$$

The answer would be YES if $E$ is reflexive. What if we impose further that $E$ is strictly convex?

$2\langle f,x\rangle-\|x\|^2\le 2\|f\|\cdot \|x\|-\|x\|^2=\|f\|^2-(\|f\|-\|x\|)^2\le \|f\|^2.$
For any $r\in (0,1)$ take  $y_r\in E$  with $0\ne y_r$ and $\langle f,y_r\rangle\ge \|f\|\cdot \|y_r\|\cdot (1-r/2).$ Let $x_r=y_r\frac {\|f\|}{\|y_r\|}.$ Then $2\langle f,x_r\rangle-\|x_r\|^2\ge \|f\|^2(1-r).$
