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

Question

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?

Answer 1

$2\langle f,x\rangle-\|x\|^2\le 2\|f\|\cdot \|x\|-\|x\|^2=\|f\|^2-(\|f\|-\|x\|)^2\le \|f\|^2.$

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 $2\langle f,x_r\rangle-\|x_r\|^2\ge \|f\|^2(1-r).$

Answer 2

To prove the other direction, we fix $\lambda \in (0, 1)$ and will look for some $x\in E$ such that $2 \langle f, x \rangle - |x|^2 \ge \lambda \|f\|^2$. Given $\eta \in (0 , 1)$, there is $x\in E$ such that $\frac{\langle f, x \rangle}{|x|} \ge \eta \|f\|$. The problem then boils down to find $\eta$ such that $2 \eta\|f\| |x| - |x|^2 \ge \lambda \|f\|^2$ or equivalently $|x|^2 - 2 \eta\|f\||x| + \lambda \|f\|^2 \le 0$. We have $\Delta' = (\eta\|f\|)^2 - \lambda \|f\|^2 = (\eta^2-\lambda)\|f\|^2$. For the inequality to have solutions, we just pick $\eta$ such that $\Delta' \ge 0$, i.e., $\eta \ge \sqrt{\lambda}$.