3
$\begingroup$

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?

$\endgroup$

2 Answers 2

Reset to default
5
$\begingroup$

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

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

$\endgroup$
3
  • $\begingroup$ I have just found another solution posted as an answer below. Could you please have a check on my attempt? $\endgroup$ Commented Mar 15, 2022 at 9:24
  • 1
    $\begingroup$ BTW even for a Banach space $E$ there might not exist $y$ such that $<f,y>=\|f\|\cdot \|y\|\ne 0.$ $\endgroup$ Commented Mar 16, 2022 at 9:00
  • $\begingroup$ Sure, in particular, the linear functional attains its norm in reflexive spaces. $\endgroup$ Commented Mar 16, 2022 at 9:02
1
$\begingroup$

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., $\eta \ge \sqrt{\lambda}$.

$\endgroup$
1
  • 1
    $\begingroup$ A little hard to follow, but OK. $\endgroup$ Commented Mar 16, 2022 at 8:53

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.