1
$\begingroup$

Let $E$ be a normed vector space. Prove that the duality map $$F(x) = \{f \in E^*; \|f\| = \|x\| \text{ and } \langle f, x\rangle = \|x\|^2\}$$ is equal to $$\tilde{F}(x) = \{f \in E^* : \frac{1}{2}\|y\|^2 - \frac{1}{2}\|x\|^2 \geq \langle f ,y-x\rangle \quad\forall y \in E\}.$$ Suppose $f \in F(x)$ and $y \in E$, then $$\langle f ,y-x \rangle \leq \|x\|\|y\|| - \|x\|^2$$ and so $$\frac{1}{2}\|y\|^2 - \frac{1}{2}\|x\|^2 - \langle f ,y-x\rangle \geq \frac{1}{2}\|y\|^2 - \frac{1}{2}\|x\|^2 -\|x\|\|y\| + \|x\|^2 = \frac{1}{2}(\|y\|-\|x\|)^2 \geq 0$$

which proves $F(x) \subset \tilde{F}(x)$.

I'm stuck on showing $\tilde{F}(x) \subset F(x)$. Suppose $f \in \tilde{F}(x)$, then I tried choosing $y = 0$ $$\frac{1}{2}\|x\|^2 \leq \langle f, x\rangle$$ but this didn't lead anywhere. How can I show that $\|f\| = \|x\|$ and $\langle f, x\rangle = \|x\|^2$?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Given $f \in \tilde F(x)$ you have that $$f(y-x) \le \frac 12 (\|y\|^2-\|x\|^2), \quad \forall y \in E.$$ For arbitrary $z\in E$ and $\varepsilon>0$ taking $ y = x + \varepsilon z $ gives that \begin{align*} f(z) &\le \tfrac 12 (\|εz+x\|^2-\|x\|^2) \\ & \le \tfrac 12 (\|εz\|^2+\|x\|^2 + 2 \|εz\| \|x\| - \|x\|^2) \\ &= \tfrac 12 ε^2 \|z\|^2 + ε\|z\| \|x\|. \end{align*} Dividing by $ε>0$ and then taking $\epsilon \to 0$ gives $f(z) \le \|z\| \|x\|$ and since $z$ is arbitrary, $$ \|f\| \le \|x\| \quad \text{ and } \quad f(x) \le \|x\|^2.$$ Now, taking $y=λx$ with $λ<1$ gives $(λ-1)f(x) \le \tfrac 12 \|x\|^2 (λ^2-1)$ or equivalently, $$ f(x) \ge \tfrac 12 \|x\|^2 (λ+1) .$$ Taking $λ\to 1^-$ gives $$f(x) \ge \|x\|^2$$ and we are done.

$\endgroup$

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.