Reference for $X' = \{ \lambda ||\cdot||'(x) : \lambda \in \mathbb R,\quad x \in X \setminus \{ 0 \} \}$

Question

I would like to find a reference for the equality $$ X' = \{ \lambda ||\cdot||'(x) : \lambda \in \mathbb R,\quad x \in X \setminus \{ 0 \} \} $$ when $X$ is a real uniformly convex Banach space whose norm is differentiable outside of $0$. I came to this result reading Villani's lecture notes on integration.

Answer 1

Taking the idea of @gerw here is my solution. Taking the usefull informations of @Jochen I give two distinct solutions.

Suppose $X$ is reflexive and not $\{ 0 \}$, so that its unit sphere $\mathbb S_X$ is non void and its unit closed ball $\overline{\mathbb B}_X$ is weakly quasi compact by Kakutani. Also suppose that the norm on $X$ is Gâteaux differentiable on $X \setminus \{ 0 \}$.

Sequential solution : By Krein Smulian $\overline{\mathbb B}_X$ is in fact sequentially compact. Let $x^* \in X' \setminus \{ 0 \}$, let $(x_j)$ be a minimizing sequence of $x^*$ on $\mathbb S_X \subset \overline{\mathbb B}_X$. By sequential compactness let $\sigma$ extraction and $x_\infty \in \overline{\mathbb B}_X$ such that $(x_{\sigma(j)})$ goes weakly to $x_\infty$. Observe that since $x^* \neq 0$ and since $x^*$ is weakly continuous $$ 0 > \inf_{\mathbb S_X} x^* = \lim_j \langle x^* , x_{\sigma(j)} \rangle = \langle x^* , x_\infty \rangle $$ so that $x_\infty$ is non zero. Then the optimality property of $x_\infty$ gives $$ \frac{1}{||x_\infty||} \langle x^* , x_\infty \rangle = \left\langle x^* , \frac{x_\infty}{||x_\infty||} \right\rangle \geq \langle x^* , x_\infty \rangle $$ so that $x_\infty \in \mathbb S_X$. Observe that then $x_\infty$ is a minimizer of $\varphi(x) = \frac{\langle x^*,x \rangle}{||x||}$ on $\mathbb B(x_\infty, 1)$ so that $$ 0 = \varphi'(x_\infty) = x^* - \langle x^* x_\infty \rangle ||\cdot ||'(x_\infty). $$

Topological solution : The map $x^*$ defined on $\overline{\mathbb B}_X$ endowed of the weak topology of $X$ is continuous on a non empty quasi compact space so it has a minimizer $x_\infty$. The rest of the proof is as above.