Let $X$ be a Banach space. For every $x\in X,$ the non-empty duality set $\mathcal{J}(x)$ is defined as:$$\mathcal{J}(x):= \left\{j(x) \in X': \langle x, j(x)\rangle = \|x\|^{2} = \|j(x)\|^{2} \right\}$$

where $X'$ is the dual of $X$. I want to ask if $\langle x, j(\alpha x)\rangle =\alpha\langle x, j(x)\rangle$ for all positive scalar $\alpha?$

  • $\begingroup$ It should follow from the fact that $\mathcal{J}(\alpha x) = \alpha \mathcal{J}(x)$ $\endgroup$ – Kore-N Jul 28 '16 at 17:38

This is trivial. From the definition, if $j(x) \in \mathcal J(x)$ you have $\langle x, j(x) \rangle = \|x\|^2$, so for $\alpha > 0$, $$\langle x, j(\alpha x) \rangle = \alpha^{-1} \langle \alpha x, j(\alpha x)\rangle = \alpha^{-1} \|\alpha x\|^2 = \alpha \|x\|^2 = \alpha \langle x, j(x)\rangle$$

BTW, it's true for all real scalars $\alpha$, not just positive ones.

What would be a bit less trivial would be $j(\alpha x) = \alpha j(x)$. The problem here is that you're pretending $j$ is a function when you haven't defined it as such: $j(x)$ could be any member of $\mathcal J(x)$, which might have cardinality $> 1$. But if you fix any choice of $j(x)$ for, say, $x$ in the unit sphere, then you can define $j(x) = \|x\| j(x/\|x\|)$ for $x \ne 0$, and then you will have $j(\alpha x) = \alpha j(x)$ for $\alpha > 0$.


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