Let $X$ be a Banach space. For every $x\in X,$ the non-empty dual 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 $X$. From this definition, what's the properties of $j$? it seems that is an isometry, but can I say that $j^{*}j = I $ (the identity) or $j^{*}j = j$?

  • $\begingroup$ Related. Note that the mapping is typically nonlinear. It reduces to the Riesz isomorphism when $X$ is Hilbert, but I suspect that it is never linear on non-Hilbert spaces. $\endgroup$ – Giuseppe Negro May 24 '16 at 15:24
  • $\begingroup$ In Hilbert space $j(x) = x$. $\endgroup$ – salma May 24 '16 at 15:26
  • $\begingroup$ Yes, that's what I said above speaking of Riesz isomorphism. What I mean is that $j$ is typically nonlinear, so what do you mean with $j^\star$? $\endgroup$ – Giuseppe Negro May 24 '16 at 15:29
  • 1
    $\begingroup$ Is $x \mapsto j(x)$ acutally a mapping? It seems to me, from this definition, that $j(x)$ is not necessarily uniquely determined, especially in view of the related question and answer linked by Guiseppe Negro. $\endgroup$ – Roland May 24 '16 at 15:34
  • 1
    $\begingroup$ @salma how do you know that an adjoint of a non-linear map actually exists? Also $j$ is not well defined and even if it were, for most banach spaces $X$ and $X^*$ are not isometric; just to mention $\ell_p$ for $p\neq 2$. $\endgroup$ – Tomek Kania May 24 '16 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.