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In the area of functional analysis, nonlinear operator theory, the normalized duality mapping on a Banach space $X$ is a set valued map from $J:X\rightarrow 2^{X^*}$ given by \begin{align} J(x)=\{j(x)\in X^*: \langle j(x), x\rangle=\|x\|^2=\|j(x)\|^2\}. \end{align} I know that if $X$ is uniformly convex and uniformly smooth that $J$ is invertible and $J^{-1}=J^*$ the duality mapping from $X^* $ to $X^{**}$.

Question: Assuming that the Banach space $X$ is uniformly convex and uniformly smooth(e.g $L_p$ spaces, $1<p<\infty$), is the normalized duality mapping symmetric? i.e Is $\langle x, j(y)\rangle = \langle y, j(x)\rangle$ for any $x,y\in X$.

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No, it is not symmetric unless $X$ is a Hilbert space. In the conjectural formula $$\langle x, j(y)\rangle = \langle y, j(x)\rangle\quad (?)$$ the left hand side is linear in $x$, but the right hand side is not, unless $j$ is a linear map.

Concrete example: in $\ell^4$, the duality map $j:\ell^4\to \ell^{4/3}$ is $$ j(x) = (x_1^3,x_2^3,x_3^3,\dots) $$ Thus, $$\langle x, j(y)\rangle = \sum_i x_iy_i^3$$ which of course need not be the same as $$\langle y, j(x)\rangle = \sum_i x_i^3y_i$$

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  • $\begingroup$ Thanks Yes! Thats correct and timely too. $\endgroup$ May 8, 2015 at 18:45
  • $\begingroup$ @user147263 This $j$ is not the duality mapping, because $\|jx\|^2\neq\|x\|^2$, but $\|jx\|^{4/3}=\|x\|^4$. $\endgroup$
    – Infinite
    Dec 10, 2018 at 4:38

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