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I am not very sure if the following problem has been treated in the literature and if so, whether it always holds:

A Banach space $X$ is isomorphic to a Hilbert space if the 'norm attaining' map $F$ from $S_{X^*}$ into $S_X$ satisfying $x^*(F(x^*)) = 1$ is a bilipschitz equivalence. Here $S_Z$ denotes the unit sphere of the Banach space $Z$.

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I think this will be correct, if I interpret your assumption correctly.

The assumption implies that the duality mapping on $X$ with gauge $t\mapsto t^2$, $$j_X(x) := \bigl\{ x^* \in X^* \colon x^*(x) = \|x^*\|\,\|x\|,~\|x^*\| = \|x\|\bigr\},$$ is bijective and Lipschitz continuous via $j_X(x) = \|x\|\, F^{-1}(x/\|x)$. Hence $X$ is $2$-smooth [Remark 5, XR]), so of type $2$ [Prop. IV.5.10, DGZ]. In particular, it is reflexive and $F$ induces the duality mapping $j_{X^*}$ on $X^*$ in the same way as $F^{-1}$ did on $X$. Analogously, $X^*$ is $2$-smooth, and this implies that $X$ is of cotype $2$ [Prop. IV.5.12, DGZ]. Finally, a Banach space which is both of type and cotype $2$ is isomorphic to a Hilbert space by Kwapien's theorem.

[XR] Xu, Zong-Ben; Roach, G. F., Characteristic inequalities of uniformly convex and uniformly smooth Banach spaces, J. Math. Anal. Appl. 157, No. 1, 189-210 (1991)

[DGZ] Deville, Robert; Godefroy, Gilles; Zizler, Václav, Smoothness and renormings in Banach spaces, Pitman Monographs and Surveys in Pure and Applied Mathematics. 64. Harlow: Longman Scientific & Technical. New York: John Wiley & Sons, Inc.. 376 p. (1993).

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  • $\begingroup$ The argument looks plausible. Shall have to check more carefuly. $\endgroup$
    – M A Sofi
    Commented Feb 21, 2020 at 11:19

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