My question was prompted by an earlier MO by @Daniel:


    [***Duality map in strictly convex Banach spaces***](https://mathoverflow.net/questions/125765/duality-map-in-strictly-convex-banach-spaces/130226#130226)


I will even use his symbol   $\phi$   below.


Let   $B$   be an arbitrary Banach space. Let   $S := \{x\in B:\|x\|=1\}$   be its unit sphere. Let   $\Gamma := \{f\in B^\*: \|f\|=1\}$   be the unit sphere in the dual space $B^\*$.


**QUESTION**   Are the following two conditions on $B$ equivalent:


1. $B$   is isometric to a Hilbert space.
2. There exists an isometry   $\phi: \Gamma \rightarrow S$   such that   $\forall\_{f\in\Gamma}\ f(\phi(f))=1$.


**?**


The finite-dimensional case is especially basic.


**REMARK 0**   Perhaps similar questions were asked in the past (on MO too?)--please, let me know.


**REMARK 2** The case of   $\mathbb R^2$   and its two dual but isometric norms   $L\_\infty\quad L\_1$   is interesting. The general question related to the one above is to describe all Banach spaces which are isometric to their dual space. Is there any beside the Hilbert spaces and   $\mathbb R^2$   with the norm(s) just mentioned above?



A characterization of Hilbert spaces?
Yes it is true. Let me show that the existence of $\phi$ implies that the norm of $B^\*$ is associated to an inner product. Then it follows easily that the spaces are Hilbert.


It suffices to verify that the norm is an inner product one on every two-dimensional subspace. (Indeed, this property is equivalent to the parallelogram law, which involves only two-dimensional configurations.)


Let $V$ be a two-dimensional subspace of $B^\*$ (equipped with the restriction of the norm of $B^\*$) and $V^\*$ the dual of $V$ (equipped with the norm dual to this restriction). There is a natural map $\pi:B\to V^\*$ dual to the inclusion $V\to B^\*$. Namely $\pi(x)(f)=f(x)$ for $x\in B$, $f\in V$. Note that $\pi$ does not increase the norm: $\|\pi(x)\|\le\|x\|$ for all $x\in B$.


Let $D$ be the unit ball of $V$ and $E\subset V$ the maximum-area ellipse contained in $D$ (i.e., the John ellipsoid of $D$). Let $\Sigma$ be the set of points where the boundaries of $D$ and $E$ meet. It is easy to see that $\Sigma$ contains at least two pairs of opposite points. The ellipse $E$ is a unit ball of a Euclidean norm $\|\cdot\|\_E$ on $V$. Since $E\subset D$, we have $\|f\|\_E\ge\|f\|$ for all $f\in V$ and equality is attained only if $x$ is proportional to an element of $\Sigma$.


On $V^\*$, there is a dual Euclidean norm, denoted by $\|\cdot\|^\*\_E$. There we have $\|y\|^\*\_E\le \|y\|$ for all $y\in V^\*$. The norm $\|\cdot\|\_E$ is associated to an inner product, which defines an isomorphism $I:V\to V^\*$ preserving the Euclidean norm.


For every $f\in\Sigma$, we have $\pi(\phi(f))=I(f)$. Indeed, $\|\pi(\phi(f))\|\le\|\phi(f)\|=1$, hence $\|\pi(\phi(f))\|^\*\_E\le 1$. On the other hand, $\pi(\phi(f))(f)=f(\phi(f))=1$. Since $\|f\|\_E=1$, this is possible only for $\pi(\phi(f))=I(f)$.


Now consider two linearly independent vectors $f,g\in\Sigma$ and look at the distance between $f$ and $-g$. Since $\phi$ is an isometry, we have 
$$
\|f+g\|=\|\phi(f)-\phi(-g)\| \ge \|\pi(\phi(f)-\phi(-g))\| .
$$
The r.h.s equals
$$
 \|\pi(\phi(f))-\pi(\phi(-g))\| = \|I(f)-I(-g)\|=\|I(f+g)\|
$$
since $\pi$ and $I$ are linear and $f,-g\in\Sigma$.


Thus $\|I(f+g)\|\le \|f+g\|$ and therefore
$$
\|I(f+g)\|\_E^\*\le\|I(f+g)\|\le \|f+g\| \le \|f+g\|\_E .
$$
But $I$ preserves the Euclidean norm, so the inequalities turn into equalities. In particular, $\|f+g\|= \|f+g\|\_E$, hence $f+g$ is proportional to an element of $\Sigma$.


Thus we have proved that, for every $f,g\in\Sigma$, the normalized bisector $\frac{f+g}{\|f+g\|}$ also belongs to $\Sigma$. Since $\Sigma$ is a closed subset of an ellipse and contains more than two points, it follows that $\Sigma$ is the whole ellipse. This means that $D=E$, so the norm on $V$ is Euclidean. Q.E.D.


**Remark.** The proof would be much easier (in fact, nearly trivial) if we assumed in advance that $\phi$ is a restriction of an isometry between $B^\*$ and $B$. Then it would be linear by Mazur-Ulam and one could just define the inner product of $f,g\in B^\*$ by $2\langle f,g\rangle=g(\phi(f))+f(\phi(g))$.


