Given a Banach space $X$ and a functional $f:X\rightarrow \mathbb R$, let


$$ X\_f := \{x\in X : f(x)\ge 0\} $$


("*functional*" means "*non-zero linear functional*"). Also, given a topological space $E$ and its topological subspace $A$, a retraction $r:E\rightarrow A$ is defined as a continuous map such that $r(x)=x$ for every $x\in A$.


**CONJECTURE:**


Let $\,X$ be an arbitrary Banach space such that for every functional $\ f:X\rightarrow \mathbb R\ $ there is a retraction $\ r:X\rightarrow X\_f\ $ such that


$$ \forall\_{x\ y\ \in\ X\setminus X\_f}\ \ \ |r(x)-r(y)| = |x-y| $$


Then $X$ is isometric to a Hilbert space.



> 
> **REMARK:** I think that questions of this type were popular in the past in the case of finite dimensional spaces (mostly 3-dim?). I am not aware of the general case (I am not a specialist thus I have to ask :-). I still believe that my usage of the ***retraction*** language here is new (even in the finite-dimensional case).
> 
> 
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A Hilbert space characterization via retractions--a conjecture
I think that the desired result can be proved on the following
lines if the dimension is at least $3$.


(1) Consider such maps for $X\_f$ and $X\_{-f}$. Denote them $r$ and
$r'$ respectively. One can show that $f(r(x))=-f(x)$ for $x\in
X\backslash X\_f$. Similarly one can show that $f(r'(x))=-f(x)$ for
$x\in X\_f$. Let $Ax=(r(x)+r'(x))/2$. Then $A$ is a $1$-Lipschitz
retraction onto $H=\{x\in X: f(x)=0\}$.


(2) Using the result of Lindenstrauss (Corollary 1 on page 270 in
Michigan Math. J. 11 (1964), 263-287), we get that there is a
linear projection of norm $1$ onto $H$.


(3) Using one of the known characterization of the Hilbert space,
see (12.8) in Amir (Characterizations of inner product spaces,
Birkhauser Verlag, Basel, 1986), one gets that the space $X$ is
Hilbert.


P.S. (1) One can replace usage of the Lindenstrauss result by the
usage of the result of Mankiewicz (Bull. Acad. Polon. Sci. Ser. Sci. Math.
Astronom. Phys. 20 (1972), 367-371) implying that the restriction
of $r$ to $X\backslash X\_f$ is a restriction of a linear isometry
$\widetilde r:X\to X$ to $X\backslash X\_f$. Then we observe that
$\frac12(I+\widetilde r)$ is a norm-$1$ projection onto $H$.


(2) One can use this observation to complete the $2$-dimensional
case which we divide into two subcases:


(2a) The unit ball is a polygon: this can be done by hand.


(2b) The unit ball has infinitely many strongly exposed points.
For each such point $x$ we consider the corresponding (exposing)
norm-one functional $f\_x$, and the half-space $H\_{f\_x}$. The
corresponding $r$ maps $x$ to $-x$. This implies that
$\hbox{ker}f\_x$ and $\mathbb{R}x$ satisfy the James orthogonality
relation (see Amir, page 24). Since there are infinitely many
strongly exposed points $x$, the assumption of (6.12'') (Amir,
page 53) is satisfied and the space is Euclidean.


