# A Hilbert space characterization via retractions--a conjecture

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).

• Your retraction condition is wrong, you probably mean 1-Lipschitz (inequality instead of the equality). – Misha Dec 28 '15 at 12:51
• @Misha He does not, his retraction is by a "folding" map that acts as a reflection on $X \ X_f$. – Will Sawin Dec 28 '15 at 13:12
• So is this clear (or known) for, say, $X=\mathbb R^2$ with a $p$ norm? – Christian Remling Dec 28 '15 at 21:52
• Whenever you have retractions for both $f$ and $-f$, there is an isometric reflection fixing the hyperplane $ker(f)$. Now, I think, it is known that if a Banach space admits an isometric reflection in every hyperplane, then it is Hilbert. (Even an open nonempty subset of hyperplanes should suffice.) – Misha Dec 29 '15 at 6:36
• Related questions went back--I think--to Hugo Steinhaus, in $\mathbb R^3$. A few dozens years ago specialists in functional analysis (e.g. Pełczyński) considered such questions but always in finite-dimensional case, in the geometric language: you had a centrally symmetric convex body, you assumed some extra property, then you ask: is it a (rounded) ball. (I don't remember who asked what). I needed my new twist with retraction to get to the problem more in full. Otherwise, in some questions you neglect to talk about, say, ellipsoid + for arbitrary B-spaces it's hard to talk about f-symmetry. – Włodzimierz Holsztyński Dec 29 '15 at 8:24

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.

• We are talking here of course about the senior Lindenstrauss (Yoram) since the Fields was awarded to Lindenstrauss (Elon) in 2010. – Włodzimierz Holsztyński Jan 1 '16 at 20:53
• Thank you, Mikhal--what a great New Year gift! Best wishes to you & to all MOers. – – Włodzimierz Holsztyński Jan 1 '16 at 20:55
• Yes, I meant Joram Lindenstrauss (Elon was born in 1970). – Mikhail Ostrovskii Jan 1 '16 at 22:41
• "Joram" --(not "Yoram"), I am very sorry for my spelling error. – Włodzimierz Holsztyński Jan 1 '16 at 23:52