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This question is related to If the answer to this question is yes, it would solve the abovementioned other MO question.

We equip ${\mathbb R}^3$ with the $\ell_3$ norm $||(x,y,z)||=(|x|^3+|y|^3+|z|^3)^{\frac{1}{3}}$. Is it true that, for any vectorial plane $V$ in ${\mathbb R}^3$, we can find a vector $w$ not in $V$ such that $||w+v|| \geq ||v||$ for all $v \in V$ ? It is easily seen that this property holds for some norms (such as the $\ell_2$ norm) and fails for others, such as the $\ell_{\infty}$ norm.

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up vote 4 down vote accepted

Here is a simple proof that the property holds only for Euclidean norms, at least if the norm in question is $C^1$ smooth and strictly convex. Surely it was known way before Gromov was born.

Let $S$ denote the unit sphere of the norm. First observe that, if $v\in S$ and $w$ are such that $\|w+tv\|\ge \|tv\|$ for all $t\in\mathbb R$, than $w$ is parallel to the tangent plane to $S$ at $v$.

So we have a map $F$ from the set of 2-dimensional linear subspaces to the set of 1-dimensional subspaces such that, for every linear plane $P$, the line $F(P)$ is parallel to the tangent planes at all points of $S\cap P$. Smoothness and strict convexity imply that this map is continuous and injective, hence it is bijective.

If a collection of planes contain a common line (or, equivalently, a common unit vector $v$), then their $F$-images belong to one plane (namely the one parallel to the tangent plane to $S$ at $v$). Now consider the set of planes and the set of lines as models of the real projective plane (they are the projectivizations of $(\mathbb R^3)^*$ and $\mathbb R^3$, resp.) Now we have a bijection from the projective plane to itself that sends lines to lines, hence a projective map.

Thus there is a linear map $L:(\mathbb R^3)^*\to\mathbb R^3$ whose projectivization is $F$. This means that, if a plane $P$ is the kernel of some element $p\in(\mathbb R^3)^*$, then $L(p)$ is a vector from the line $F(P)$. This map $L$ defines a non-degenerate bilinear form on $\mathbb R^3$ in a standard way, and $F(P)$ is just the orthogonal complement to $P$ w.r.t. this form. Since $F(P)$ obviously never belongs to $P$, the bilinear form is sign-definite (assume positive), so it is a Euclidean structure. In this Euclidean structure, the tangent planes to $S$ are orthogonal to their corresponding radial vectors. It follows that $S$ is a sphere of this Euclidean structure.

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Nice, but I don't see how to reduce the Kakutani theorem to the special case you treat. IIRC (it has been so long that I did not even remember that Kakutani proved it!) the proof is not very difficult, but I can't get his paper and also don't have access to Amir's book here. – Bill Johnson Feb 8 '11 at 14:46
@Bill: This works for $\ell_3$ and the like, and yes, running this argument for non-smooth norms would be difficult. I don't know how Kakutani proved it either. – Sergei Ivanov Feb 8 '11 at 15:34

Another way to state your condition is to say that every 2 dimensional subspace is norm one complemented. This implies that the space is isometrically isomorphic to a Hilbert space. Kakutani proved this in Some characterizations of Euclidean space, Jap. J. Math. 16, (1939). 93–97. It is also a remark at the end of Bruck's paper

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