# Distance of a point from a subspace vs. diameter

Let X = $(\Bbb R^N, \|\cdot\|)$ be a Banach space. Let $x_0 \in S^{N-1} = \{x \in \Bbb R^N : \sqrt{x_1^2+...+x_n^2}=1\}$. Denote $B^N_2 = \{x \in \Bbb R^N : \sqrt{x_1^2+...+x_N^2} \le 1 \}$. Define for $A \subset \Bbb R^N$, $\mathrm{diam} (A) := \mathrm sup_{x,y \in A}\|x - y\|$ (Diameter by $\|\cdot\|$, which is not necessarily Euclidean).

Is the following claim true?

Claim: $\forall (N-n)$-dimensional subspace $V \subset \Bbb R^N \space \exists$ an $n$-dimensional subspace $L \subset \Bbb R^N$ and $z \in L$ such that $\inf\limits_{y\in L}\|x_0-y\|= C \cdot \mathrm{diam} (B^N_2 \cap (z + V))$, ($C>0$ is a constant).

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Is $C$ a given, fixed constant? –  40 votes Jul 24 '13 at 21:42
$C$ may be any constant, but must not depend on the choice of $V$. –  Fernandez Jul 25 '13 at 13:09