Theorem: Kottman Constant of Infinite-Dimensional Hilbert Space is Root 2

Let $\GF \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be an infinite-dimensional Hilbert space over $\GF$.
Let $\map K \HH$ be the Kottman constant of $\HH$.
Then $\map K \HH = \sqrt 2$.

Proof:
First consider the case $\GF = \R$.
Let $\norm {\, \cdot \,}$ be the inner product norm on $\HH$.
Let $B_\HH^-$ be the closed unit ball of $\struct {\HH, \norm {\, \cdot \,} }$. 
We first show that $\map K \HH \ge \sqrt 2$.
Since $\HH$ is infinite-dimensional, there exists a countable linearly independent subset $S$ of $\HH$.
By Gram-Schmidt Orthogonalization, there exists an orthonormal subset $\sequence {e_n}_{n \mathop \in \N}$ such that:
:$\map \span S = \map \span {\set {e_n : n \in \N} }$
From Pythagoras's Theorem (Inner Product Space), we have:
:$\norm {e_i - e_j}^2 = \norm {e_i}^2 + \norm {e_j}^2 = 2$ for $i \ne j$
since:
:$\innerprod {e_i} {-e_j} = 0$ for $i \ne j$. 
Hence $\sequence {e_n}_{n \mathop \in \N}$ is a sequence in $B_\HH^-$ such that:
:$\norm {e_i - e_j} = \sqrt 2$ for $i \ne j$.
Hence $\map K \HH \ge \sqrt 2$.
We now show that $\map K \HH \le \sqrt 2$.
{{AimForCont}} $\map K \HH > \sqrt 2$.
Then there exists $\epsilon > 0$ and a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $B_\HH^-$ such that:
:$\norm {x_i - x_j}^2 \ge 2 + 2 \epsilon$ for $i \ne j$.
Let $i \ne j$.
We can write:
{{begin-eqn}}
{{eqn	| l = \norm {x_i - x_j}^2
		| r = \innerprod {x_i - x_j} {x_i - x_j}
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \innerprod {x_i} {x_i} - \innerprod {x_j} {x_i} - \innerprod {x_i} {x_j} + \innerprod {x_j} {x_j} 
}}
{{eqn	| r = \norm {x_i}^2 - 2 \innerprod {x_i} {x_j} + \norm {x_j}^2
}}
{{eqn	| r = 2 - 2 \innerprod {x_i} {x_j}
}}
{{eqn	| r = \norm {x_i}^2 - 2 \innerprod {x_i} {x_j} + \norm {x_j}^2
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = 2 \paren {1 - \innerprod {x_i} {x_j} }
}}
{{end-eqn}}
Hence:
:$2 \paren {1 - \innerprod {x_i} {x_j} } \ge 2 + 2 \epsilon$
so that:
:$1 - \innerprod {x_i} {x_j} \ge 1 + \epsilon$
so:
:$\innerprod {x_i} {x_j} \le -\epsilon < 0$ for each $i \ne j$.
Now let $n \in \N$.
Write:
{{begin-eqn}}
{{eqn	| l = \norm {\sum_{i \mathop = 1}^n x_i}^2
		| r = \innerprod {\sum_{i \mathop = 1}^n x_i} {\sum_{j \mathop = 1}^n x_j}
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \sum_{1 \le i, j \le n} \innerprod {x_i} {x_j}
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| r = \sum_{1 \le i = j \le n} \innerprod {x_i} {x_i} + \sum_{1 \le i \ne j \le n} \innerprod {x_i} {x_j}
}}
{{eqn	| r = \sum_{i \mathop = 1}^n \norm {x_i}^2 + \sum_{1 \le i \ne j \le n} \innerprod {x_i} {x_j}
		| c = {{Defof|Inner Product Norm}}
}}
{{eqn	| o = \le
		| r = n - \epsilon \sum_{1 \le i \ne j \le n} 1
		| c = since $\norm {x_i}^2 = 1$ for all $i$, and $\innerprod {x_i} {x_j} \le -\epsilon$ for $i \ne j$
}}
{{end-eqn}}
Note that there are $n \paren {n - 1}$ tuples $\tuple {i, j}$ with $1 \le i \ne j \le n$, since:
:there are $n$ choices for $i$
:then, once $i$ is fixed, there are $n - 1$ choices for $j$. 
Hence:
:$\ds \epsilon \sum_{1 \le i \ne j \le n} 1 = n \paren {n - 1}$
Hence:
:$\ds \norm {\sum_{i \mathop = 1}^n x_i}^2 \le n - \epsilon n \paren {n - 1}$
On the other hand, we have:
:$\ds 0 \le \norm {\sum_{i \mathop = 1}^n x_i}^2$
so that:
:$\epsilon n \paren {n - 1} \le n$ 
Dividing by $n$ we have:
:$\epsilon \paren {n - 1} \le 1$
Hence:
:$\ds n \le 1 + \frac 1 \epsilon$
This is a contradiction to $n \in \N$ being arbitrary.
Hence, we have a contradiction and there exists no such sequence $\sequence {x_n}_{n \mathop \in \N}$.
Hence $\map K \HH \le \sqrt 2$.
Since $\map K \HH \ge \sqrt 2$ and $\map K \HH \le \sqrt 2$, we have $\map K \HH = \sqrt 2$.
{{qed}}
Category:Hilbert Spaces
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