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Let H be a separable and infinite-dimensional Hilbert space which we can consider to be a vector space with real scalars and one of its points h as origin (null vector). Let S be an infinite set of unit length vectors of H, all originating from h. Must S always contain pairs of distinct vectors which subtend an angle less than or equal to pi/2 at h? If the answer to this question is negative and S is a counterexample, must S nevertheless always contain pairs of distinct vectors that subtend obtuse angles at h which are arbitrarily close to pi/2?

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1 Answer 1

I'll use $(u,v)$ to denote the inner product. If $u_1, \ldots, u_n$ are unit vectors, $0 \le \|u_1 + \ldots + u_n\|^2 = \sum_{i,j=1}^n (u_i, u_j)$. There are $n$ terms $(u_i, u_i) = 1$ and $n^2-n$ with $i \ne j$, so the average of the $(u_i, u_j)$ with $i \ne j$ must be at least $-1/(n-1)$, and therefore at least one pair has $(u_i, u_j) \ge -1/(n-1)$. In particular, for an infinite set of vectors the infimum of the angles must be at most $\pi/2$.

For an example where all angles are obtuse, if $e_1,\ e_2, \ldots$ are orthonormal vectors consider the unit vectors $v_n =\sqrt{3/4} \left(e_n - \sum_{i=n+1}^\infty 2^{n-i} e_i\right)$ and note that for $n < m$, $(v_n, v_m) = (3/4) 2^{n-m} \left( 1 - \sum_{i=1}^\infty 2^{-2i} \right) < 0$.

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Bravo! This is a beautiful answer. –  Garabed Gulbenkian Dec 14 '11 at 20:57
    
I assume that by "supremum" you mean "Limit Superior" and not "Least Upper Bound" –  Garabed Gulbenkian Dec 14 '11 at 21:22
    
Oops, I meant infimum. I'll edit –  Robert Israel Dec 14 '11 at 22:12

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