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Let $H$ be a separable Hilbert space, $\Pi:H\to L$ the orthogonal projection to a linear subspace of finite dimension $p$, and $U$ the open cone of vectors $u\in H$ such that $\langle u,\Pi u\rangle>\|u\|^2/\sqrt{2}$. Which is the maximum number of orthonormal vectors contained in $U$?

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and the motivation for the question is?... – Anthony Quas Mar 1 '12 at 15:02
Is it a convex cone? Otherwise, what do you mean by dimension of cone? – Uday Mar 1 '12 at 15:31
About the motivation, I arrived to this question studying growth conditions for the eigenvalues of self-adjoint operators with discrete spectrum. Suggestions about this kind of study are also welcome. As an example, it is easy to see that the maximum number is 1 if $p=1$, and it is $\ge 3$ if $p=2$. The dimension of U is not mentioned in the question, but its definition uses a linear subspace of finite dimension. – Jesús Álvarez Mar 1 '12 at 16:06
oops! I interpreted the 'number of orthonormal vectors' as dimension. I thought that is the natural thing to do. In the convex cone case, the answer must be p (dimension of the range). – Uday Mar 1 '12 at 16:16
For the maximum number with $p=2$ to be $\ge 3$ you need the right side to be at most $(2/3) \|u\|^2$, not $\|u\|^2/\sqrt{2}$. I suspect you're thinking about $\|u\|^2/2$. – Robert Israel Mar 1 '12 at 17:06
up vote 3 down vote accepted

Suppose $v_1, \ldots, v_p$ is an orthonormal basis of $\Pi H$. If $u_1, \ldots, u_k$ are orthonormal vectors in $U$, we have $\frac{1}{\sqrt{2}} \le \langle u_j, \Pi u_j \rangle = \sum_{i=1}^p \left|\langle v_i, u_j \rangle\right|^2$. Adding these inequalities for $j = 1 \ldots k$, $\frac{k}{\sqrt{2}} \le \sum_{i=1}^p \sum_{j=1}^k \left|\langle v_i, u_j \rangle\right|^2 \le \sum_{i=1}^p 1 = p$. So $k \le p \sqrt{2}$.

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