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 $U$?



Maximum number of orthonormal vectors contained in an open cone
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 $k \le p \sqrt{2}$. 


