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 $U$?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $k \le p \sqrt{2}$. 

