Let $\{e_i\}$ be an orthonormal set. Then prove that $S_n=\{e_i:|\langle x,e_i \rangle|^2>\lVert x\rVert^2/ n\}$ has at most $n-1$ vectors.

Question

Let $\{e_i\}$ be an orthonormal set in a Hilbert space $H$, $x$ be any vector in $H$ and $n \in N$, then prove that $S_n=\{e_i:\lvert\langle x,e_i \rangle\rvert^2 > \lVert x \rVert^2/ n\}$ has at most $n-1$ vectors.

In the book it is written, by Bessel's inequality the result follows. But I could not prove it by using Bessel's inequality or by another way.

Answer 1

Suppose we have $r$ vectors in $S_n$. Then $$\sum_{k=1}^{r}|\langle x,e_k \rangle|^2 > \frac{r}{n}\|x\|^2.$$ If $r \ge n$ then $\frac{r}{n}\|x\|^2 \ge \|x\|^2$. Consequently, $$\sum_{k=1}^{r}| \langle x,e_k\rangle|^2 > \|x\|^2,$$ which is incorrect due to Bessel's inequality that states $$\sum_{k=1}^{r}|\langle x,e_k\rangle|^2 \le \|x\|^2.$$

Answer 2

For the finite case, we have by Bessel's inequality, $\sum_{i=1}^k|\langle x,e_i\rangle|^2\leq \sum_{i=1}^\infty|\langle x,e_i\rangle|^2\leq ||x||^2$. If $|S_n|=k$, then taking their sum, we have $$ \sum_{i=1}^k |\langle x,e_i\rangle|^2>k||x||^2/n\geq ||x||^2, $$ a contradiction.

For the infinite case, we have by Bessel's inequality, $\sum_{i=1}^\infty|\langle x,e_{j_i}\rangle|^2\leq \sum_{i=1}^\infty|\langle x,e_i\rangle|^2\leq ||x||^2$. If $|S_n|=\infty$, then taking their sum, we have $$ \sum_{i=1}^\infty |\langle x,e_{j_i}\rangle|^2>\infty. $$ implying the sum is unbounded, but by Bessel's inequality, the sum is bounded, a contradiction.