Sequences in Hilbert spaces We have a sequence $(x_{n})_{n=1}^\infty$ in a Hilbert space $H$ and we know that:


*

*for every $n \in \mathbb{N}$ holds $\|x_{n}\|\le1$

*for every $m\ne n$ holds $\|x_{n}-x_{m}\|\ge r > 0$


How I can show that $r \le \sqrt2$ ?
 A: Notice that whenever $m \neq n$, we have
$$ r^2 \leq \|x_n - x_m\|^2 \leq 2 - 2\operatorname{Re} \langle x_m, x_n \rangle
\quad \Rightarrow \quad
2\operatorname{Re} \langle x_m, x_n \rangle \leq 2 - r^2. \tag{*}$$
Then for any positive integer $N \geq 2$,
\begin{align*}
0
\leq \left\| \sum_{n=1}^{N} x_n \right\|^2
&= \sum_{m,n = 1}^{N} \langle x_m, x_n \rangle \\
&= \sum_{n=1}^{N} \langle x_n, x_n \rangle + \sum_{1 \leq m < n \leq N} 2 \operatorname{Re}\langle x_m, x_n \rangle \\
&\stackrel{(*)}{\leq} N + \frac{N(N-1)}{2}(2-r^2).
\end{align*}
So it follows that $ 2 - r^2 \geq -\frac{2}{N-1} $ and letting $N \to \infty$ yields the desired claim.
A: The unit ball $B$ of a Hilbert space is weakly sequentially compact so the sequence $(x_n)_{n=1}^\infty$ has a weakly-convergent subsequence with limit in $B$.
WLOG we can assume that $x_n \rightharpoonup x$ weakly. Then for every $m\ne n$ we have
$$r^2 \leq \|x_n - x_m\|^2 \leq 2 - 2\operatorname{Re} \langle x_m, x_n \rangle \xrightarrow{m,n\to\infty} 2-\|x\|^2 \le 2$$
so $r \le \sqrt{2}$.
