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 $r \le \sqrt2$ ?
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mechanodroid
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flyaway07
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I thought a bounded sequence must have a convergent subsequence... – Kenny Lau Jan 06 '18 at 19:36
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If the vectors form an orthogonal set, then square, throw away the inner product term and take the square root again. Otherwise, I don't know - it does look like the Riesz-Lemma though, see https://math.stackexchange.com/questions/163500/an-application-of-riesz-lemma – max_zorn Jan 06 '18 at 19:36
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@Kenny, so $H$ is infinite-dimensional. – max_zorn Jan 06 '18 at 19:36
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Use the triangle inequality. – kimchi lover Jan 06 '18 at 19:37
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@KennyLau Not necessarily in infinite-dimensional spaces. Consider the canonical vectors $(e_n)_{n=1}^\infty$ in $\ell^2$. It is bounded but $|e_m - e_n| = \sqrt{2}$ for $m \ne n$. – mechanodroid Jan 06 '18 at 20:43
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@kimchilover: the assertion is not true in finite-dimensional spaces, so I fail to see how your suggestion might work. – Martin Argerami Jan 06 '18 at 22:16
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@KennyLau. That is true in $\Bbb R^n$. It is not true in Hilbert space – DanielWainfleet Jan 07 '18 at 02:49
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@DanielWainfleet I know. I don't need three people telling me the same thing. – Kenny Lau Jan 07 '18 at 02:50
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@KennyLau. Sorry. I didn't see the others. – DanielWainfleet Jan 07 '18 at 02:57
2 Answers
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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.
Sangchul Lee
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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 $r \le \sqrt{2}$.
mechanodroid
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