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Let $E$ be a complex Hilbert space. Let $(x_n)_{n}\subseteq E$ and $(y_n)_{n}\subseteq E$ such that $\|x_n\|=\|y_n\|=1$. Assume that $\forall\, \theta<1$, we have $|\langle x_n\; ,\;y_n\rangle|> \theta$.

Why $$\displaystyle\lim_{n\longrightarrow\infty}|\langle x_{n}\; ,\;y_{n}\rangle|=1?$$

Student
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  • If $\forall, \theta<1$, |$\langle x_n; ,;y_n\rangle|> \theta$ then $|\langle x_n; ,;y_n\rangle| = 1$ – Burrrrb Nov 25 '17 at 16:12

2 Answers2

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By Cauchy-Schwarz inequality $|<x_n,y_n>|\leq ||x_n||\cdot ||y_n|| = 1$. Now choose for every $k\in \mathbb{N}$ an index $n_k$ such that $|<x_{n_k},y_{n_k}>|>1-1/k$. Then we have $1-1/k<|<x_{n_k},y_{n_k}>|\leq 1$, so $\lim_{k\to \infty}|<x_{n_k},y_{n_k}>|=1$.

Nightgap
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  • Thank you, but my problem why can I choose $n_k$ such that $n_{k+1}>n_k$? – Student Nov 25 '17 at 16:18
  • How exactly do you mean $\forall \theta < 1: |<x_n,y_n>|>\theta$? Because if this is for example for all $n\in \mathbb{N}$ we would have $|<x_n,y_n>|=1$ for all $n\in \mathbb{N}$ as mentioned by PotatoHead47. Then there would be no need to pass to a subsequence. – Nightgap Nov 25 '17 at 16:21
  • Yes, thank you very much – Student Nov 25 '17 at 16:24
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We have from Cauchy-Swartz we have that $$\theta <|<x_n,y_n>|\leq ||x_n|| ||y_n||=1,\forall \theta<1,\forall n \in \Bbb{N}$$

Thus $$\lim_{\theta \to 1} |<x_n,y_n>|=1 \Rightarrow |<x_n,y_n>|=1,\forall n \in \Bbb{N}$$ from uniqueness of limits.

Thus the sequence $b_n=<x_n,y_n>$ is bounded in $\Bbb{R}($or $\Bbb{C})$

So you can find from bolzano-Weirestrass a convergent subsequence $b_{n_k}$ such that $|b_{n_k}|=1$ again form uniqueness of limits.

Marios Gretsas
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