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Let $(x_n)_{n\geq1}\subseteq \mathcal{H}$ and $(y_n)_{n\geq1}\subseteq \mathcal{H}$ such that $\|x_n\|=\|y_n\|=1$. I see in a paper the following statements: We claim that there exists a constant $\theta\in \mathbb{R}$ such that $|\langle x_n\; |\;y_n\rangle|\leq \theta< 1$ for all $n$ sufficiently large. To see this, suppose otherwise. By replacing $(x_n)_{n\geq1}\subseteq \mathcal{H}$ and $(y_n)_{n\geq1}\subseteq \mathcal{H}$ with subsequences, we may assume that $\lim_{n\longrightarrow\infty}|\langle x_n\; |\;y_n\rangle|=1$. I don't understand how to prove that $$\lim_{n\longrightarrow\infty}|\langle x_n\; |\;y_n\rangle|=1.$$ Where $\mathcal{H}$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$. And thank you.

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  • $\begingroup$ What does $\langle x_n|y_n\rangle$ denote? $\endgroup$ – 5xum Aug 2 '17 at 12:16
  • $\begingroup$ $\mathcal{H}$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$. $\endgroup$ – Student Aug 2 '17 at 12:20
  • $\begingroup$ You have not given all informations !! It is given that $|\langle x_n\; |\;y_n\rangle|\leq 1$ for all $n$ ? $\endgroup$ – Fred Aug 2 '17 at 12:22
  • $\begingroup$ Yes because $\|x_n\|=\|y_n\|=1$ so by cauchy Schwartz we get the inequality . Thank you $\endgroup$ – Student Aug 2 '17 at 12:28
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Suppose that it is not true that there is a constant $\theta<1$ such that $|\langle x_n|y_n\rangle|\leq\theta <1 $ for all $n$ sufficiently large. Therfore, for every positive integer $k$, the number $\theta=1-1/k$ does not satisfy $|\langle x_n|y_n\rangle|\leq 1-1/k $ for all $n$ sufficiently large, which means that it is not true that there exists an $N$ such that for all $n>N$ this inequality is valid. Therefore, there exists some $n_k$ such that $1\geq |\langle x_{n_k}|y_{n_k}\rangle|> 1-1/k $ where the l.h.s inequality from from the fact that $x_n,y_n$ are unit vectors, and in fact we can always choose $n_{k+1}>n_k$ (why?). It follows that $$\lim_{k\to\infty}|\langle x_{n_k}|y_{n_k}\rangle|=1$$ as was declared.

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  • $\begingroup$ It is not very clear for me why we can always choose $n_{k+1}>n_k$??. What means by l.h.s inequality? And thank you $\endgroup$ – Student Aug 5 '17 at 8:59
  • $\begingroup$ I can explain it to you but I think it is better that you figure it out yourself. You are welcome. $\endgroup$ – uniquesolution Aug 5 '17 at 11:13

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