Let $(x_n)_{n\geq1}\subseteq \mathcal{H}$ and $(y_n)_{n\geq1}\subseteq \mathcal{H}$ such that $\|x_n\|=\|y_n\|=1$, Where $\mathcal{H}$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$. We assume 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. Let $u_n\in span\{x_n,y_n\}$ for all $n\in \mathbb{N}$, then there exists scalars $\alpha_n,\beta_n\in \mathbb{K}$ such that $u_n=\alpha_nx_n+\beta_ny_n$. Assume that $\|u_n\|=1$. Let $M:=\frac{1}{\sqrt{1-\theta^2}}$. Why we have $|\alpha_n|\leq M$ and $|\beta_n|\leq M$ for sufficiently large $n$?? and thank you.
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$\begingroup$ There doesn't? Take $u_n = nx_n$, then $\alpha_n = n$ grows without bound. $\endgroup$– NealAug 2, 2017 at 14:24
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$\begingroup$ Sorry I forget the assumption $u_n$ is a unit vector $\endgroup$– StudentAug 2, 2017 at 14:32
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3 Answers
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Notice that for sufficiently large $n\in \mathbb{N}$, we have that $$1= \lvert\lvert{u_{n}\rvert\rvert}^{2} = \lvert {\alpha_{n}\rvert}^{2} + \lvert{\beta_{n}\rvert}^{2} + 2 \Re\left(\alpha_{n}\bar{\beta_{n}}\langle x_{n}, \, y_{n} \rangle\right) \geq \lvert {\alpha_{n}\rvert}^{2} + \lvert{\beta_{n}\rvert}^{2} - 2\lvert\alpha_{n}\rvert \lvert \beta_{n} \rvert \theta = $$ $$\left(\lvert \alpha_{n} \rvert - \lvert \beta_{n} \rvert\right)^{2} + 2 \lvert\alpha_{n}\rvert \lvert \beta_{n} \rvert\left(1-\theta\right) \geq 2 \lvert\alpha_{n}\rvert \lvert \beta_{n} \rvert\left(1-\theta\right)\geq 0$$
Thus shows that the $\alpha_{n}$'s and the $\beta_{n}$'s must be bounded.
Inspired by DominikS, we have that $$ 1 \geq \lvert {\alpha_{n}\rvert}^{2} + \lvert{\beta_{n}\rvert}^{2} - 2\lvert\alpha_{n}\rvert \lvert \beta_{n} \rvert \theta = (1-\theta^{2})\lvert{\alpha_{n}\rvert}^{2} + \left( \lvert \beta_{n} \rvert - \theta \lvert \alpha_{n} \rvert \right)^{2} \geq (1-\theta^{2})\lvert{\alpha_{n}\rvert}^{2}$$
By symmetry you can adapt the same argument to the $\beta_{n}$'s.
answered Aug 2, 2017 at 14:48
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$\begingroup$ But why $\alpha_n$ and $\beta_n$ are bounded by the same constant $M:=\frac{1}{\sqrt{1-\theta^2}}$??Thank you $\endgroup$– StudentAug 2, 2017 at 14:51
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$\begingroup$ Is there a reason to believe that $M$ has precisely this shape? $\endgroup$– DominikSAug 2, 2017 at 14:53
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$\begingroup$ I don't know but I see in a paper that $M:=\frac{1}{\sqrt{1-\theta^2}}$. $\endgroup$– StudentAug 2, 2017 at 14:57
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Using $$1 = \|u_n\|^2 \ge |\alpha_n|^2 + |\beta_n|^2 -2|\alpha_n||\beta_n|\theta = \theta (|\alpha_n|-|\beta_n|)^2 + (1-\theta)(|\alpha_n|^2 + |\beta_n|^2)\ge (1-\theta)(|\alpha_n|^2 + |\beta_n|^2)$$ would imply that $$|\alpha_n|, |\beta_n|\le \frac 1{\sqrt{1-\theta}},$$ which is unfortunately a weaker statement then the desired one.
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Let's get rid of the $n$, because it does not matter. Now let $e_1, e_2$ be an ONB of $\mathrm{span}\{x, y\}$ with $e_1=x$. Then $$ 1\ge|\langle e_2, u\rangle|=|\langle e_2, \alpha x+\beta y\rangle|=|\beta|\cdot|\langle e_2, y\rangle| $$ and $$ 1=\|y\|^2=|\langle e_1, y\rangle|^2+|\langle e_2, y\rangle|^2 \le \theta^2 + |\langle e_2, y\rangle|^2, $$ hence, combining these, $$ |\beta|^2\le\frac1{|\langle e_2, y\rangle|^2}\le\frac1{1-\theta^2}. $$ The inequality for $\alpha$ follows by symmetry.