2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ There doesn't? Take $u_n = nx_n$, then $\alpha_n = n$ grows without bound. $\endgroup$
    – Neal
    Aug 2, 2017 at 14:24
  • $\begingroup$ Sorry I forget the assumption $u_n$ is a unit vector $\endgroup$
    – Student
    Aug 2, 2017 at 14:32

3 Answers 3

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ But why $\alpha_n$ and $\beta_n$ are bounded by the same constant $M:=\frac{1}{\sqrt{1-\theta^2}}$??Thank you $\endgroup$
    – Student
    Aug 2, 2017 at 14:51
  • $\begingroup$ Is there a reason to believe that $M$ has precisely this shape? $\endgroup$
    – DominikS
    Aug 2, 2017 at 14:53
  • $\begingroup$ I don't know but I see in a paper that $M:=\frac{1}{\sqrt{1-\theta^2}}$. $\endgroup$
    – Student
    Aug 2, 2017 at 14:57
  • $\begingroup$ I have edited my post! $\endgroup$ Aug 2, 2017 at 15:10
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.