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I'd appreciate if someone can provide the best way to deal with this problem.

Let $\{\alpha_n\}$ be an orthonormal sequence for a Hilbert space H and let $\{\beta_n\}$ be an orthonormal sequence such that $$\beta_n=\frac{c_n\alpha_{2n-1}+\alpha_{2n}}{\sqrt{1+c_n^2}}$$ where $c_n\in(0,\infty)$ and $c_n\to0$. If $M=\overline{\text{span}\{\alpha_{2n}\}}$ and $N=\overline{\text{span}\{\beta_n\}}$, then $M\cap N=\{0\}$.

I have tried the following approaches but I always find myself at a dead end: (1) If I could show that $M^\perp=N$(assuming it's the right direction), then I'm done. (2) If $x\in M\cap N$, then I can express $x$ as the series $$x=\sum_{n=1}^\infty\lambda_n\alpha_{2n}\text{ and }x=\sum_{n=1}^\infty\phi_n\beta_n$$ where $\{\lambda_n\},\{\phi_n\}\subseteq\mathbb{C}$ such that $\sum_{n=1}^\infty|\lambda_n|^2$ and $\sum_{n=1}^\infty|\phi_n|^2$ are both finite. I tried to show that $\langle x,x\rangle=\|x\|^2=0$ but I failed. Is there any other way to deal with this problem?

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  • $\begingroup$ The two spaces aren't orthogonal, so (1) doesn't work. Let $x \in M \setminus \{0\}$. Show that $x\notin N$. Let $k = \min \{ n : \langle x, \alpha_{2n}\rangle \neq 0\}$. What can you say about $\langle x, \beta_k\rangle$ and $\langle x , \alpha_{2k-1}\rangle$? $\endgroup$ May 11 '16 at 18:37
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First you can show that $M=\left\{a_{2n-1}:n\right\}^\perp$, so $N\cap M=\left\{x\in N:x\perp\alpha_{2n-1}\forall n\right\}$.

Now note that $\alpha_{2n-1}$ is orthogonal to all $\beta_k$, $k\neq n$. Now given $x=\sum_k\lambda_k\beta_k$ take the inner product with $\alpha_{2n-1}$ and use the previous facts to conclude that $\lambda_n=0$. Therefore $x=0$.

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  • $\begingroup$ Thank you very much for your help. $\endgroup$
    – user338853
    May 11 '16 at 19:10

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