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Let $\alpha, \beta \in \mathbb C - \{0\}$ such that $0 < |\frac{\beta}{\alpha}| < 1$ and $$ x_1 = (\alpha, \beta, 0,...), x_2 = (0, \alpha, \beta, 0,...), x_3 = (0, 0, \alpha, \beta, 0,...),... $$ Show that $(x_n)_{n \ge 1}$ is a Schauder basis of $\ell^2$.

I know that each $x_i$ could be written as follows $$ x_i = \sum^{\infty}_{k=1} \alpha e_i + \beta e_{i+1} $$ where $e_i$ represents the canonical vector with input 1 in the $i$ th coordinate and 0 in the rest, but I don't know how to use the condition of $0 < |\frac{\beta}{\alpha}| < 1$.

JuanDa
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1 Answers1

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Hint:

Let $\mathbf{e}_j=\delta_j$ ($1$ in the $j$-th component, zero otherwise). Since for any $\mathbf{x}\in\ell_2$, $\mathbf{x}=\sum^\infty_{j=1}\langle \mathbf{x},\mathbf{e}_j\rangle \mathbf{e}_i$, it is enough to show that each $\mathbf{e}_j$ is of the form $\sum_k a_{jk}\mathbf{x}_k$.

\begin{aligned} \mathbf{e}_1&=\frac{1}{\alpha}\mathbf{x}_1-\frac{\beta}{\alpha}\mathbf{e}_2\\ \mathbf{e}_2&=\frac{1}{\alpha}\mathbf{x}_2-\frac{\beta}{\alpha}\mathbf{e}_3\\ &\ldots\\ \mathbf{e}_j&=\frac{1}{\alpha}\mathbf{x}_j-\frac{\beta} {\alpha}\mathbf{e}_{j+1}\\ \ldots \end{aligned}

By induction

$$ \mathbf{e}_1 = \frac{1}{\alpha}\sum^N_{j=1}\big(-\frac{\beta}{\alpha}\big)^{j-1}\mathbf{x}_j +\big(-\frac{\beta}{\alpha}\big)^N\mathbf{e}_{N+1} $$

From this and the assumption $0<|\beta/\alpha|<1$, one can see that $\mathbf{e}_1$ has the desired expression. Similar procedure for other $\mathbf{e}_j$'s.

Mittens
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    I understand that this is just a hint, but simply showing that each $e_i$ can be expressed in terms of the $x_i$ does not make ${x_i}_i$ a Schauder basis. In other words, a lot more work is needed. – Ruy Sep 17 '20 at 01:39
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    forget the previous comment.The next step would be to find the norm of $e_1$ minus the sum, which gives zero, but since you find the scalars so that for $ v \in \ell^2$ the two norm of $v$ minus the sum gives zero? – JuanDa Sep 18 '20 at 03:05