Schauder basis in $\ell^2(\mathbb{N})$.

Question

I have a question about this problem posted here. Essentially they are trying to prove that the sequence $\{x_n\}_{k = 1}^\infty = \{\alpha e_n + \beta e_{n+1}\}_{n=1}^\infty$ (here $\{e_n\}_{n = 1}^\infty$ is the canonical orthonormal basis for $\ell^2(\mathbb{N})$) is a Schauder basis for $\ell^2(\mathbb{N})$ whenever $|\beta| < |\alpha|.$

The accepted answer on here gives a hint on how to write the orthonormal basis in terms of this sequence. Following this hint gives us that we can write (uniquely) $$e_k = \frac{1}{\alpha}\sum_{n=k}^\infty \left(-\frac{\beta}{\alpha}\right)^{n-1}x_n.$$ However, why is this enough to conclude that $\{x_n\}_{n=1}^\infty$ is a Schauder basis? In order for this to happen for each $y \in \ell^2(\mathbb{N})$ there must be a unique way to write $$y = \sum_{n=1}^\infty c_k x_k$$ for some scalars $\{c_n\}_{n=1}^\infty$. We already know we have the decomposition of $y$ in terms of the orthonormal basis and each element of the orthonormal basis has a unique decomposition. Is this the argument? I am having trouble believing it works rigorously.

Edit: For every $y$ we have $$y = \sum_{k=1}^\infty \langle y, e_k\rangle e_k = \sum_{k=1}^\infty \sum_{n=k}^\infty \frac{\langle y , e_k\rangle}{\alpha}\left(-\frac{\beta}{\alpha}\right)^{n-1}x_n.$$ If we can swap the order of the sum we are done.

Answer 1

Observe that for an invertible linear operator $T$ and a Schauder basis $\{e_n\}_{n=1}^\infty$ the sequence $\{Te_n\}_{n=1}^\infty$ consitutes the Schauder basis as well. Indeed for any element $x\in H$ we have $$T^{-1}x=\sum_{n=1}^\infty b_ne_n \iff x=\sum_{n=1}^\infty b_nTe_n$$ Consider the operator $T=\alpha I+\beta S=\alpha(I+\gamma S)$, where $\gamma=\beta/\alpha$ and $S$ is the shift operator defined by $Se_n=e_{n+1}.$ As $|\gamma|<1$ and $\|S\|=1,$ the operator $ T$ is invertible.

Answer 2

As you observed, it suffices to check existence.

We first note a truncation estimate: $$e_k = \frac{1}{\alpha}\sum_{n=k}^M \left(-\frac{\beta}{\alpha}\right)^{n-1} x_n + O\left(\left(\frac{\beta}{\alpha}\right)^M \right)$$ as $M \to \infty$.

Suppose $y = \sum_{i=1}^{\infty} a_i e_i$. Let $\epsilon > 0$. Take $M$ large such that $O\left(\left(\frac{\beta}{\alpha}\right)^M \right) = O(\epsilon)$, and take $N > M$ such that $$y = \sum_{i=1}^N a_i e_i + O(\epsilon).$$ Substitute the truncated expression of $e_i$'s in the sum; all sums are finite, so swapping sums have no issues, and we see that \begin{align*} y = \, & \sum_{i=1}^N a_i e_i + O(\epsilon) \\ = \, & \frac{1}{\alpha}\sum_{i=1}^N a_i \sum_{n=i}^M \left(-\frac{\beta}{\alpha}\right)^{n-1} x_n + O(\epsilon) \\ = \, & \frac{1}{\alpha}\sum_{n=1}^M \left(\sum_{i \leq n} a_i\right) \left(-\frac{\beta}{\alpha}\right)^{n-1} x_n + O(\epsilon) + O\left(\left(\frac{\beta}{\alpha}\right)^M\right) \\ = \, & \frac{1}{\alpha}\sum_{n=1}^M \left(\sum_{i \leq n} a_i\right) \left(-\frac{\beta}{\alpha}\right)^{n-1} x_n + O(2\epsilon). \end{align*} Hence $$y = \frac{1}{\alpha}\sum_{n=1}^{\infty} \left(\sum_{i \leq n} a_i\right) \left(-\frac{\beta}{\alpha}\right)^{n-1} x_n.$$