Almost Mathieu operator I'm having trouble showing that the almost Mathieu operator given by
$$(Hu)_n = u_{n+1} + u_{n-1} + 2\lambda \cos \ [2 \pi (w + n\alpha)]u_n$$ 
Where $\lambda, \in \mathbb{R}$$, \alpha, w \in \mathbb{R/Z}$, $\alpha \not \in \mathbb{Q}$ are parameters, and $\lambda \not= 0$ 
where $H$ is the almost mathieu operator on  the Hilbert space $\mathcal l^{2}(\mathbb{Z})=\left \{ a=\left \{ a_{n} \right \}_{n=1}^{\infty}:a_{n}  \in  \mathbb{Z}, 1\leqslant n\leqslant \infty , \sum_{n=1}^{\infty}\left | a_{n} \right |^{2}< \infty  \right \}$ is bounded.
I tried to do the inner product to show that $\langle Ax,y\rangle =\langle x,Ay\rangle$ since according to Hellinger-Toeplitz theorem then the operator is bounded. But I end up with an odd answer.
In order for $\langle Ax,y\rangle =\langle x,Ay\rangle $ to be correct then $\sum_{n=1}^{\infty}[x_{n}\cdot y_{n+1}+x_{n}\cdot y_{n-1}]$ has to be equal to $\sum_{n=1}^{\infty}[x_{n+1}\cdot y_{n}+x_{n-1}y_{n}]$ which it's clearly not for whenever $x$ is different from $y$??
Since
$\langle Ax,y\rangle $=$\sum_{n=1}^{\infty}[(x_{n+1}+x_{n-1}+2\lambda \cos\ [2 \pi (w + n\alpha)]x_n)\cdot y_{n}]$
and
$\langle x,Ay\rangle$=$\sum_{n=1}^{\infty}[x_{n}\cdot( y_{n+1}+ y_{n-1}+2\lambda \cos \ [2 \pi (w + n\alpha)]y_n)]$
Am I miscalculating or misunderstanding the inner product in $\mathcal l^{2}(\mathbb{Z})$?
According to articles etc. the Almost Mathieu operator is bounded and hence self-adjoint, but they don't prove it.
I also tried to calculate $||Hu||$ since if $0 <||Hu||< \infty \implies H$ is bounded.
This leads me to some ugly sums which I'm not sure whether all of them converge. If these sums converge however since we're in an Hilbert space, then it's bounded... But if it's bounded it should also be self-adjoint which I tried to show with no luck! All articles claim the operator is bounded and self-adjoint so what am I doing wrong here? :( 
I should add I also try to show that the linear operator is continuous at a point (if it is then it's bounded), but I haven't managed yet.
 A: First, be aware that you've misunderstood the definition of the Hilbert space $\ell^2(\mathbb{Z})$; it's the Hilbert space of all square-summable complex sequences indexed by $\mathbb{Z}$, i.e.,
$$
 \ell^2(\mathbb{Z}) := \Bigg\{a : \mathbb{Z} \to \mathbb{C} \; \Bigg\vert \; \sum_{n = -\infty}^\infty \lvert a_n \rvert^2 < +\infty \Bigg\}
$$
endowed with the inner product
$$
 \langle a, b \rangle = \sum_{n = -\infty}^\infty \overline{a_n} b_n,
$$
if you take inner products to be linear in the second argument, or
$$
 \langle a, b \rangle = \sum_{n = -\infty}^\infty a_n\overline{b_n}
$$
if you take inner products to be linear in the first; this is a matter of convention and taste.
Now, the Hellinger–Toeplitz theorem says that an everywhere defined symmetric operator on a Hilbert space is actually bounded (and hence, in particular, self-adjoint). Linearity of your almost-Mathieu operator $H$ is trivial, so you really have two tasks:


*

*Show that $Hu \in \ell^2(\mathbb{Z})$ for all $u \in \ell^2(\mathbb{Z})$, so that $H$ is defined everywhere—this is not the same as showing it's bounded.

*Given that $H$ is everywhere defined, show that $H$ is symmetric, i.e., $\langle H a, b \rangle = \langle a, H b \rangle$ for all $a$, $b \in \ell^2(\mathbb{Z})$.


With regard to the first task, observe that for $u \in \ell^2(\mathbb{Z})$,
$$
 Hu = Au + Bu + Cu, \quad (Au)_n := u_{n+1}, \quad (Bu)_n := u_{n-1}, \quad (Cu)_n := 2\lambda \cos(2\pi\omega(n+\alpha))u_n
$$
(in fact, it turns out that $A$ defines a well-known unitary operator on $\ell^2(\mathbb{Z})$ called the left shift, whilst $B = A^\ast$ is the right shift); hence it suffices to show that $Au$, $Bu$, and $Cu \in \ell^2(\mathbb{Z})$, i.e., that
$$
 \sum_{n = -\infty}^\infty \lvert u_{n+1} \rvert^2 < +\infty, \quad \sum_{n = -\infty}^\infty \lvert u_{n-1} \rvert^2 < +\infty, \quad \sum_{n=-\infty}^\infty \lvert2\lambda \cos(2\pi\omega(n+\alpha))u_n\rvert^2 < +\infty.
$$
Given that $u \in \ell^2(\mathbb{Z})$, why is this a triviality? Remember, if you're summing from $-\infty$ to $\infty$, you're completely free to reindex such a sum by $m = n-1$ or $m=n+1$ and get one and the same sum.
With regard to the second task, redoing you computation to take into account the fact that our sequences are complex valued and that the summation is over $\mathbb{Z}$ and not $\mathbb{N}$, you'll have to show equalities similar to what you're worried about, e.g.,
$$
 \sum_{n=-\infty}^\infty \overline{a_{n+1}}b_n = \sum_{n=-\infty}^\infty \overline{a_n} b_{n-1}, \quad \sum_{n=-\infty}^\infty \overline{a_{n-1}}b_n = \sum_{n=-\infty}^\infty \overline{a_n}b_{n+1};
$$
the key is the observation above about summing from $-\infty$ to $\infty$.
