Operator norm of almost mathieu operator The almost Mathieu operator has become famous since it is the central object of the ten martini problem. 
In this paper here a bound on the operator norm is given. Although the bound is of course great, the calculation is rather cumbersome and hard to summarize. 
I was unable to find a simple proof of the far(!) weaker result that the operator norm of
$$H^{\alpha} u(n) = u(n+1) + u(n-1) + 2  \cos(2\pi (n\alpha)) u(n)$$
is strictly smaller than $4$ if $\alpha \in (0,1)$. If anybody has a good idea how to show this, I would be greatful. Maybe, this was also proved in some other context which I am not aware of. In any case, thank you for thinking about this.
 A: Yes, this has an easy proof. Essentially, the argument is that if we replace $2\cos(\ldots)$ by $\pm 2$, then the operator norm is exactly $4$, but the actual potential is close to $\pm 2$ only occasionally, and it's not possible to have an approximate eigenfunction supported by those $n$.
Here's a sketch how to do this more formally: Notice that $(H_0u)=u(n+1)+u(n-1)+2u(n)$ has spectrum $[0,4]$ (for example, take Fourier transforms to see this), and $H=H_0 + V-2$, with $V(n)=2\cos (2\pi n\alpha)$, so $V(n)-2\le 0$ and in fact $V(n)-2\le-\delta<0$ unless $n\alpha\simeq 0\bmod 1$.
So if we wanted to get $\langle u, Hu\rangle$ close to $4$ with a normalized test function $\|u\|=1$, since $\langle u, H_0 u\rangle <4$, we'd have to make $$
\langle u, (V-2) u\rangle = \sum (V(n)-2)|u(n)|^2 \simeq 0 .
$$
By the properties of $V-2$ we observed, this means that up to a small error, $u$ must be supported by those $n$ for which $n\alpha\simeq 0\bmod 1$. More precisely, if we write $S=\{ n: V(n)-2\le -\delta\}$, then $\sum_{n\in S} |u_n|^2\simeq 0$. Also, if $\delta>0$ was chosen small enough, then $n\notin S$ will imply that $n\pm 1\in S$. This means that $\sum |u_n u_{n\pm 1}|$ will stay small, and thus $\langle u, H_0 u\rangle$ can actually not be close to $4$ for such a $u$.
This shows that $H\le 4-\epsilon$, and $H\ge -4+\epsilon$ can be shown in the same way.
