Question on equivalent norm in sequence space I have a question that: Given $\alpha$ is an irrational number and $\{a_n\}_{n \in \mathbb{Z}} $ is belong to $l^2$, the sequence space i.e: $\sum_{n}a_n^2 < +\infty$. Hence, does there exist a constant $c > 0$ such that
$\sum_{n}a_n^2(1-\cos(2n\pi\alpha)) \geq c \sum_{n}a_n^2$
 A: So the answer is no. Assume for contradiction $c > 0$ is such a constant. The key fact here is that $(\cos(2 \pi n \alpha))_{n}$ is dense in $[-1, 1]$. Then there exist infinitely many $n$ such that $\cos (2 \pi n \alpha) \geq 1 - \frac{c}{2}$. Since it's infinite, let $(n_k)_k$ be a sequence such that $\cos (2 \pi n_{k} \alpha) \geq 1 - c/2$ for all $k$. Let $(a_n)_n$ be defined such that $a_n \neq 0$ only if $n$ is in $(n_k)_k$. Then $\sum_n a_{n}^{2} (1 - \cos( 2 \pi n \alpha )) = \sum_k a_{n_{k}}^{2} (1 - \cos( 2 \pi n_{k} \alpha )) \leq \frac{c}{2} \sum_k a_{n_{k}}^{2} = (c / 2) \sum_n a_{n}^{2}$.
A: Assume there was such a constant $c$ and we will reach a contradiction.
We will use the fact that $1- \cos ( 2 n \pi \alpha)  $ is dense in $[0,1]$. 
Choose $k_n$ such $1- \cos ( 2 k_n \pi \alpha)  \leq c/2$.
Now let $a_{k_n}=\frac{1}{k_n}$ and zero everywhere else.
Therefore 
\begin{align*} 
\sum_{n}a_{n}^2(1- \cos ( 2 n \pi \alpha)) &= \sum_{n}a_{k_n}^2(1- \cos ( 2 k_n \pi \alpha))\\
&\leq c/2\sum_{n} a_{k_n}^2
\end{align*}
Contradicting the fact that $\sum_{n}a_{n}^2(1- \cos ( 2 n \pi \alpha))\geq c \sum_{n}a_{n}^2.$
A: Ok, now if we have $\epsilon >0$ given. Does there exist $c_{\epsilon} >0$ such that 
$\sum_{n}a_n^2(1-\cos(2n\pi\alpha)) + \epsilon \geq c_{\epsilon} \sum_{n}a_n^2$
