A postitive sequence $\{x_k\}$ converges Q-linearly to $0$ if there is some $\mu\in(0,1)$ and $K\in\mathbb{N}$ such that
$$ x_{k+1}\le \mu x_k \text{ for }k\ge K.$$
If $\{x_k\}$ converges linearly to $0$ and $0< y_k\le x_k$ for all $k$, does it mean that $y_k$ converges Q-linearly to $0?$
The answer is no since we can choose $x^k=\mu^k$ for all $k\ge 1$ and $y^{2k-1}=y^{2k}=\mu^{2k}$ for all $k\ge 1$. Then for $\mu<1$, it is clear that $x^k$ converges $Q$-linearly to $0$, $y^k\le x^k$ for all $k$ but $y^k$ does not converges $Q$-linearly to $0.$