Let's say we have a sequence $X_{n}(k)$, where $n$ and $k$ are $1, 2\ldots$ Next, let's fix $k$ and assume that there exists function $g$ $$ X_{n}(k) \leq k^{3}g(n) $$ with $g$ some strictly positive monotone decreasing function of $n.$
Therefore, for a fixed $k$ we can say that $$ X_{n} = \mathcal{O}(g(n)) $$ as $n\to\infty,$ where $\mathcal{O}(\cdot)$ means Big–O notation.
Does this meant then that
$$ X_{n} = \mathcal{O}(k^{3} g(n)) $$ as $n, k \to\infty$?