Let $(R,\mathfrak{m},k)$ be a local domain of Krull dimension $d$ with $K=\operatorname{Frac}(R)$, the field of fractions of $R$. A rank one discrete $K$-valuation $\nu$ is an $\mathfrak{m}$-valuation if $\nu(R) \ge 0$, $\nu(\mathfrak{m}) >0$, and if $(V_\nu,\mathfrak{m}_\nu,k_\nu)$ is the valuation ring of $\nu$, then $k_\nu$ has transcendence degree $d-1$ over $k$.
Let $I_n(\nu) = \{x \in R \mid \nu(x) \ge n \text{ or } x = 0\}$. Then $I_n(\nu)$ is an integrally closed ideal of $R$, and $\overline{\mathfrak{m}^n} \subset I_n(\nu)$. This I can show without too much difficulty.
Say I also happen to know in this particular case, that there is an $l$ such that for all $n$, $I_{ln}(\nu)\subset \mathfrak{m}^n$. Now let $w$ be another $\mathfrak{m}$-valuation. I want to show that $\nu(x) \le (2l-1)w(x)$ for all $x \in R-\{0\}$.
My problem is, I have difficulty when working with valuations since everything seems built to only give lower bounds. How do I find an upper bound for $v(x)$ in this way? Of course, $w(x^{2l-1}) = (2l-1)w(x)$ but I can't seem to make that work for me. For any $x$, $x$ must be in some maximal power of $\mathfrak{m}$ by Krull's Intersection Theorem, so can I use that to my advantage? I.e., if $x \in \mathfrak{m}^m \setminus \mathfrak{m}^{m+1}$, do I know $\nu(x) < (m+1)\nu(\mathfrak{m})$?
I think I have a solution now. If someone can verify, please let me know.
Say $w(x)=n$. Then $x \in I_n(w) \setminus I_{n+1}(w)$, and since $I_{n+1}(w) \supset I_{ln+l}(\nu)$, then $\nu(x) < ln+l$. Hence $\nu(x) \le ln+l-1$, and so since $n \ge 1$, $\nu(x) \le ln+(l-1)n=(2l-1)n=(2l-1)w(x)$.