At least they are equivalent on every real (or complex) finite dimensional normed vector space.
Proposition.
Let \(\norm{.}\) be an arbitrary norm on \(\R ^n\).Then \(\norm{.}\) is equivalent to \(\norm{.}_\infty\).
Proof.
Let \(\{e_i: i = 1,\dots, n \}\) be the standard basis on \(\R ^n\). Then for any \(v\in \R ^n\) \[\norm{v} = \norm{\sum_{i=1}^n \alpha_i e_i} \le \sum_{i=1}^n |\alpha_i| \norm{ e_i} \le n \sum_{i=1}^n\norm{ e_i}\norm{v}_\infty \le C_1 \norm{v}_\infty.\]
As we have \(|\norm{v} - \norm{w}| \le \norm{v-w}\) the function \begin{align*} f : \left< \R ^n, \norm{.}_\infty\right> &\to \left< \R , |.|\right> \\ v &\mapsto \norm{v} \end{align*} is continuous.
As \(S^{n-1} = \{v \in \R^n : \norm{v}_\infty = 1 \}\) is compact \(f\) attains a minimum on that set
\[C_2 := \min_{v \in S^{n-1}} f(v) = \min_{v \in S^{n-1}} \norm{v}.\]
As \(0 \notin S^{n-1}\) it holds that \(C_2 > 0\).
For arbitrary \(v \in \R^d\) define
\(w =\frac{1}{\norm{v}_\infty}v\). Since \(w \in S^{n-1}\) we have \(\norm{w} \ge C_2\).
But this implies
\[ C_2 \norm{v}_\infty \le \norm{v}.\]
\(\Box\)
Proposition.
Let \(V\) be an arbitrary finite dimensional vector space. Then every norm is equivalent.
Proof.
For basis \(\{b_i : i = 1,\dots,n\}\) define \[\norm{v}_\infty = \norm{\sum_{i=1}^n \alpha_i b_i}_\infty = \max_{i=1,\dots, n} |\alpha_i|.\] Show that every norm is equivalent to \(\norm{.}_\infty\) as above. As the equivalence relation is transitive every norm is equivalent.
\(\Box\)