$L^\infty$ into $\mathbb{R}^n$, equivalent norms Consider $f:\Omega\rightarrow \mathbb{R}^n$ measurable. I am interested in knowing if $|| |u|_n ||_\infty$ is an equivalent norm to $| ||u||_\infty |_n$. Here $|\cdot |_n$ is any norm on $n$ dimensional vectors, and $||u||_\infty=(||u_i||_\infty)_i$.
One direction is quite easy: $|| |u|_n ||_\infty\leq C|| \sum_i|u_i| ||_\infty \leq C \sum_i || |u_i| ||_\infty =C \sum_i || u_i ||_\infty \leq CC'| ||u||_\infty |_n$.
The other direction I am having quite some difficulty, one could maybe try to see that both spaces are Banach and then conclude, but maybe there is an easier purely algebraic version?
 A: The equivalence is clear if the norm $\|\cdot\|$  (which you denoted $| \cdot|_n$) that you use on $n$-dimensional vectors is itself the $\ell^\infty$ norm, as in that case both sides are equal.
In the general case, just use the fact that for any norm $\|\cdot\|$ on $\mathbb R^n$, there are constants $C_1,C_2 >0 $ such that for all $v \in \mathbb R^n$, we have
$$\|v\|_\infty \le C_1 \|v\| \quad \text{and} \quad  \|v\| \le C_2  \|v\|_\infty \,.$$
Addendum: Supremum and maximum can indeed be exchanged. More generally,  for any function $g: X \times Y \to [0,\infty)$, we have $$\sup_{x \in X} \sup_{y \in Y} g(x,y)=\sup_{y \in Y} \sup_{x \in X} g(x,y) \,.$$
Addendum 2: As requested by the OP in a comment, let's verify that given a measure  space $(\Omega,\mathcal F,\mu)$ and measurable functions  $g_i:\Omega \to \mathbb R$, we have
$$\max_{i \le n} \;  \text{ess-}\!\sup g_i=  \text{ess-}\!\sup  g \,, \tag{*}$$
where $g(x):=\displaystyle \max_{i \le n} g_i(x)$ for all $x \in \Omega$.
First, since $g_i \le g$ pointwise for each $i$, we obtain
$$ \text{ess-}\!\sup g_i \le \text{ess-}\!\sup  g \,, $$
so the LHS of $(*)$ is at most the RHS.
Second, suppose that $$b>\max_{i \le n} \;  \text{ess-}\!\sup g_i\,. \tag{**}$$
Then $\mu\{x \in \Omega: g_i(x)>b\}=0$ for each $i$, so
$$\mu\{x \in \Omega: g(x)>b\}= \mu\Bigl(\cup_{i \le n} \{x \in \Omega: g_i(x)>b\}\Bigr)=0 \,.$$
Thus $\text{ess-}\!\sup  g \le b$. Since this holds for every $b$ satisfying $(**)$, we have verified that the RHS of $(*)$ is at most the LHS.
https://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf
https://en.wikipedia.org/wiki/Essential_infimum_and_essential_supremum
