For two finite sets $A$ and $B$, let $x_{i,j}$ be in both $\ell^2(A)$ and $\ell^\infty(B)$. Then is it possible that $$ \| \|x_{i,j} \|_{\ell^\infty(B)}\|_{\ell^2(A)} = \| \|x_{i,j} \|_{\ell^2(A)}\|_{\ell^\infty(B)} $$ or inequality $$ \| \|x_{i,j} \|_{\ell^\infty(B)}\|_{\ell^2(A)} \le C\, \| \|x_{i,j} \|_{\ell^2(A)}\|_{\ell^\infty(B)} $$ holds for some $C$?
I tried to use the $\ell^p$ inclusion for counting measure, which is $\ell^\infty \subset\ell^1 \subset \ell^2$ but I cannot conclude it. Can I have some hint or counterexample (if it is false)?
EDIT: The $x_{i,j}$ are decomposed vectors of some vector $x\in \mathbb{R^n}$ by $$x=\sum_{i\in A} \sum_{j\in B} x_{i,j}.$$
$C=\sqrt n$ works. Just use the fact that $\|y\|_2\leq \sqrt n \|y\|_{\infty}$ to see that LHS is dominatted by $\sqrt n$ times the same expression where both norms are replaced by $\|.\|_{\infty}$. The inequality then becomes obvious.