On the proof that $\|\cdot\|_1,\|\cdot\|_2,\|\cdot\|_\infty$ are equivalent on $\mathbb R^n$ On looking for the proof that $\|\cdot\|_1,\|\cdot\|_2,\|\cdot\|_\infty$ are equivalent on $\mathbb R^n$, I found the following question on MSE: 1 and 2 norm inequality.
In the accepted answer it is stated that $\|x \|_{\infty} \leq \|x \|_{2} \leq \|x \|_{1} \leq n \|x \|_{\infty}\ $ for $x\in\mathbb R^n$, and then proceeds to show so. At first I wasn't convinced that this statement showed that $\|\cdot\|_1$ and $\|\cdot\|_2$ were equivalent, but instead that they are both equivalent to $\|\cdot\|_\infty.$ In particular, there isn't provided some numbers $a,b\gt0:a\|x\|_1\le\|x\|_2\le b\|x\|_1$ holds.
Is the reasoning that since both $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent to $\|\cdot\|_\infty$ they must be equivalent to each other? More formally, is it correct to justify this reasoning by saying that the equivalence of norms is a equivalence relation, wherein we make use of transitivity? 
 A: Yes it is an equivalent relation. Say you have $3$ norms $||\cdot||_1,||\cdot||_2$ and $||\cdot||_3$ such that both $3,2$ are equivalent to $1$. That means that there exists positive constants $a,b,A,B$ such that 
$$a||\cdot||_1\leqslant ||\cdot||_2\leqslant b||\cdot||_1$$
and $$A||\cdot||_1\leqslant ||\cdot||_3\leqslant B||\cdot||_1$$
from these you can infer that 
$$\frac{A}{b}||\cdot||_2\leqslant ||\cdot||_3\leqslant\frac{B}{a}||\cdot||_2$$
hence $2$ and $3$ are equivalent. 
A: You can proceed as follows:
$$\|x \|_{\infty} \leq \|x \|_{2} \leq \|x \|_{1} \leq n \|x \|_{\infty}\le n\|x\|_2\le n\|x\|_1\le n^2\|x\|_\infty.$$
You can prove transitivity similarly.
A: If $||\cdot ||_1$ and $||\cdot ||_2$ are both equivalent to $||\cdot ||_\infty$, then we can find positive constants $c_1,c_2,d_1,d_2$, such that the following hold everywhere
$$
c_1||\cdot ||_1\leq ||\cdot||_\infty\leq c_2||\cdot||_1\\
d_1||\cdot ||_2\leq ||\cdot||_\infty\leq d_2||\cdot||_2
$$
We can then see that
$$
\frac{d_1}{c_2}||\cdot||_2\leq ||\cdot||_1\leq\frac{d_2}{c_1}||\cdot||_2
$$
and thus the 1 and 2 norms are equivalent.
