Is there a pair of norms such that they are not equivalent, but for any sequence of $(x_k)$, we have $\left \| x_k-x \right \|_1\to 0$ if and only if $\left \| x_k-x \right \|_2\rightarrow 0$?
I know if the norms are equivalent then it follows, but it's not stated as a necessary and sufficient condition for norms to be equivalent, I got curious. Many thanks in advance!
Let $V$ be a real/complex vector space and $\|\cdot\|_1$ and $\|\cdot\|_2$ two norms, and denote the corresponding normed vector spaces as $V_1,V_2$. Let $I:V\to V$ denote the identity map on the set $V$.
If each convergent sequence in $V_1$ converges in $V_2$ to the same element then it is saying $\|x_k-x\|_1\to 0$ implies $\|I(x_k)-I(x)\|_2=\|x_k-x\|_2\to 0$. Thus $I:V_1\to V_2$ is continuous. Reversing the roles of $1,2$ gives the other statement that $I:V_2\to V_1$ is continuous. Therefore, with your hypothesis, the identity map $I:V_1\to V_2$ is a homeomorphism. This means the topology on $V_1$ is equal to the topology on $V_2$, and thus by definition, the two norms are equivalent.
Assume $\|x_k\|_1\to 0$ iff $\|x_k\|_2\to 0.$ Then the norms are equivalent. Indeed, assume for a contradiction that there is no constant such that $\|x\|_2\le c\|x\|_1.$ Hence for any $n$ there is $x_n$ such that $\|x_n\|_2> n^2\|x_n\|_1.$ WLOG (by the homogeneity of the norms) we may assume $\|x_n\|_1=n^{-1}.$ Then $\|x_n\|_1\to 0$ but $\|x_n\|_2>n,$ i.e. $\|x_n\|_2\not\to 0,$ a contradiction. Reveresing the roles of $\|\cdot\|_1$ and $\|\cdot\|_2$ we get $ \|x\|_1\le d\|x\|_2$ for a positive constant $d.$
