# Convergence of sequence vs NOT equivalent norms

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 $$d.$$