Question: Why was this marked as a duplicate? The referenced question asked nearly a year later than this question. In fact I'm not even sure that they are identical at all.
I was working on a homework problem which stated:
Let $X$ be a normed linear space with two norms $\|\cdot\|_1$ and $\|\cdot\|_2$. Assume that $\|x_n - x^1\|_1\to 0$ for some $x^1\in X$ if and only if $\|x_n - x^2\|_2\to 0$ for some $x^2\in X$. Show that $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.
At first I errantly misinterpreted the question as $x_n\to x$ in $\|\cdot\|_1$ if and only if $x_n\to x$ in $\|\cdot\|_2$.
UPDATE: I found a way to go directly from the convergence condition to equivalence of norms. So I don't need to prove that the sequences converge to the same limit in this way (since it now follows from equivalence of norms). I won't erase anything though, as the answer from Andres below may be helpful to someone else.
Now that I realize my mistake I'm trying to fix the proof. Since the conclusion I'm trying to prove (that the norms are equivalent) implies that the limits in either norm must agree, I know that must also be true, even though it's not given in the question. After failing to prove this extra condition, I am getting suspicious that it might be needed as a hypothesis after all. Can anyone confirm or refute this?
My attempt:
Taking $\alpha = \dfrac12\min\left(\|x^1 - x^2\|_1, \|x^1 - x^2\|_2\right)$,
I let $B_{1} = \{x\in X : ||x - x^1||_{1} < \alpha\}$ and let $B_{2} = \{x\in X : ||x - x^{2}||_{2} < \alpha\}$.
Now suppose for a contradiction that there is some $x\in B_{1}\cap B_{2}$. Then $||x - x^{1}||_{1} < \alpha$. And $||x - x^{2}||_{2} < \alpha$. I want to use this to prove $||x^{1} - x^{2}||$ (in either norm) is less than $2\alpha$ (which would give me a contradiction). But I keep ending up with these floating $||x - x^{2}||_{1}$ or $||x - x^{1}||_{2}$ terms that I can't do anything with:
$ \begin{eqnarray*} ||x^{1} - x^{2}||_{1} &\leq& ||x^{1} - x||_{1} + ||x - x^{2}||_{1}\\ &<& \alpha + ||x - x^{2}||_{1} \end{eqnarray*} $
and
$ \begin{eqnarray*} ||x^{1} - x^{2}||_{2} &\leq& ||x^{1} - x||_{2} + ||x - x^{2}||_{2}\\ &<& ||x^{1} - x||_{2} + \alpha \end{eqnarray*} $