Do there exist a real vector space $X$ *complete* with respect to norms $|\cdot|$ and $\|\cdot\|$ and a sequence $(x\_n)\_{n\in \mathbb N} \subset X$ such that there exist $x,y\in X$: $x\ne y$, $|x\_n - x|\to 0$ and $\|x\_n -y\|\to 0$ as $n\to \infty$?


Without requiring $X$ to be complete with respect to $|\cdot|$ and $\|\cdot\|$ this is possible. In particular one can repeat [Bill Johnson's construction](https://mathoverflow.net/a/53978/44463): let $(e\_n)\_{n\in\mathbb N}$ be the standard orthonormal basis of $\ell\_2$. Let $A\colon \ell\_2 \to \ell\_2$ be a linear operator such that $A(e\_1 + \frac{1}{n} e\_n) = \frac{2}{n}e\_n$ for any $n\in \mathbb N$ (it can be constructed by extending the linearly independent sets $(e\_1 + \frac{1}{n} e\_n)\_n$ and $(\frac{2}{n} e\_n)\_n$ to Hamel bases of $\ell\_2$ and extending the bijection between these bases to a linear isomorphism). Let $\|\cdot\|$ be the standard norm of $\ell\_2$ and let $|x| := \|Ax\|$, $x\in \ell\_2$. Then clearly $x\_n = e\_1 + \frac{1}{n} e\_n$ is the desired sequence (with $x=0$ and $y=e\_1$).


In this example clearly $(X, \|\cdot\|)$ is Banach, but it is not clear whether $(X, |\cdot|)$ is Banach, as [Christopher A. Wong](https://math.stackexchange.com/questions/202493/example-of-a-sequence-that-converges-to-two-different-limits-with-respect-to-two?rq=1#comment6133149_205042) commented in the discussion of a [similar question on MSE](https://math.stackexchange.com/q/202493/116974).



Sequence converging to different limits with respect to two different _complete_ norms
$|\cdot|$ is complete. Indeed, suppose $(x\_n)$ is a Cauchy sequence in $(X,|\cdot|)$. Then $(Ax\_n)$ is a Cauchy sequence in $(X, \|\cdot\|)$ which is complete and hence has a limit $y$. Since $A$ was constructed to be a linear isomorphism, $y=Ax$ for some $x\in X$. Therefore $|x\_n-x|=\|A(x\_n-x)\|=\|Ax\_n-y\|$ tends to $0$, proving $(x\_n)$ is convergent.


