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: 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 commented in the discussion of a similar question on MSE.


$|\cdot|$ 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.

  • $\begingroup$ You are perfectly right, so simple! Thanks! $\endgroup$ – Skeeve Dec 9 '19 at 13:46

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