TITLE: Example of sequences with different limits for two norms
QUESTION [33 upvotes]: I was explaining to my students that if there is an inequality between two norms, then there is an inclusion between their spaces of convergent sequences, with matching limits. I then proceeded to show examples of such inequalities on the normed spaces they knew, and counterexamples of sequences which converge for a norm and not for another, stating the equivalence of norms in finite dimension, etc.
It is then that I wondered about the following : does there exist a vector space, two norms on that vector space and a single sequence which converges for both norms, but with different  limits?
The first remark is that such a counter-example cannot exist in finite dimension ; and one first has to find "really inequivalent norms", which do exist : consider the space of polynomials in one variable, and define norms on it by summing the absolute values of the coefficients :

 first with a weight $1$ for every coefficient ;
 second with $2^n$ or $2^{-n}$ depending on the parity of the degree $n$.

It's now easy to find a sequence going to zero for the first and not for the second, and a sequence going to zero for the second and not for the first - so there can't be an inequality between those.
Notice this is all over the real or complex numbers, though the question could be amusing in a more general setting.

REPLY [4 votes]: I'm not sure if you would consider this an "example", but it seems this is not as widely known as I would expect.
Theorem. Let $X$ be a Banach space and let $||\cdot||_1, ||\cdot||_2$ be non-equivalent norms on $X$. Then there exists a sequence $(x_n)$ in $X$ and $x\neq y \in X$ such that $x_n \to x$ with respect to  $||\cdot||_1$ and $x_n \to y$ with respect to  $||\cdot||_2$.
Proof. By the bounded inverse theorem it must be that the identity map $\iota: (X, ||\cdot||_1) \to (X, ||\cdot||_2)$ is discontinuous. Thus by the closed graph theorem it must be that the graph of $\iota$ is not closed in $(X, ||\cdot||_1) \times (X, ||\cdot||_2)$. Since the graph is not closed, we may choose a sequence $(x_n, \iota(x_n))$ in the graph converging in $(X, ||\cdot||_1) \times (X, ||\cdot||_2)$ to some $(x,y)$ such that $(x,y)$ is not in the graph. Converging in $(X, ||\cdot||_1) \times (X, ||\cdot||_2)$ means that $x_n \to x$ in $(X, ||\cdot||_1)$ and $\iota(x_n) = x_n \to y$ in $(X, ||\cdot||_2)$. But $(x,y)$ not being in the graph means that $y \neq x$.