# Example of a sequence that converges to two different limits with respect to two complete norms

I've wondered about the following question :

Is there an (explicit?) example of a vector space $X$, two complete norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on $X$, and a sequence $(x_n) \subseteq X$ such that $x_n$ converges to $x$ with respect to $\|\cdot\|_1$, $x_n$ converges to $y$ with respect to $\|\cdot\|_2$, but $x \neq y$?

Obviously, this would imply that $\|\cdot\|_1$ and $\|\cdot\|_2$ are not equivalent. In fact, these two statements are equivalent, which is a consequence of the Open Mapping Theorem.

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Are those norms the $L_1$ and $L_2$ norms, or placeholders for arbitrary norms? – templatetypedef Sep 25 '12 at 23:39
No, they are arbitrary norms. Would it be better to use another notation? – Malik Younsi Sep 25 '12 at 23:39

Bill Johnson's example in MathOverflow seems answer the question.

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Yes, you're right, it does! Thank you. It relies on the construction of a Hamel basis though, which requires the axiom of choice. Out of curiosity, is it known if the axiom of choice is necessary to obtain the existence of two inequivalent complete norms? – Malik Younsi Oct 1 '12 at 12:45

This is not possible! Obviuosly, all norm topologies are equivalent for findim linear spaces, so you must search for an example in infinite-dim spaces. Assuming $x_n$ converges to $x$ in $\| \cdot \|_1$ but converges to $y$ in $\|\cdot \|_2$, and choose $a = \min\{ \| x-y\|_1, \| x-y \|_2 \}$. Then you can easily get an counterexample by choosing $n$ so large that $x_n$ are within $a/3$ of $x$ measured by the first norm, and within $a/3$ of $y$ measured with the second norm. Then use the triangle inequality for the distance between $x$ and $y$, write $x-y = (x-x_n) - (y - x_n)$. Done.

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I dont think this is correct. When you use the triangle inequality, which norm do you use, the first one or the other one? – Malik Younsi Sep 25 '12 at 23:50
You have only two norms! The argument works with both, by the way we did choose $a$. – kjetil b halvorsen Sep 25 '12 at 23:57
Let me be more explicit : you have that $\|x_n-x\|_1<a/3$ and $\|x_n-y\|_2<a/3$. So, for example, $$\|x-y\|_1 = \|(x-x_n)-(y-x_n)\|_1 \leq \|x-x_n\|_1 + \|y-x_n\|_1 < a/3 + \|y-x_n\|_1.$$ But how do you estimate $\|y-x_n\|_1$? – Malik Younsi Sep 26 '12 at 0:04
googling, I found these, which can be relevant: mathoverflow.net/questions/53971/… – kjetil b halvorsen Sep 26 '12 at 0:53
This is also relevant: math.stackexchange.com/questions/159810/… – kjetil b halvorsen Sep 26 '12 at 0:56

Not sure but I would guess no. Maybe see if $(X, ||.||_1 + ||.||_2)$ is also a Banach space then you'll probably get a contradiction if you assume x is not y.

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But how do you prove that $(X, \|\cdot\|_1+\|\cdot\|_2)$ is Banach without assuming the uniqueness of limit property, as in the question? – Malik Younsi Sep 25 '12 at 23:40
Yeah, I don't think you're going to get any easy example. Looking here homepage.ntlworld.com/ivan.wilde/notes/fa1/fa1.pdf it seems like any construction of a Banach space with two non equivalent norms requires some abstraction (here the existence of a Hamel basis) – Blitzer Sep 26 '12 at 1:30
In other words, you're probably gonna have to use choice somewhere – Blitzer Sep 26 '12 at 1:31
Yes, perhaps choice is unavoidable here. I'm curious to see if it's the case! – Malik Younsi Sep 26 '12 at 1:50