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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.

Davide Giraudo
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Malik Younsi
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2 Answers2

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

Community
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Davide Giraudo
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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 01 '12 at 12:45
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    I recently stumbled upon this question and very interesting example. However, it does not seem to me that in the example given, the second norm is complete. Does anybody have verification of this? – Christopher A. Wong Oct 25 '18 at 21:22
  • @ChristopherA.Wong I am also interested in this completeness question, and posted it on MO – Skeeve Dec 09 '19 at 13:33
  • @ChristopherA.Wong and in fact this norm turns out to be complete. – Skeeve Dec 09 '19 at 13:48
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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.

Blitzer
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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
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    Yeah, I don't think you're going to get any easy example. Looking here http://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 01:30
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    In other words, you're probably gonna have to use choice somewhere – Blitzer Sep 26 '12 at 01:31
  • Yes, perhaps choice is unavoidable here. I'm curious to see if it's the case! – Malik Younsi Sep 26 '12 at 01:50