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Does there exist a vector space $X$ and two norms $\Vert \cdot \Vert$ and $\Vert \cdot \Vert_1$ on $X$ such that both spaces $(X, \Vert \cdot \Vert)$ and $(X, \Vert \cdot \Vert_1)$ are complete, but the two norms $\Vert \cdot \Vert$ and $\Vert \cdot \Vert_1$ are not equivalent?

A fact: When one of these two norms is stronger than the other one, then they are equivalent. So I want to find a counter example, where the above condition does not hold.

Jean Marie
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Lei Zhang
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  • I have modified your title in order for it to be more explicit 1) do you agree 2) there is a problem, you say "compact" in your title and you say "complete" in your text... – Jean Marie May 03 '17 at 23:07
  • That is, Can a vector space be complete for two incomparable norms? – Lei Zhang May 03 '17 at 23:19
  • Thus "compact" in your title should be replaced by "complete", all right ? – Jean Marie May 03 '17 at 23:20
  • @JeanMarie Yes, Thanks. – Lei Zhang May 03 '17 at 23:21
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    https://math.stackexchange.com/questions/613034/can-a-vector-space-be-complete-for-two-non-compatible-norms I found an answer, but it is not very clear for me. – Lei Zhang May 03 '17 at 23:24
  • The key part of Daniel Fischer's answer is the bit about the open mapping theorem. See also flytothesurface's answer to a question I asked a while ago on math.SE: https://math.stackexchange.com/questions/2158628/applications-of-the-open-mapping-theorem-for-banach-spaces – Kenny Wong May 03 '17 at 23:30
  • Motivated by the Daniel Fischer's answer, I found an answer. Let $X= C(\overline{\Omega})$ with $\Vert u \Vert = \sup_{ x \in \mathbb{R}} \vert u(x) \vert$ and $\Vert u \Vert_1 = \sup_{ x \in \mathbb{R}} \vert x u(x) \vert$.

    And I am now understand Daniel Fischer's answer.

    – Lei Zhang May 03 '17 at 23:43

1 Answers1

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Let $X$ be a separable infinite-dimensional Banach space (for instance $X=\ell_p$ for some $p\in [1,\infty]$). Every infinite-dimensional separable Banach space has Hamel basis of cardinality continuum; let us choose any such space $Y$ that is not isomorphic to $X$ (for instance, $c_0$ if $X=\ell_p$). As $X$ and $Y$ have bases of the same cardinality, there exists a bijective linear map $T\colon X\to Y$ which sends one basis onto the other. Define

$$\|x\|^\prime = \|Tx\|\quad (x\in X).$$

Then this is essentially the norm in $Y$, hence inequivalent as we have taken $Y$ to be non-isomorphic to $X$.

Tomasz Kania
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