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## Comparable Norm on Banach [closed]

Let $\mathcal{X}$ be a vector space, $| \cdot |$ and $\| \cdot \|$ be two norms on which $\mathcal{X}$ is complete with respect to both. Can the two norms be not equivalent? Please just give me some hint, I want to be able to do it myself. Thx!

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 Please don't give it away, just hint please – Li Oct 7 at 0:22 Bill Johnson gave a hint (which might be a giveaway...). – Todd Trimble Oct 7 at 0:31

## closed as not a real question by Bill Johnson, Qiaochu Yuan, Deane Yang, Chris Godsil, algoriOct 7 at 0:39

Take $\mathcal{X}=\ell_1(\mathbb{N})$. Then $\mathcal{X}$ is complete under $\ell_1$ and $\ell_\infty$, while the two norms are not equivalent.

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 $l^1(\mathbb{N})$ is not complete under the sup norm... Just take the sequence $x_n = (1,1/2,...,1/n,0,...)$. Then it is Cauchy under the sup norm, but can't converge to anything in $l^1(\mathbb{N}$ – Li Oct 7 at 0:16 I have thought about giving a wrong answer to a nonsense question but never had the guts to do so. Congratulations, Aryeh. – Bill Johnson Oct 7 at 0:17 Oh, right. My mistake. – Aryeh Kontorovich Oct 7 at 0:18 Why is this question nonsense? – Todd Trimble Oct 7 at 0:20 Any two separable infinite dimensional Banach spaces have the same Hamel dimension, Todd. – Bill Johnson Oct 7 at 0:26
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Let's see. Do two infinite dimensional Banach spaces that have the same density character also have the same Hamel dimension? If so, the cited result is trivial (for then every infinite dimensional Banach space obviously has a norm on it that makes it isomorphic to a Hilbert space direct summed with $\ell_p$, $1\le p < \infty$). Of course, the cited article contains other theorems. – Bill Johnson Oct 7 at 0:50