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I'm self teaching Metric Spaces and came across this question:

Let X denote the vector space of sequences $x = (x_n)$ with finite sum of squares. Explain why $||x|| := sup|x_n|$ is a well-defined norm on X. Is the metric induced by this norm equivalent to the metric induced by the $l_2$-norm?

I've shown that the norm is well defined. From my understanding, these two norms are equivalent. My definition of equivalent metrics is that they are equivalent if the identity map is a homeomorphism. Would appreciate some help, thanks!

yw_2003
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    For infinite sequence space these two norms are not equivalent. You can easily show this by the fact that $X$ is not complete w.r.t. the sup-norm but complete w.r.t. the $l^2$-norm. – Zerox Oct 17 '22 at 13:33
  • @Zerox Can I ask why? – yw_2003 Oct 17 '22 at 13:35
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    $x_n=(1,\dfrac{1}{\sqrt{2}},\cdots,\dfrac{1}{\sqrt{n}},0,0,\cdots) \in X$ is a Cauchy sequence w.r.t. the sup-norm but it doesn't converge in the $l^2$-norm. – Zerox Oct 17 '22 at 13:45
  • @Zerox thank you! – yw_2003 Oct 17 '22 at 13:47

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This is false.

Take $(x_{n})$ with $x_{n}=1$. Then $\|x\|_{l^{2}}=\infty$ but $\|x\|_{l^{\infty}}=1$.

Medo
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For a finite dimensional vector space, such as $\mathbb R^n$, all norms are equivalent. See Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces for a proof.

For an infinite dimensional vector space, such as $L^p$, the norms are not equivalent. Counter examples were given in the other answers.

Syd Amerikaner
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