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Consider two norms in the space $c_0$: $$\lVert x \rVert = \sup \lvert x_i \rvert$$ and $$\lVert x \rVert _0=\sum 2^{-i} \lvert x_i \rvert$$ Prove that above two norms are not equivalent.


I know $c_0$ with norm $\lVert \cdot \rVert$ is Banach space. And I guess that $c_0$ with the latter norm is not Banach. However, I can't find any example to show that.

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Think about sequence $e_n \in c_0$, such that $e_n$ has $1$ on $n^{th}$ place and zero otherwise. Then $\Vert e_n \Vert_0 \rightarrow 0$, but $\Vert e_n \Vert = 1$ for all $n \in \mathbb N$.

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  • $\begingroup$ Do you have any example that shows $c_0$ with norm $\lVert . \rVert _0$ is not Banach space? $\endgroup$ – RuaSun Oct 27 '15 at 6:51
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It is clear that $\|x\|_0\le\|x\|$. To show that they are not equivalent you need to show that $\|x\|/\|x\|_0$ is unbounded. For this, try to choose sequences $a_n\in c_0$ such that $\|a_n\|/\|a_n\|_0\to\infty$ as $n\to\infty$. There is a pretty obvious choice.

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  • $\begingroup$ Many thanks. I see. But can we prove that $c_0$ with second norm is not Banach space? And therefore those norms are not equivalent. I'm finding a Cauchy sequence in $c_0$ converging to an element that not in $c_0$. I have no result for the time being. $\endgroup$ – RuaSun Oct 27 '15 at 6:56
  • $\begingroup$ No, $c_0$ with $\|\cdot\|_0$ is not complete. If $a_n=(1,\dots,1,0,\dots)$ has $n$ $1$'s, then $\{a_n\}$ is Cauchy for the norm $\|\cdot\|_0$, but does not converge in $c_0$. $\endgroup$ – Julián Aguirre Oct 27 '15 at 9:49

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