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I'm confused about the common proof of the equivalence of the norms in Banach spaces. It was very clear and nice answered here equivalent norms in Banach spaces of infinite dimension, but i still don't understand, why we can use identity map?

More precisely: Let $X$ be a Banach space and $i: (X, \|\cdot\|_2) \rightarrow (X, \|\cdot\|_1), x \mapsto x$ be the identity map on $X$ (i.e. $i(x)=x$ for all $x \in X$). It is given, that $\|x\|_2 \le C \|x\|_1, C > 0$. Then $i$ is continuous, since $\|i(x)\|_1=\|x\|_2 \le C \|x\|_1$ (by assumption).

The only question is: why it does not means, that $\|x\|_1=\|i(x)\|_1=\|x\|_2$?

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Why should you be able to conclude that $||x||_1=||x||_2$? This is just not true. Consider $X=\mathbb{R}^n$ with $||\cdot||_1=||\cdot||_{L^1}$ and $||\cdot||_2=||\cdot||_{L^\infty}$. Then these norms are equivalent (in particular, both identity maps are continuous) but not identically equal.

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  • $\begingroup$ My confusion comes from the fact, that $i(x)=x$. Then if we know that $\|i(x)\|_1=\|x\|_2$ and if we can write instead of $i(x)$ just its value ($x$) then $\|x\|_1=\|i(x)\|_1=\|x\|_2$. Where am I wrong? $\endgroup$
    – Thomas J
    Dec 12 '19 at 19:32
  • $\begingroup$ How do you know that $||i(x)||_1 = ||x||_2$? $\endgroup$ Dec 12 '19 at 20:11
  • $\begingroup$ @KeeperOfSecrets, I know it, since by definition $i$ maps from one normed space (with the norm $\| \cdot \|_2$) to another normed space (with the norm $\| \cdot \|_1$). But why we can not take a value of $i(x)$ inside of $\| \cdot \|_1$, when $i$ is identity? I know, that it is maybe a very stupid argument, but I don't see any contrdiction, or an argument, why we can not to do so... $\endgroup$
    – Thomas J
    Dec 12 '19 at 21:20
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    $\begingroup$ Don't think of these as the same space. They are not. Just because $i$ maps identically on the elements, does not mean that $i(x)\in (X, ||\cdot||_2)$. This is not true. What is true is that $i(x)\in (X, ||\cdot||_1)$ and $x \in (X, ||\cdot||_2)$. Think of it like this: $i((x, ||\cdot||_2))=(x, ||\cdot||_1)$ $\endgroup$ Dec 12 '19 at 21:57

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