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 $\|x\|_1=\|i(x)\|_1=\|x\|_2$?
Best Answer
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.