Is possible to find an example of a Banach space $E_1$ and normed vector space $E_2$ and bijection operator $A \in \mathcal{L} (E_1, E_2)$, that $A^{-1}$ will not be bounded?
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$\begingroup$ See this. $\endgroup$– David MitraCommented Jun 3, 2014 at 16:34
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$\begingroup$ @DavidMitra That one goes in the other direction, though. Here the domain is assumed complete, not the codomain. $\endgroup$– user147263Commented Jun 5, 2014 at 13:37
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$\begingroup$ @wordsthatendinGRY The question contains the standard example. $\endgroup$– David MitraCommented Jun 5, 2014 at 13:59
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You can even make $E_1$ a Hilbert space, say $\ell^2$. Define another norm on $\ell^2$ by $$\|x\|_*^2 = \sum_n \frac{|x_n|^2}{n^2}$$ The identity map $\operatorname{id}:\ell^2\to (\ell^2,\|\cdot\|_*)$ is bounded, but its inverse is not. Indeed, for the standard basis vectors we have $\|e_n\|_*= 1/n$ while $\|e_n\|_{\ell^2}=1$.