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I have just been introduced to the concept of invertibility for bounded linear operators. Specifically, we defined a bounded operator $A$ to be invertible if there exists a bounded $A^{-1}$ which is its right and left inverse, i.e. $AA^{-1}=\mathrm{id}_{\mathrm{Im}A},A^{-1}A=\mathrm{id}_{\mathrm{Dom}A}$. So I was wondering: is the requirement of boundedness (or equivalently of continuity) of the inverse important? Or, is it asking more than is granted by the invertibility? The open mapping theorem states a continuous linear surjective operator between Banach spaces is an open map, and thus if it has an inverse, that inverse is necessarily continuous. So for Banach spaces, we could avoid requiring this continuity explicitly, as it is automatic. But what about non-Banach spaces?

So the question is: do there exist normed spaces $X,Y$ and operators between them which are linear and bounded but have un-bounded inverses?

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Yes, unbounded inverses exist. Here is an example where $X$ and $Y$ are normed spaces.

For more examples, see Mohamed's answer here, where $X$ is actually Banach.

And for examples with $Y$ Banach, see this overflow post.

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Yes, bounded inverses of unbounded operators are very important. Let $H$ be an unbounded self-adjoint operator acting in a Hilbert space $\mathcal{H}$ . Then $z-H$ with $Imz\neq 0$ is also unbounded but $[z-H]^{-1},$the resolvent of $H$ is bounded. The set $\{z\}$ for which the resolvent exists and is bounded is known as the resolvent set. Its complement is the spectrum of $H$. Resolvents play an essential role in the study of Schrödinger operators.

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  • $\begingroup$ The question was about unbounded inverses of bounded operators. $\endgroup$ – Alex M. Jul 21 '18 at 6:42
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The easier example is the identity with different norms in each side, the norm in the right strictly finer than the norm in the left.

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