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I was wondering whether the inverse Operator $A\mapsto A^{-1}$ is continuous on the set of continuously invertible linear operators $G\subset L(X)$, where X is supposed to be a Banach space (Can this may be weakend?). If this is not the case, do anyone has a counterexample. Thanks in advance.

Best Aristo

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Yes. In fact if $A$ is a Banach algebra with identity and $G$ is the group of invertible elements then the map $x\to x^{-1}$ is continuous on $G$:

Say $x\in G$. If $||y||<1/||x^{-1}||$ then $||yx^{-1}||<1$, hence $x-y=(e-yx^{-1})x\in G$ and in fact $$(x-y)^{-1}=((e-yx^{-1})x)^{-1}=x^{-1}(e-yx^{-1})^{-1} =x^{-1}(e+yx^{-1}+(yx^{-1})^2+\dots).$$This shows that $(x-y)^{-1}\to x^{-1}$ as $y\to 0$: $$||(x-y)^{-1}-x^{-1}||\le||x^{-1}||(||x^{-1}||(||y||\,||x^{-1}||+||y||^2||x^{-1}||^2+\dots).$$

(If you don't know anything about Banach algebras you can find an answer to your question above by assuming $x$ and $y$ are operators and $e=I$. Note that we use the fact that $||xy||\le||x||\,||y||$ in various places.)

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  • $\begingroup$ Thank you very much David for your nice answer. Notice that you forgot a facor $\Vert x^{-1}\Vert$ on the right hand side of the ineqality, which does not make the proof false. Best Aristo $\endgroup$ – Aristo Nov 25 '17 at 15:38

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