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Let $A_n \overset{SOT}{\to} A$ where $A$ is invertible. Does $A_n^{-1} \overset{SOT}{\to} A^{-1}$? Does $A_n^{-1} \overset{WOT}{\to} A^{-1}$?


EDIT: Forgot to mention $\{A,A_n\}\in\mathscr{B(H)}$ where $\mathscr{H}$ is a separable Hilbert space.

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    What have you tried? On which space are you working? Normed spaces? Topological vector spaces?2012-05-20
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    I tried mucking around with the definitions... Wouldn't call anything a "lead", I'm very thoroughly stumped. Forgot to mention I'm working in $B(H)$ where $H$ is a separable Hilbert space.2012-05-20
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    Have you tried writing $$A_n^{-1}-A^{-1}=A_n^{-1}(A-A_n)A^{-1}?$$2012-05-20
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    [...]However to make this work we need some upper bound for $\lVert A_n^{-1}\rVert$. The existence of such a bound seems reasonable to me because $A$ is a bounded operator, but unfortunately I don't know if there is one or not.2012-05-20
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    I tried to infer one using the uniform boundedness principle. Guess I'll try again :/2012-05-21
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    I had not noticed the "Linked" on the right. There is a complete (and interesting) answer by t.b., in case you had not noticed it as well.2012-05-21

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