Let $H$ a Hilbert space and $A_n, A$ bounded and invertible operators such that $A_n\to A$ in the strong sense, that is for any $x\in H$, $$ ||A_n x - Ax||\to 0. $$ Is it true that $A_n^{-1}\to A^{-1}$ in the strong sense? Thank you for your help.
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A necessary condition would be that $\lVert A_n^{-1}\rVert$ be bounded. Does that follow? – Daniel Fischer Oct 28 '16 at 15:08
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Yes you are right. So the question could be, is the fact that $A$ is invertible forcing $A_n^{-1}$ to be uniformly bounded? – Alarico_folko Oct 28 '16 at 15:14
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Do you expect that to be the case? – Daniel Fischer Oct 28 '16 at 15:15
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I don't think so. But I can't find any counterexample. – Alarico_folko Oct 28 '16 at 15:16
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Let $A = \operatorname{id}$. Suppose that $H$ is separable, and let ${e_k : k \in \mathbb{N}}$ be an orthonormal basis. Let $A_n$ change only one coordinate. – Daniel Fischer Oct 28 '16 at 15:20
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Why $A_n^{-1}$ is not uniformly bounded? – Alarico_folko Oct 28 '16 at 15:26
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That depends on how you define $A_n$. It may be uniformly bounded, but it need not be. – Daniel Fischer Oct 28 '16 at 15:27
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Excuse me, I don't understand. Can you give me a specific example? – Alarico_folko Oct 28 '16 at 15:42
1 Answers
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It need not be the case that $A_n^{-1} \to A_n$ in the strong operator topology.
Let $H = \ell^2(\mathbb{N})$, and for every $n\in \mathbb{N}$ define
$$A_nx = x - \frac{n}{n+1}\langle x, e_n\rangle e_n.$$
Then $A_n \to A = \operatorname{id}$ strongly, for
$$\lVert x - A_n x\rVert = \frac{n}{n+1}\lvert \langle x, e_n\rangle\rvert \to 0,$$
but we have $A_n^{-1}x = x + n\langle x, e_n\rangle e_n$ and therefore $\lVert A_n^{-1}\rVert = n+1$, so $\{ A_n^{-1} : n \in \mathbb{N}\}$ is not uniformly bounded.
Specifically, for
$$x = \sum_{n = 0}^{\infty} \frac{1}{n+1}e_n$$
we have
$$\lVert A_n^{-1} x - x\rVert = \frac{n}{n+1}\to 1.$$
Daniel Fischer
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If the inverses are uniformly bounded, is the original question then true? – Rick Sanchez Oct 06 '18 at 21:33