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This seems to be very interesting result , ie

If operators $\{A_n\}$ in Banach space $B(X)$ and if $A_n \to A$ , $A \in B(X)$in operator norm then $\lambda_n \in \sigma(A_n)$ ie spectrum of $A_n$ then $\lambda_n$ converges to $\lambda$ and $\lambda$ is spectrum of $A$ .

Can you help me to prove that the above statements is infact true ? Thank you .

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To be precise, are you saying that if $A_n \to A$ in the operator norm, and if there exists a sequence $\{\lambda_n\}$ with $\lambda_n \in \sigma(A_n)$ that happens to converge to some $\lambda$, then $\lambda \in \sigma(A)$? –  Branimir Ćaćić Jan 20 '13 at 17:43
    
@BranimirĆaćić : Yes , exactly . –  Theorem Jan 20 '13 at 17:45

1 Answer 1

  • If $\lVert A_n-A\rVert_{\mathcal B(X)}\to 0$ and $\lambda_n\to \lambda$, then $\lVert (A_n-\lambda_nI)-(A-\lambda I)\rVert_{\mathcal B(X)}\to 0$.
  • The set of non-invertible operators is closed for the norm topology of $\mathcal B(X)$.
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