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If a sequence of operator $A_n$ converges in norm to $A$, i.e. $\lim \lVert A_n-A\rVert=0$)where $A_n$ and $A\in B(H)$ ($H$ is the Hilbert space). Is it true that $A_n^*$ converges in norm to $A^*$?

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  • $\begingroup$ So you mean norm convergence? $\endgroup$ – JSchlather Dec 9 '12 at 20:25
  • $\begingroup$ Yes I mean norm convergence $\endgroup$ – 89085731 Dec 9 '12 at 20:25
  • $\begingroup$ Actually, an operator has the same norm as the norm of its adjoint. $\endgroup$ – Davide Giraudo Dec 9 '12 at 20:34
  • $\begingroup$ @DavideGiraudo Get it. $\endgroup$ – 89085731 Dec 9 '12 at 20:41
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Since $(A_n-A)^*=A_n^*-A^*$ and an operator has the same norm with the adjoint operator,so $||A_n-A||=||A_n^*-A^*||$,then we get the answer.

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  • $\begingroup$ The step "an operator has the same norm as its adjoint" deserves more details. $\endgroup$ – Davide Giraudo Dec 9 '12 at 20:55

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