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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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So you mean norm convergence? –  JSchlather Dec 9 '12 at 20:25
Yes I mean norm convergence –  Joe Dec 9 '12 at 20:25
Actually, an operator has the same norm as the norm of its adjoint. –  Davide Giraudo Dec 9 '12 at 20:34
@DavideGiraudo Get it. –  Joe Dec 9 '12 at 20:41

1 Answer 1

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

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