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Let $T_n$ be a sequence of bounded normal operators on a Hilbert space which converges to a normal operator $T$ in the strong operator topology. Show that $T_n^*$ also converges to $T^*$ in SOT.

I have tried to telescope the terms but it leads to nowhere.

ML So
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    I think one has $|T_n^\ast x - T^\ast x|=|T_n x - T x|$. This would then easily imply the result. – PhoemueX Apr 09 '22 at 07:30
  • @PhoemueX This formula was my first idea too, but it seems that for a proof $T_n T^\ast =T^\ast T_n$ is needed. I dont't see this. – Gerd Apr 09 '22 at 08:32
  • See also here https://math.stackexchange.com/questions/3309734/normal-operator-iff-norm-on-v-equivalent-to-that-of-adjoint – PhoemueX Apr 09 '22 at 09:24
  • @PhoemueX I agree that $|Tx|= |T^\ast x|$ for a normal operator $T$, but the sum (or difference) of normal operators is not normal, in general, see https://math.stackexchange.com/questions/368757/is-the-sum-of-two-normal-operators-normal – Gerd Apr 09 '22 at 11:46
  • @Gerd: Oh, right, damn. I knew this at some point :) Thanks for reminding me. – PhoemueX Apr 09 '22 at 12:51

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We have $$ \|T_n^\ast x-T^\ast x\|^2 = \langle T_n^\ast x,T_n^\ast x \rangle -\langle T_n^\ast x,T^\ast x \rangle -\langle T^\ast x,T_n^\ast x \rangle +\langle T^\ast x,T^\ast x \rangle $$ $$ =\langle T_n x,T_n x \rangle -\langle x,T_nT^\ast x \rangle -\langle T_nT^\ast x,x \rangle +\langle T x,T x \rangle $$ $$ \to \langle T x,T x \rangle -\langle x,TT^\ast x \rangle -\langle TT^\ast x,x \rangle +\langle T x,T x \rangle $$ $$ =\langle T x,T x \rangle -\langle x,T^\ast T x \rangle -\langle T^\ast T x,x \rangle +\langle T x,T x \rangle $$ $$ =\langle T x,T x \rangle -\langle Tx,T x \rangle -\langle T x,Tx \rangle +\langle T x,T x \rangle =0. $$

Gerd
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