SOT convergence of normal operators

Question

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.

Answer 1

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