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Let $(T_n)\subset B(H)$ be a sequence of operators such that $T_n\to 0$ in strong operator topology. Show that $\|T_nK\|\to 0$ and $\|KT_n\|\to 0$ for every compact operator $K$.

Let $f,g \neq 0 \in H$, and define $f\otimes g=\langle .,g \rangle h$. It's a rank one operator. Since finite rank operators are dense in $\mathcal{K}(H)$ (the subspace of compact operators) it is sufficient to show that the result is true for rank one operators.

We have $\|T_nf\otimes g \|=\|T_n(f)\otimes g\|=\|T_n(f)\|\|g\|\to 0$. That show $\|T_nK\|\to 0$. But for $\|Kt_n\|$ it seems a bit more difficult since $KT_n=f\otimes T^*_n(g)$ and there is no reason that $\|T^*_n(g)\|\to 0$ because the involution $*$ is not continuous for the strong operator topology. How can I avoid this problem ?

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The assertion $\|KT_n\|\to0$ does not hold in general. Let $H=\ell^2(\mathbb N)$, and $$ T_n(a_1,a_2,\ldots,)=(a_n,a_{n+1},\ldots). $$ Then $T_n\to0$ in the strong operator topology.

Consider the rank-one operator $P$ given by $P(a_1,a_2,\ldots)=(a_1,0,0,\ldots)$. Then $$ PT_n(a_1,a_2,\ldots)=(a_n,0,0,\ldots). $$ If $\delta_n$ is the sequence with $1$ in the $n^{\rm th}$ position and zeroes elsewhere, $$ \|PT_n\delta_n\|=1, $$ so $\|PT_n\|=1$ for all $n$.

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