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 ?


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 $n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.