A basic question on the induced SOT topology on a subspace I am trying to understand the concepts of operator topologies and I am struggling with the technical definitions. I asked myself a basic question: Suppose H is a Hilbert Space and consider the space B(H) with the Strong Operator Topology. Let U be a subspace of H and now consider B(U) with the SOT on it. Denote the topology T_u. If we regard B(U) as a subset of B(H) (we identify every operator on U with an operator coinciding on U and giving zero outside U), is the subspace topology (induced from the SOT on B(H)) equals T_u? Thanks in advance!

You cannot just define $T=0$ 'outside' $U$. That would destroy linearity. Assuming that $U$ is a closed subspace we can write $H=U+U^{\perp}$ and any $T$ in $B(U)$ gives rise to an operator $T'$ on $H$ by defining $T'(u+v)=T(u)+0$ for $u \in U, v \in V$. What you are asking is: does $T_n(x)\to T(x)$ for all $x \in U$ imply $T_n'(z) \to T'(z)$ for all $z \in H$. This is true as you can see by just writing $z$as $u+v$ with $u \in U, v \in V$.