Standard form of $B(H)$ [duplicate]

Question

For a Hilbert space $H$, consider the representation $$ \pi_{l}:B(H) \to B(L^2(H)): \pi_{l}(x)(y):=xy$$ where $L^2(H)$ denotes the Hilbert Schmidt class operators in $B(H)$.

I want to show that this defines an injective $*$-homomorphism. In the provided solution, as $\pi_{l}(x)(y)=xy=0$ they simply take $y=x^*$ and thus $xx^*=0$ and conclude that $x=0$. I do not understand why they can just take $y=x^*$ as for an arbitrary $x\in B(H)$, $x^*$ need not be in $L^2(H)$ and $\pi_{l}(x)$ is defined on $L^2(H)$.

Answer 1

As you say, you cannot take $x^*$ because it might not be Hilbert-Schmidt. But the finite-rank (and hence the Hilbert-Schmidt) operators are dense in $B(H)$, so you can write $x^*=\lim_j z_j$, where $z_j$ is finite-rank and the limit is in the weak or strong operator topology. Then $$ xx^*=\lim_j xz_j=0. $$