Definitions:
1 - An operator $y\in B(H)$ is said to be of trace class if $y$ is compact, and also $\sum|\alpha_n| <\infty$ where $\alpha_n \in \sigma(y)$ and $y$ has a representation $\sum \alpha_n \xi_n\otimes \eta_n $ for orthonormal systems $\{\xi_n\} , \{\eta_n\}$.
2 - An operator $x\in B(H)$ is said to be of Hilbert-Schmidt class if $x^*x$ is trace class.
By above definitions I can not show that $yx \in L^2(H) $ (Hilbert-Schmdt class), where $y\in B(H)$, and $x\in L^2(H)$. Please give me a hint. Thanks.
I am assuming $H$ is a separable Hilbert space. Let $(e_k)$ be an orthonormal basis of $H$ of eigenvectors of $x^*x$. Then $$ \sum \langle (yx)^*(yx) e_k,e_k\rangle =\sum \|yxe_k\|^2 \le \|y\|^2 \sum \|xe_k\|^2 =\|y\|^2_\infty \sum \langle x^*xe_k,e_k\rangle <\infty $$ because $x^*x$ is trace class and $y$ is bounded. Since the left-hand side is independent of the choice of basis (see Problem 5.64 page 433 here), we can replace the basis $(e_k)$ by a basis of eigenvectors of $(yx)^*yx$, which shows that it is trace class.