Because $C_i=L_iL_i^*$ is a covariance operator,
$$
\lVert L_i\rVert_{\rm HS}^2
=\operatorname{tr}(L_iL_i^*)=\operatorname{tr}C_i<\infty.
$$
Thus both factors are <Hilbert-Schmidt operator>[Hilbert-Schmidt]. The product of two Hilbert-Schmidt operators is a <trace-class operator>, with
$$
\lVert L_2^*L_1\rVert_1
\leq\lVert L_2\rVert_{\rm HS}\lVert L_1\rVert_{\rm HS}<\infty.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 225 / 3 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-225/3/ii/solution