Proving that $||AB||_{HS}\leq ||B||_H||A||_{HS}$ for Hilbert-Schmidt operator $A$ and a bounded linear operator $B$

Question

The trick I am looking for is likely some well-known trick in functional analysis that I have either not learned or have completely forgotten. I am trying to show that

$$||AB||_{HS}\leq ||B||_H ||A||_{HS}$$

for Hilbert-Schmidt operator $A$ and a bounded linear operator $B$ in a Hilbert space $H$ over the field $\mathbb{C}$ with a Hilbert basis $\{e_i\}$. By definition of a HS-operator,

$$||AB||_{HS}^2 = \sum_{i=1}^\infty ||ABe_i||_H^2 \leq \sum_{i=1}^\infty ||B||_H^2||A\left(\frac{Be_i}{||Be_i||}\right)||_H^2$$

which is almost what I need, except that I have to still argue why

$$\sum_{i=1}^\infty ||A\left(\frac{Be_i}{||Be_i||}\right)||_H^2\leq \sum_{i=1}^\infty ||Ae_i||_H^2 = ||A||_{HS}^2$$

which is something I don't know how to do. So what tool(s) do I need to conclude the claim?

Answer 1

Let $T:=A^*$ and $S:=B^*$. Using that $\|T\|_{HS}=\|A\|_{HS}$ (and similarly, $\|ST\|_{HS}=\|(AB)^*\|_{HS}=\|AB\|_{HS}$), and $\|S\|_H=\|B\|_H$, your goal boils down to check that $$\|ST\|_{HS}\le\|S\|_H\|T\|_{HS},$$ which is much easier.