 # For A &#x2208;<!-- ∈ --> B ( H ) where H is a separable Hilbert space (so a coun 2nalfq8 2022-06-30 Answered
For $A\in B\left(H\right)$ where $H$ is a separable Hilbert space (so a countable orthonormal basis exists or also known as a maximal orthonormal system) where $A$ is also a Hilbert-Schmidt operator and $B\in B\left(H\right)$ (so $B$ does not need to be a Hilbert-Schmidt operator. It is just linear and bounded.) How do I show that $||AB|{|}_{HS}\le ||A|{|}_{HS}||B|{|}_{H}$ and $||BA|{|}_{HS}\le ||A|{|}_{HS}||B|{|}_{H}$ where $HS$ is the Hilbert Schmidt norm and $H$ is just the standard operator norm. I have never been introduced to the trace operator so I can't use that. I can only use the standard Hilbert space inequalities like Cauchy-Schwarz, Parseval's and Bessel's inequality as well as the fact that $‖A{‖}_{H}\le ‖A{‖}_{HS}$ I have not been able to turn the left hand side anywhere close to the right hand side so I'm not sure if there is some trick I'm missing or some inequality I haven't thought of.
You can still ask an expert for help

## Want to know more about Inequalities systems and graphs?

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it lofoptiformfp
Hint: Use $‖BA{e}_{i}{‖}_{H}\le ‖B{‖}_{H}‖A{e}_{i}{‖}_{H}$ to show $‖BA{‖}_{HS}\le ‖B{‖}_{H}‖A{‖}_{HS}$. For the other inequality it is useful to first show $‖{B}^{\ast }{‖}_{H}=‖B{‖}_{H}$ and $‖{A}^{\ast }{‖}_{HS}=‖A{‖}_{HS}.$