2nalfq8

2022-06-30

For $A\in B(H)$ 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 $B\in B(H)$ (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 $\Vert A{\Vert}_{H}\le \Vert A{\Vert}_{HS}$ 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.

lofoptiformfp

Beginner2022-07-01Added 16 answers

Hint: Use $\Vert BA{e}_{i}{\Vert}_{H}\le \Vert B{\Vert}_{H}\Vert A{e}_{i}{\Vert}_{H}$ to show $\Vert BA{\Vert}_{HS}\le \Vert B{\Vert}_{H}\Vert A{\Vert}_{HS}$. For the other inequality it is useful to first show $\Vert {B}^{\ast}{\Vert}_{H}=\Vert B{\Vert}_{H}$ and $\Vert {A}^{\ast}{\Vert}_{HS}=\Vert A{\Vert}_{HS}.$

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