How to prove that $A B A^* \leq \|B\| A A^*$ for operators A,B?

179 Views Asked by At

Let $A$, $B$ bounded operators on a Hilbert space $H$. Further let $B$ be self-adjoint. Then we have that $A B A^* \leq \|B\| A A^*$. I wanted to ask how to prove this inequality or where I can find a proof if it is a very common ineuqality.

1

There are 1 best solutions below

4
On BEST ANSWER

Proof: For all $h \in H$, $\langle A^* BA h, h \rangle = \langle BA h, A h \rangle \leq \|B\| \cdot \langle A h, A h \rangle = \langle \|B\|A^*A h, h \rangle$.