• May 20th 2010, 07:08 PM
ejgmath
If $S,T:H\rightarrow H$ are bounded operators, show that $(ST)^*=T^*S^*$.

I'm assuming that $H$ is a Hilbert space, although it doesn't say this in the question. I'm really not sure where to start with this. All I have is that since $S,T$ are bounded there adjoints $S^*$ and $T^*$ exist and that:

$=$ for all $x\in H$, $y\in H$
• May 20th 2010, 07:19 PM
Focus
Start by applying the adjoint principal to Tx, i.e.
$
\langle S(Tx),y \rangle=\langle Tx,S^*y \rangle
$

I hope you can now see the next step.
• May 20th 2010, 07:20 PM
mabruka
This kind of identities are usually proved as follows:

For any x,y in the hilbert space
$\langle x,(ST)^* y \rangle=\langle STx,y\rangle =\langle Tx,S^*y\rangle=\langle x,T^*S^*y\rangle$

From where

$\langle x,(ST)^* y \rangle -\langle x,T^*S^*y\rangle = 0$ for all $x,y \in \mathcal H$

Therefore $(ST)^*=T^*S^*$

Here it is used that if $\langle x,(B-C)y \rangle = 0$ for all $x,y \in \mathcal H$ then B=C