Adjoint of Bounded Operators

**ejgmath** · May 20th 2010, 06:08 PM

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:

$<Tx,y>=<x,T^*y>$ for all $x\in H$ , $y\in H$

**Focus** · May 20th 2010, 06:19 PM

Start by applying the adjoint principal to Tx, i.e.
$<br /> \langle S(Tx),y \rangle=\langle Tx,S^*y \rangle<br />$

I hope you can now see the next step.

**mabruka** · May 20th 2010, 06:20 PM

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

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