Adjoint of Bounded Operators

Jan 2010
32
1
If \(\displaystyle S,T:H\rightarrow H\) are bounded operators, show that \(\displaystyle (ST)^*=T^*S^*\).

I'm assuming that \(\displaystyle H\) 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 \(\displaystyle S,T\) are bounded there adjoints \(\displaystyle S^*\) and \(\displaystyle T^*\) exist and that:

\(\displaystyle <Tx,y>=<x,T^*y>\) for all \(\displaystyle x\in H\),\(\displaystyle y\in H\)
 
Aug 2009
228
80
Start by applying the adjoint principal to Tx, i.e.
\(\displaystyle
\langle S(Tx),y \rangle=\langle Tx,S^*y \rangle
\)

I hope you can now see the next step.
 
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Jan 2010
150
29
Mexico City
This kind of identities are usually proved as follows:

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


From where

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

Therefore \(\displaystyle (ST)^*=T^*S^*\)

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