#### ejgmath

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 $$\displaystyle x\in H$$,$$\displaystyle y\in H$$

#### Focus

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.

• ejgmath

#### mabruka

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

• ejgmath