Proving Equality Regarding the Adjoint of Bounded Linear Operator

Question

I am proving the Proposition 2.13 from Elementary Functional Analysis by MacCluer, mainly on c) and d)

We're given that For any $A,B \in \mathscr{B}(\mathscr{H})$, we have \begin{align*} (\alpha A)^* &= \overline{\alpha}A^*, \alpha \in \mathbb{C} \tag*{c)}\\ (AB)^* &= B^*A^* \tag*{d)} \end{align*} where '$*$' stands for adjoint.

My attempt for c) shows something weird: \begin{align*} \langle (\alpha A)^*x,y\rangle = \langle x,(\alpha A)^{**}y\rangle=\langle x,\alpha Ay\rangle =\alpha\langle x,Ay\rangle \end{align*} whereas \begin{align*} \langle \overline{\alpha} A^*x,y\rangle &=\langle \overline{\alpha}x,A^{**}y\rangle\\ &=\langle \overline{\alpha}x,Ay\rangle\\ &=\overline{\alpha}\langle x,Ay\rangle \end{align*} How can this be?

For d), so far I have got: \begin{align*} \langle x,B^*A^*y \rangle &= \langle Bx,A^*y\rangle\\ &=\langle ABx,y\rangle\\ & =\langle x, (AB)^* y\rangle \end{align*} This shows that $(AB)^* = B^*A^*$

Answer 1

$\langle x,\alpha Ay\rangle =\overline{\alpha}\langle x,Ay\rangle$ and not $\alpha\langle x,Ay\rangle$ . That's how the inner product is defined. It is not symmetric but conjugate symmetric.

Proof of $d$ is correct .

Answer 2

You wrote: My attempt for c) shows something weird: \begin{align*} \langle (\alpha A)^*x,y\rangle = \langle x,(\alpha A)^{**}y\rangle=\langle x,\alpha Ay\rangle =\alpha\langle x,Ay\rangle \end{align*}

The last equality is wrong. It should be

\begin{align*} \langle (\alpha A)^*x,y\rangle = \langle x,(\alpha A)^{**}y\rangle=\langle x,\alpha Ay\rangle =\overline {\alpha}\langle x,Ay\rangle \end{align*}