Ask a question on adjoint operator?

Question

Let $H_1,H_2$ be Hilbert spaces with inner products $\langle \cdot, \cdot \rangle_1$ and $\langle \cdot, \cdot \rangle_2$. Corresponding to every $T \in \mathcal B(H_1,H_2)$, there is a unique element $T^*$ of $B(H_2,H_1)$ determined by the relation

$$\langle T x_1, x_2 \rangle_2=\langle x_1, T^* x_2 \rangle_1$$

for all $x_1 \in H_1, x_2 \in H_2$.

Proof: Consider $\langle T x_1, x_2 \rangle_2$ as a function of $x_1$ for fixed $x_2$. It is bounded and linear so that the Riesz representation theorem may be applied to see that there exists a unique $y \in H_1$ such that $\langle T x_1, x_2 \rangle_2=\langle x_1, y \rangle_1$. Thus, we take $T^* x_2=y$. This definition gives us a linear mapping. To see that it bounded first note that $T$ is necessarily the adjoint of $T^*$. Then

$$|| T^* x_2 ||^2_1= | \langle x_2，T T^* x_2 \rangle_2 |$$

$$\leq ||T|| || T^* x_2 ||_1 ||x_2||_2$$

Question: Why $|| T^* x_2 ||^2_1= | \langle x_2，T T^* x_2 \rangle_2 |$?

Answer 1

Question: Why $\|T^*x_2\|^2_1=|\langle x_2，TT^*x_2\rangle_2|$?

Because, as just noted, $T$ is necessarily the adjoint of $T^*$, i.e. $$\langle T^*x_2,x_1\rangle_1= \langle x_2,Tx_1\rangle_2$$ for all $x_1\in H_1,x_2\in H_2$, hence $$\Bbb R_{\ge0}\ni \|T^*x_2\|_1^2=\langle T^*x_2,T^*x_2\rangle_1=\langle x_2,TT^*x_2\rangle_2. $$

Answer 2

Question: Why $\| T^* x_2 \|^2_1= | \langle x_2，T T^* x_2 \rangle_2 |$?

Because you just proved that $T^*$ exists and it is the adjoint of $T$. Then $$ \|T^*x_2\|_1^2=\langle T^*x_2,T^*x_2\rangle_1=\langle T(T^*x_2),x_2\rangle_2 =\overline{\langle x_2,TT^*x_2\rangle_2}=\langle x_2,TT^*x_2\rangle_2, $$ where the conjugation is not necessary because it is applied to a real number.