The Hilbert adjoint operator $T: \mathcal{H_1} \to \mathcal{H_2}$ is $T^*: \mathcal{H_2} \to \mathcal{H_1}$ such that for all $x \in \mathcal{H_1}$ and $y \in \mathcal{H_2}$ the inner product
$$\langle Tx, y\rangle = \langle x, T^*y \rangle $$
where $\mathcal{H_1}, \, \mathcal{H_2}$ are two different Hilbert spaces.
My question what the difference between the two theorems is? The first theorem tells us that the Hilbert adjoint operator if it does exist, it is a unique bounded linear operator with norm $\|T\| = \|T^*\|$. And the second theorem tells us that $T: \mathcal{H} \to \mathcal{H}$ is normal if and only if $\|Tx\| = \|T^*x\|$ for all $x \in \mathcal{H}$. I mean what is the normality for when we already known that the norm of an operator and its adoint are equal in Hilbert spaces?
Hilbert adjoint operator
functional-analysis
Best Answer
Normality is a far stronger information than $||T||=||T^*||$.
Consider the function $\varphi_T : x \mapsto \frac{||T(x)||}{||x||}$ defined for all nonzero $x$. A consequence of normality is that $\varphi_T (x)=\varphi_{T^*}(x)$ for ALL $x\neq 0$. This need not be true at all when you only know that the operator norms of (T) and $T^*$ are equal! By definition, $||T||:= sup \ \varphi_T$. Intuitively, the amount of maximum stretching that the two operators do will be the same, while you could have that there are vectors that get stretched by $T$ and squished by $T^*$, or vice versa.
Example: consider $T: \mathbb{C}^2 \rightarrow \mathbb{C}^2$ which defined by the matrix $$ \begin{pmatrix} 1 & 1\\ 0 & 1\\ \end{pmatrix} $$ Thus $T^*$ is given by $$ \begin{pmatrix} 1 & 0\\ 1 & 1\\ \end{pmatrix} $$ Clearly $||T\begin{pmatrix} 1\\ 0\\ \end{pmatrix}||=1\neq2=||T^*\begin{pmatrix} 1\\ 0\\ \end{pmatrix}||$ so $T$ is not normal.