# For adjoint operators, $\|T^*\|=\|T\|$, so $\|T^*f\|=\|T f\|$ for all $f$?

Suppose $H$ is a Hilbert space and let $T \in B(H,H)$ where in our notation $B(H,H)$ denotes the set of all linear continuous operators $H \to H$.

We defined the adjoint of $T$ as the unique $T^* \in B(H,H)$ such that $\langle T x, y \rangle = \langle x, T^* y \rangle$ for all $x,y$ in $H$.

Since $\| T^* \| = \| T \|$, can I write

$$\| T^* f \| = \| T f \| \text{ for all } f \in H?$$

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Certainly not. Let $H = \mathbb{R}^2$ with the standard inner product and let $T$ be the matrix $$T = \begin{pmatrix} 0 & 1 \\ -2 & 0\end{pmatrix}$$ Its adjoint operator, as $H$ is a finite dimensional real vector space, is just the transpose of the matrix $$T^* = \begin{pmatrix} 0 & -2 \\ 1 & 0\end{pmatrix}$$

Let $v = \begin{pmatrix} 1 \\ 0\end{pmatrix}$ the standard unit vector in the $x_1$ direction, you have that

$$\|Tv\| = 2 \neq 1 = \|T^* v\|$$

A direct generalisation of the above example provides a counterexample for all Hilbert spaces of dimension $> 1$. In one dimension linear operations are scalar multiplications, and hence all elements of $B(H,H)$ are self-adjoint, whence your claim is trivially true.

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If we are talking about complex Hilbert space, then this property characterizes normal operators.

If $\|Tx\|=\|T^*x\|$ for all $x$, then $\langle T^*Tx,x\rangle=\|Tx\|^2=\|T^*x\|^2=\langle TT^*x,x\rangle$ for all $x$, and this question gives a hint for how to finish showing that $T$ is normal. (The converse is more direct.)

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