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