# Operator norm and Action

The sup in the equation is a supremum over all states out of the Hilbert space $$\mathcal{H}$$. In other words, you pick the state $$|\psi\rangle$$ out of $$\mathcal{H}$$ for which the number $$\frac{\mid\mid A|\psi\rangle\mid\mid}{\mid\mid |\psi\rangle\mid\mid}$$ becomes the largest. It can be shown that the operator norm of $$A$$ corresponds to the largest eigenvalue of $$\sqrt{A^{\dagger}A}$$. As an operator acting on $$\mathcal{H}$$, $$A$$ maps one state out of it to another, which may be not normalized since you have no guarantee on $$A$$ being unitary. So you can think of the action of $$A$$ as a measure for “how much” non-normalized $$A|\psi\rangle$$ is, given that $$|\psi\rangle$$ is, i.e. $$\langle\psi | \psi \rangle=1$$.

An operator is unbounded if its supremum norm is infinite. This follows trivially by negating the definition of a bounded operator in a preBanach space.

"A linear operator $$A :D(A) \rightarrow X$$ is called bounded, iff

$$\forall \psi\in D(A), \exists k\in \mathbb R, \text{so that} ||A\psi|| \leq k ||\psi||.$$

The maximum value of $$k$$ over the domain of A is called the supremum norm of A.