I am thinking the sentence which may be about the Banach algebra function:
Let $A \in \mathcal{B}(H)$ where $H$ is Hilbert. \begin{equation} \| A \| = \sup_{ \| u \|, \| v \| \leq 1} | \langle Au, v \rangle | \end{equation}
It is the norm but with continuous functions $u,v \in H$, I think. I am thinking which properties it holds to prove it. Apparently, completeness, continuity.
My proposal
I think we need here the definition of the norm in Hilbert space i.e. vector's norm: \begin{equation} \|x\| = \sqrt{ \langle x, x \rangle } \end{equation} where I think the square root should be disappear through the Cauchy-Schwarz inequality such that the operator norm equals the right-hand-side.
Let's consider $A$ is continuous, by provig the inequality \begin{equation*} | A | \leq M, \forall u,v \in H, \forall u,v \leq 1. \end{equation*} Start with some $u \in H$. Note $u^{\pm} = |u| \pm u \in H$, so $u \geq 0$ so $\|A\| \geq 0$. So we get $| A | \leq M$, and since $u,v \leq 1$, we have the inequality. $\square$
Comments
- I really think that the proof of this statement should be more straight-forward. Cauchy-Schwarz should be enough but I have apparently mixed up some symbols in my attempt.
- I am not sure if the part starting from A is continuous, by proving ... is necessary for the proof.
- I am not sure about $u^{\pm}$. I tried to imply a similar proof of linear functionals from here.
How can you prove more rigorously the norm in Hilbert space?