norm of bounded operator on a hilbert space

Question

If we have a bounded linear operator on a complex Hilbert space, we can define the operator norm by the scalar product.

I have a doubt about the last expression of the following equality:

\begin{equation} \lVert A \rVert=\sup_{{\lVert x \rVert}\ne 0,\,{\lVert y \rVert}\ne 0} \frac{|\langle y,Ax \rangle|}{{\lVert y \rVert}{\lVert x \rVert}}=\sup_{{\lVert x \rVert}\ne 0,{\lVert y \rVert}\ne 0} \frac{\operatorname{Re}\langle y,Ax \rangle}{{\lVert y \rVert}{\lVert x \rVert}} \end{equation}

where $x,y$ $\in$ $D(A)$, the domain of the operator $A$.

In my view a factor phase is missing in the last expression, but i'm not sure. It should be:

\begin{equation} \sup_{{\lVert x \rVert}\ne 0,\,{\lVert y \rVert}\ne 0} \frac{|\langle y,Ax \rangle|}{{\lVert y \rVert}{\lVert x \rVert}}=\sup_{{\lVert x \rVert}\ne 0,{\lVert y \rVert}\ne 0} \frac{\operatorname{Re}(e^{i\theta}\langle y,Ax \rangle)}{{\lVert y \rVert}{\lVert x \rVert}} \end{equation}

Is it right?

Any suggestion?

Thanks

Answer 1

You write $$|\langle y,Ax\rangle|=\lambda\,\langle y,Ax\rangle$$ for an appropriate $\lambda\in\mathbb C$ with $|\lambda|=1$. Then $$ |\langle y,Ax\rangle|=\lambda\,\langle y,Ax\rangle =\langle \lambda y,Ax\rangle=\text{Re}\,\langle \lambda y,Ax\rangle. $$ As you go through every possible nonzero $y$, the $\lambda$ is absorbed by $y$.