Numerical radius of the absolute value of an operator

Question

Let $T$ be a compact operator on a (complex) Hilbert space $H$ and $|T| \equiv (T^* T)^{1/2}$ the absolute value of $T$. (So, $|T|$ is compact as well and also self-adjoint.)

If the numerical radius $\nu (T)$ of $T$ is $1$, that is, $$ \nu (T) \equiv \sup \{ | \langle Tx, x \rangle | : \|x\|=1\} = 1, $$ then it is easy to see that $\nu(|T|) = \|T\|$ and $\|T \| \geq \nu (T) = 1$.

Is there an example of a compact operator $T$ with $\nu(T) = 1$ and $\nu (|T|) > 1$?

I tried to prove that the numerical radius $\nu (|T|)$ of $|T|$ is $1$ as well, but I wonder whether this is the case or not.

Any help will be appreciated!

Answer 1

Let $\mathcal H=\ell_2(\mathbb N)$ where $(e_n)_{n\in\mathbb N}$ denotes the standard basis of $\ell_2(\mathbb N)$. As notation we use $x=(x_1,x_2,\ldots)$ so $x_j=\langle e_j,x\rangle$.

Choose $T=2\langle e_1,\cdot\rangle e_2$. As $T$ is finite-rank, it is compact and

$$ \nu(T)=2\sup_{\Vert x\Vert=1}|x_1||x_2|=2\sup_{|x_1|^2+|x_2|^2=1}|x_1||x_2|=2\cdot\frac12=1 $$

but $|T|=\sqrt{(2\langle e_2,\cdot\rangle e_1)(2\langle e_1,\cdot\rangle e_2)}=\sqrt{4\Vert e_2\Vert^2\langle e_1,\cdot\rangle e_1}=2\langle e_1,\cdot\rangle e_1$ satisfies

$$ \nu(|T|)=2\sup_{\Vert x\Vert=1}|x_1|^2=2>1. $$

Answer 2

It is not an answer; it is rather a simplification:

define the operator $ x\otimes y $ by

$$ (x\otimes y)(t)=\langle t ,y\rangle x $$

we have $$ \Vert x\otimes y \Vert= \Vert x\Vert\Vert y\Vert$$

$$ \vert x\otimes y \vert= \Vert x\Vert y\otimes y $$

$$ (x\otimes y)^{*}= y \otimes x $$

Then $T= 2 e_{2}\otimes e_{1}$ and $\vert T \vert= 2\Vert e_{2}\Vert e_{1}\otimes e_{1} = 2 e_{1}\otimes e_{1}$ and since $ \vert T \vert$ is selfadjoint, we obtain

$$ \nu(\vert T \vert)=2 \Vert e_{1}\otimes e_{1} \Vert = 2 \Vert e_{1} \Vert \Vert e_{1}\Vert =2$$