Question related to the numerical radius

In this question $$\mathcal{H}$$ stands for a Hilbert space over $$\mathbb{K}=\mathbb{R}$$ or $$\mathbb{C}$$, with inner product $$\langle\cdot\;| \;\cdot\rangle$$ and the norm $$\|\cdot\|$$ and let $$\mathcal{B}(\mathcal{H})$$ the algebra of all bounded linear operators from $$\mathcal{H}$$ to $$\mathcal{H}$$.

If $${\bf T} = (T_1,\cdots,T_d)\in\mathcal{B}(\mathcal{H})^d$$, then $$\omega({\bf T})$$ is given by

$$\omega({\bf T}) =\sup\{\bigg(\displaystyle\sum_{k=1}^d|\langle T_kx\;|\;x\rangle|^2\bigg)^{1/2},\;x\in \mathcal{H},\;\|x\|=1\;\}.$$

It is well known that for each $$k\in\{1,\cdots,d\}$$, we have $$\begin{eqnarray*} \omega(T_k) &=&\sup\left\{|\langle T_kx\;|\;x\rangle|,\;x\in \mathcal{H},\;\|x\|=1\;\right\}. \end{eqnarray*}$$

Is $$\omega({\bf T})=\bigg(\displaystyle\sum_{k=1}^d\omega(T_k)^2\bigg)^{1/2}\;??$$

1 Answer

The statement is false in general for $d > 1$. To see this, define the operators $T_i : \ell^2 \to \ell^2$ as $T_ix = x_ie_i$, like in my answer to one of your previous questions.

For $x = (x_n)_{n=1}^\infty \in \ell^2$ have:

$$\left|\langle T_ix\,|\,x\rangle\right| = \left|\langle x_ie_i\,|\,x\rangle\right| = \left|x_i\right|^2 \le \|x\|_2^2$$

so $\omega(T_i) = \sup\limits_{\|x\|=1} \left|\langle T_ix\,|\,x\rangle\right| \le 1$. On the other hand, for $x = e_i$ we have $\left|\langle T_ie_i\,|\,e_i\rangle\right| = \left|\langle e_i\,|\,e_i\rangle\right|=1$.

Thus, $\omega(T_i) = 1$. Now we get:

$$\sqrt{\sum_{k=1}^d\omega(T_k)^2} = \sqrt{\sum_{k=1}^d 1} = \sqrt{d}$$

However, for $\mathbf{T} = (T_1, \ldots, T_d) \in \mathcal{B}\left(\ell^2\right)^d$ we have:

$$\omega(\mathbf{T}) = \sup_{\|x\|_2=1}\sqrt{\sum_{k=1}^d\left|\langle T_kx\,|\,x\rangle\right|^2} = \sup_{\|x\|_2=1}\sqrt{\sum_{k=1}^d\left|x_k\right|^2} \le \sup_{\|x\|_2=1}\sqrt{\sum_{k=1}^\infty\left|x_k\right|^2} = \sup_{\|x\|_2 = 1}\|x\|_2 = 1$$

Thus for $d > 1$ we cannot have $\omega(\mathbf{T}) = \sqrt{\sum_{k=1}^d\omega(T_k)^2}$.

In general it certainly holds $\omega(\mathbf{T}) \le \sqrt{\sum_{k=1}^d\omega(T_k)^2}$

• Thank you very much for your answer. I want to get a closed formula if there exist $\omega({\bf T})$ and $\omega(T_k)$. – Student Sep 27 '17 at 19:33
• @Student In that case, perhaps don't accept my answer yet as less people will look at your question. I don't think the answer is anything simple. – mechanodroid Sep 27 '17 at 19:34
• Thank you but your answer is very helpful for me. – Student Sep 27 '17 at 19:36
• @Student I saw your question. Unfortunately I have no idea. I only know that for self-adjoint (and perhaps normal) operators $T \in \mathcal{B}(\mathcal{H})$ we have $$\|T\| = \sup_{\|x\| = 1} |\langle Tx, x\rangle|$$ But this leads back to the issue that the supremum for different $T_k$ may be attained at different $x \in \mathcal{H}$. – mechanodroid Oct 2 '17 at 8:05