A numerical radius inequality

Question

Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{B}(E)$ the algebra of all bounded linear operators from $E$ to $E$.

I want to show that for $(T_1,...,T_d) \in \mathcal{B}(E)^d$ we have: $$\displaystyle\frac{1}{2}\left\|\displaystyle\sum_{k=1}^dT_k^*T_k \right\|^{1/2}\leq \omega(T_1,\cdots,T_d) \leq \left\|\displaystyle\sum_{k=1}^dT_k^*T_k \right\|^{1/2},$$ where $$\omega(T_1,\cdots,T_d)=\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{k=1}^d|\langle T_kx\;|\;x\rangle|^2\bigg)^{1/2}.$$

And you for you help.

Answer 1

The inequality you have written cannot be proven.

Gelu Popescu has a big paper on multivariable generalizations of the numerical radius (Memoirs of the AMS, arXiv). He calls what you have above the Euclidian operator radius of a tuple: $$ \omega_e(T_1,\dots, T_d) = \sup_{\|x\|=1} \left(\sum_{k=1}^d |\langle T_kx,x\rangle|^2\right)^{1/2} $$ and he proves many things about this in section 2 of his paper. Moreover, Proposition 2.4 establishes that $$ \frac{1}{2\sqrt n} \|[T_1, \dots, T_d]\| \leq \omega_e(T_1,\dots, T_d) \leq \|[T_1, \dots, T_d]\| $$ and the inequality is sharp, noting that $\|[T_1, \dots, T_d]\| = \|\sum_{k=1}^d T_k^*T_k\|^{1/2}$. The proof is developed throughout section 2 but is not too difficult to follow.