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Let $E$ be a complex Hilbert space.

In (arXiv) it was shown that for $A=(A_1,...,A_n) \in \mathcal{B}(E)^n$ we have,

$$\displaystyle\frac{1}{2\sqrt{n}}\|A\|\leq \omega(A) \leq \|A\|,$$ where $$ \omega(A) = \sup_{\|x\|=1} \left(\sum_{k=1}^n |\langle A_kx,x\rangle|^2\right)^{1/2}, $$ and $$\|A\|= \left\|\displaystyle\sum_{k=1}^nA_k^*A_k \right\|^{1/2}.$$

How can we prove that $\frac{1}{2\sqrt{n}}$ is optimal?

If we assume $\dim_\mathbb{C}(E)\geq n+1$, then it suffices to consider the case $E=\mathbb{C}^{n+1}$ with the standard scalar product (otherwise choose a subspace of dimension $n+1$ and identify it with $\mathbb{C}^{n+1}$). We choose $A_k: \mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n+1}$ such that $$ A_k (x_1, \dots, x_{n+1})= (0, \dots,0 , x_1, 0, \dots, 0) $$ where $x_1$ is in the kth slot. The corresponding matrix consists of all zeros and a one in the kth row of the first column, e.g. for $n=3, k=2$ $$ A_2 = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ Now we set $$ A=(A_2, \dots, A_{n+1} )$$ We have $$ \Vert A_k x \Vert^2 = \langle (0, \dots, x_1, \dots, 0),(0, \dots, x_1, \dots, 0)\rangle = \vert x_1 \vert^2 $$ Thus, we get $$ \Vert A \Vert = \sup_{\Vert x \Vert = 1} \sqrt{n} \vert x_1 \vert = \sqrt{n} $$ On the other hand we have $$ \vert \langle A_k x, x \rangle \vert^2 = \vert \langle (0, \dots, x_1, \dots, 0), (x_1, \dots, x_{n+1}) \rangle \vert^2 = \vert x_1 \vert^2 \cdot \vert x_k \vert^2 $$ And hence, for $\Vert x \Vert=1$ $$ \left(\sum_{k=2}^{n+1} \vert \langle A_k x, x \rangle \vert^2\right)^\frac{1}{2} = \left(\vert x_1 \vert^2 \cdot \sum_{k=2}^{n+1} \vert x_k \vert^2 \right)^\frac{1}{2} = \left(\vert x_1 \vert^2 \cdot (1- \vert x_1 \vert^2)\right)^\frac{1}{2} = \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2}$$ Hence, we get $$ \omega(A) = \sup_{\Vert x \Vert=1} \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2} = \sup_{\vert x_1 \vert\leq 1} \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2} = \frac{1}{2} $$ Thus, we finally get $$ \frac{1}{2\sqrt{n}} \Vert A \Vert = \frac{1}{2\sqrt{n}} \cdot \sqrt{n} = \frac{1}{2} = \omega(A) $$

The optimality was proven under the additional assumption $\dim_\mathbb{C}(E)\geq n+1$. What about lower dimension? More precisely, how can we prove the optimality when $\dim_\mathbb{C}(E)\geq 2$?.

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    $\begingroup$ well, in dimension 1 it is not optimal (we always have $\omega(A)=\|A\|$) $\endgroup$ – Fedor Petrov Jan 13 at 9:19
  • $\begingroup$ @FedorPetrov For $dim E\geq2$? $\endgroup$ – Student Jan 13 at 9:24
  • $\begingroup$ you mean for $\dim\ge 2$? $\endgroup$ – Fedor Petrov Jan 13 at 9:27
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    $\begingroup$ To which extent is this different from the question on the Mathematics site: Find an example which shows that the following inequality is sharp? It seems that some of the information you provide here is taken from the answer posted there - probably it would be reasonable to add at least a link. $\endgroup$ – Martin Sleziak Jan 13 at 10:50
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    $\begingroup$ A more informative title would be helpful. $\endgroup$ – YCor Jan 15 at 9:13

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