# inequality involving tuple of operators on Hilbert spaces

Let $E$ be a complex Hilbert space.

Let $(A_1,...,A_n) \in \mathcal{L}(E)^n$, could you please help me to show that $$\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{i=1}^n\| A_i^*x\|^2\bigg)\leq (4n)\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{i=1}^n|\langle A_ix\;,\;x\rangle|^2\bigg).$$

And you for you help.

• As usual with your questions: what partial results have you obtained? Why do you suppose this might be true? What evidence is there to support this claim? What, in short, have you actually tried? – Yemon Choi Dec 13 '17 at 12:31
• This follows immediately from my answer to your other question mathoverflow.net/questions/285471/a-numerical-radius-inequality/… – Chris Ramsey Dec 13 '17 at 15:35
• Are you sure that you are happy just with the factor $4n$ in front of the supremum on the right? If you are, you only need to consider $n=1$, don't you? – fedja Dec 17 '17 at 1:55

Notice that $$\|[A_1 \dots A_n]\| = \|[A_1 \dots A_n]^*\| = \left\| \left[\begin{smallmatrix} A_1^* \\ \\\vdots \\ A_n^* \end{smallmatrix}\right] \right\| = \sup_{\|x\|=1} \left( \sum_{i=1}^n \|A_i^*x\|^2 \right)^{1/2}.$$
By the answer to this question we have that $$\frac{1}{2\sqrt n} \|[A_1 \dots A_n]\| \leq \sup_{\|x\|=1} \left(\sum_{i=1}^n |\langle A_ix,x\rangle|^2\right)^{1/2}.$$