# 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? Dec 13, 2017 at 12:31
• This follows immediately from my answer to your other question mathoverflow.net/questions/285471/a-numerical-radius-inequality/… Dec 13, 2017 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? Dec 17, 2017 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 $$\frac{1}{2\sqrt n} \|[A_1 \dots A_n]\| \leq \sup_{\|x\|=1} \left(\sum_{i=1}^n |\langle A_ix,x\rangle|^2\right)^{1/2}.$$