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.

  • 2
    $\begingroup$ 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? $\endgroup$
    – Yemon Choi
    Dec 13, 2017 at 12:31
  • 1
    $\begingroup$ This follows immediately from my answer to your other question mathoverflow.net/questions/285471/a-numerical-radius-inequality/… $\endgroup$ Dec 13, 2017 at 15:35
  • $\begingroup$ 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? $\endgroup$
    – fedja
    Dec 17, 2017 at 1:55

1 Answer 1


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 $$ \frac{1}{2\sqrt n} \|[A_1 \dots A_n]\| \leq \sup_{\|x\|=1} \left(\sum_{i=1}^n |\langle A_ix,x\rangle|^2\right)^{1/2}. $$

Combine these two statements to get your desired inequality.

  • $\begingroup$ When I have some time I'll look over the result. $\endgroup$ Dec 13, 2017 at 17:11

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