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Let $(a_n)_n$ be sequence of bounded linear operator on a Hilbert space $E$ and $b$ be a positive operator on $E$, Why $$\left\|\displaystyle\sum_{n=1}^da_n^*ba_n\right\|\leq\|b\|\left\|\displaystyle\sum_{n=1}^da_n^*a_n\right\|\;??$$ Thank you for your help..

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    $\begingroup$ If you assume $b$ is a positive operator, then $a_n^*b a_n$ is positive. You've got: $$\langle x,\sum_n a_n^* b a_n x\rangle =\sum_n \langle a_n x, b a_nx\rangle≤\|b\|\,\sum_n\langle x,a_n^*a_nx\rangle ≤\|b\|\,\left\|\sum a_n^*a_n\right\|\,\|x\|^2$$ Taking the supremum over $\|x\|=1$ gives the desired inequality. I think you might be able to start from this case and see the general case. $\endgroup$ – s.harp Jun 30 '17 at 14:12
  • $\begingroup$ Thank you. I modify the question. Please write your comment as an answer in order to accept it. $\endgroup$ – Student Jul 1 '17 at 9:01
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If $b$ is positive it follows that $a_n^* b a_n$ is positive. As a sum of positives $\sum a_n^* ba_n$ is positve. One can then see

$$\langle x, \sum_n a_n^* b a_nx \rangle  = \sum_n \langle a_n x, b a_nx \rangle ≤\|b\|\sum_n\langle a_n x,a_nx\rangle =\|b\|\,\langle x, \sum_n a_n^* a_n x\rangle ≤ \|b\|\ \left\|\sum_n a_n^*a_n\right\|\,\|x\|$$ Taking the supremum over $\|x\|=1$ gives you the desired inequality in the case that $b$ is positive.

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