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In this question $E$ stands for a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$.

Let $T= (T_1,...,T_d) \in \mathcal{L}(E)^d$ .

We can establish that, $\|T\|:=\bigg(\displaystyle\sum_{k=1}^d\|T_k\|^2\bigg)^{1/2}\geq\left\|\displaystyle\sum_{k=1}^dT_kT_k^* \right\|^{1/2}$. Is the converse inequality true?

Thank you for your help.

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The converse is not true. Take $E=\mathbb R^d$, $d>1$, and let $T_k$ be the matrix with all entries zero except the $(k,k)$ entry, which is one. Then $\|T_k\|=1$, $\sum_{k=1}^d T_kT_k^*=I$, and $$\|T\| = \sqrt d \not\le \|\sum_{k=1}^d T_kT_k^*\|^{1/2}=1.$$

One can however prove the converse inequality involving the factor $\sqrt d$: $$ \|\sum_{k=1}^d T_kT_k^*\| = \sup_{\|x\|\le1}\langle x, \sum_{k=1}^d T_kT_k^*x\rangle =\sup_{\|x\|\le1}\sum_{k=1}^d\|T_k^*x\|^2. $$ This shows $\|\sum_{k=1}^d T_kT_k^*\| \ge \|T_jT_j^*\|^2$ for all $j$, hence $$ \|\sum_{k=1}^d T_kT_k^*\| \ge \max_{k=1\dots d} \|T_kT_k^*\|^2 \ge d^{-1} \sum_{k=1}^d \|T_kT_k^*\|^2 = d^{-1} \sum_{k=1}^d \|T_k\|^2=d^{-1} \|T\|^2. $$

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  • $\begingroup$ @Student thanks for correcting it. $\endgroup$ – daw Oct 20 '17 at 18:24

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