Let $E$ be a complex Hilbert space, and $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.

For ${\bf A}:=(A_1,...,A_n) \in \mathcal{L}(E)^n$ we recall the definitions of the following two norms on $\mathcal{L}(E)^n$: $$\|{\bf A}\|=\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{k=1}^n\|A_kx\|^2\bigg)^{\frac{1}{2}},$$

and $$\omega_e({\bf A})=\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{k=1}^n|\langle A_kx\;|\;x\rangle|^2\bigg)^{1/2}.$$

It's not difficult to prove that $\omega_e({\bf A}) \leq \|{\bf A}\|$. Are $\|\cdot\|$ and $\omega_e(\cdot)$ two equivalent norms on $\mathcal{L}(E)^n\,?$ If the answer is true, I hope to find $\alpha$ such that $$\alpha \|{\bf A}\|\leq \omega_e({\bf A}) \leq \|{\bf A}\|.$$ Note that if $n=1$, it is well known that $$\displaystyle\frac{1}{2}\|A\|\leq \omega(A)\leq\|A\|.$$

And you for you help.


Yes, these two norms are equivalent. Recall that the following two norms are equivalent:

$$ \Vert \cdot \Vert_{\max}: \mathbb{R}^n \rightarrow \mathbb{R}, \ \Vert (x_1, \dots, x_n )\Vert_{\max} = \max_{j\in \{1, \dots, n \}} \vert x_j \vert $$


$$ \Vert \cdot \Vert_{2}: \mathbb{R}^n \rightarrow \mathbb{R}, \ \Vert (x_1, \dots, x_n )\Vert_{2} = \left( \sum_{j=1}^n \vert x_j \vert^2 \right)^{1/2}.$$

In fact, we have that for all $y\in \mathbb{R}^n$ holds (this are the optimal constants)

$$ \Vert y \Vert_{\max} \leq \Vert y \Vert_2 \leq \sqrt{n} \Vert y \Vert_{\max}.$$

Note that

$$ \Vert A \Vert = \sup_{\Vert x \Vert = 1} \Vert (A_1x, \dots , A_n x)\Vert_2 $$


$$ \omega_e(A) = \sup_{\Vert x \Vert = 1} \Vert (\vert \langle A_1 x,x\rangle \vert \dots , \vert \langle A_n x, x \rangle\vert )\Vert_2 $$

Furthermore, as pointed out by @Student we have

$$ \Vert A \Vert \leq 2 \sup_{\Vert x \Vert =1} \vert \langle Ax , x \rangle \vert .$$

This follows from the fact that for normal operators $B$ we have $\Vert B \Vert = \sup_{\Vert x \Vert=1} \vert \langle Bx, x \rangle \vert$ and the following computation

$$ \Vert A \Vert \leq \frac{1}{2} ( \Vert A + A^\star \Vert + \Vert A - A^\star \Vert ) = \frac{1}{2} ( \sup_{\Vert x \Vert = 1} \vert \langle ( A + A^\star)x, x \rangle \vert + \sup_{\Vert x \Vert = 1} \vert \langle ( A - A^\star)x, x \rangle \vert) \leq 2 \sup_{\Vert x \Vert = 1} \vert \langle Ax, x \rangle \vert.$$

Putting everything together yields

$$ \Vert A \Vert = \sup_{\Vert x \Vert = 1} \Vert (A_1x, \dots , A_n x)\Vert_2 \leq \sqrt{n} \sup_{\Vert x \Vert = 1} \Vert (A_1x, \dots , A_n x)\Vert_{\max} \leq 2 \sqrt{n} \sup_{\Vert x \Vert = 1} \Vert (\vert \langle A_1x, x \rangle \vert, \dots , \vert \langle A_n x, x \rangle \vert)\Vert_{\max} \leq 2 \sqrt{n} \sup_{\Vert x \Vert = 1} \Vert (\vert \langle A_1x, x \rangle \vert, \dots , \vert \langle A_n x, x \rangle \vert)\Vert_{2} = 2\sqrt{n} \omega_e(A). $$

Therefore, $\alpha= \frac{1}{2\sqrt{n}}$ will do the job.

Added: I was asked to adress the optimality of the constants in the inequality. The estimate $\omega_e(A) \leq \Vert A \Vert$ is sharp, as we can take $A=(Id, \dots, Id)$ and for this choice we get equality.

The other estimate is more difficult. I only managed to prove optimality under the additional assumption $dim_\mathbb{C}(E)\geq n+1$. In case someone has some insight for lower dimension please let me know.

If we assume $dim_\mathbb{C}(E)\geq n+1$, then it suffices to consider the case $E=\mathbb{C}^{n+1}$ with the standard scalar product (otherwise choose a subspace of dimension $n+1$ and identify it with $\mathbb{C}^{n+1}$). We choose $A_k: \mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n+1}$ such that $$ A_k (x_1, \dots, x_{n+1})= (0, \dots,0 , x_1, 0, \dots, 0) $$ where $x_1$ is in the kth slot. The corresponding matrix consists of all zeros and a one in the kth row of the first column, e.g. for $n=3, k=2$ $$ A_2 = \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$ Now we set $$ A=(A_2, \dots, A_{n+1} )$$ We have $$ \Vert A_k x \Vert^2 = \langle (0, \dots, x_1, \dots, 0),(0, \dots, x_1, \dots, 0)\rangle = \vert x_1 \vert^2 $$ Thus, we get $$ \Vert A \Vert = \sup_{\Vert x \Vert = 1} \sqrt{n} \vert x_1 \vert = \sqrt{n} $$ On the other hand we have $$ \vert \langle A_k x, x \rangle \vert^2 = \vert \langle (0, \dots, x_1, \dots, 0), (x_1, \dots, x_{n+1}) \rangle \vert^2 = \vert x_1 \vert^2 \cdot \vert x_k \vert^2 $$ And hence, for $\Vert x \Vert=1$ $$ \left(\sum_{k=2}^{n+1} \vert \langle A_k x, x \rangle \vert^2\right)^\frac{1}{2} = \left(\vert x_1 \vert^2 \cdot \sum_{k=2}^{n+1} \vert x_k \vert^2 \right)^\frac{1}{2} = \left(\vert x_1 \vert^2 \cdot (1- \vert x_1 \vert^2)\right)^\frac{1}{2} = \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2}$$ Hence, we get $$ \omega_e(A) = \sup_{\Vert x \Vert=1} \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2} = \sup_{\vert x_1 \vert\leq 1} \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2} = \frac{1}{2} $$ Thus, we finally get $$ \frac{1}{2\sqrt{n}} \Vert A \Vert = \frac{1}{2\sqrt{n}} \cdot \sqrt{n} = \frac{1}{2} = \omega_e(A) $$

  • $\begingroup$ @Student I added some details. Is it fine or should I elaborate a bit more? $\endgroup$ – Severin Schraven Dec 20 '17 at 11:06
  • $\begingroup$ @Student Thank you for pointing out my mistakes. I'll correct them as soon as possible. $\endgroup$ – Severin Schraven Dec 21 '17 at 11:38
  • $\begingroup$ Dear Professor Severin. I think that since the inequalities $$ \Vert y \Vert_{\max} \leq \Vert y \Vert_2 \leq \sqrt{n} \Vert y \Vert_{\max},$$ are sharp, i.e. $1,\sqrt{n}$ are the optimal constants, then the inequalities $$\frac{1}{2\sqrt{n}}\|\mathbf{A}\|\leq \omega_e(\mathbf{A})\leq \|\mathbf{A}\|,$$ are sharp (the constants $\frac{1}{2\sqrt{n}}$ and $1$ are the best). Do you agree with me? Thanks a lot. $\endgroup$ – Student Jan 10 at 10:15
  • 1
    $\begingroup$ Dear Severin, I would like to thank you for your effort in order to help me. Since a period I'm trying to find an example. According to ([arXiv][1], page 25), if we take $A_k=\frac{1}{\sqrt{n}}T$ for all $k\in \{1,\cdots,n\}$ with $$T=\begin{pmatrix}0&0\\1&0\end{pmatrix}.$$ [1]: arxiv.org/pdf/math/0410492.pdf $\endgroup$ – Student Jan 11 at 6:05
  • 1
    $\begingroup$ We obtain: Hence, $$w(A_1,\cdots,A_n)=\sqrt{n}w(\frac{1}{\sqrt{n}}T)=w(T)=\frac{1}{2},$$ however, $$\displaystyle\frac{1}{2\sqrt{n}}\left\|\displaystyle\sum_{k=1}^nA_k^*A_k \right\|^{1/2}=\frac{1}{2\sqrt{n}}\sqrt{n}\|\frac{1}{\sqrt{n}}T\|=\frac{1}{2\sqrt{n}}.$$ $\endgroup$ – Student Jan 11 at 6:06

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