7
$\begingroup$

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.

$\endgroup$
2
+50
$\begingroup$

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

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

and

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

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

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

and

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

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 $$ \frac{1}{2\sqrt{n}} \Vert A \Vert = \frac{1}{2\sqrt{n}} \cdot \sqrt{n} = \frac{1}{2} = \omega_e(A) $$

$\endgroup$
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.