Two equivalent norms on $\mathcal{L}(E)^n$ 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.
 A: 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
$$ \frac{1}{2\sqrt{n}} \Vert A \Vert = \frac{1}{2\sqrt{n}} \cdot \sqrt{n} = \frac{1}{2} = \omega_e(A) $$
