Dtyjlui

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 $mathcal{L}(E)^n$:
$$|{bf A}|=displaystylesup_{|x|=1}bigg(displaystylesum_{k=1}^n|A_kx|^2bigg)^{frac{1}{2}},$$

and
$$omega_e({bf A})=displaystylesup_{|x|=1}bigg(displaystylesum_{k=1}^n|langle A_kx;|;xrangle|^2bigg)^{1/2}.$$

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
$$displaystylefrac{1}{2}|A|leq omega(A)leq|A|.$$

And you for you help.

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

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

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 $yin mathbb{R}^n$ 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

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

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

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

Therefore, $alpha= frac{1}{2sqrt{n}}$ 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 $Vert x Vert=1$
$$ left(sum_{k=2}^{n+1} vert langle A_k x, x rangle vert^2right)^frac{1}{2}
= left(vert x_1 vert^2 cdot sum_{k=2}^{n+1} vert x_k vert^2 right)^frac{1}{2}
= left(vert x_1 vert^2 cdot (1- vert x_1 vert^2)right)^frac{1}{2}
= vert x_1 vert cdot sqrt{1- vert x_1 vert^2}$$
Hence, we get
$$ omega_e(A) = sup_{Vert x Vert=1} vert x_1 vert cdot sqrt{1- vert x_1 vert^2}
= sup_{vert x_1 vertleq 1} vert x_1 vert cdot sqrt{1- vert x_1 vert^2}
= frac{1}{2} $$
Thus, we finally get
$$ frac{1}{2sqrt{n}} Vert A Vert = frac{1}{2sqrt{n}} cdot sqrt{n} = frac{1}{2} = omega_e(A) $$