Mixing
Find an example which shows that the following inequality is sharp
$begingroup$
Let $E$ be a complex Hilbert space.
In (arXiv) it was shown that for $A=(A_1,...,A_n) in mathcal{B}(E)^n$ we have,
$$displaystylefrac{1}{2sqrt{n}}|A|leq omega(A) leq |A|,$$
where
$$
omega(A) = sup_{|x|=1} left(sum_{k=1}^n |langle A_kx,xrangle|^2right)^{1/2},
$$
and
$$|A|= left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$
How can we prove that $frac{1}{2sqrt{n}}$ is optimal?
linear-algebra functional-analysis
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
$endgroup$
add a comment |
$begingroup$
Let $E$ be a complex Hilbert space.
In (arXiv) it was shown that for $A=(A_1,...,A_n) in mathcal{B}(E)^n$ we have,
$$displaystylefrac{1}{2sqrt{n}}|A|leq omega(A) leq |A|,$$
where
$$
omega(A) = sup_{|x|=1} left(sum_{k=1}^n |langle A_kx,xrangle|^2right)^{1/2},
$$
and
$$|A|= left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$
How can we prove that $frac{1}{2sqrt{n}}$ is optimal?
linear-algebra functional-analysis
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
$endgroup$
add a comment |
$begingroup$
Let $E$ be a complex Hilbert space.
In (arXiv) it was shown that for $A=(A_1,...,A_n) in mathcal{B}(E)^n$ we have,
$$displaystylefrac{1}{2sqrt{n}}|A|leq omega(A) leq |A|,$$
where
$$
omega(A) = sup_{|x|=1} left(sum_{k=1}^n |langle A_kx,xrangle|^2right)^{1/2},
$$
and
$$|A|= left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$
How can we prove that $frac{1}{2sqrt{n}}$ is optimal?
linear-algebra functional-analysis
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
$endgroup$
Let $E$ be a complex Hilbert space.
In (arXiv) it was shown that for $A=(A_1,...,A_n) in mathcal{B}(E)^n$ we have,
$$displaystylefrac{1}{2sqrt{n}}|A|leq omega(A) leq |A|,$$
where
$$
omega(A) = sup_{|x|=1} left(sum_{k=1}^n |langle A_kx,xrangle|^2right)^{1/2},
$$
and
$$|A|= left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$
How can we prove that $frac{1}{2sqrt{n}}$ is optimal?
linear-algebra functional-analysis
linear-algebra functional-analysis
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
edited Jan 13 at 9:15
Student
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
asked Dec 27 '18 at 13:11
StudentStudent
2,4242524
2,4242524
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
We now 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
$$ frac{1}{2sqrt{n}} Vert A Vert = frac{1}{2sqrt{n}} cdot sqrt{n} = frac{1}{2} = omega_e(A) $$
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
$endgroup$
add a comment |
$begingroup$
As stated, the example cannot be right. The three expressions in the inequality spit coefficients, so if $T/sqrt n$ works, so does $T$.
Also the left inequalities is not sharp. For instance, if $E=mathbb C$, then
$$
omega(A_1,ldots,A_n)=left|sum_{j=1}^n A_j^*A_jright|^{1/2}.
$$
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
$begingroup$
Thank you Professor for your answer. However, please I don't understand why the left inequality cannot be sharp? The example taken in the paper is not correct. However, why we cannot find another example? By the way I think that the inequality $$omega(A_1,cdots,A_n) leq left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}$$ is sharp. Indeed, if one takes $A_k=S=begin{pmatrix}1&0\0&0end{pmatrix}$ for all $k$, we obtain equality in last inequality.
$endgroup$
– Student
Dec 28 '18 at 5:17
$begingroup$
It's not about having equality in one case, but in all of them. The example I'm giving you is that when $dim E=1$, your left inequality is always strict.
$endgroup$
– Martin Argerami
Dec 28 '18 at 5:29
$begingroup$
If I understand you, you mean by not ''sharp'' that is, it is impossible to find a same example $(A_1,cdots,A_n)$ such that $$displaystylefrac{1}{2sqrt{n}}left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}= omega(A_1,cdots,A_n)=left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$ Please correct me if I made wrong. Thank you.
$endgroup$
– Student
Dec 28 '18 at 15:38
$begingroup$
That's what I said. And the the case where $dim E=1$ gives an example where that inequality is never sharp. But I don't know what happens when $dim Egeq2$.
$endgroup$
– Martin Argerami
Dec 28 '18 at 15:59
$begingroup$
I'm puzzled. I don't see anything wrong with your computation.
$endgroup$
– Martin Argerami
Dec 28 '18 at 16:08
|
show 1 more comment
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2 Answers
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$begingroup$
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.
We now 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
$$ frac{1}{2sqrt{n}} Vert A Vert = frac{1}{2sqrt{n}} cdot sqrt{n} = frac{1}{2} = omega_e(A) $$
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
$endgroup$
add a comment |
$begingroup$
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.
We now 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
$$ frac{1}{2sqrt{n}} Vert A Vert = frac{1}{2sqrt{n}} cdot sqrt{n} = frac{1}{2} = omega_e(A) $$
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
$endgroup$
add a comment |
$begingroup$
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.
We now 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
$$ frac{1}{2sqrt{n}} Vert A Vert = frac{1}{2sqrt{n}} cdot sqrt{n} = frac{1}{2} = omega_e(A) $$
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
$endgroup$
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.
We now 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
$$ frac{1}{2sqrt{n}} Vert A Vert = frac{1}{2sqrt{n}} cdot sqrt{n} = frac{1}{2} = omega_e(A) $$
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
edited Jan 13 at 10:53
Martin Sleziak
44.7k9117272
44.7k9117272
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
answered Jan 12 at 13:57
Severin SchravenSeverin Schraven
6,1281934
6,1281934
add a comment |
add a comment |
$begingroup$
As stated, the example cannot be right. The three expressions in the inequality spit coefficients, so if $T/sqrt n$ works, so does $T$.
Also the left inequalities is not sharp. For instance, if $E=mathbb C$, then
$$
omega(A_1,ldots,A_n)=left|sum_{j=1}^n A_j^*A_jright|^{1/2}.
$$
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
$begingroup$
Thank you Professor for your answer. However, please I don't understand why the left inequality cannot be sharp? The example taken in the paper is not correct. However, why we cannot find another example? By the way I think that the inequality $$omega(A_1,cdots,A_n) leq left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}$$ is sharp. Indeed, if one takes $A_k=S=begin{pmatrix}1&0\0&0end{pmatrix}$ for all $k$, we obtain equality in last inequality.
$endgroup$
– Student
Dec 28 '18 at 5:17
$begingroup$
It's not about having equality in one case, but in all of them. The example I'm giving you is that when $dim E=1$, your left inequality is always strict.
$endgroup$
– Martin Argerami
Dec 28 '18 at 5:29
$begingroup$
If I understand you, you mean by not ''sharp'' that is, it is impossible to find a same example $(A_1,cdots,A_n)$ such that $$displaystylefrac{1}{2sqrt{n}}left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}= omega(A_1,cdots,A_n)=left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$ Please correct me if I made wrong. Thank you.
$endgroup$
– Student
Dec 28 '18 at 15:38
$begingroup$
That's what I said. And the the case where $dim E=1$ gives an example where that inequality is never sharp. But I don't know what happens when $dim Egeq2$.
$endgroup$
– Martin Argerami
Dec 28 '18 at 15:59
$begingroup$
I'm puzzled. I don't see anything wrong with your computation.
$endgroup$
– Martin Argerami
Dec 28 '18 at 16:08
|
show 1 more comment
$begingroup$
As stated, the example cannot be right. The three expressions in the inequality spit coefficients, so if $T/sqrt n$ works, so does $T$.
Also the left inequalities is not sharp. For instance, if $E=mathbb C$, then
$$
omega(A_1,ldots,A_n)=left|sum_{j=1}^n A_j^*A_jright|^{1/2}.
$$
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
$begingroup$
Thank you Professor for your answer. However, please I don't understand why the left inequality cannot be sharp? The example taken in the paper is not correct. However, why we cannot find another example? By the way I think that the inequality $$omega(A_1,cdots,A_n) leq left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}$$ is sharp. Indeed, if one takes $A_k=S=begin{pmatrix}1&0\0&0end{pmatrix}$ for all $k$, we obtain equality in last inequality.
$endgroup$
– Student
Dec 28 '18 at 5:17
$begingroup$
It's not about having equality in one case, but in all of them. The example I'm giving you is that when $dim E=1$, your left inequality is always strict.
$endgroup$
– Martin Argerami
Dec 28 '18 at 5:29
$begingroup$
If I understand you, you mean by not ''sharp'' that is, it is impossible to find a same example $(A_1,cdots,A_n)$ such that $$displaystylefrac{1}{2sqrt{n}}left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}= omega(A_1,cdots,A_n)=left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$ Please correct me if I made wrong. Thank you.
$endgroup$
– Student
Dec 28 '18 at 15:38
$begingroup$
That's what I said. And the the case where $dim E=1$ gives an example where that inequality is never sharp. But I don't know what happens when $dim Egeq2$.
$endgroup$
– Martin Argerami
Dec 28 '18 at 15:59
$begingroup$
I'm puzzled. I don't see anything wrong with your computation.
$endgroup$
– Martin Argerami
Dec 28 '18 at 16:08
|
show 1 more comment
$begingroup$
As stated, the example cannot be right. The three expressions in the inequality spit coefficients, so if $T/sqrt n$ works, so does $T$.
Also the left inequalities is not sharp. For instance, if $E=mathbb C$, then
$$
omega(A_1,ldots,A_n)=left|sum_{j=1}^n A_j^*A_jright|^{1/2}.
$$
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
$endgroup$
As stated, the example cannot be right. The three expressions in the inequality spit coefficients, so if $T/sqrt n$ works, so does $T$.
Also the left inequalities is not sharp. For instance, if $E=mathbb C$, then
$$
omega(A_1,ldots,A_n)=left|sum_{j=1}^n A_j^*A_jright|^{1/2}.
$$
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
answered Dec 28 '18 at 3:43
Martin ArgeramiMartin Argerami
126k1182180
126k1182180
$begingroup$
Thank you Professor for your answer. However, please I don't understand why the left inequality cannot be sharp? The example taken in the paper is not correct. However, why we cannot find another example? By the way I think that the inequality $$omega(A_1,cdots,A_n) leq left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}$$ is sharp. Indeed, if one takes $A_k=S=begin{pmatrix}1&0\0&0end{pmatrix}$ for all $k$, we obtain equality in last inequality.
$endgroup$
– Student
Dec 28 '18 at 5:17
$begingroup$
It's not about having equality in one case, but in all of them. The example I'm giving you is that when $dim E=1$, your left inequality is always strict.
$endgroup$
– Martin Argerami
Dec 28 '18 at 5:29
$begingroup$
If I understand you, you mean by not ''sharp'' that is, it is impossible to find a same example $(A_1,cdots,A_n)$ such that $$displaystylefrac{1}{2sqrt{n}}left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}= omega(A_1,cdots,A_n)=left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$ Please correct me if I made wrong. Thank you.
$endgroup$
– Student
Dec 28 '18 at 15:38
$begingroup$
That's what I said. And the the case where $dim E=1$ gives an example where that inequality is never sharp. But I don't know what happens when $dim Egeq2$.
$endgroup$
– Martin Argerami
Dec 28 '18 at 15:59
$begingroup$
I'm puzzled. I don't see anything wrong with your computation.
$endgroup$
– Martin Argerami
Dec 28 '18 at 16:08
|
show 1 more comment
$begingroup$
Thank you Professor for your answer. However, please I don't understand why the left inequality cannot be sharp? The example taken in the paper is not correct. However, why we cannot find another example? By the way I think that the inequality $$omega(A_1,cdots,A_n) leq left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}$$ is sharp. Indeed, if one takes $A_k=S=begin{pmatrix}1&0\0&0end{pmatrix}$ for all $k$, we obtain equality in last inequality.
$endgroup$
– Student
Dec 28 '18 at 5:17
$begingroup$
It's not about having equality in one case, but in all of them. The example I'm giving you is that when $dim E=1$, your left inequality is always strict.
$endgroup$
– Martin Argerami
Dec 28 '18 at 5:29
$begingroup$
If I understand you, you mean by not ''sharp'' that is, it is impossible to find a same example $(A_1,cdots,A_n)$ such that $$displaystylefrac{1}{2sqrt{n}}left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}= omega(A_1,cdots,A_n)=left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$ Please correct me if I made wrong. Thank you.
$endgroup$
– Student
Dec 28 '18 at 15:38
$begingroup$
That's what I said. And the the case where $dim E=1$ gives an example where that inequality is never sharp. But I don't know what happens when $dim Egeq2$.
$endgroup$
– Martin Argerami
Dec 28 '18 at 15:59
$begingroup$
I'm puzzled. I don't see anything wrong with your computation.
$endgroup$
– Martin Argerami
Dec 28 '18 at 16:08
$begingroup$
Thank you Professor for your answer. However, please I don't understand why the left inequality cannot be sharp? The example taken in the paper is not correct. However, why we cannot find another example? By the way I think that the inequality $$omega(A_1,cdots,A_n) leq left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}$$ is sharp. Indeed, if one takes $A_k=S=begin{pmatrix}1&0\0&0end{pmatrix}$ for all $k$, we obtain equality in last inequality.
$endgroup$
– Student
Dec 28 '18 at 5:17
$begingroup$
Thank you Professor for your answer. However, please I don't understand why the left inequality cannot be sharp? The example taken in the paper is not correct. However, why we cannot find another example? By the way I think that the inequality $$omega(A_1,cdots,A_n) leq left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}$$ is sharp. Indeed, if one takes $A_k=S=begin{pmatrix}1&0\0&0end{pmatrix}$ for all $k$, we obtain equality in last inequality.
$endgroup$
– Student
Dec 28 '18 at 5:17
$begingroup$
It's not about having equality in one case, but in all of them. The example I'm giving you is that when $dim E=1$, your left inequality is always strict.
$endgroup$
– Martin Argerami
Dec 28 '18 at 5:29
$begingroup$
It's not about having equality in one case, but in all of them. The example I'm giving you is that when $dim E=1$, your left inequality is always strict.
$endgroup$
– Martin Argerami
Dec 28 '18 at 5:29
$begingroup$
If I understand you, you mean by not ''sharp'' that is, it is impossible to find a same example $(A_1,cdots,A_n)$ such that $$displaystylefrac{1}{2sqrt{n}}left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}= omega(A_1,cdots,A_n)=left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$ Please correct me if I made wrong. Thank you.
$endgroup$
– Student
Dec 28 '18 at 15:38
$begingroup$
If I understand you, you mean by not ''sharp'' that is, it is impossible to find a same example $(A_1,cdots,A_n)$ such that $$displaystylefrac{1}{2sqrt{n}}left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}= omega(A_1,cdots,A_n)=left|displaystylesum_{k=1}^nA_k^*A_k right|^{1/2}.$$ Please correct me if I made wrong. Thank you.
$endgroup$
– Student
Dec 28 '18 at 15:38
$begingroup$
That's what I said. And the the case where $dim E=1$ gives an example where that inequality is never sharp. But I don't know what happens when $dim Egeq2$.
$endgroup$
– Martin Argerami
Dec 28 '18 at 15:59
$begingroup$
That's what I said. And the the case where $dim E=1$ gives an example where that inequality is never sharp. But I don't know what happens when $dim Egeq2$.
$endgroup$
– Martin Argerami
Dec 28 '18 at 15:59
$begingroup$
I'm puzzled. I don't see anything wrong with your computation.
$endgroup$
– Martin Argerami
Dec 28 '18 at 16:08
$begingroup$
I'm puzzled. I don't see anything wrong with your computation.
$endgroup$
– Martin Argerami
Dec 28 '18 at 16:08
|
show 1 more comment
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