Numerical radius of a pair of operators in Hilbert spaces
$begingroup$
Let $(C,D)$ be a pair of bounded linear operators on a complex Hilbert space $E$. The Euclidean operator radius is defined by
$$w_e(C,D)=displaystylesup_{|x|=1}left(|langle Cx,x rangle|^2+|langle Dx,x rangle|^2right)^{1/2}.$$
Moreover, the following inequality holds:
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}leq w_e(C,D)leq |C^*C+D^*D|^{1/2}.$$
I want to show that the constants $frac{sqrt{2}}{4}$ and $1$ in the above inequalities are the best possible.
For the second inequality, the following example show that we have equality:
Let $(C,D)=(B,B)$, with $B=begin{pmatrix}1&0\0&0end{pmatrix}$ (operator on $(mathbb{C}^2,|cdot|)$). Hence, I get
$w_e(C,D)=sqrt{2}$ and $|C^*C+D^*D|^{1/2}=sqrt{2}.$
I want to find $(C,D)$ such that
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D).$$
For a single operator, we have the following theorem:
Do you think that if $text{Im}(C)perp text{Im}(C^*)$ and $text{Im}(D)perp text{Im}(D^*)$ we have
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D),?$$
Thank you in advance.
functional-analysis
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
$endgroup$
add a comment |
$begingroup$
Let $(C,D)$ be a pair of bounded linear operators on a complex Hilbert space $E$. The Euclidean operator radius is defined by
$$w_e(C,D)=displaystylesup_{|x|=1}left(|langle Cx,x rangle|^2+|langle Dx,x rangle|^2right)^{1/2}.$$
Moreover, the following inequality holds:
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}leq w_e(C,D)leq |C^*C+D^*D|^{1/2}.$$
I want to show that the constants $frac{sqrt{2}}{4}$ and $1$ in the above inequalities are the best possible.
For the second inequality, the following example show that we have equality:
Let $(C,D)=(B,B)$, with $B=begin{pmatrix}1&0\0&0end{pmatrix}$ (operator on $(mathbb{C}^2,|cdot|)$). Hence, I get
$w_e(C,D)=sqrt{2}$ and $|C^*C+D^*D|^{1/2}=sqrt{2}.$
I want to find $(C,D)$ such that
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D).$$
For a single operator, we have the following theorem:
Do you think that if $text{Im}(C)perp text{Im}(C^*)$ and $text{Im}(D)perp text{Im}(D^*)$ we have
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D),?$$
Thank you in advance.
functional-analysis
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
$endgroup$
add a comment |
$begingroup$
Let $(C,D)$ be a pair of bounded linear operators on a complex Hilbert space $E$. The Euclidean operator radius is defined by
$$w_e(C,D)=displaystylesup_{|x|=1}left(|langle Cx,x rangle|^2+|langle Dx,x rangle|^2right)^{1/2}.$$
Moreover, the following inequality holds:
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}leq w_e(C,D)leq |C^*C+D^*D|^{1/2}.$$
I want to show that the constants $frac{sqrt{2}}{4}$ and $1$ in the above inequalities are the best possible.
For the second inequality, the following example show that we have equality:
Let $(C,D)=(B,B)$, with $B=begin{pmatrix}1&0\0&0end{pmatrix}$ (operator on $(mathbb{C}^2,|cdot|)$). Hence, I get
$w_e(C,D)=sqrt{2}$ and $|C^*C+D^*D|^{1/2}=sqrt{2}.$
I want to find $(C,D)$ such that
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D).$$
For a single operator, we have the following theorem:
Do you think that if $text{Im}(C)perp text{Im}(C^*)$ and $text{Im}(D)perp text{Im}(D^*)$ we have
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D),?$$
Thank you in advance.
functional-analysis
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
$endgroup$
Let $(C,D)$ be a pair of bounded linear operators on a complex Hilbert space $E$. The Euclidean operator radius is defined by
$$w_e(C,D)=displaystylesup_{|x|=1}left(|langle Cx,x rangle|^2+|langle Dx,x rangle|^2right)^{1/2}.$$
Moreover, the following inequality holds:
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}leq w_e(C,D)leq |C^*C+D^*D|^{1/2}.$$
I want to show that the constants $frac{sqrt{2}}{4}$ and $1$ in the above inequalities are the best possible.
For the second inequality, the following example show that we have equality:
Let $(C,D)=(B,B)$, with $B=begin{pmatrix}1&0\0&0end{pmatrix}$ (operator on $(mathbb{C}^2,|cdot|)$). Hence, I get
$w_e(C,D)=sqrt{2}$ and $|C^*C+D^*D|^{1/2}=sqrt{2}.$
I want to find $(C,D)$ such that
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D).$$
For a single operator, we have the following theorem:
Do you think that if $text{Im}(C)perp text{Im}(C^*)$ and $text{Im}(D)perp text{Im}(D^*)$ we have
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}= w_e(C,D),?$$
Thank you in advance.
functional-analysis
functional-analysis
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
edited Jan 17 '18 at 15:29
Student
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
asked Jan 11 '18 at 12:31
StudentStudent
2,4172524
2,4172524
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For your first question, consider
$$ C = D = begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}| = frac{sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = sqrt{2} sup_{|x|=1} |langle begin{pmatrix}0 \ x_1 end{pmatrix}, begin{pmatrix} x_1\x_2 end{pmatrix}rangle| = sqrt{2} sup_{|x|=1} |x_1x_2| = frac{sqrt{2}}{2}.$$
For your second question: Consider
$$ C = begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}, D = begin{pmatrix} 0 & 0 \ 1 & 0 end{pmatrix},$$
then $Im(C) perp Im(C^ast)$ and $Im(D) perp Im(D^ast)$ but
$$ frac{sqrt{2}}{4} | CC^ast + DD^ast|^{1/2} = frac{sqrt{2}}{4}, quad w_e(C,D)) = frac{1}{sqrt{2}}.$$
answered Jan 17 '18 at 11:46
agbagb
1,069415
$endgroup$
$begingroup$
I think there exits a wrong because we have $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|^{1/2}$$ and not $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|. $$
$endgroup$
– Student
Dec 22 '18 at 13:34
$begingroup$
Do you agree with me?
$endgroup$
– Student
Jan 11 at 8:46
add a comment |
$begingroup$
In view of (arXiv):
Let $(C,D)=(frac{1}{sqrt{2}}A,frac{1}{sqrt{2}}A)$, with $A=begin{pmatrix}0&0\1&0end{pmatrix}$. Hence,
$$w_e(C,D)=w(A)=frac{1}{2},$$
but if I don't make wrong in calculus I get
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}=frac{sqrt{2}}{4}.$$
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For your first question, consider
$$ C = D = begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}| = frac{sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = sqrt{2} sup_{|x|=1} |langle begin{pmatrix}0 \ x_1 end{pmatrix}, begin{pmatrix} x_1\x_2 end{pmatrix}rangle| = sqrt{2} sup_{|x|=1} |x_1x_2| = frac{sqrt{2}}{2}.$$
For your second question: Consider
$$ C = begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}, D = begin{pmatrix} 0 & 0 \ 1 & 0 end{pmatrix},$$
then $Im(C) perp Im(C^ast)$ and $Im(D) perp Im(D^ast)$ but
$$ frac{sqrt{2}}{4} | CC^ast + DD^ast|^{1/2} = frac{sqrt{2}}{4}, quad w_e(C,D)) = frac{1}{sqrt{2}}.$$
answered Jan 17 '18 at 11:46
agbagb
1,069415
$endgroup$
$begingroup$
I think there exits a wrong because we have $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|^{1/2}$$ and not $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|. $$
$endgroup$
– Student
Dec 22 '18 at 13:34
$begingroup$
Do you agree with me?
$endgroup$
– Student
Jan 11 at 8:46
add a comment |
$begingroup$
For your first question, consider
$$ C = D = begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}| = frac{sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = sqrt{2} sup_{|x|=1} |langle begin{pmatrix}0 \ x_1 end{pmatrix}, begin{pmatrix} x_1\x_2 end{pmatrix}rangle| = sqrt{2} sup_{|x|=1} |x_1x_2| = frac{sqrt{2}}{2}.$$
For your second question: Consider
$$ C = begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}, D = begin{pmatrix} 0 & 0 \ 1 & 0 end{pmatrix},$$
then $Im(C) perp Im(C^ast)$ and $Im(D) perp Im(D^ast)$ but
$$ frac{sqrt{2}}{4} | CC^ast + DD^ast|^{1/2} = frac{sqrt{2}}{4}, quad w_e(C,D)) = frac{1}{sqrt{2}}.$$
answered Jan 17 '18 at 11:46
agbagb
1,069415
$endgroup$
$begingroup$
I think there exits a wrong because we have $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|^{1/2}$$ and not $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|. $$
$endgroup$
– Student
Dec 22 '18 at 13:34
$begingroup$
Do you agree with me?
$endgroup$
– Student
Jan 11 at 8:46
add a comment |
$begingroup$
For your first question, consider
$$ C = D = begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}| = frac{sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = sqrt{2} sup_{|x|=1} |langle begin{pmatrix}0 \ x_1 end{pmatrix}, begin{pmatrix} x_1\x_2 end{pmatrix}rangle| = sqrt{2} sup_{|x|=1} |x_1x_2| = frac{sqrt{2}}{2}.$$
For your second question: Consider
$$ C = begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}, D = begin{pmatrix} 0 & 0 \ 1 & 0 end{pmatrix},$$
then $Im(C) perp Im(C^ast)$ and $Im(D) perp Im(D^ast)$ but
$$ frac{sqrt{2}}{4} | CC^ast + DD^ast|^{1/2} = frac{sqrt{2}}{4}, quad w_e(C,D)) = frac{1}{sqrt{2}}.$$
answered Jan 17 '18 at 11:46
agbagb
1,069415
$endgroup$
For your first question, consider
$$ C = D = begin{pmatrix} 0 & 0 \ 1 & 0end{pmatrix}.$$
Then the LHS of the first inequality reads
$$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}| = frac{sqrt{2}}{2}. $$
For the RHS it is
$$ w_e(C,D) = sqrt{2} sup_{|x|=1} |langle begin{pmatrix}0 \ x_1 end{pmatrix}, begin{pmatrix} x_1\x_2 end{pmatrix}rangle| = sqrt{2} sup_{|x|=1} |x_1x_2| = frac{sqrt{2}}{2}.$$
For your second question: Consider
$$ C = begin{pmatrix} 0 & 1 \ 0 & 0 end{pmatrix}, D = begin{pmatrix} 0 & 0 \ 1 & 0 end{pmatrix},$$
then $Im(C) perp Im(C^ast)$ and $Im(D) perp Im(D^ast)$ but
$$ frac{sqrt{2}}{4} | CC^ast + DD^ast|^{1/2} = frac{sqrt{2}}{4}, quad w_e(C,D)) = frac{1}{sqrt{2}}.$$
answered Jan 17 '18 at 11:46
agbagb
1,069415
answered Jan 17 '18 at 11:46
agbagb
1,069415
answered Jan 17 '18 at 11:46
agbagb
1,069415
answered Jan 17 '18 at 11:46
agbagb
1,069415
1,069415
$begingroup$
I think there exits a wrong because we have $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|^{1/2}$$ and not $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|. $$
$endgroup$
– Student
Dec 22 '18 at 13:34
$begingroup$
Do you agree with me?
$endgroup$
– Student
Jan 11 at 8:46
add a comment |
$begingroup$
I think there exits a wrong because we have $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|^{1/2}$$ and not $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|. $$
$endgroup$
– Student
Dec 22 '18 at 13:34
$begingroup$
Do you agree with me?
$endgroup$
– Student
Jan 11 at 8:46
$begingroup$
I think there exits a wrong because we have $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|^{1/2}$$ and not $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|. $$
$endgroup$
– Student
Dec 22 '18 at 13:34
$begingroup$
I think there exits a wrong because we have $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|^{1/2}$$ and not $$ frac{sqrt{2}}{4}| 2begin{pmatrix} 0 & 0 \ 0 & 1end{pmatrix}|. $$
$endgroup$
– Student
Dec 22 '18 at 13:34
$begingroup$
Do you agree with me?
$endgroup$
– Student
Jan 11 at 8:46
$begingroup$
Do you agree with me?
$endgroup$
– Student
Jan 11 at 8:46
add a comment |
$begingroup$
In view of (arXiv):
Let $(C,D)=(frac{1}{sqrt{2}}A,frac{1}{sqrt{2}}A)$, with $A=begin{pmatrix}0&0\1&0end{pmatrix}$. Hence,
$$w_e(C,D)=w(A)=frac{1}{2},$$
but if I don't make wrong in calculus I get
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}=frac{sqrt{2}}{4}.$$
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
$endgroup$
add a comment |
$begingroup$
In view of (arXiv):
Let $(C,D)=(frac{1}{sqrt{2}}A,frac{1}{sqrt{2}}A)$, with $A=begin{pmatrix}0&0\1&0end{pmatrix}$. Hence,
$$w_e(C,D)=w(A)=frac{1}{2},$$
but if I don't make wrong in calculus I get
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}=frac{sqrt{2}}{4}.$$
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
$endgroup$
add a comment |
$begingroup$
In view of (arXiv):
Let $(C,D)=(frac{1}{sqrt{2}}A,frac{1}{sqrt{2}}A)$, with $A=begin{pmatrix}0&0\1&0end{pmatrix}$. Hence,
$$w_e(C,D)=w(A)=frac{1}{2},$$
but if I don't make wrong in calculus I get
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}=frac{sqrt{2}}{4}.$$
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
$endgroup$
In view of (arXiv):
Let $(C,D)=(frac{1}{sqrt{2}}A,frac{1}{sqrt{2}}A)$, with $A=begin{pmatrix}0&0\1&0end{pmatrix}$. Hence,
$$w_e(C,D)=w(A)=frac{1}{2},$$
but if I don't make wrong in calculus I get
$$frac{sqrt{2}}{4}|C^*C+D^*D|^{1/2}=frac{sqrt{2}}{4}.$$
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
edited Jan 10 at 19:11
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
answered Jan 14 '18 at 15:56
StudentStudent
2,4172524
2,4172524
add a comment |
add a comment |
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