Mixing
Bound operator norm
Clash Royale CLAN TAG#URR8PPP
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Consider the Hilbert space $ell^2(mathbbN)$ and let $P:ell^2(mathbbN)toell^2(mathbbN)$ be an orthogonal projection. Let $(delta_i)_iinmathbbN$ denote the canonical orthonormal basis of $ell^2(mathbbN)$ and define for $CsubsetmathbbN$
$$
Q_C=sum_kin CQ_k,
$$
where $Q_kx=langledelta_k,xrangledelta_k$. Put differently, $Q_C$ is the orthogonal projection on $ell^2(C)=overlinemathrmspandelta_k_kin C$.
Now I want to prove:
$$
|Q_CP(mathrmId-Q_C)PQ_C|leq frac14.
$$
My attemp: By the min-max theorem, $|Q_CP(mathrmId-Q_C)PQ_C|=sup_xlangle x,(Q_CP(mathrmId-Q_C)PQ_C)xrangle$ for $|x|=1$. By the outermost $Q_C$'s, we may restrict from the beginning to $xinoverlinemathrmspandelta_k_kin C$. In summary,
$$
|Q_CP(mathrmId-Q_C)PQ_C|=sup_substackxlangle x,P(mathrmId-Q_C)Pxrangle=sup_substackxleft[|Px|^2-sum_kin C|langle Px,delta_krangle|^2right].
$$
However, this approach does not look very fruitful...
functional-analysis operator-theory hilbert-spaces norm projection
edited Aug 1 at 14:37
Davide Giraudo
121k15146249
asked Jul 31 at 14:59
julian
312110
add a comment |Â
up vote
3
down vote
favorite
Consider the Hilbert space $ell^2(mathbbN)$ and let $P:ell^2(mathbbN)toell^2(mathbbN)$ be an orthogonal projection. Let $(delta_i)_iinmathbbN$ denote the canonical orthonormal basis of $ell^2(mathbbN)$ and define for $CsubsetmathbbN$
$$
Q_C=sum_kin CQ_k,
$$
where $Q_kx=langledelta_k,xrangledelta_k$. Put differently, $Q_C$ is the orthogonal projection on $ell^2(C)=overlinemathrmspandelta_k_kin C$.
Now I want to prove:
$$
|Q_CP(mathrmId-Q_C)PQ_C|leq frac14.
$$
My attemp: By the min-max theorem, $|Q_CP(mathrmId-Q_C)PQ_C|=sup_xlangle x,(Q_CP(mathrmId-Q_C)PQ_C)xrangle$ for $|x|=1$. By the outermost $Q_C$'s, we may restrict from the beginning to $xinoverlinemathrmspandelta_k_kin C$. In summary,
$$
|Q_CP(mathrmId-Q_C)PQ_C|=sup_substackxlangle x,P(mathrmId-Q_C)Pxrangle=sup_substackxleft[|Px|^2-sum_kin C|langle Px,delta_krangle|^2right].
$$
However, this approach does not look very fruitful...
functional-analysis operator-theory hilbert-spaces norm projection
edited Aug 1 at 14:37
Davide Giraudo
121k15146249
asked Jul 31 at 14:59
julian
312110
Do you know theorem 2 in Halmos' paper here : ams.org/journals/tran/1969-144-00/S0002-9947-1969-0251519-5/… ?
– Julien
Aug 1 at 12:53
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider the Hilbert space $ell^2(mathbbN)$ and let $P:ell^2(mathbbN)toell^2(mathbbN)$ be an orthogonal projection. Let $(delta_i)_iinmathbbN$ denote the canonical orthonormal basis of $ell^2(mathbbN)$ and define for $CsubsetmathbbN$
$$
Q_C=sum_kin CQ_k,
$$
where $Q_kx=langledelta_k,xrangledelta_k$. Put differently, $Q_C$ is the orthogonal projection on $ell^2(C)=overlinemathrmspandelta_k_kin C$.
Now I want to prove:
$$
|Q_CP(mathrmId-Q_C)PQ_C|leq frac14.
$$
My attemp: By the min-max theorem, $|Q_CP(mathrmId-Q_C)PQ_C|=sup_xlangle x,(Q_CP(mathrmId-Q_C)PQ_C)xrangle$ for $|x|=1$. By the outermost $Q_C$'s, we may restrict from the beginning to $xinoverlinemathrmspandelta_k_kin C$. In summary,
$$
|Q_CP(mathrmId-Q_C)PQ_C|=sup_substackxlangle x,P(mathrmId-Q_C)Pxrangle=sup_substackxleft[|Px|^2-sum_kin C|langle Px,delta_krangle|^2right].
$$
However, this approach does not look very fruitful...
functional-analysis operator-theory hilbert-spaces norm projection
edited Aug 1 at 14:37
Davide Giraudo
121k15146249
asked Jul 31 at 14:59
julian
312110
Consider the Hilbert space $ell^2(mathbbN)$ and let $P:ell^2(mathbbN)toell^2(mathbbN)$ be an orthogonal projection. Let $(delta_i)_iinmathbbN$ denote the canonical orthonormal basis of $ell^2(mathbbN)$ and define for $CsubsetmathbbN$
$$
Q_C=sum_kin CQ_k,
$$
where $Q_kx=langledelta_k,xrangledelta_k$. Put differently, $Q_C$ is the orthogonal projection on $ell^2(C)=overlinemathrmspandelta_k_kin C$.
Now I want to prove:
$$
|Q_CP(mathrmId-Q_C)PQ_C|leq frac14.
$$
My attemp: By the min-max theorem, $|Q_CP(mathrmId-Q_C)PQ_C|=sup_xlangle x,(Q_CP(mathrmId-Q_C)PQ_C)xrangle$ for $|x|=1$. By the outermost $Q_C$'s, we may restrict from the beginning to $xinoverlinemathrmspandelta_k_kin C$. In summary,
$$
|Q_CP(mathrmId-Q_C)PQ_C|=sup_substackxlangle x,P(mathrmId-Q_C)Pxrangle=sup_substackxleft[|Px|^2-sum_kin C|langle Px,delta_krangle|^2right].
$$
However, this approach does not look very fruitful...
functional-analysis operator-theory hilbert-spaces norm projection
edited Aug 1 at 14:37
Davide Giraudo
121k15146249
asked Jul 31 at 14:59
julian
312110
edited Aug 1 at 14:37
Davide Giraudo
121k15146249
edited Aug 1 at 14:37
Davide Giraudo
121k15146249
edited Aug 1 at 14:37
Davide Giraudo
121k15146249
121k15146249
asked Jul 31 at 14:59
julian
312110
asked Jul 31 at 14:59
julian
312110
asked Jul 31 at 14:59
julian
312110
312110
Do you know theorem 2 in Halmos' paper here : ams.org/journals/tran/1969-144-00/S0002-9947-1969-0251519-5/… ?
– Julien
Aug 1 at 12:53
add a comment |Â
Do you know theorem 2 in Halmos' paper here : ams.org/journals/tran/1969-144-00/S0002-9947-1969-0251519-5/… ?
– Julien
Aug 1 at 12:53
Do you know theorem 2 in Halmos' paper here : ams.org/journals/tran/1969-144-00/S0002-9947-1969-0251519-5/… ?
– Julien
Aug 1 at 12:53
Do you know theorem 2 in Halmos' paper here : ams.org/journals/tran/1969-144-00/S0002-9947-1969-0251519-5/… ?
– Julien
Aug 1 at 12:53
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The way I would see your inequality is as
$$
|(I-Q_C)PQ_C|leqfrac12,
$$
since $$|Q_CP(I-Q_C)PQ_C|=|Q_CP(I-Q_C)^*(I-Q_C)PQ_C|=|(I-Q_C)PQ_C|^2.$$
And this is basically saying that the off diagonal entries of a projection cannot contribute more than $1/2$ to the norm. In spirit, this happens because the equality $P^2=P$ gives, for every diagonal entry, $$ P_kk-P_kk^2=sum_jne k |P_kj|^2.$$
To make this into a proof, we write $P$ in terms of the decomposition $H=Q_CHoplus (I-Q_C)H$,
$$
P=beginbmatrix A&B\ B^*&Cendbmatrix,
$$
with $A,Cgeq0$ (since $P$ is positive). If we now look at the equality $P^2=P$ at the $1,1$ entry, we get
$$
A^2+BB^*=A.
$$
So $BB^*=A-A^2$. As $A$ is a positive contraction (from $P$ being a projection), we have
$$
sigma(A-A^2)subsetlambda-lambda^2: lambdain[0,1].
$$
For all such $lambda$ we have $lambda-lambda^2leq1/4$, so $$|B|^2=|BB^*|=|A-A^2|leqfrac14,$$ so $|B|leq1/2$. Thus
$$
|(I-Q_C)PQ_C|=|B|leqfrac12.
$$
answered Aug 3 at 1:24
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
I give the argument based on Halmos' paper as suggested by Julien in his comment. All credits are due to him for pointing out this neat result.
Let us first cite the theorem of Halmos:
If $M$ and $N$ are subspaces in generic position in a Hilbert space $H$, with respective projections $tildeP$ and $Q$, then there exists a Hilbert space $K$, and there exist positive contractions $S$ and $C$ on $K$, with $S^2 + C^2 = 1$ and $ker S=ker C=0$, such that $tildeP$ and $Q$ are unitarily equivalent to
$$
beginpmatrix1& 0\0&0endpmatrix,qquad beginpmatrixC^2&CS\CS&S^2endpmatrix
$$
respectively.
What's not written in the theorem but follows by the proof is that $K=mathrmranP$ and that $tildeP$ and $Q$ admit the same unitary $U:Hto Koplus K$. Moreover, $C$ and $S$ commute. Then using Halmos' theorem with $tildeP=Q_C$, $Q=P$, we see by a little computation
$$
|Q_CP(mathrmId-Q_C)PQ_C|=left|beginpmatrixC^2S^2&0\0&0endpmatrixright|=|C^2-C^4|.
$$
An easy exercise in spectral calculus shows $sigma(C^2-C^4)subset [0,1/4]$ from which the result follows.
answered Aug 1 at 14:13
julian
312110
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The way I would see your inequality is as
$$
|(I-Q_C)PQ_C|leqfrac12,
$$
since $$|Q_CP(I-Q_C)PQ_C|=|Q_CP(I-Q_C)^*(I-Q_C)PQ_C|=|(I-Q_C)PQ_C|^2.$$
And this is basically saying that the off diagonal entries of a projection cannot contribute more than $1/2$ to the norm. In spirit, this happens because the equality $P^2=P$ gives, for every diagonal entry, $$ P_kk-P_kk^2=sum_jne k |P_kj|^2.$$
To make this into a proof, we write $P$ in terms of the decomposition $H=Q_CHoplus (I-Q_C)H$,
$$
P=beginbmatrix A&B\ B^*&Cendbmatrix,
$$
with $A,Cgeq0$ (since $P$ is positive). If we now look at the equality $P^2=P$ at the $1,1$ entry, we get
$$
A^2+BB^*=A.
$$
So $BB^*=A-A^2$. As $A$ is a positive contraction (from $P$ being a projection), we have
$$
sigma(A-A^2)subsetlambda-lambda^2: lambdain[0,1].
$$
For all such $lambda$ we have $lambda-lambda^2leq1/4$, so $$|B|^2=|BB^*|=|A-A^2|leqfrac14,$$ so $|B|leq1/2$. Thus
$$
|(I-Q_C)PQ_C|=|B|leqfrac12.
$$
answered Aug 3 at 1:24
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
The way I would see your inequality is as
$$
|(I-Q_C)PQ_C|leqfrac12,
$$
since $$|Q_CP(I-Q_C)PQ_C|=|Q_CP(I-Q_C)^*(I-Q_C)PQ_C|=|(I-Q_C)PQ_C|^2.$$
And this is basically saying that the off diagonal entries of a projection cannot contribute more than $1/2$ to the norm. In spirit, this happens because the equality $P^2=P$ gives, for every diagonal entry, $$ P_kk-P_kk^2=sum_jne k |P_kj|^2.$$
To make this into a proof, we write $P$ in terms of the decomposition $H=Q_CHoplus (I-Q_C)H$,
$$
P=beginbmatrix A&B\ B^*&Cendbmatrix,
$$
with $A,Cgeq0$ (since $P$ is positive). If we now look at the equality $P^2=P$ at the $1,1$ entry, we get
$$
A^2+BB^*=A.
$$
So $BB^*=A-A^2$. As $A$ is a positive contraction (from $P$ being a projection), we have
$$
sigma(A-A^2)subsetlambda-lambda^2: lambdain[0,1].
$$
For all such $lambda$ we have $lambda-lambda^2leq1/4$, so $$|B|^2=|BB^*|=|A-A^2|leqfrac14,$$ so $|B|leq1/2$. Thus
$$
|(I-Q_C)PQ_C|=|B|leqfrac12.
$$
answered Aug 3 at 1:24
Martin Argerami
115k1071164
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The way I would see your inequality is as
$$
|(I-Q_C)PQ_C|leqfrac12,
$$
since $$|Q_CP(I-Q_C)PQ_C|=|Q_CP(I-Q_C)^*(I-Q_C)PQ_C|=|(I-Q_C)PQ_C|^2.$$
And this is basically saying that the off diagonal entries of a projection cannot contribute more than $1/2$ to the norm. In spirit, this happens because the equality $P^2=P$ gives, for every diagonal entry, $$ P_kk-P_kk^2=sum_jne k |P_kj|^2.$$
To make this into a proof, we write $P$ in terms of the decomposition $H=Q_CHoplus (I-Q_C)H$,
$$
P=beginbmatrix A&B\ B^*&Cendbmatrix,
$$
with $A,Cgeq0$ (since $P$ is positive). If we now look at the equality $P^2=P$ at the $1,1$ entry, we get
$$
A^2+BB^*=A.
$$
So $BB^*=A-A^2$. As $A$ is a positive contraction (from $P$ being a projection), we have
$$
sigma(A-A^2)subsetlambda-lambda^2: lambdain[0,1].
$$
For all such $lambda$ we have $lambda-lambda^2leq1/4$, so $$|B|^2=|BB^*|=|A-A^2|leqfrac14,$$ so $|B|leq1/2$. Thus
$$
|(I-Q_C)PQ_C|=|B|leqfrac12.
$$
answered Aug 3 at 1:24
Martin Argerami
115k1071164
The way I would see your inequality is as
$$
|(I-Q_C)PQ_C|leqfrac12,
$$
since $$|Q_CP(I-Q_C)PQ_C|=|Q_CP(I-Q_C)^*(I-Q_C)PQ_C|=|(I-Q_C)PQ_C|^2.$$
And this is basically saying that the off diagonal entries of a projection cannot contribute more than $1/2$ to the norm. In spirit, this happens because the equality $P^2=P$ gives, for every diagonal entry, $$ P_kk-P_kk^2=sum_jne k |P_kj|^2.$$
To make this into a proof, we write $P$ in terms of the decomposition $H=Q_CHoplus (I-Q_C)H$,
$$
P=beginbmatrix A&B\ B^*&Cendbmatrix,
$$
with $A,Cgeq0$ (since $P$ is positive). If we now look at the equality $P^2=P$ at the $1,1$ entry, we get
$$
A^2+BB^*=A.
$$
So $BB^*=A-A^2$. As $A$ is a positive contraction (from $P$ being a projection), we have
$$
sigma(A-A^2)subsetlambda-lambda^2: lambdain[0,1].
$$
For all such $lambda$ we have $lambda-lambda^2leq1/4$, so $$|B|^2=|BB^*|=|A-A^2|leqfrac14,$$ so $|B|leq1/2$. Thus
$$
|(I-Q_C)PQ_C|=|B|leqfrac12.
$$
answered Aug 3 at 1:24
Martin Argerami
115k1071164
edited Aug 3 at 6:16
answered Aug 3 at 1:24
Martin Argerami
115k1071164
answered Aug 3 at 1:24
Martin Argerami
115k1071164
answered Aug 3 at 1:24
Martin Argerami
115k1071164
115k1071164
add a comment |Â
add a comment |Â
up vote
1
down vote
I give the argument based on Halmos' paper as suggested by Julien in his comment. All credits are due to him for pointing out this neat result.
Let us first cite the theorem of Halmos:
If $M$ and $N$ are subspaces in generic position in a Hilbert space $H$, with respective projections $tildeP$ and $Q$, then there exists a Hilbert space $K$, and there exist positive contractions $S$ and $C$ on $K$, with $S^2 + C^2 = 1$ and $ker S=ker C=0$, such that $tildeP$ and $Q$ are unitarily equivalent to
$$
beginpmatrix1& 0\0&0endpmatrix,qquad beginpmatrixC^2&CS\CS&S^2endpmatrix
$$
respectively.
What's not written in the theorem but follows by the proof is that $K=mathrmranP$ and that $tildeP$ and $Q$ admit the same unitary $U:Hto Koplus K$. Moreover, $C$ and $S$ commute. Then using Halmos' theorem with $tildeP=Q_C$, $Q=P$, we see by a little computation
$$
|Q_CP(mathrmId-Q_C)PQ_C|=left|beginpmatrixC^2S^2&0\0&0endpmatrixright|=|C^2-C^4|.
$$
An easy exercise in spectral calculus shows $sigma(C^2-C^4)subset [0,1/4]$ from which the result follows.
answered Aug 1 at 14:13
julian
312110
add a comment |Â
up vote
1
down vote
I give the argument based on Halmos' paper as suggested by Julien in his comment. All credits are due to him for pointing out this neat result.
Let us first cite the theorem of Halmos:
If $M$ and $N$ are subspaces in generic position in a Hilbert space $H$, with respective projections $tildeP$ and $Q$, then there exists a Hilbert space $K$, and there exist positive contractions $S$ and $C$ on $K$, with $S^2 + C^2 = 1$ and $ker S=ker C=0$, such that $tildeP$ and $Q$ are unitarily equivalent to
$$
beginpmatrix1& 0\0&0endpmatrix,qquad beginpmatrixC^2&CS\CS&S^2endpmatrix
$$
respectively.
What's not written in the theorem but follows by the proof is that $K=mathrmranP$ and that $tildeP$ and $Q$ admit the same unitary $U:Hto Koplus K$. Moreover, $C$ and $S$ commute. Then using Halmos' theorem with $tildeP=Q_C$, $Q=P$, we see by a little computation
$$
|Q_CP(mathrmId-Q_C)PQ_C|=left|beginpmatrixC^2S^2&0\0&0endpmatrixright|=|C^2-C^4|.
$$
An easy exercise in spectral calculus shows $sigma(C^2-C^4)subset [0,1/4]$ from which the result follows.
answered Aug 1 at 14:13
julian
312110
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I give the argument based on Halmos' paper as suggested by Julien in his comment. All credits are due to him for pointing out this neat result.
Let us first cite the theorem of Halmos:
If $M$ and $N$ are subspaces in generic position in a Hilbert space $H$, with respective projections $tildeP$ and $Q$, then there exists a Hilbert space $K$, and there exist positive contractions $S$ and $C$ on $K$, with $S^2 + C^2 = 1$ and $ker S=ker C=0$, such that $tildeP$ and $Q$ are unitarily equivalent to
$$
beginpmatrix1& 0\0&0endpmatrix,qquad beginpmatrixC^2&CS\CS&S^2endpmatrix
$$
respectively.
What's not written in the theorem but follows by the proof is that $K=mathrmranP$ and that $tildeP$ and $Q$ admit the same unitary $U:Hto Koplus K$. Moreover, $C$ and $S$ commute. Then using Halmos' theorem with $tildeP=Q_C$, $Q=P$, we see by a little computation
$$
|Q_CP(mathrmId-Q_C)PQ_C|=left|beginpmatrixC^2S^2&0\0&0endpmatrixright|=|C^2-C^4|.
$$
An easy exercise in spectral calculus shows $sigma(C^2-C^4)subset [0,1/4]$ from which the result follows.
answered Aug 1 at 14:13
julian
312110
I give the argument based on Halmos' paper as suggested by Julien in his comment. All credits are due to him for pointing out this neat result.
Let us first cite the theorem of Halmos:
If $M$ and $N$ are subspaces in generic position in a Hilbert space $H$, with respective projections $tildeP$ and $Q$, then there exists a Hilbert space $K$, and there exist positive contractions $S$ and $C$ on $K$, with $S^2 + C^2 = 1$ and $ker S=ker C=0$, such that $tildeP$ and $Q$ are unitarily equivalent to
$$
beginpmatrix1& 0\0&0endpmatrix,qquad beginpmatrixC^2&CS\CS&S^2endpmatrix
$$
respectively.
What's not written in the theorem but follows by the proof is that $K=mathrmranP$ and that $tildeP$ and $Q$ admit the same unitary $U:Hto Koplus K$. Moreover, $C$ and $S$ commute. Then using Halmos' theorem with $tildeP=Q_C$, $Q=P$, we see by a little computation
$$
|Q_CP(mathrmId-Q_C)PQ_C|=left|beginpmatrixC^2S^2&0\0&0endpmatrixright|=|C^2-C^4|.
$$
An easy exercise in spectral calculus shows $sigma(C^2-C^4)subset [0,1/4]$ from which the result follows.
answered Aug 1 at 14:13
julian
312110
answered Aug 1 at 14:13
julian
312110
answered Aug 1 at 14:13
julian
312110
answered Aug 1 at 14:13
julian
312110
312110
add a comment |Â
add a comment |Â
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Do you know theorem 2 in Halmos' paper here : ams.org/journals/tran/1969-144-00/S0002-9947-1969-0251519-5/… ?
– Julien
Aug 1 at 12:53