Mixing
Let $H$ be a Hilbert space with projections $P,Q$, then $lim_{ntoinfty}(PQ)^nx= Rx$ for all $xin H$, where...
Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.
We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.
functional-analysis operator-theory hilbert-spaces
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
add a comment |
Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.
We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.
functional-analysis operator-theory hilbert-spaces
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 '18 at 15:42
1
(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 '18 at 15:53
1
Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 '18 at 15:54
add a comment |
Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.
We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.
functional-analysis operator-theory hilbert-spaces
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.
We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.
functional-analysis operator-theory hilbert-spaces
functional-analysis operator-theory hilbert-spaces
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
asked Nov 20 '18 at 15:32
Monstrous Moonshine
2,6941630
2,6941630
Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 '18 at 15:42
1
(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 '18 at 15:53
1
Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 '18 at 15:54
add a comment |
Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 '18 at 15:42
1
(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 '18 at 15:53
1
Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 '18 at 15:54
Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 '18 at 15:42
Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 '18 at 15:42
1
1
(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 '18 at 15:53
(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 '18 at 15:53
1
1
Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 '18 at 15:54
Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 '18 at 15:54
add a comment |
1 Answer
1
active
oldest
votes
I work in your reduced case and consider instead the operator $A = QPQ$ and note that
$$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
so it will suffice to show that $A^n to 0$ strongly as $n to infty$.
The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.
Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.
It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.
Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
$$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
$$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.
In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.
We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.
$$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006463%2flet-h-be-a-hilbert-space-with-projections-p-q-then-lim-n-to-inftypqn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I work in your reduced case and consider instead the operator $A = QPQ$ and note that
$$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
so it will suffice to show that $A^n to 0$ strongly as $n to infty$.
The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.
Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.
It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.
Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
$$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
$$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.
In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.
We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.
$$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
add a comment |
I work in your reduced case and consider instead the operator $A = QPQ$ and note that
$$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
so it will suffice to show that $A^n to 0$ strongly as $n to infty$.
The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.
Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.
It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.
Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
$$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
$$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.
In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.
We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.
$$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
add a comment |
I work in your reduced case and consider instead the operator $A = QPQ$ and note that
$$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
so it will suffice to show that $A^n to 0$ strongly as $n to infty$.
The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.
Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.
It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.
Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
$$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
$$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.
In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.
We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.
$$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
I work in your reduced case and consider instead the operator $A = QPQ$ and note that
$$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
so it will suffice to show that $A^n to 0$ strongly as $n to infty$.
The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.
Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.
It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.
Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
$$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
$$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.
In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.
We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.
$$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
edited Nov 20 '18 at 16:50
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
answered Nov 20 '18 at 16:43
Rhys Steele
6,0531829
6,0531829
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006463%2flet-h-be-a-hilbert-space-with-projections-p-q-then-lim-n-to-inftypqn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Do you assume your projections to be orthogonal?
– Rhys Steele
Nov 20 '18 at 15:42
1
(I've just realised there may be some confusion. What I'm asking is if $P$ is the orthogonal projection onto $Pmathcal{H}$ and similarly for $Q$. Not if you are assuming that $P$ is orthogonal to $Q$)
– Rhys Steele
Nov 20 '18 at 15:53
1
Oh I see, then I’m considering non-commuting orthogonal projections, using your terminology.
– Monstrous Moonshine
Nov 20 '18 at 15:54