Mixing
Representation of linear operator between $L^p$ spaces.
$begingroup$
I was wondering where I could find a reference to the a characterization of continuous linear operators:
$$T:L^p(X,mu)to L^q(Y,eta)$$
of the form $T(f)(y)=int_{X} k(x,y)f(x)dmu$ for some $k$ satisfying some properties.
functional-analysis reference-request banach-spaces normed-spaces lp-spaces
asked Jan 23 at 3:32
LolmanLolman
1,000613
$endgroup$
add a comment |
$begingroup$
I was wondering where I could find a reference to the a characterization of continuous linear operators:
$$T:L^p(X,mu)to L^q(Y,eta)$$
of the form $T(f)(y)=int_{X} k(x,y)f(x)dmu$ for some $k$ satisfying some properties.
functional-analysis reference-request banach-spaces normed-spaces lp-spaces
asked Jan 23 at 3:32
LolmanLolman
1,000613
$endgroup$
$begingroup$
The question is not properly stated. The form you are looking for is not sated clearly, but I guess you want $T$ to an integral operator. The answer depends on what type of properties you are interested in. If $p=q=2$, for example, not every operator on $L^{2}$ is compact and integral operators of interest are usually compact operators.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:55
add a comment |
$begingroup$
I was wondering where I could find a reference to the a characterization of continuous linear operators:
$$T:L^p(X,mu)to L^q(Y,eta)$$
of the form $T(f)(y)=int_{X} k(x,y)f(x)dmu$ for some $k$ satisfying some properties.
functional-analysis reference-request banach-spaces normed-spaces lp-spaces
asked Jan 23 at 3:32
LolmanLolman
1,000613
$endgroup$
I was wondering where I could find a reference to the a characterization of continuous linear operators:
$$T:L^p(X,mu)to L^q(Y,eta)$$
of the form $T(f)(y)=int_{X} k(x,y)f(x)dmu$ for some $k$ satisfying some properties.
functional-analysis reference-request banach-spaces normed-spaces lp-spaces
functional-analysis reference-request banach-spaces normed-spaces lp-spaces
asked Jan 23 at 3:32
LolmanLolman
1,000613
asked Jan 23 at 3:32
LolmanLolman
1,000613
edited Jan 26 at 17:00
Lolman
asked Jan 23 at 3:32
LolmanLolman
1,000613
asked Jan 23 at 3:32
LolmanLolman
1,000613
asked Jan 23 at 3:32
LolmanLolman
1,000613
1,000613
$begingroup$
The question is not properly stated. The form you are looking for is not sated clearly, but I guess you want $T$ to an integral operator. The answer depends on what type of properties you are interested in. If $p=q=2$, for example, not every operator on $L^{2}$ is compact and integral operators of interest are usually compact operators.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:55
add a comment |
$begingroup$
The question is not properly stated. The form you are looking for is not sated clearly, but I guess you want $T$ to an integral operator. The answer depends on what type of properties you are interested in. If $p=q=2$, for example, not every operator on $L^{2}$ is compact and integral operators of interest are usually compact operators.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:55
$begingroup$
The question is not properly stated. The form you are looking for is not sated clearly, but I guess you want $T$ to an integral operator. The answer depends on what type of properties you are interested in. If $p=q=2$, for example, not every operator on $L^{2}$ is compact and integral operators of interest are usually compact operators.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:55
$begingroup$
The question is not properly stated. The form you are looking for is not sated clearly, but I guess you want $T$ to an integral operator. The answer depends on what type of properties you are interested in. If $p=q=2$, for example, not every operator on $L^{2}$ is compact and integral operators of interest are usually compact operators.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not every bounded linear operator from $L^p$ to $L^q$ has an integral kernel. Schachermeyer (Integral Operators on $L^p$ Spaces, Indiana Uni. Math. J.) gave the following characterization of integral operators:
Let $pin[1,infty)$ and $qin [0,infty]$. A linear operator $Tcolon
L^p(X,mu)to L^q(Y,nu)$ has an integral kernel if and only if it
maps order intervals to equi-measurable sets.
An order interval is a set of the form ${fin L^p(X,mu)mid gleq fleq h}$ for $g,hin L^p$. A subset $mathfrak{F}$ of $L^q(Y,nu)$ is called equi-measurable if for every $epsilon>0$ there exists a measurable subset $Y_epsilon$ of $Y$ such that $nu(Ysetminus Y_epsilon)<epsilon$ and $mathfrak{F}|_{Y_epsilon}$ is relatively norm compact in $L^infty(Y_epsilon,mu|_{Y_epsilon})$.
Related results can be found in the articles Kernel operators and Compactness properties of an operator which imply that it is an integral operator by Schep, and The integral representation of linear operators by Buhvalov.
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
$endgroup$
$begingroup$
Thanks, that's exactly one type of result I was searching for! Do you know how not integral operators look like?
$endgroup$
– Lolman
Jan 23 at 16:09
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%2f3084053%2frepresentation-of-linear-operator-between-lp-spaces%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
$begingroup$
Not every bounded linear operator from $L^p$ to $L^q$ has an integral kernel. Schachermeyer (Integral Operators on $L^p$ Spaces, Indiana Uni. Math. J.) gave the following characterization of integral operators:
Let $pin[1,infty)$ and $qin [0,infty]$. A linear operator $Tcolon
L^p(X,mu)to L^q(Y,nu)$ has an integral kernel if and only if it
maps order intervals to equi-measurable sets.
An order interval is a set of the form ${fin L^p(X,mu)mid gleq fleq h}$ for $g,hin L^p$. A subset $mathfrak{F}$ of $L^q(Y,nu)$ is called equi-measurable if for every $epsilon>0$ there exists a measurable subset $Y_epsilon$ of $Y$ such that $nu(Ysetminus Y_epsilon)<epsilon$ and $mathfrak{F}|_{Y_epsilon}$ is relatively norm compact in $L^infty(Y_epsilon,mu|_{Y_epsilon})$.
Related results can be found in the articles Kernel operators and Compactness properties of an operator which imply that it is an integral operator by Schep, and The integral representation of linear operators by Buhvalov.
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
$endgroup$
$begingroup$
Thanks, that's exactly one type of result I was searching for! Do you know how not integral operators look like?
$endgroup$
– Lolman
Jan 23 at 16:09
add a comment |
$begingroup$
Not every bounded linear operator from $L^p$ to $L^q$ has an integral kernel. Schachermeyer (Integral Operators on $L^p$ Spaces, Indiana Uni. Math. J.) gave the following characterization of integral operators:
Let $pin[1,infty)$ and $qin [0,infty]$. A linear operator $Tcolon
L^p(X,mu)to L^q(Y,nu)$ has an integral kernel if and only if it
maps order intervals to equi-measurable sets.
An order interval is a set of the form ${fin L^p(X,mu)mid gleq fleq h}$ for $g,hin L^p$. A subset $mathfrak{F}$ of $L^q(Y,nu)$ is called equi-measurable if for every $epsilon>0$ there exists a measurable subset $Y_epsilon$ of $Y$ such that $nu(Ysetminus Y_epsilon)<epsilon$ and $mathfrak{F}|_{Y_epsilon}$ is relatively norm compact in $L^infty(Y_epsilon,mu|_{Y_epsilon})$.
Related results can be found in the articles Kernel operators and Compactness properties of an operator which imply that it is an integral operator by Schep, and The integral representation of linear operators by Buhvalov.
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
$endgroup$
$begingroup$
Thanks, that's exactly one type of result I was searching for! Do you know how not integral operators look like?
$endgroup$
– Lolman
Jan 23 at 16:09
add a comment |
$begingroup$
Not every bounded linear operator from $L^p$ to $L^q$ has an integral kernel. Schachermeyer (Integral Operators on $L^p$ Spaces, Indiana Uni. Math. J.) gave the following characterization of integral operators:
Let $pin[1,infty)$ and $qin [0,infty]$. A linear operator $Tcolon
L^p(X,mu)to L^q(Y,nu)$ has an integral kernel if and only if it
maps order intervals to equi-measurable sets.
An order interval is a set of the form ${fin L^p(X,mu)mid gleq fleq h}$ for $g,hin L^p$. A subset $mathfrak{F}$ of $L^q(Y,nu)$ is called equi-measurable if for every $epsilon>0$ there exists a measurable subset $Y_epsilon$ of $Y$ such that $nu(Ysetminus Y_epsilon)<epsilon$ and $mathfrak{F}|_{Y_epsilon}$ is relatively norm compact in $L^infty(Y_epsilon,mu|_{Y_epsilon})$.
Related results can be found in the articles Kernel operators and Compactness properties of an operator which imply that it is an integral operator by Schep, and The integral representation of linear operators by Buhvalov.
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
$endgroup$
Not every bounded linear operator from $L^p$ to $L^q$ has an integral kernel. Schachermeyer (Integral Operators on $L^p$ Spaces, Indiana Uni. Math. J.) gave the following characterization of integral operators:
Let $pin[1,infty)$ and $qin [0,infty]$. A linear operator $Tcolon
L^p(X,mu)to L^q(Y,nu)$ has an integral kernel if and only if it
maps order intervals to equi-measurable sets.
An order interval is a set of the form ${fin L^p(X,mu)mid gleq fleq h}$ for $g,hin L^p$. A subset $mathfrak{F}$ of $L^q(Y,nu)$ is called equi-measurable if for every $epsilon>0$ there exists a measurable subset $Y_epsilon$ of $Y$ such that $nu(Ysetminus Y_epsilon)<epsilon$ and $mathfrak{F}|_{Y_epsilon}$ is relatively norm compact in $L^infty(Y_epsilon,mu|_{Y_epsilon})$.
Related results can be found in the articles Kernel operators and Compactness properties of an operator which imply that it is an integral operator by Schep, and The integral representation of linear operators by Buhvalov.
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
answered Jan 23 at 15:26
MaoWaoMaoWao
3,793617
3,793617
$begingroup$
Thanks, that's exactly one type of result I was searching for! Do you know how not integral operators look like?
$endgroup$
– Lolman
Jan 23 at 16:09
add a comment |
$begingroup$
Thanks, that's exactly one type of result I was searching for! Do you know how not integral operators look like?
$endgroup$
– Lolman
Jan 23 at 16:09
$begingroup$
Thanks, that's exactly one type of result I was searching for! Do you know how not integral operators look like?
$endgroup$
– Lolman
Jan 23 at 16:09
$begingroup$
Thanks, that's exactly one type of result I was searching for! Do you know how not integral operators look like?
$endgroup$
– Lolman
Jan 23 at 16:09
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.
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%2f3084053%2frepresentation-of-linear-operator-between-lp-spaces%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
$begingroup$
The question is not properly stated. The form you are looking for is not sated clearly, but I guess you want $T$ to an integral operator. The answer depends on what type of properties you are interested in. If $p=q=2$, for example, not every operator on $L^{2}$ is compact and integral operators of interest are usually compact operators.
$endgroup$
– Kavi Rama Murthy
Jan 23 at 5:55