Lp convergence involving powers
$begingroup$
Let $(X,mu)$ be a finite measure space. Suppose $int_Xlvert f_n-frvert^p,dmuto0$ for $pgeq1$. For $qin(0,p]$ does it hold that $$int_Xlvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}},dmuto0quad?$$
measure-theory
asked May 11 '17 at 13:34
user375366user375366
885138
$endgroup$
add a comment |
$begingroup$
Let $(X,mu)$ be a finite measure space. Suppose $int_Xlvert f_n-frvert^p,dmuto0$ for $pgeq1$. For $qin(0,p]$ does it hold that $$int_Xlvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}},dmuto0quad?$$
measure-theory
asked May 11 '17 at 13:34
user375366user375366
885138
$endgroup$
add a comment |
$begingroup$
Let $(X,mu)$ be a finite measure space. Suppose $int_Xlvert f_n-frvert^p,dmuto0$ for $pgeq1$. For $qin(0,p]$ does it hold that $$int_Xlvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}},dmuto0quad?$$
measure-theory
asked May 11 '17 at 13:34
user375366user375366
885138
$endgroup$
Let $(X,mu)$ be a finite measure space. Suppose $int_Xlvert f_n-frvert^p,dmuto0$ for $pgeq1$. For $qin(0,p]$ does it hold that $$int_Xlvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}},dmuto0quad?$$
measure-theory
measure-theory
asked May 11 '17 at 13:34
user375366user375366
885138
asked May 11 '17 at 13:34
user375366user375366
885138
edited May 11 '17 at 13:52
user375366
asked May 11 '17 at 13:34
user375366user375366
885138
asked May 11 '17 at 13:34
user375366user375366
885138
asked May 11 '17 at 13:34
user375366user375366
885138
885138
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes. I will assume that we work on a probability space.
From item 6 of this answer, we know that convergence to $0$ in $mathbb L^1$ of a sequence of random variables $left(X_nright)_{ngeqslant 1}$ is equivalent to its convergence in probability combined with uniform integrability. Define
$$
X_n:= lvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}}.
$$
By the assumption, $f_nto f$ in probability hence by item 3 of this answer,
$X_nto 0$ in probability. For uniform integrability, we bound $X_n$ using the inequality
$$
lvertlvert arvert^q-lvert brvert^qrvert^{frac{p}{q}}
leqslant lvertlvert arvert^q+lvert brvert^qrvert^{frac{p}{q}}
leqslant 2^{p/q-1}left(lvertlvert arvert^p+lvertlvert brvert^pright).
$$
In this way, we are reduced to show the uniform integrability of the sequence $left(lvertlvert f_nrvert^pright)_{ngeqslant 1}$, which is a consequence of $int_Xlvert f_n-frvert^p,dmuto0$.
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
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%2f2276380%2flp-convergence-involving-powers%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
$begingroup$
Yes. I will assume that we work on a probability space.
From item 6 of this answer, we know that convergence to $0$ in $mathbb L^1$ of a sequence of random variables $left(X_nright)_{ngeqslant 1}$ is equivalent to its convergence in probability combined with uniform integrability. Define
$$
X_n:= lvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}}.
$$
By the assumption, $f_nto f$ in probability hence by item 3 of this answer,
$X_nto 0$ in probability. For uniform integrability, we bound $X_n$ using the inequality
$$
lvertlvert arvert^q-lvert brvert^qrvert^{frac{p}{q}}
leqslant lvertlvert arvert^q+lvert brvert^qrvert^{frac{p}{q}}
leqslant 2^{p/q-1}left(lvertlvert arvert^p+lvertlvert brvert^pright).
$$
In this way, we are reduced to show the uniform integrability of the sequence $left(lvertlvert f_nrvert^pright)_{ngeqslant 1}$, which is a consequence of $int_Xlvert f_n-frvert^p,dmuto0$.
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
Yes. I will assume that we work on a probability space.
From item 6 of this answer, we know that convergence to $0$ in $mathbb L^1$ of a sequence of random variables $left(X_nright)_{ngeqslant 1}$ is equivalent to its convergence in probability combined with uniform integrability. Define
$$
X_n:= lvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}}.
$$
By the assumption, $f_nto f$ in probability hence by item 3 of this answer,
$X_nto 0$ in probability. For uniform integrability, we bound $X_n$ using the inequality
$$
lvertlvert arvert^q-lvert brvert^qrvert^{frac{p}{q}}
leqslant lvertlvert arvert^q+lvert brvert^qrvert^{frac{p}{q}}
leqslant 2^{p/q-1}left(lvertlvert arvert^p+lvertlvert brvert^pright).
$$
In this way, we are reduced to show the uniform integrability of the sequence $left(lvertlvert f_nrvert^pright)_{ngeqslant 1}$, which is a consequence of $int_Xlvert f_n-frvert^p,dmuto0$.
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
add a comment |
$begingroup$
Yes. I will assume that we work on a probability space.
From item 6 of this answer, we know that convergence to $0$ in $mathbb L^1$ of a sequence of random variables $left(X_nright)_{ngeqslant 1}$ is equivalent to its convergence in probability combined with uniform integrability. Define
$$
X_n:= lvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}}.
$$
By the assumption, $f_nto f$ in probability hence by item 3 of this answer,
$X_nto 0$ in probability. For uniform integrability, we bound $X_n$ using the inequality
$$
lvertlvert arvert^q-lvert brvert^qrvert^{frac{p}{q}}
leqslant lvertlvert arvert^q+lvert brvert^qrvert^{frac{p}{q}}
leqslant 2^{p/q-1}left(lvertlvert arvert^p+lvertlvert brvert^pright).
$$
In this way, we are reduced to show the uniform integrability of the sequence $left(lvertlvert f_nrvert^pright)_{ngeqslant 1}$, which is a consequence of $int_Xlvert f_n-frvert^p,dmuto0$.
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
$endgroup$
Yes. I will assume that we work on a probability space.
From item 6 of this answer, we know that convergence to $0$ in $mathbb L^1$ of a sequence of random variables $left(X_nright)_{ngeqslant 1}$ is equivalent to its convergence in probability combined with uniform integrability. Define
$$
X_n:= lvertlvert f_nrvert^q-lvert frvert^qrvert^{frac{p}{q}}.
$$
By the assumption, $f_nto f$ in probability hence by item 3 of this answer,
$X_nto 0$ in probability. For uniform integrability, we bound $X_n$ using the inequality
$$
lvertlvert arvert^q-lvert brvert^qrvert^{frac{p}{q}}
leqslant lvertlvert arvert^q+lvert brvert^qrvert^{frac{p}{q}}
leqslant 2^{p/q-1}left(lvertlvert arvert^p+lvertlvert brvert^pright).
$$
In this way, we are reduced to show the uniform integrability of the sequence $left(lvertlvert f_nrvert^pright)_{ngeqslant 1}$, which is a consequence of $int_Xlvert f_n-frvert^p,dmuto0$.
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
answered Dec 11 '18 at 21:01
Davide GiraudoDavide Giraudo
128k17156268
128k17156268
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.
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%2f2276380%2flp-convergence-involving-powers%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
