Mixing
How do I induce this Hardy-Littlewood-Sobolev inequality
$begingroup$
The Hardy-Littlewood-Sobolev inequality in my book,
Let $0<alpha < d$, $1<p<q<infty$, and $frac{1}{q} +1 = frac {1}{p} + frac {alpha}{d}$. Then
$$ lVert f ast |cdot|^{alpha} rVert_{L^{q}(mathbb{R}^d )} leq C_{p,q} lVert f rVert_{L^{p}(mathbb{R}^d)}.$$
But I saw another one,
$$int_{mathbb{R}^d} int_{mathbb{R}^d} frac {f(x)g(y)}{|x-y|^{alpha}}dxdy
leq C_{alpha, p, d }lVert f rVert_{L^{p}(mathbb{R}^d)}lVert g rVert_{L^{q}(mathbb{R}^d)}. $$
for all $f in L^p, g in L^q,quad 1<p,q< infty,quad frac{1}{p} + frac{1}{q} + frac {alpha}{d} = 2$ and $0<alpha <d $.
I want to show bottom one but I am failing it.
fourier-analysis harmonic-analysis
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
$endgroup$
add a comment |
$begingroup$
The Hardy-Littlewood-Sobolev inequality in my book,
Let $0<alpha < d$, $1<p<q<infty$, and $frac{1}{q} +1 = frac {1}{p} + frac {alpha}{d}$. Then
$$ lVert f ast |cdot|^{alpha} rVert_{L^{q}(mathbb{R}^d )} leq C_{p,q} lVert f rVert_{L^{p}(mathbb{R}^d)}.$$
But I saw another one,
$$int_{mathbb{R}^d} int_{mathbb{R}^d} frac {f(x)g(y)}{|x-y|^{alpha}}dxdy
leq C_{alpha, p, d }lVert f rVert_{L^{p}(mathbb{R}^d)}lVert g rVert_{L^{q}(mathbb{R}^d)}. $$
for all $f in L^p, g in L^q,quad 1<p,q< infty,quad frac{1}{p} + frac{1}{q} + frac {alpha}{d} = 2$ and $0<alpha <d $.
I want to show bottom one but I am failing it.
fourier-analysis harmonic-analysis
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
$endgroup$
$begingroup$
I think a direct use of the definition of convolution and Holder Inequality should give you this result, have you tried this already?
$endgroup$
– Zixiao_Liu
Jan 13 at 9:12
$begingroup$
@Zixiao_Liu Oh my days, I'm such an idiot. Thank you so much. I couldn't think about elementary method. Can't I vote here?
$endgroup$
– Idkwhat
Jan 13 at 9:30
$begingroup$
Maybe you would like to print a sketch of proof and admit it as the answer yourself ;) It is quite lengthy and not interesting, but still a good exercise anyway. LOL
$endgroup$
– Zixiao_Liu
Jan 13 at 10:33
add a comment |
$begingroup$
The Hardy-Littlewood-Sobolev inequality in my book,
Let $0<alpha < d$, $1<p<q<infty$, and $frac{1}{q} +1 = frac {1}{p} + frac {alpha}{d}$. Then
$$ lVert f ast |cdot|^{alpha} rVert_{L^{q}(mathbb{R}^d )} leq C_{p,q} lVert f rVert_{L^{p}(mathbb{R}^d)}.$$
But I saw another one,
$$int_{mathbb{R}^d} int_{mathbb{R}^d} frac {f(x)g(y)}{|x-y|^{alpha}}dxdy
leq C_{alpha, p, d }lVert f rVert_{L^{p}(mathbb{R}^d)}lVert g rVert_{L^{q}(mathbb{R}^d)}. $$
for all $f in L^p, g in L^q,quad 1<p,q< infty,quad frac{1}{p} + frac{1}{q} + frac {alpha}{d} = 2$ and $0<alpha <d $.
I want to show bottom one but I am failing it.
fourier-analysis harmonic-analysis
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
$endgroup$
The Hardy-Littlewood-Sobolev inequality in my book,
Let $0<alpha < d$, $1<p<q<infty$, and $frac{1}{q} +1 = frac {1}{p} + frac {alpha}{d}$. Then
$$ lVert f ast |cdot|^{alpha} rVert_{L^{q}(mathbb{R}^d )} leq C_{p,q} lVert f rVert_{L^{p}(mathbb{R}^d)}.$$
But I saw another one,
$$int_{mathbb{R}^d} int_{mathbb{R}^d} frac {f(x)g(y)}{|x-y|^{alpha}}dxdy
leq C_{alpha, p, d }lVert f rVert_{L^{p}(mathbb{R}^d)}lVert g rVert_{L^{q}(mathbb{R}^d)}. $$
for all $f in L^p, g in L^q,quad 1<p,q< infty,quad frac{1}{p} + frac{1}{q} + frac {alpha}{d} = 2$ and $0<alpha <d $.
I want to show bottom one but I am failing it.
fourier-analysis harmonic-analysis
fourier-analysis harmonic-analysis
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
asked Jan 13 at 8:55
IdkwhatIdkwhat
236
236
$begingroup$
I think a direct use of the definition of convolution and Holder Inequality should give you this result, have you tried this already?
$endgroup$
– Zixiao_Liu
Jan 13 at 9:12
$begingroup$
@Zixiao_Liu Oh my days, I'm such an idiot. Thank you so much. I couldn't think about elementary method. Can't I vote here?
$endgroup$
– Idkwhat
Jan 13 at 9:30
$begingroup$
Maybe you would like to print a sketch of proof and admit it as the answer yourself ;) It is quite lengthy and not interesting, but still a good exercise anyway. LOL
$endgroup$
– Zixiao_Liu
Jan 13 at 10:33
add a comment |
$begingroup$
I think a direct use of the definition of convolution and Holder Inequality should give you this result, have you tried this already?
$endgroup$
– Zixiao_Liu
Jan 13 at 9:12
$begingroup$
@Zixiao_Liu Oh my days, I'm such an idiot. Thank you so much. I couldn't think about elementary method. Can't I vote here?
$endgroup$
– Idkwhat
Jan 13 at 9:30
$begingroup$
Maybe you would like to print a sketch of proof and admit it as the answer yourself ;) It is quite lengthy and not interesting, but still a good exercise anyway. LOL
$endgroup$
– Zixiao_Liu
Jan 13 at 10:33
$begingroup$
I think a direct use of the definition of convolution and Holder Inequality should give you this result, have you tried this already?
$endgroup$
– Zixiao_Liu
Jan 13 at 9:12
$begingroup$
I think a direct use of the definition of convolution and Holder Inequality should give you this result, have you tried this already?
$endgroup$
– Zixiao_Liu
Jan 13 at 9:12
$begingroup$
@Zixiao_Liu Oh my days, I'm such an idiot. Thank you so much. I couldn't think about elementary method. Can't I vote here?
$endgroup$
– Idkwhat
Jan 13 at 9:30
$begingroup$
@Zixiao_Liu Oh my days, I'm such an idiot. Thank you so much. I couldn't think about elementary method. Can't I vote here?
$endgroup$
– Idkwhat
Jan 13 at 9:30
$begingroup$
Maybe you would like to print a sketch of proof and admit it as the answer yourself ;) It is quite lengthy and not interesting, but still a good exercise anyway. LOL
$endgroup$
– Zixiao_Liu
Jan 13 at 10:33
$begingroup$
Maybe you would like to print a sketch of proof and admit it as the answer yourself ;) It is quite lengthy and not interesting, but still a good exercise anyway. LOL
$endgroup$
– Zixiao_Liu
Jan 13 at 10:33
add a comment |
0
active
oldest
votes
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%2f3071822%2fhow-do-i-induce-this-hardy-littlewood-sobolev-inequality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3071822%2fhow-do-i-induce-this-hardy-littlewood-sobolev-inequality%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
$begingroup$
I think a direct use of the definition of convolution and Holder Inequality should give you this result, have you tried this already?
$endgroup$
– Zixiao_Liu
Jan 13 at 9:12
$begingroup$
@Zixiao_Liu Oh my days, I'm such an idiot. Thank you so much. I couldn't think about elementary method. Can't I vote here?
$endgroup$
– Idkwhat
Jan 13 at 9:30
$begingroup$
Maybe you would like to print a sketch of proof and admit it as the answer yourself ;) It is quite lengthy and not interesting, but still a good exercise anyway. LOL
$endgroup$
– Zixiao_Liu
Jan 13 at 10:33