Does the 'continuous convolution relation' imply a discrete version?
$begingroup$
In particular, I want to know if $L^p(mathbb{R}^d)*L^q(mathbb{R}^d)subseteq L^r(mathbb{R}^d)$ implies $ell^{p}(mathbb{Z}^d)*ell^{q}(mathbb{Z}^d)subseteq ell^{r}(mathbb{Z}^d)$? My attempt is to try using the hypothesis on functions defined via sequences. That is, for any $a={a_n}_{nin mathbb{Z}^d} in ell^p(mathbb{Z}^d),$ $b={b_n }_{nin mathbb{Z}^d}in ell^{q}(mathbb{Z}^d)$, define $f_a =sum_{nin mathbb{Z}^d} a_nchi_{n}$, and $f_b=sum_{nin mathbb{Z}^d}b_n chi_{n}$ where for any $nin mathbb{Z}^d,$ $chi_{n}=chi_{n+[0,1)^{d}}$. I naively tried to show that $||a*b||_{ell^r}leq ||f_a*f_b||_{L^r},$ but failed miserably.
functional-analysis convolution
asked Dec 8 '18 at 12:08
KuromeKurome
378114
$endgroup$
add a comment |
$begingroup$
In particular, I want to know if $L^p(mathbb{R}^d)*L^q(mathbb{R}^d)subseteq L^r(mathbb{R}^d)$ implies $ell^{p}(mathbb{Z}^d)*ell^{q}(mathbb{Z}^d)subseteq ell^{r}(mathbb{Z}^d)$? My attempt is to try using the hypothesis on functions defined via sequences. That is, for any $a={a_n}_{nin mathbb{Z}^d} in ell^p(mathbb{Z}^d),$ $b={b_n }_{nin mathbb{Z}^d}in ell^{q}(mathbb{Z}^d)$, define $f_a =sum_{nin mathbb{Z}^d} a_nchi_{n}$, and $f_b=sum_{nin mathbb{Z}^d}b_n chi_{n}$ where for any $nin mathbb{Z}^d,$ $chi_{n}=chi_{n+[0,1)^{d}}$. I naively tried to show that $||a*b||_{ell^r}leq ||f_a*f_b||_{L^r},$ but failed miserably.
functional-analysis convolution
asked Dec 8 '18 at 12:08
KuromeKurome
378114
$endgroup$
add a comment |
$begingroup$
In particular, I want to know if $L^p(mathbb{R}^d)*L^q(mathbb{R}^d)subseteq L^r(mathbb{R}^d)$ implies $ell^{p}(mathbb{Z}^d)*ell^{q}(mathbb{Z}^d)subseteq ell^{r}(mathbb{Z}^d)$? My attempt is to try using the hypothesis on functions defined via sequences. That is, for any $a={a_n}_{nin mathbb{Z}^d} in ell^p(mathbb{Z}^d),$ $b={b_n }_{nin mathbb{Z}^d}in ell^{q}(mathbb{Z}^d)$, define $f_a =sum_{nin mathbb{Z}^d} a_nchi_{n}$, and $f_b=sum_{nin mathbb{Z}^d}b_n chi_{n}$ where for any $nin mathbb{Z}^d,$ $chi_{n}=chi_{n+[0,1)^{d}}$. I naively tried to show that $||a*b||_{ell^r}leq ||f_a*f_b||_{L^r},$ but failed miserably.
functional-analysis convolution
asked Dec 8 '18 at 12:08
KuromeKurome
378114
$endgroup$
In particular, I want to know if $L^p(mathbb{R}^d)*L^q(mathbb{R}^d)subseteq L^r(mathbb{R}^d)$ implies $ell^{p}(mathbb{Z}^d)*ell^{q}(mathbb{Z}^d)subseteq ell^{r}(mathbb{Z}^d)$? My attempt is to try using the hypothesis on functions defined via sequences. That is, for any $a={a_n}_{nin mathbb{Z}^d} in ell^p(mathbb{Z}^d),$ $b={b_n }_{nin mathbb{Z}^d}in ell^{q}(mathbb{Z}^d)$, define $f_a =sum_{nin mathbb{Z}^d} a_nchi_{n}$, and $f_b=sum_{nin mathbb{Z}^d}b_n chi_{n}$ where for any $nin mathbb{Z}^d,$ $chi_{n}=chi_{n+[0,1)^{d}}$. I naively tried to show that $||a*b||_{ell^r}leq ||f_a*f_b||_{L^r},$ but failed miserably.
functional-analysis convolution
functional-analysis convolution
asked Dec 8 '18 at 12:08
KuromeKurome
378114
asked Dec 8 '18 at 12:08
KuromeKurome
378114
edited Dec 8 '18 at 12:43
Kurome
asked Dec 8 '18 at 12:08
KuromeKurome
378114
asked Dec 8 '18 at 12:08
KuromeKurome
378114
asked Dec 8 '18 at 12:08
KuromeKurome
378114
378114
add a comment |
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%2f3031037%2fdoes-the-continuous-convolution-relation-imply-a-discrete-version%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%2f3031037%2fdoes-the-continuous-convolution-relation-imply-a-discrete-version%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
