Lebesgue Density $f$ with compact support, polynomial $p$ of degree $k$. Then convolution $fstar p$ is...
Let $f(.)$ be a continuous Lebesgue-density-function with compact support $mathrm{supp}f={x:f(x)neq0}$. Let $p(.)$ be a polynomial with degree $k$.
Then the convolution $h:=fstar p$ is an integrable function and can be written as polynomial.
What I tried:
$fstar p(y)=int_{-infty}^{+infty}f(x-y)p(y)dy=int_{-infty}^{+infty}f(y)p(x-y)dy$.
$p=sum_{m=0}^{k}alpha_mt^m$. We only consider $p(t)=t^m forall m=0,ldots,k$.
So $fstar p(x) =int_{-infty}^{+infty}f(y)(x-y)^m$ which is by the binomial theorem $sum_{i=1}^{m}x^i{mchoose i}int_{-infty}^{infty}f(y)(-y)^{m-i}$ Where the product of the terms after $x^i$ denote the polynomial coefficient. So $f star p$ is a polynomial.
Now I want to show that $f star p$ is integrable.
The support is compact so we can reach the maximum of p$max_{x in mathrm{supp} f}p(x)=:K<infty$
$int_{-infty}^{+infty}int_{-infty}^{+infty}f(x-y)p(y)dydx=int_{-infty}^{infty}int_{supp f}f(x-y)p(y)dydxleint_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)Kdydx$.
Since $f$ is a density function
$int_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)dydxle1$ (I'm not sure about that)
So $int_{-infty}^{+infty}(fstar p)(x)dxle K<infty$
real-analysis analysis measure-theory convolution
asked Nov 27 '18 at 21:57
conradconrad
757
add a comment |
Let $f(.)$ be a continuous Lebesgue-density-function with compact support $mathrm{supp}f={x:f(x)neq0}$. Let $p(.)$ be a polynomial with degree $k$.
Then the convolution $h:=fstar p$ is an integrable function and can be written as polynomial.
What I tried:
$fstar p(y)=int_{-infty}^{+infty}f(x-y)p(y)dy=int_{-infty}^{+infty}f(y)p(x-y)dy$.
$p=sum_{m=0}^{k}alpha_mt^m$. We only consider $p(t)=t^m forall m=0,ldots,k$.
So $fstar p(x) =int_{-infty}^{+infty}f(y)(x-y)^m$ which is by the binomial theorem $sum_{i=1}^{m}x^i{mchoose i}int_{-infty}^{infty}f(y)(-y)^{m-i}$ Where the product of the terms after $x^i$ denote the polynomial coefficient. So $f star p$ is a polynomial.
Now I want to show that $f star p$ is integrable.
The support is compact so we can reach the maximum of p$max_{x in mathrm{supp} f}p(x)=:K<infty$
$int_{-infty}^{+infty}int_{-infty}^{+infty}f(x-y)p(y)dydx=int_{-infty}^{infty}int_{supp f}f(x-y)p(y)dydxleint_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)Kdydx$.
Since $f$ is a density function
$int_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)dydxle1$ (I'm not sure about that)
So $int_{-infty}^{+infty}(fstar p)(x)dxle K<infty$
real-analysis analysis measure-theory convolution
asked Nov 27 '18 at 21:57
conradconrad
757
No non-constant polynomial is integrable on $mathbb R$.
– Kavi Rama Murthy
Nov 27 '18 at 23:48
add a comment |
Let $f(.)$ be a continuous Lebesgue-density-function with compact support $mathrm{supp}f={x:f(x)neq0}$. Let $p(.)$ be a polynomial with degree $k$.
Then the convolution $h:=fstar p$ is an integrable function and can be written as polynomial.
What I tried:
$fstar p(y)=int_{-infty}^{+infty}f(x-y)p(y)dy=int_{-infty}^{+infty}f(y)p(x-y)dy$.
$p=sum_{m=0}^{k}alpha_mt^m$. We only consider $p(t)=t^m forall m=0,ldots,k$.
So $fstar p(x) =int_{-infty}^{+infty}f(y)(x-y)^m$ which is by the binomial theorem $sum_{i=1}^{m}x^i{mchoose i}int_{-infty}^{infty}f(y)(-y)^{m-i}$ Where the product of the terms after $x^i$ denote the polynomial coefficient. So $f star p$ is a polynomial.
Now I want to show that $f star p$ is integrable.
The support is compact so we can reach the maximum of p$max_{x in mathrm{supp} f}p(x)=:K<infty$
$int_{-infty}^{+infty}int_{-infty}^{+infty}f(x-y)p(y)dydx=int_{-infty}^{infty}int_{supp f}f(x-y)p(y)dydxleint_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)Kdydx$.
Since $f$ is a density function
$int_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)dydxle1$ (I'm not sure about that)
So $int_{-infty}^{+infty}(fstar p)(x)dxle K<infty$
real-analysis analysis measure-theory convolution
asked Nov 27 '18 at 21:57
conradconrad
757
Let $f(.)$ be a continuous Lebesgue-density-function with compact support $mathrm{supp}f={x:f(x)neq0}$. Let $p(.)$ be a polynomial with degree $k$.
Then the convolution $h:=fstar p$ is an integrable function and can be written as polynomial.
What I tried:
$fstar p(y)=int_{-infty}^{+infty}f(x-y)p(y)dy=int_{-infty}^{+infty}f(y)p(x-y)dy$.
$p=sum_{m=0}^{k}alpha_mt^m$. We only consider $p(t)=t^m forall m=0,ldots,k$.
So $fstar p(x) =int_{-infty}^{+infty}f(y)(x-y)^m$ which is by the binomial theorem $sum_{i=1}^{m}x^i{mchoose i}int_{-infty}^{infty}f(y)(-y)^{m-i}$ Where the product of the terms after $x^i$ denote the polynomial coefficient. So $f star p$ is a polynomial.
Now I want to show that $f star p$ is integrable.
The support is compact so we can reach the maximum of p$max_{x in mathrm{supp} f}p(x)=:K<infty$
$int_{-infty}^{+infty}int_{-infty}^{+infty}f(x-y)p(y)dydx=int_{-infty}^{infty}int_{supp f}f(x-y)p(y)dydxleint_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)Kdydx$.
Since $f$ is a density function
$int_{-infty}^{+infty}int_{mathrm{supp} f}f(x-y)dydxle1$ (I'm not sure about that)
So $int_{-infty}^{+infty}(fstar p)(x)dxle K<infty$
real-analysis analysis measure-theory convolution
real-analysis analysis measure-theory convolution
asked Nov 27 '18 at 21:57
conradconrad
757
asked Nov 27 '18 at 21:57
conradconrad
757
asked Nov 27 '18 at 21:57
conradconrad
757
asked Nov 27 '18 at 21:57
conradconrad
757
asked Nov 27 '18 at 21:57
conradconrad
757
757
No non-constant polynomial is integrable on $mathbb R$.
– Kavi Rama Murthy
Nov 27 '18 at 23:48
add a comment |
No non-constant polynomial is integrable on $mathbb R$.
– Kavi Rama Murthy
Nov 27 '18 at 23:48
No non-constant polynomial is integrable on $mathbb R$.
– Kavi Rama Murthy
Nov 27 '18 at 23:48
No non-constant polynomial is integrable on $mathbb R$.
– Kavi Rama Murthy
Nov 27 '18 at 23:48
add a comment |
1 Answer
1
active
oldest
votes
No, $f*g$ need not be integrable. Let $f=1$ on $[0,1],$ $f=0$ elsewhere. Let $p(y)=y.$ Then
$$f*p(x) = int_0^1 1cdot(x-y),dy = x-1/2.$$
The function on the right is not integrable on $mathbb R.$
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
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%2f3016357%2flebesgue-density-f-with-compact-support-polynomial-p-of-degree-k-then-co%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
No, $f*g$ need not be integrable. Let $f=1$ on $[0,1],$ $f=0$ elsewhere. Let $p(y)=y.$ Then
$$f*p(x) = int_0^1 1cdot(x-y),dy = x-1/2.$$
The function on the right is not integrable on $mathbb R.$
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
add a comment |
No, $f*g$ need not be integrable. Let $f=1$ on $[0,1],$ $f=0$ elsewhere. Let $p(y)=y.$ Then
$$f*p(x) = int_0^1 1cdot(x-y),dy = x-1/2.$$
The function on the right is not integrable on $mathbb R.$
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
add a comment |
No, $f*g$ need not be integrable. Let $f=1$ on $[0,1],$ $f=0$ elsewhere. Let $p(y)=y.$ Then
$$f*p(x) = int_0^1 1cdot(x-y),dy = x-1/2.$$
The function on the right is not integrable on $mathbb R.$
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
No, $f*g$ need not be integrable. Let $f=1$ on $[0,1],$ $f=0$ elsewhere. Let $p(y)=y.$ Then
$$f*p(x) = int_0^1 1cdot(x-y),dy = x-1/2.$$
The function on the right is not integrable on $mathbb R.$
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
answered Nov 27 '18 at 22:42
zhw.zhw.
71.8k43075
71.8k43075
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%2f3016357%2flebesgue-density-f-with-compact-support-polynomial-p-of-degree-k-then-co%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
No non-constant polynomial is integrable on $mathbb R$.
– Kavi Rama Murthy
Nov 27 '18 at 23:48