Commutative Convolution. Problem 26 Royden 2 ed.












1












$begingroup$


Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.



Why $fast g=gast f$?



I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$






real-analysis measure-theory convolution







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See this question, or perhaps this one?
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:35








  • 1




    $begingroup$
    when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
    $endgroup$
    – Dunham
    Jan 7 at 20:36






  • 1




    $begingroup$
    Okay, but what happens to the limits of integration at the same time?
    $endgroup$
    – Doug M
    Jan 7 at 20:38
















1












$begingroup$


Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.



Why $fast g=gast f$?



I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$






real-analysis measure-theory convolution







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    See this question, or perhaps this one?
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:35








  • 1




    $begingroup$
    when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
    $endgroup$
    – Dunham
    Jan 7 at 20:36






  • 1




    $begingroup$
    Okay, but what happens to the limits of integration at the same time?
    $endgroup$
    – Doug M
    Jan 7 at 20:38














1












1








1


1



$begingroup$


Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.



Why $fast g=gast f$?



I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$






real-analysis measure-theory convolution







share|cite|improve this question









$endgroup$




Let $f$ and $g$ be functions in $L^1(—infty,infty)$, and define $fast g$ to be the
function $h$ defined by $h(y) = int f(y — x)g(x) dx$.



Why $fast g=gast f$?



I have this:
If $y-x=z$ then $int f(y-x)g(x)dx=int f(z)g(y-z)(-dz)=-gast f$






real-analysis measure-theory convolution




real-analysis measure-theory convolution






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 7 at 20:32









eraldcoileraldcoil

395211




395211








  • 1




    $begingroup$
    See this question, or perhaps this one?
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:35








  • 1




    $begingroup$
    when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
    $endgroup$
    – Dunham
    Jan 7 at 20:36






  • 1




    $begingroup$
    Okay, but what happens to the limits of integration at the same time?
    $endgroup$
    – Doug M
    Jan 7 at 20:38














  • 1




    $begingroup$
    See this question, or perhaps this one?
    $endgroup$
    – Dietrich Burde
    Jan 7 at 20:35








  • 1




    $begingroup$
    when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
    $endgroup$
    – Dunham
    Jan 7 at 20:36






  • 1




    $begingroup$
    Okay, but what happens to the limits of integration at the same time?
    $endgroup$
    – Doug M
    Jan 7 at 20:38








1




1




$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35






$begingroup$
See this question, or perhaps this one?
$endgroup$
– Dietrich Burde
Jan 7 at 20:35






1




1




$begingroup$
when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36




$begingroup$
when you changed variables, you didn't change the interval of integration. $infty -> - infty$ and vice versa
$endgroup$
– Dunham
Jan 7 at 20:36




1




1




$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38




$begingroup$
Okay, but what happens to the limits of integration at the same time?
$endgroup$
– Doug M
Jan 7 at 20:38










1 Answer
1






active

oldest

votes


















2












$begingroup$

The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$



Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The substitution of variable is valid in Lebesgue integral?
    $endgroup$
    – eraldcoil
    Jan 9 at 3:39










  • $begingroup$
    @eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
    $endgroup$
    – NicNic8
    Jan 9 at 4:59











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065463%2fcommutative-convolution-problem-26-royden-2-ed%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









2












$begingroup$

The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$



Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The substitution of variable is valid in Lebesgue integral?
    $endgroup$
    – eraldcoil
    Jan 9 at 3:39










  • $begingroup$
    @eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
    $endgroup$
    – NicNic8
    Jan 9 at 4:59
















2












$begingroup$

The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$



Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The substitution of variable is valid in Lebesgue integral?
    $endgroup$
    – eraldcoil
    Jan 9 at 3:39










  • $begingroup$
    @eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
    $endgroup$
    – NicNic8
    Jan 9 at 4:59














2












2








2





$begingroup$

The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$



Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$






share|cite|improve this answer









$endgroup$



The indefinite integral is not a value by a function (the anti-derivative of the integrand). Convolution is defined with a definite integral as follows:
$$ (fast g)(y) = int_{-infty}^{infty} f(y-x) g(x) , dx. $$



Now let's do the substitution where $z=y-x$. Then
$$ (fast g)(y) = int_{-infty}^{infty} f(z) , g(y-z) , (-dz). $$
Bringing out the negative sign and reversing the limits of integration yields
$$ (fast g)(y) = int_{-infty}^{infty} g(y-z) , f(z) , dz = (gast f)(y). $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 7 at 20:45









NicNic8NicNic8

4,60531123




4,60531123












  • $begingroup$
    The substitution of variable is valid in Lebesgue integral?
    $endgroup$
    – eraldcoil
    Jan 9 at 3:39










  • $begingroup$
    @eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
    $endgroup$
    – NicNic8
    Jan 9 at 4:59


















  • $begingroup$
    The substitution of variable is valid in Lebesgue integral?
    $endgroup$
    – eraldcoil
    Jan 9 at 3:39










  • $begingroup$
    @eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
    $endgroup$
    – NicNic8
    Jan 9 at 4:59
















$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39




$begingroup$
The substitution of variable is valid in Lebesgue integral?
$endgroup$
– eraldcoil
Jan 9 at 3:39












$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59




$begingroup$
@eraldcoil The substitution is valid as a Riemann integral. Doesn't that make it automatically valid as a Lebesgue integral? I'm not sure.
$endgroup$
– NicNic8
Jan 9 at 4:59


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065463%2fcommutative-convolution-problem-26-royden-2-ed%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Human spaceflight

Image
Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
Read more

張江高科駅

Image
張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
Read more