What is $intdelta(x-y)delta(y-z)f(y):{rm d}y$?
$begingroup$
Let $(Omega,mathcal A,mu)$ be a measurable space and $delta$ denote the Dirac delta function. If $finmathcal L^1(mu)$ and $x,zinOmega$, what is $$intdelta(x-y)delta(y-z)f(y):mu({rm d}y)?$$ I've found that in a paper, but isn't that undefined?
real-analysis functional-analysis dirac-delta
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
$endgroup$
add a comment |
$begingroup$
Let $(Omega,mathcal A,mu)$ be a measurable space and $delta$ denote the Dirac delta function. If $finmathcal L^1(mu)$ and $x,zinOmega$, what is $$intdelta(x-y)delta(y-z)f(y):mu({rm d}y)?$$ I've found that in a paper, but isn't that undefined?
real-analysis functional-analysis dirac-delta
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
$endgroup$
$begingroup$
Indeed, I would say it's undefined as well.
$endgroup$
– Crostul
Dec 30 '18 at 14:48
1
$begingroup$
Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $delta(x-z)f(x)$
$endgroup$
– Winther
Dec 30 '18 at 14:50
$begingroup$
Unfortunately "a paper" is not enough for us to find it. Some papers are pure nonsense. Other papers have definitions in them revealing what their notation means.
$endgroup$
– GEdgar
Dec 30 '18 at 14:51
$begingroup$
@Winther It's occuring implicitly in the definition of $kappa_n^circ$ on page 4 here.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 14:52
$begingroup$
Don't know how to rigorously define and prove the results above (products of distributions are troublesome, but my guess is that this one can make sense). One way of justifying a formula like the one above is to consider a smooth approximation to the Dirac $delta$ like for example $delta_epsilon(x) equiv frac{1}{sqrt{2pi epsilon}}e^{-frac{x^2}{2epsilon}}$. In this case it's easy to check that $int delta_epsilon(x-y)delta_epsilon(y-z){rm d}y = delta_{epsilon/2}(x-z)$ so atleast for cases where the delta is just an approximation for a sharp pulse the formula above should hold.
$endgroup$
– Winther
Dec 30 '18 at 15:15
add a comment |
$begingroup$
Let $(Omega,mathcal A,mu)$ be a measurable space and $delta$ denote the Dirac delta function. If $finmathcal L^1(mu)$ and $x,zinOmega$, what is $$intdelta(x-y)delta(y-z)f(y):mu({rm d}y)?$$ I've found that in a paper, but isn't that undefined?
real-analysis functional-analysis dirac-delta
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
$endgroup$
Let $(Omega,mathcal A,mu)$ be a measurable space and $delta$ denote the Dirac delta function. If $finmathcal L^1(mu)$ and $x,zinOmega$, what is $$intdelta(x-y)delta(y-z)f(y):mu({rm d}y)?$$ I've found that in a paper, but isn't that undefined?
real-analysis functional-analysis dirac-delta
real-analysis functional-analysis dirac-delta
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
1,64441534
1,64441534
$begingroup$
Indeed, I would say it's undefined as well.
$endgroup$
– Crostul
Dec 30 '18 at 14:48
1
$begingroup$
Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $delta(x-z)f(x)$
$endgroup$
– Winther
Dec 30 '18 at 14:50
$begingroup$
Unfortunately "a paper" is not enough for us to find it. Some papers are pure nonsense. Other papers have definitions in them revealing what their notation means.
$endgroup$
– GEdgar
Dec 30 '18 at 14:51
$begingroup$
@Winther It's occuring implicitly in the definition of $kappa_n^circ$ on page 4 here.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 14:52
$begingroup$
Don't know how to rigorously define and prove the results above (products of distributions are troublesome, but my guess is that this one can make sense). One way of justifying a formula like the one above is to consider a smooth approximation to the Dirac $delta$ like for example $delta_epsilon(x) equiv frac{1}{sqrt{2pi epsilon}}e^{-frac{x^2}{2epsilon}}$. In this case it's easy to check that $int delta_epsilon(x-y)delta_epsilon(y-z){rm d}y = delta_{epsilon/2}(x-z)$ so atleast for cases where the delta is just an approximation for a sharp pulse the formula above should hold.
$endgroup$
– Winther
Dec 30 '18 at 15:15
add a comment |
$begingroup$
Indeed, I would say it's undefined as well.
$endgroup$
– Crostul
Dec 30 '18 at 14:48
1
$begingroup$
Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $delta(x-z)f(x)$
$endgroup$
– Winther
Dec 30 '18 at 14:50
$begingroup$
Unfortunately "a paper" is not enough for us to find it. Some papers are pure nonsense. Other papers have definitions in them revealing what their notation means.
$endgroup$
– GEdgar
Dec 30 '18 at 14:51
$begingroup$
@Winther It's occuring implicitly in the definition of $kappa_n^circ$ on page 4 here.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 14:52
$begingroup$
Don't know how to rigorously define and prove the results above (products of distributions are troublesome, but my guess is that this one can make sense). One way of justifying a formula like the one above is to consider a smooth approximation to the Dirac $delta$ like for example $delta_epsilon(x) equiv frac{1}{sqrt{2pi epsilon}}e^{-frac{x^2}{2epsilon}}$. In this case it's easy to check that $int delta_epsilon(x-y)delta_epsilon(y-z){rm d}y = delta_{epsilon/2}(x-z)$ so atleast for cases where the delta is just an approximation for a sharp pulse the formula above should hold.
$endgroup$
– Winther
Dec 30 '18 at 15:15
$begingroup$
Indeed, I would say it's undefined as well.
$endgroup$
– Crostul
Dec 30 '18 at 14:48
$begingroup$
Indeed, I would say it's undefined as well.
$endgroup$
– Crostul
Dec 30 '18 at 14:48
1
1
$begingroup$
Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $delta(x-z)f(x)$
$endgroup$
– Winther
Dec 30 '18 at 14:50
$begingroup$
Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $delta(x-z)f(x)$
$endgroup$
– Winther
Dec 30 '18 at 14:50
$begingroup$
Unfortunately "a paper" is not enough for us to find it. Some papers are pure nonsense. Other papers have definitions in them revealing what their notation means.
$endgroup$
– GEdgar
Dec 30 '18 at 14:51
$begingroup$
Unfortunately "a paper" is not enough for us to find it. Some papers are pure nonsense. Other papers have definitions in them revealing what their notation means.
$endgroup$
– GEdgar
Dec 30 '18 at 14:51
$begingroup$
@Winther It's occuring implicitly in the definition of $kappa_n^circ$ on page 4 here.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 14:52
$begingroup$
@Winther It's occuring implicitly in the definition of $kappa_n^circ$ on page 4 here.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 14:52
$begingroup$
Don't know how to rigorously define and prove the results above (products of distributions are troublesome, but my guess is that this one can make sense). One way of justifying a formula like the one above is to consider a smooth approximation to the Dirac $delta$ like for example $delta_epsilon(x) equiv frac{1}{sqrt{2pi epsilon}}e^{-frac{x^2}{2epsilon}}$. In this case it's easy to check that $int delta_epsilon(x-y)delta_epsilon(y-z){rm d}y = delta_{epsilon/2}(x-z)$ so atleast for cases where the delta is just an approximation for a sharp pulse the formula above should hold.
$endgroup$
– Winther
Dec 30 '18 at 15:15
$begingroup$
Don't know how to rigorously define and prove the results above (products of distributions are troublesome, but my guess is that this one can make sense). One way of justifying a formula like the one above is to consider a smooth approximation to the Dirac $delta$ like for example $delta_epsilon(x) equiv frac{1}{sqrt{2pi epsilon}}e^{-frac{x^2}{2epsilon}}$. In this case it's easy to check that $int delta_epsilon(x-y)delta_epsilon(y-z){rm d}y = delta_{epsilon/2}(x-z)$ so atleast for cases where the delta is just an approximation for a sharp pulse the formula above should hold.
$endgroup$
– Winther
Dec 30 '18 at 15:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Talking non-rigorously, $delta(x-y) delta(y-z) f(y)$ will be non-zero only when $x-y=0$ and $y-z=0$, i.e. when $x=y=z.$ Therefore the integral over $y$ would be non-zero only when $x=z.$ We can thus expect the integral to be a multiple of $delta(x-z).$
So, let $phi$ be a nice function and study the formal integral
$$
int left( int delta(x-y) delta(y-z) f(y) , dy right) phi(z) , dz.
$$
Swapping the order of integration gives
$$
int delta(x-y) left( int delta(y-z) phi(z) , dz right) f(y) , dy
= int delta(x-y) phi(y) f(y) , dy
= phi(x) f(x).
$$
Thus,
$$
int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
$$
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
$endgroup$
add a comment |
$begingroup$
Recall the usual $intdelta(x-y)g(y)dy=g(x)$ (assuming it's a definite integral over $A$), in this case with $g(y):=f(y)delta(y-z)$, giving $color{blue}{f(x)delta(x-z)}$ for your in-title integral (again, if it's definite over $A$). Here we've eliminated the $delta$ containing $x$, effectively imposing $y=x$. Of course we could have used the other $delta$ to impose $y=z$ instead, obtaining $color{blue}{f(z)delta(x-z)}$. But by inspection, these two answers are equivalent.
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
$endgroup$
add a comment |
$begingroup$
$$color{green}{delta(x!-!z)}tag{1a}$$and$$color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)}tag{1b}$$ are informal notations for the distributions
$u,vin D^{prime}(mathbb{R}^2)$ given by$^1$
$$u[varphi]~:=~int_{mathbb{R}} !mathrm{d}y~varphi(y,y),tag{2a}$$
$$v[varphi]~:=~int_{mathbb{R}} !mathrm{d}y ~f(y)~varphi(y,y),tag{2b} $$
respectively, where $varphiin D(mathbb{R}^2)$ is a test function. The distributions are written in the informal notation as
$$u[varphi]~=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{green}{delta(x!-!z)}~ varphi(x,z), tag{3a}$$
$$v[varphi]~:=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)} ~varphi(x,z),tag{3b} $$
respectively.
--
$^1$Disclaimer: Here we assume for simplicity that OP's space is $Omega=mathbb{R}$ and the measure $mu$ is the Lebesque measure. Note that e.g. boundaries $partial Omega$ often requires extra attention in distribution theory.
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
$endgroup$
add a comment |
Your Answer
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%2f3056887%2fwhat-is-int-deltax-y-deltay-zfy-rm-dy%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Talking non-rigorously, $delta(x-y) delta(y-z) f(y)$ will be non-zero only when $x-y=0$ and $y-z=0$, i.e. when $x=y=z.$ Therefore the integral over $y$ would be non-zero only when $x=z.$ We can thus expect the integral to be a multiple of $delta(x-z).$
So, let $phi$ be a nice function and study the formal integral
$$
int left( int delta(x-y) delta(y-z) f(y) , dy right) phi(z) , dz.
$$
Swapping the order of integration gives
$$
int delta(x-y) left( int delta(y-z) phi(z) , dz right) f(y) , dy
= int delta(x-y) phi(y) f(y) , dy
= phi(x) f(x).
$$
Thus,
$$
int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
$$
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
$endgroup$
add a comment |
$begingroup$
Talking non-rigorously, $delta(x-y) delta(y-z) f(y)$ will be non-zero only when $x-y=0$ and $y-z=0$, i.e. when $x=y=z.$ Therefore the integral over $y$ would be non-zero only when $x=z.$ We can thus expect the integral to be a multiple of $delta(x-z).$
So, let $phi$ be a nice function and study the formal integral
$$
int left( int delta(x-y) delta(y-z) f(y) , dy right) phi(z) , dz.
$$
Swapping the order of integration gives
$$
int delta(x-y) left( int delta(y-z) phi(z) , dz right) f(y) , dy
= int delta(x-y) phi(y) f(y) , dy
= phi(x) f(x).
$$
Thus,
$$
int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
$$
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
$endgroup$
add a comment |
$begingroup$
Talking non-rigorously, $delta(x-y) delta(y-z) f(y)$ will be non-zero only when $x-y=0$ and $y-z=0$, i.e. when $x=y=z.$ Therefore the integral over $y$ would be non-zero only when $x=z.$ We can thus expect the integral to be a multiple of $delta(x-z).$
So, let $phi$ be a nice function and study the formal integral
$$
int left( int delta(x-y) delta(y-z) f(y) , dy right) phi(z) , dz.
$$
Swapping the order of integration gives
$$
int delta(x-y) left( int delta(y-z) phi(z) , dz right) f(y) , dy
= int delta(x-y) phi(y) f(y) , dy
= phi(x) f(x).
$$
Thus,
$$
int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
$$
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
$endgroup$
Talking non-rigorously, $delta(x-y) delta(y-z) f(y)$ will be non-zero only when $x-y=0$ and $y-z=0$, i.e. when $x=y=z.$ Therefore the integral over $y$ would be non-zero only when $x=z.$ We can thus expect the integral to be a multiple of $delta(x-z).$
So, let $phi$ be a nice function and study the formal integral
$$
int left( int delta(x-y) delta(y-z) f(y) , dy right) phi(z) , dz.
$$
Swapping the order of integration gives
$$
int delta(x-y) left( int delta(y-z) phi(z) , dz right) f(y) , dy
= int delta(x-y) phi(y) f(y) , dy
= phi(x) f(x).
$$
Thus,
$$
int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
$$
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,65411129
8,65411129
add a comment |
add a comment |
$begingroup$
Recall the usual $intdelta(x-y)g(y)dy=g(x)$ (assuming it's a definite integral over $A$), in this case with $g(y):=f(y)delta(y-z)$, giving $color{blue}{f(x)delta(x-z)}$ for your in-title integral (again, if it's definite over $A$). Here we've eliminated the $delta$ containing $x$, effectively imposing $y=x$. Of course we could have used the other $delta$ to impose $y=z$ instead, obtaining $color{blue}{f(z)delta(x-z)}$. But by inspection, these two answers are equivalent.
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
$endgroup$
add a comment |
$begingroup$
Recall the usual $intdelta(x-y)g(y)dy=g(x)$ (assuming it's a definite integral over $A$), in this case with $g(y):=f(y)delta(y-z)$, giving $color{blue}{f(x)delta(x-z)}$ for your in-title integral (again, if it's definite over $A$). Here we've eliminated the $delta$ containing $x$, effectively imposing $y=x$. Of course we could have used the other $delta$ to impose $y=z$ instead, obtaining $color{blue}{f(z)delta(x-z)}$. But by inspection, these two answers are equivalent.
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
$endgroup$
add a comment |
$begingroup$
Recall the usual $intdelta(x-y)g(y)dy=g(x)$ (assuming it's a definite integral over $A$), in this case with $g(y):=f(y)delta(y-z)$, giving $color{blue}{f(x)delta(x-z)}$ for your in-title integral (again, if it's definite over $A$). Here we've eliminated the $delta$ containing $x$, effectively imposing $y=x$. Of course we could have used the other $delta$ to impose $y=z$ instead, obtaining $color{blue}{f(z)delta(x-z)}$. But by inspection, these two answers are equivalent.
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
$endgroup$
Recall the usual $intdelta(x-y)g(y)dy=g(x)$ (assuming it's a definite integral over $A$), in this case with $g(y):=f(y)delta(y-z)$, giving $color{blue}{f(x)delta(x-z)}$ for your in-title integral (again, if it's definite over $A$). Here we've eliminated the $delta$ containing $x$, effectively imposing $y=x$. Of course we could have used the other $delta$ to impose $y=z$ instead, obtaining $color{blue}{f(z)delta(x-z)}$. But by inspection, these two answers are equivalent.
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
answered Dec 30 '18 at 15:12
J.G.J.G.
33.7k23252
33.7k23252
add a comment |
add a comment |
$begingroup$
$$color{green}{delta(x!-!z)}tag{1a}$$and$$color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)}tag{1b}$$ are informal notations for the distributions
$u,vin D^{prime}(mathbb{R}^2)$ given by$^1$
$$u[varphi]~:=~int_{mathbb{R}} !mathrm{d}y~varphi(y,y),tag{2a}$$
$$v[varphi]~:=~int_{mathbb{R}} !mathrm{d}y ~f(y)~varphi(y,y),tag{2b} $$
respectively, where $varphiin D(mathbb{R}^2)$ is a test function. The distributions are written in the informal notation as
$$u[varphi]~=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{green}{delta(x!-!z)}~ varphi(x,z), tag{3a}$$
$$v[varphi]~:=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)} ~varphi(x,z),tag{3b} $$
respectively.
--
$^1$Disclaimer: Here we assume for simplicity that OP's space is $Omega=mathbb{R}$ and the measure $mu$ is the Lebesque measure. Note that e.g. boundaries $partial Omega$ often requires extra attention in distribution theory.
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
$endgroup$
add a comment |
$begingroup$
$$color{green}{delta(x!-!z)}tag{1a}$$and$$color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)}tag{1b}$$ are informal notations for the distributions
$u,vin D^{prime}(mathbb{R}^2)$ given by$^1$
$$u[varphi]~:=~int_{mathbb{R}} !mathrm{d}y~varphi(y,y),tag{2a}$$
$$v[varphi]~:=~int_{mathbb{R}} !mathrm{d}y ~f(y)~varphi(y,y),tag{2b} $$
respectively, where $varphiin D(mathbb{R}^2)$ is a test function. The distributions are written in the informal notation as
$$u[varphi]~=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{green}{delta(x!-!z)}~ varphi(x,z), tag{3a}$$
$$v[varphi]~:=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)} ~varphi(x,z),tag{3b} $$
respectively.
--
$^1$Disclaimer: Here we assume for simplicity that OP's space is $Omega=mathbb{R}$ and the measure $mu$ is the Lebesque measure. Note that e.g. boundaries $partial Omega$ often requires extra attention in distribution theory.
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
$endgroup$
add a comment |
$begingroup$
$$color{green}{delta(x!-!z)}tag{1a}$$and$$color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)}tag{1b}$$ are informal notations for the distributions
$u,vin D^{prime}(mathbb{R}^2)$ given by$^1$
$$u[varphi]~:=~int_{mathbb{R}} !mathrm{d}y~varphi(y,y),tag{2a}$$
$$v[varphi]~:=~int_{mathbb{R}} !mathrm{d}y ~f(y)~varphi(y,y),tag{2b} $$
respectively, where $varphiin D(mathbb{R}^2)$ is a test function. The distributions are written in the informal notation as
$$u[varphi]~=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{green}{delta(x!-!z)}~ varphi(x,z), tag{3a}$$
$$v[varphi]~:=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)} ~varphi(x,z),tag{3b} $$
respectively.
--
$^1$Disclaimer: Here we assume for simplicity that OP's space is $Omega=mathbb{R}$ and the measure $mu$ is the Lebesque measure. Note that e.g. boundaries $partial Omega$ often requires extra attention in distribution theory.
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
$endgroup$
$$color{green}{delta(x!-!z)}tag{1a}$$and$$color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)}tag{1b}$$ are informal notations for the distributions
$u,vin D^{prime}(mathbb{R}^2)$ given by$^1$
$$u[varphi]~:=~int_{mathbb{R}} !mathrm{d}y~varphi(y,y),tag{2a}$$
$$v[varphi]~:=~int_{mathbb{R}} !mathrm{d}y ~f(y)~varphi(y,y),tag{2b} $$
respectively, where $varphiin D(mathbb{R}^2)$ is a test function. The distributions are written in the informal notation as
$$u[varphi]~=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{green}{delta(x!-!z)}~ varphi(x,z), tag{3a}$$
$$v[varphi]~:=~int_{mathbb{R}^2} !mathrm{d}x~mathrm{d}z~color{red}{int_{mathbb{R}} !mathrm{d}y ~f(y)delta(x!-!y)delta(y!-!z)} ~varphi(x,z),tag{3b} $$
respectively.
--
$^1$Disclaimer: Here we assume for simplicity that OP's space is $Omega=mathbb{R}$ and the measure $mu$ is the Lebesque measure. Note that e.g. boundaries $partial Omega$ often requires extra attention in distribution theory.
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
answered Jan 5 at 12:10
QmechanicQmechanic
5,20711859
5,20711859
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%2f3056887%2fwhat-is-int-deltax-y-deltay-zfy-rm-dy%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
$begingroup$
Indeed, I would say it's undefined as well.
$endgroup$
– Crostul
Dec 30 '18 at 14:48
1
$begingroup$
Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $delta(x-z)f(x)$
$endgroup$
– Winther
Dec 30 '18 at 14:50
$begingroup$
Unfortunately "a paper" is not enough for us to find it. Some papers are pure nonsense. Other papers have definitions in them revealing what their notation means.
$endgroup$
– GEdgar
Dec 30 '18 at 14:51
$begingroup$
@Winther It's occuring implicitly in the definition of $kappa_n^circ$ on page 4 here.
$endgroup$
– 0xbadf00d
Dec 30 '18 at 14:52
$begingroup$
Don't know how to rigorously define and prove the results above (products of distributions are troublesome, but my guess is that this one can make sense). One way of justifying a formula like the one above is to consider a smooth approximation to the Dirac $delta$ like for example $delta_epsilon(x) equiv frac{1}{sqrt{2pi epsilon}}e^{-frac{x^2}{2epsilon}}$. In this case it's easy to check that $int delta_epsilon(x-y)delta_epsilon(y-z){rm d}y = delta_{epsilon/2}(x-z)$ so atleast for cases where the delta is just an approximation for a sharp pulse the formula above should hold.
$endgroup$
– Winther
Dec 30 '18 at 15:15