What is $intdelta(x-y)delta(y-z)f(y):{rm d}y$?
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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
2,01241532
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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
2,01241532
$endgroup$
$begingroup$
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 $delta(x-z)f(x)$
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– Winther
Dec 30 '18 at 14:50
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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
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@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
2,01241532
$endgroup$
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
2,01241532
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
2,01241532
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
2,01241532
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
2,01241532
asked Dec 30 '18 at 14:46
0xbadf00d0xbadf00d
2,01241532
2,01241532
$begingroup$
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
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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
8,06111028
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add a comment |
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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.
28k22844
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add a comment |
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$$color{green}{delta(x!-!z)}tag{1a}$$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
5,09211856
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add a comment |
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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
8,06111028
$endgroup$
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
8,06111028
$endgroup$
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
8,06111028
$endgroup$
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
8,06111028
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,06111028
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,06111028
answered Dec 30 '18 at 15:09
md2perpemd2perpe
8,06111028
8,06111028
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.
28k22844
$endgroup$
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.
28k22844
$endgroup$
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.
28k22844
$endgroup$
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.
28k22844
answered Dec 30 '18 at 15:12
J.G.J.G.
28k22844
answered Dec 30 '18 at 15:12
J.G.J.G.
28k22844
answered Dec 30 '18 at 15:12
J.G.J.G.
28k22844
28k22844
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
5,09211856
$endgroup$
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
5,09211856
$endgroup$
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
5,09211856
$endgroup$
$$color{green}{delta(x!-!z)}tag{1a}$$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
5,09211856
answered Jan 5 at 12:10
QmechanicQmechanic
5,09211856
answered Jan 5 at 12:10
QmechanicQmechanic
5,09211856
answered Jan 5 at 12:10
QmechanicQmechanic
5,09211856
5,09211856
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$begingroup$
Indeed, I would say it's undefined as well.
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– Crostul
Dec 30 '18 at 14:48
1
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Well providing the reference (paper) would be useful! If well-defined then I would guess this should equal to something like $delta(x-z)f(x)$
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– Winther
Dec 30 '18 at 14:50
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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.
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– GEdgar
Dec 30 '18 at 14:51
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@Winther It's occuring implicitly in the definition of $kappa_n^circ$ on page 4 here.
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– 0xbadf00d
Dec 30 '18 at 14:52
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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.
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– Winther
Dec 30 '18 at 15:15