What is $intdelta(x-y)delta(y-z)f(y):{rm d}y$?












2












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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







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  • $begingroup$
    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


















2












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












2








2


0



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 30 '18 at 14:46









0xbadf00d0xbadf00d

1,88241533




1,88241533












  • $begingroup$
    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






  • 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












3 Answers
3






active

oldest

votes


















4












$begingroup$

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,
$$
int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      $$color{green}{delta(x!-!z)}tag{1a}$$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.






      share|cite|improve this answer









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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        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,
        $$
        int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
        $$






        share|cite|improve this answer









        $endgroup$


















          4












          $begingroup$

          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,
          $$
          int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
          $$






          share|cite|improve this answer









          $endgroup$
















            4












            4








            4





            $begingroup$

            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,
            $$
            int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
            $$






            share|cite|improve this answer









            $endgroup$



            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,
            $$
            int delta(x-y) delta(y-z) f(y) , dy = f(x) delta(z-x).
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 30 '18 at 15:09









            md2perpemd2perpe

            8,31111028




            8,31111028























                0












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 30 '18 at 15:12









                    J.G.J.G.

                    31.8k23250




                    31.8k23250























                        0












                        $begingroup$

                        $$color{green}{delta(x!-!z)}tag{1a}$$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.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          $$color{green}{delta(x!-!z)}tag{1a}$$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.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            $$color{green}{delta(x!-!z)}tag{1a}$$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.






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                            $endgroup$



                            $$color{green}{delta(x!-!z)}tag{1a}$$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.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 12:10









                            QmechanicQmechanic

                            5,17711858




                            5,17711858






























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