Partial Derivative of Multivariate Indicator Function in the Distributional Sense

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Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.



Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.



I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.










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  • Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
    – Kavi Rama Murthy
    Sep 7 at 6:19














up vote
2
down vote

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Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.



Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.



I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.










share|cite|improve this question





















  • Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
    – Kavi Rama Murthy
    Sep 7 at 6:19












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.



Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.



I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.










share|cite|improve this question













Define $f:mathbbR^2tomathbbR$ by $left(x,yright)mapsto1$ if $x^2+y^2leq1$ and $left(x,yright)mapsto0$ otherwise.



Is there a definition of $partial f/partial x$ in the distributional sense? If so, what is it? References are welcome.



I know that $partial f/partial x=0$ on $leftleft(x,yright)inmathbbR^2:x^2+y^2neq1right$, and my intuition tells me that $partial f/partial x$ behaves like the Dirac delta function on $leftleft(x,yright)inmathbbR^2:x^2+y^2=1right$, but I do not know how to express this rigorously.







distribution-theory






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asked Sep 7 at 6:08









Cleric

3,08432464




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  • Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
    – Kavi Rama Murthy
    Sep 7 at 6:19
















  • Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
    – Kavi Rama Murthy
    Sep 7 at 6:19















Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
– Kavi Rama Murthy
Sep 7 at 6:19




Any locally integrable function has a derivative in the sense of distributions/ generalized functions because every distribution has a derivative.
– Kavi Rama Murthy
Sep 7 at 6:19










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let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
$$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
$$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then



$$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$



i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
in the sense of distributions.



I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.






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    1 Answer
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    up vote
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    accepted










    let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
    $$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
    so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
    $$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
    then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
    The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then



    $$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$



    i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
    in the sense of distributions.



    I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.






    share|cite|improve this answer


























      up vote
      2
      down vote



      accepted










      let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
      $$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
      so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
      $$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
      then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
      The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then



      $$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$



      i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
      in the sense of distributions.



      I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.






      share|cite|improve this answer
























        up vote
        2
        down vote



        accepted







        up vote
        2
        down vote



        accepted






        let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
        $$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
        so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
        $$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
        then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
        The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then



        $$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$



        i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
        in the sense of distributions.



        I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.






        share|cite|improve this answer














        let $vec x = (x_1,dots,x_n)$, and let $vecphi:mathbb R^n to mathbb R^n $ be a vector valued test function, extend the pairing between smooth functions of compact support $langle vec f,vec grangle := int_mathbb R^n vec fcdotvec g$ to distributions analagously to the scalar case. By extending Gauss's theorem, the natural generalisation of integration by parts to higher dimensions, we have the natural generalisation of the distributional derivative,
        $$langle nabla f, vecphirangle := -langle f,nablacdot vecphi rangle = -int_<1 nabla cdot vecphi(vec x) dvec x = -int_ vecphicdot vec n dl$$
        so $nabla f$ is a vector valued distribution supported on $mathbb S^1 = $ which points in the direction of the outward normal from $mathbb S^1$. If you wanted to, you could define for any test function $varphi : mathbb R^n to mathbb R$,
        $$langle delta_mathbb S^1,varphirangle := int_ varphi dl,$$
        then (somewhat abusively since $vec n$ is only defined on $mathbb S^1$), $$nabla f = -vec n delta_mathbb S^1$$
        The components $partial_i f$ are related as follows. If $varphi : mathbb R^n to mathbb R$ is a test function, set $vecphi = varphi vec e_i$, where $vec e_i$ is the $i$th standard basis vector of $vec R^n$. Then



        $$langle partial_i f, varphirangle := - langle f , partial_i varphirangle = - langle f,nabla cdot vecphirangle = langle nabla f,vecphirangle = langle -vec n delta_mathbb S^1 , vecphirangle = langle -n_idelta_mathbb S^1, varphirangle$$



        i.e. $$ fracpartial fpartial x_i = -n_idelta_mathbb S^1$$
        in the sense of distributions.



        I don't have any references for the above computation; it is not difficult, but I have never seen it before. It would be interesting to have a theory of generalised $n$-forms that would allow both the above and the usual computations with generalised functions, but I don't know of it. So I would also greatly appreciate some references.







        share|cite|improve this answer














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        edited Sep 7 at 14:25

























        answered Sep 7 at 6:49









        Calvin Khor

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