Taking a functional derivative












0














I am following the derivation here. I reproduce a smaller part of it here:



Consider the functional



$J(p) = eta_0int_{-infty}^{infty}p(x)dx$



for some probability distribution $p(x)$. It is then stated that



$frac{delta J}{delta p} = eta_0$.



I'm not sure what exactly happened in the step before. What I think is going on is the following analogue of discrete case. If



$J = eta_0sum_i p_i(x)$, I can see that



$frac{delta J}{delta p_j} = eta_0sum_i frac{partial p_i(x)}{partial p_j(x)} = eta_0sum_i delta_{ij} = eta_0$.



Is this what is happening in the continuous case as well? How is the operation $frac{delta J}{delta p}$ defined exactly?






real-analysis calculus probability







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    0














    I am following the derivation here. I reproduce a smaller part of it here:



    Consider the functional



    $J(p) = eta_0int_{-infty}^{infty}p(x)dx$



    for some probability distribution $p(x)$. It is then stated that



    $frac{delta J}{delta p} = eta_0$.



    I'm not sure what exactly happened in the step before. What I think is going on is the following analogue of discrete case. If



    $J = eta_0sum_i p_i(x)$, I can see that



    $frac{delta J}{delta p_j} = eta_0sum_i frac{partial p_i(x)}{partial p_j(x)} = eta_0sum_i delta_{ij} = eta_0$.



    Is this what is happening in the continuous case as well? How is the operation $frac{delta J}{delta p}$ defined exactly?






    real-analysis calculus probability







    share|cite|improve this question



























      0












      0








      0







      I am following the derivation here. I reproduce a smaller part of it here:



      Consider the functional



      $J(p) = eta_0int_{-infty}^{infty}p(x)dx$



      for some probability distribution $p(x)$. It is then stated that



      $frac{delta J}{delta p} = eta_0$.



      I'm not sure what exactly happened in the step before. What I think is going on is the following analogue of discrete case. If



      $J = eta_0sum_i p_i(x)$, I can see that



      $frac{delta J}{delta p_j} = eta_0sum_i frac{partial p_i(x)}{partial p_j(x)} = eta_0sum_i delta_{ij} = eta_0$.



      Is this what is happening in the continuous case as well? How is the operation $frac{delta J}{delta p}$ defined exactly?






      real-analysis calculus probability







      share|cite|improve this question















      I am following the derivation here. I reproduce a smaller part of it here:



      Consider the functional



      $J(p) = eta_0int_{-infty}^{infty}p(x)dx$



      for some probability distribution $p(x)$. It is then stated that



      $frac{delta J}{delta p} = eta_0$.



      I'm not sure what exactly happened in the step before. What I think is going on is the following analogue of discrete case. If



      $J = eta_0sum_i p_i(x)$, I can see that



      $frac{delta J}{delta p_j} = eta_0sum_i frac{partial p_i(x)}{partial p_j(x)} = eta_0sum_i delta_{ij} = eta_0$.



      Is this what is happening in the continuous case as well? How is the operation $frac{delta J}{delta p}$ defined exactly?






      real-analysis calculus probability




      real-analysis calculus probability






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 17 hours ago









      Scientifica

      6,37641335




      6,37641335










      asked 17 hours ago









      user1936752

      5201513




      5201513






















          1 Answer
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          Given partial derivatives $frac{partial p_i}{partial p_j}=delta_i^j$, the discrete labels $i,,j$ index variables $p_i$. With functional derivatvies we have continuous labels, viz. $frac{delta p(x)}{delta p(y)}=delta(x-y)$ (note the right-hand side is now a Dirac delta, not a Kronecker delta). So if $eta$ has at most one argument, we can write $frac{delta J}{delta p(y)}=eta_0(y)int_{Bbb R}delta(x-y)dx=eta_0(y)$.






          share|cite|improve this answer





















          • Thank you. Just to clarify, the meaning of $frac{delta J}{delta p}$ on Wikipedia is what exactly? Is it $frac{delta J}{delta p(y)}$ for specific $y$?
            – user1936752
            13 hours ago






          • 1




            @user1936752 The meaning follows from the definition I've given, together with $frac{delta}{delta p}$ commuting with integration operators, obeying the product & chain rules etc. The $y$-dependence is needed in general, because e.g. $frac{delta}{delta p(y)}int_{Bbb R}p^2(x)dx=2p(y)$ (proof is an exercise). However, you shouldn't think of it as a specific $y$; differentiating the integral with respect to a function gives a function.
            – J.G.
            13 hours ago













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          1 Answer
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          active

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          active

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          1














          Given partial derivatives $frac{partial p_i}{partial p_j}=delta_i^j$, the discrete labels $i,,j$ index variables $p_i$. With functional derivatvies we have continuous labels, viz. $frac{delta p(x)}{delta p(y)}=delta(x-y)$ (note the right-hand side is now a Dirac delta, not a Kronecker delta). So if $eta$ has at most one argument, we can write $frac{delta J}{delta p(y)}=eta_0(y)int_{Bbb R}delta(x-y)dx=eta_0(y)$.






          share|cite|improve this answer





















          • Thank you. Just to clarify, the meaning of $frac{delta J}{delta p}$ on Wikipedia is what exactly? Is it $frac{delta J}{delta p(y)}$ for specific $y$?
            – user1936752
            13 hours ago






          • 1




            @user1936752 The meaning follows from the definition I've given, together with $frac{delta}{delta p}$ commuting with integration operators, obeying the product & chain rules etc. The $y$-dependence is needed in general, because e.g. $frac{delta}{delta p(y)}int_{Bbb R}p^2(x)dx=2p(y)$ (proof is an exercise). However, you shouldn't think of it as a specific $y$; differentiating the integral with respect to a function gives a function.
            – J.G.
            13 hours ago


















          1














          Given partial derivatives $frac{partial p_i}{partial p_j}=delta_i^j$, the discrete labels $i,,j$ index variables $p_i$. With functional derivatvies we have continuous labels, viz. $frac{delta p(x)}{delta p(y)}=delta(x-y)$ (note the right-hand side is now a Dirac delta, not a Kronecker delta). So if $eta$ has at most one argument, we can write $frac{delta J}{delta p(y)}=eta_0(y)int_{Bbb R}delta(x-y)dx=eta_0(y)$.






          share|cite|improve this answer





















          • Thank you. Just to clarify, the meaning of $frac{delta J}{delta p}$ on Wikipedia is what exactly? Is it $frac{delta J}{delta p(y)}$ for specific $y$?
            – user1936752
            13 hours ago






          • 1




            @user1936752 The meaning follows from the definition I've given, together with $frac{delta}{delta p}$ commuting with integration operators, obeying the product & chain rules etc. The $y$-dependence is needed in general, because e.g. $frac{delta}{delta p(y)}int_{Bbb R}p^2(x)dx=2p(y)$ (proof is an exercise). However, you shouldn't think of it as a specific $y$; differentiating the integral with respect to a function gives a function.
            – J.G.
            13 hours ago
















          1












          1








          1






          Given partial derivatives $frac{partial p_i}{partial p_j}=delta_i^j$, the discrete labels $i,,j$ index variables $p_i$. With functional derivatvies we have continuous labels, viz. $frac{delta p(x)}{delta p(y)}=delta(x-y)$ (note the right-hand side is now a Dirac delta, not a Kronecker delta). So if $eta$ has at most one argument, we can write $frac{delta J}{delta p(y)}=eta_0(y)int_{Bbb R}delta(x-y)dx=eta_0(y)$.






          share|cite|improve this answer












          Given partial derivatives $frac{partial p_i}{partial p_j}=delta_i^j$, the discrete labels $i,,j$ index variables $p_i$. With functional derivatvies we have continuous labels, viz. $frac{delta p(x)}{delta p(y)}=delta(x-y)$ (note the right-hand side is now a Dirac delta, not a Kronecker delta). So if $eta$ has at most one argument, we can write $frac{delta J}{delta p(y)}=eta_0(y)int_{Bbb R}delta(x-y)dx=eta_0(y)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 17 hours ago









          J.G.

          23.1k22137




          23.1k22137












          • Thank you. Just to clarify, the meaning of $frac{delta J}{delta p}$ on Wikipedia is what exactly? Is it $frac{delta J}{delta p(y)}$ for specific $y$?
            – user1936752
            13 hours ago






          • 1




            @user1936752 The meaning follows from the definition I've given, together with $frac{delta}{delta p}$ commuting with integration operators, obeying the product & chain rules etc. The $y$-dependence is needed in general, because e.g. $frac{delta}{delta p(y)}int_{Bbb R}p^2(x)dx=2p(y)$ (proof is an exercise). However, you shouldn't think of it as a specific $y$; differentiating the integral with respect to a function gives a function.
            – J.G.
            13 hours ago




















          • Thank you. Just to clarify, the meaning of $frac{delta J}{delta p}$ on Wikipedia is what exactly? Is it $frac{delta J}{delta p(y)}$ for specific $y$?
            – user1936752
            13 hours ago






          • 1




            @user1936752 The meaning follows from the definition I've given, together with $frac{delta}{delta p}$ commuting with integration operators, obeying the product & chain rules etc. The $y$-dependence is needed in general, because e.g. $frac{delta}{delta p(y)}int_{Bbb R}p^2(x)dx=2p(y)$ (proof is an exercise). However, you shouldn't think of it as a specific $y$; differentiating the integral with respect to a function gives a function.
            – J.G.
            13 hours ago


















          Thank you. Just to clarify, the meaning of $frac{delta J}{delta p}$ on Wikipedia is what exactly? Is it $frac{delta J}{delta p(y)}$ for specific $y$?
          – user1936752
          13 hours ago




          Thank you. Just to clarify, the meaning of $frac{delta J}{delta p}$ on Wikipedia is what exactly? Is it $frac{delta J}{delta p(y)}$ for specific $y$?
          – user1936752
          13 hours ago




          1




          1




          @user1936752 The meaning follows from the definition I've given, together with $frac{delta}{delta p}$ commuting with integration operators, obeying the product & chain rules etc. The $y$-dependence is needed in general, because e.g. $frac{delta}{delta p(y)}int_{Bbb R}p^2(x)dx=2p(y)$ (proof is an exercise). However, you shouldn't think of it as a specific $y$; differentiating the integral with respect to a function gives a function.
          – J.G.
          13 hours ago






          @user1936752 The meaning follows from the definition I've given, together with $frac{delta}{delta p}$ commuting with integration operators, obeying the product & chain rules etc. The $y$-dependence is needed in general, because e.g. $frac{delta}{delta p(y)}int_{Bbb R}p^2(x)dx=2p(y)$ (proof is an exercise). However, you shouldn't think of it as a specific $y$; differentiating the integral with respect to a function gives a function.
          – J.G.
          13 hours ago




















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