Derivative of an L1 norm of transform of a vector.

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I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$



where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that



$$C = psi F$$



then my l1 norm is defined as:



$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,



$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$



My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?







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  • Why do you need to do this? If you're solving an optimization problem, there might be a better way.
    – littleO
    Feb 20 '15 at 3:21














up vote
1
down vote

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I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$



where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that



$$C = psi F$$



then my l1 norm is defined as:



$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,



$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$



My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?







share|cite|improve this question






















  • Why do you need to do this? If you're solving an optimization problem, there might be a better way.
    – littleO
    Feb 20 '15 at 3:21












up vote
1
down vote

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1









up vote
1
down vote

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I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$



where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that



$$C = psi F$$



then my l1 norm is defined as:



$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,



$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$



My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?







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I have to take derivative of the l-1 norm. L1 is the function R in the following expression:
$$ R(psi Fx) $$



where x is a vector, F is the inverse Fourier transform, and $psi$ is a wavelet transform. If I define a variable C such that



$$C = psi F$$



then my l1 norm is defined as:



$$||Cx||^1_1$$ I know that taking a derivative of an l1 norm is not possible. The l1 norm is defined as: $$sumnolimits|x_i^1_1$$ To take a derivative of the l1 term, I addd a small positive number, call it $epsilon$. Therefore,



$$sumnolimits|x_1^1_1 = sumnolimitssqrtx^*_ix_i + epsilon$$



My question is, what is the derivative of the l1 norm Cx and what would be the elements of the matrix C?









share|cite|improve this question













share|cite|improve this question




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edited Feb 20 '15 at 2:45









Mathemagician1234

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asked Feb 20 '15 at 2:29









user212257

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  • Why do you need to do this? If you're solving an optimization problem, there might be a better way.
    – littleO
    Feb 20 '15 at 3:21
















  • Why do you need to do this? If you're solving an optimization problem, there might be a better way.
    – littleO
    Feb 20 '15 at 3:21















Why do you need to do this? If you're solving an optimization problem, there might be a better way.
– littleO
Feb 20 '15 at 3:21




Why do you need to do this? If you're solving an optimization problem, there might be a better way.
– littleO
Feb 20 '15 at 3:21










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Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.



See also the answer to
Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.






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    Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.



    See also the answer to
    Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.






    share|cite|improve this answer


























      up vote
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      down vote













      Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.



      See also the answer to
      Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.






      share|cite|improve this answer
























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        up vote
        0
        down vote









        Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.



        See also the answer to
        Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.






        share|cite|improve this answer














        Solving in coordinates, use the formula $fracpartialpartial x_k |mathbfx|_p = fracx_kmathbfx$ for $p=1$ and with obvious existence conditions.



        See also the answer to
        Taking derivative of $L_0$-norm, $L_1$-norm, $L_2$-norm.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 13 '17 at 12:20









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        answered Apr 30 '15 at 1:31









        rych

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