How to find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?












1












$begingroup$


Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.



Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?



Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:



$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$



Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?



Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.






derivatives frechet-derivative







share|cite|improve this question











$endgroup$












  • $begingroup$
    Write $f$ as a composition of an affine function and a bilinear function.
    $endgroup$
    – Will M.
    Dec 10 '18 at 4:05
















1












$begingroup$


Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.



Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?



Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:



$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$



Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?



Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.






derivatives frechet-derivative







share|cite|improve this question











$endgroup$












  • $begingroup$
    Write $f$ as a composition of an affine function and a bilinear function.
    $endgroup$
    – Will M.
    Dec 10 '18 at 4:05














1












1








1





$begingroup$


Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.



Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?



Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:



$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$



Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?



Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.






derivatives frechet-derivative







share|cite|improve this question











$endgroup$




Given a real $m times n$ matrix $A$ and $b in mathbb{R}^m$, let $f(x)=|Ax-b|^2$ for any $x in mathbb{R}^n$.



Find Frechet derivative of $f(x)=|Ax-b|^2$ at any $x^*$?



Actually I am wondering how to use the following to find the Frechet derivative, i.e., $J$:



$$lim_{h rightarrow 0} frac{|f(x+h)-f(x)-Jh|}{|h|} =0$$



Another question is what the difference between what we would get from Frechet derivative and the gradient $nabla f(x)=2A^T(Ax-b)$?



Please explain your reasons in detail, especially, what is the difference between gradient of $f$ and Frechet derivative. Also, explain when they might be identical.






derivatives frechet-derivative




derivatives frechet-derivative






share|cite|improve this question















share|cite|improve this question













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edited Dec 9 '18 at 21:02









Saeed

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asked Dec 9 '18 at 19:37









SepideSepide

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2788












  • $begingroup$
    Write $f$ as a composition of an affine function and a bilinear function.
    $endgroup$
    – Will M.
    Dec 10 '18 at 4:05


















  • $begingroup$
    Write $f$ as a composition of an affine function and a bilinear function.
    $endgroup$
    – Will M.
    Dec 10 '18 at 4:05
















$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05




$begingroup$
Write $f$ as a composition of an affine function and a bilinear function.
$endgroup$
– Will M.
Dec 10 '18 at 4:05










1 Answer
1






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oldest

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0












$begingroup$

If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$



$$u(y) = (y mid x_u).$$



Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:



$$f'(a) cdot h = (nabla f(a) mid h).$$



In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
$$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$



If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$



    $$u(y) = (y mid x_u).$$



    Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:



    $$f'(a) cdot h = (nabla f(a) mid h).$$



    In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
    $$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$



    If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$



      $$u(y) = (y mid x_u).$$



      Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:



      $$f'(a) cdot h = (nabla f(a) mid h).$$



      In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
      $$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$



      If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$



        $$u(y) = (y mid x_u).$$



        Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:



        $$f'(a) cdot h = (nabla f(a) mid h).$$



        In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
        $$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$



        If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.






        share|cite|improve this answer









        $endgroup$



        If $mathrm{H}$ is a Hilbert space, then every continuous linear function $u:mathrm{H} to mathbf{R}$ can be represented by means of scalar product with respect to a unique vector, here denoted as $x_u:$



        $$u(y) = (y mid x_u).$$



        Hence, if $f:mathrm{H} to mathbf{R}$ is a differentiable function, then its derivative $u = f'(a)$ at $a$ is a continuous linear function. The vector $x_u$ is denoted $nabla f(a)$ in this case. And we have the fundamental relation:



        $$f'(a) cdot h = (nabla f(a) mid h).$$



        In regards to your particular $f,$ we can write $f(x) = (Ax -b mid Ax - b)$ and by the products and chain rules,
        $$f'(x) cdot h = (Ax - b mid Ah) + (Ah mid Ax - b) = 2(Ah mid Ax - b).$$



        If $mathrm{H} = mathbf{R}^d,$ and we are dealing with the standard Euclidean inner product, we can write further $f'(x) cdot h = (2A^intercal (Ax - b) mid h),$ this signifies $nabla f(x) = 2A^intercal (Ax - b).$ Q.E.D.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 5:13









        Will M.Will M.

        2,615315




        2,615315






























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