Proving the product rule for Fréchet Derivative.












1














Let $X$ be a normed vector space, $Usubset X$, and $F,G:Urightarrow mathbb{R}$ differentiable at $xin U$. Show that the map $Fcdot G:Urightarrow mathbb{R}$, $Fcdot G(x)=F(x)G(x)$ is also differentiable at $xin U$ and that
$$ D(Fcdot G)_{|_x}=F(x)DG_{|_x}+G(x)DF_{|_x} $$
AKA Prove the product rule for the Fréchet Derivative.




To be Fréchet differentiable means the following: Let $X,Y$ be normed vector spaces, U open in X, and $F:Urightarrow Y$. Let $x,hin U$ and let $T:Xrightarrow Y$ be a linear map. Then the limit
$$ underset{hrightarrow 0}{lim}frac{||F(x+h)-F(x)-Th||_Y}{||h||_X}=0$$
exists. We denote $T$ as $DF_{|_x}$.




Here is my attempt at a proof.



$Proof.$



Let
$$ S=F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h $$
By some algebraic manipulations we have,
$$ =F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h + F(x+h)G(x)-F(x+h)G(x)$$
$$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] -F(x)DG_{|_x}h - G(x)DF_{|_x}h +F(x+h)DG_{|_x}h+G(x)DF_{|_x}h $$
$$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] + DG_{|_x}hbig[F(x+h)-F(x)big] $$
Then since F and G are Fréchet differentiable at x, we have
$$ underset{hrightarrow 0}{lim}frac{|S|}{||h||_X} $$
$$ =F(x+h)(0)+G(x)(0)+underset{hrightarrow 0}{lim}frac{|DG_{|_x}||h||F(x+h)-F(x)|}{||h||_X} $$




At the end there, I'm pretty sure that the $h$'s do not cancel since it is possible that $X$ is infinite dimensional and, therefore, the norms cannot be said to be equivalent. If it was finite dimensional, I would have argued by continuity of $F$ that $F(x+h)-F(x)rightarrow 0$ as $hrightarrow 0$.



Does anyone know how to proceed? Or a different approach to the problem






real-analysis proof-verification normed-spaces frechet-derivative







share|cite|improve this question





























    1














    Let $X$ be a normed vector space, $Usubset X$, and $F,G:Urightarrow mathbb{R}$ differentiable at $xin U$. Show that the map $Fcdot G:Urightarrow mathbb{R}$, $Fcdot G(x)=F(x)G(x)$ is also differentiable at $xin U$ and that
    $$ D(Fcdot G)_{|_x}=F(x)DG_{|_x}+G(x)DF_{|_x} $$
    AKA Prove the product rule for the Fréchet Derivative.




    To be Fréchet differentiable means the following: Let $X,Y$ be normed vector spaces, U open in X, and $F:Urightarrow Y$. Let $x,hin U$ and let $T:Xrightarrow Y$ be a linear map. Then the limit
    $$ underset{hrightarrow 0}{lim}frac{||F(x+h)-F(x)-Th||_Y}{||h||_X}=0$$
    exists. We denote $T$ as $DF_{|_x}$.




    Here is my attempt at a proof.



    $Proof.$



    Let
    $$ S=F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h $$
    By some algebraic manipulations we have,
    $$ =F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h + F(x+h)G(x)-F(x+h)G(x)$$
    $$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] -F(x)DG_{|_x}h - G(x)DF_{|_x}h +F(x+h)DG_{|_x}h+G(x)DF_{|_x}h $$
    $$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] + DG_{|_x}hbig[F(x+h)-F(x)big] $$
    Then since F and G are Fréchet differentiable at x, we have
    $$ underset{hrightarrow 0}{lim}frac{|S|}{||h||_X} $$
    $$ =F(x+h)(0)+G(x)(0)+underset{hrightarrow 0}{lim}frac{|DG_{|_x}||h||F(x+h)-F(x)|}{||h||_X} $$




    At the end there, I'm pretty sure that the $h$'s do not cancel since it is possible that $X$ is infinite dimensional and, therefore, the norms cannot be said to be equivalent. If it was finite dimensional, I would have argued by continuity of $F$ that $F(x+h)-F(x)rightarrow 0$ as $hrightarrow 0$.



    Does anyone know how to proceed? Or a different approach to the problem






    real-analysis proof-verification normed-spaces frechet-derivative







    share|cite|improve this question



























      1












      1








      1


      0





      Let $X$ be a normed vector space, $Usubset X$, and $F,G:Urightarrow mathbb{R}$ differentiable at $xin U$. Show that the map $Fcdot G:Urightarrow mathbb{R}$, $Fcdot G(x)=F(x)G(x)$ is also differentiable at $xin U$ and that
      $$ D(Fcdot G)_{|_x}=F(x)DG_{|_x}+G(x)DF_{|_x} $$
      AKA Prove the product rule for the Fréchet Derivative.




      To be Fréchet differentiable means the following: Let $X,Y$ be normed vector spaces, U open in X, and $F:Urightarrow Y$. Let $x,hin U$ and let $T:Xrightarrow Y$ be a linear map. Then the limit
      $$ underset{hrightarrow 0}{lim}frac{||F(x+h)-F(x)-Th||_Y}{||h||_X}=0$$
      exists. We denote $T$ as $DF_{|_x}$.




      Here is my attempt at a proof.



      $Proof.$



      Let
      $$ S=F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h $$
      By some algebraic manipulations we have,
      $$ =F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h + F(x+h)G(x)-F(x+h)G(x)$$
      $$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] -F(x)DG_{|_x}h - G(x)DF_{|_x}h +F(x+h)DG_{|_x}h+G(x)DF_{|_x}h $$
      $$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] + DG_{|_x}hbig[F(x+h)-F(x)big] $$
      Then since F and G are Fréchet differentiable at x, we have
      $$ underset{hrightarrow 0}{lim}frac{|S|}{||h||_X} $$
      $$ =F(x+h)(0)+G(x)(0)+underset{hrightarrow 0}{lim}frac{|DG_{|_x}||h||F(x+h)-F(x)|}{||h||_X} $$




      At the end there, I'm pretty sure that the $h$'s do not cancel since it is possible that $X$ is infinite dimensional and, therefore, the norms cannot be said to be equivalent. If it was finite dimensional, I would have argued by continuity of $F$ that $F(x+h)-F(x)rightarrow 0$ as $hrightarrow 0$.



      Does anyone know how to proceed? Or a different approach to the problem






      real-analysis proof-verification normed-spaces frechet-derivative







      share|cite|improve this question















      Let $X$ be a normed vector space, $Usubset X$, and $F,G:Urightarrow mathbb{R}$ differentiable at $xin U$. Show that the map $Fcdot G:Urightarrow mathbb{R}$, $Fcdot G(x)=F(x)G(x)$ is also differentiable at $xin U$ and that
      $$ D(Fcdot G)_{|_x}=F(x)DG_{|_x}+G(x)DF_{|_x} $$
      AKA Prove the product rule for the Fréchet Derivative.




      To be Fréchet differentiable means the following: Let $X,Y$ be normed vector spaces, U open in X, and $F:Urightarrow Y$. Let $x,hin U$ and let $T:Xrightarrow Y$ be a linear map. Then the limit
      $$ underset{hrightarrow 0}{lim}frac{||F(x+h)-F(x)-Th||_Y}{||h||_X}=0$$
      exists. We denote $T$ as $DF_{|_x}$.




      Here is my attempt at a proof.



      $Proof.$



      Let
      $$ S=F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h $$
      By some algebraic manipulations we have,
      $$ =F(x+h)G(x+h)-F(x)G(x)-F(x)DG_{|_x}h - G(x)DF_{|_x}h + F(x+h)G(x)-F(x+h)G(x)$$
      $$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] -F(x)DG_{|_x}h - G(x)DF_{|_x}h +F(x+h)DG_{|_x}h+G(x)DF_{|_x}h $$
      $$ =F(x+h)big[G(x+h)-G(x)-DG_{|_x}hbig] + G(x)big[F(x+h)-F(x)-DF_{|_x}big] + DG_{|_x}hbig[F(x+h)-F(x)big] $$
      Then since F and G are Fréchet differentiable at x, we have
      $$ underset{hrightarrow 0}{lim}frac{|S|}{||h||_X} $$
      $$ =F(x+h)(0)+G(x)(0)+underset{hrightarrow 0}{lim}frac{|DG_{|_x}||h||F(x+h)-F(x)|}{||h||_X} $$




      At the end there, I'm pretty sure that the $h$'s do not cancel since it is possible that $X$ is infinite dimensional and, therefore, the norms cannot be said to be equivalent. If it was finite dimensional, I would have argued by continuity of $F$ that $F(x+h)-F(x)rightarrow 0$ as $hrightarrow 0$.



      Does anyone know how to proceed? Or a different approach to the problem






      real-analysis proof-verification normed-spaces frechet-derivative




      real-analysis proof-verification normed-spaces frechet-derivative






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Oct 29 '18 at 10:44

























      asked Oct 28 '18 at 4:21









      Joe Man Analysis

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      33419






















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          I'd prove using the following general rules of differentiation:




          1. Chain rule $(g circ f)'(x) = g'(f(x)) circ f'(x)$;


          2. Leibniz rule $B'(x,y) cdot (h, k) = B(x, k) + B(h, y)$ for continuous bilinear functions.


          3. Derivate of product spaces $(f, g)'(x) = (f'(x), g'(x)).$



          Then, you want to differentiate $psi(x) = F(x) G(x) = (B circ (F,G)) (x),$ where $B$ is the bilinear function $(x, y) mapsto xy.$






          share|cite|improve this answer





















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            I'd prove using the following general rules of differentiation:




            1. Chain rule $(g circ f)'(x) = g'(f(x)) circ f'(x)$;


            2. Leibniz rule $B'(x,y) cdot (h, k) = B(x, k) + B(h, y)$ for continuous bilinear functions.


            3. Derivate of product spaces $(f, g)'(x) = (f'(x), g'(x)).$



            Then, you want to differentiate $psi(x) = F(x) G(x) = (B circ (F,G)) (x),$ where $B$ is the bilinear function $(x, y) mapsto xy.$






            share|cite|improve this answer


























              0














              I'd prove using the following general rules of differentiation:




              1. Chain rule $(g circ f)'(x) = g'(f(x)) circ f'(x)$;


              2. Leibniz rule $B'(x,y) cdot (h, k) = B(x, k) + B(h, y)$ for continuous bilinear functions.


              3. Derivate of product spaces $(f, g)'(x) = (f'(x), g'(x)).$



              Then, you want to differentiate $psi(x) = F(x) G(x) = (B circ (F,G)) (x),$ where $B$ is the bilinear function $(x, y) mapsto xy.$






              share|cite|improve this answer
























                0












                0








                0






                I'd prove using the following general rules of differentiation:




                1. Chain rule $(g circ f)'(x) = g'(f(x)) circ f'(x)$;


                2. Leibniz rule $B'(x,y) cdot (h, k) = B(x, k) + B(h, y)$ for continuous bilinear functions.


                3. Derivate of product spaces $(f, g)'(x) = (f'(x), g'(x)).$



                Then, you want to differentiate $psi(x) = F(x) G(x) = (B circ (F,G)) (x),$ where $B$ is the bilinear function $(x, y) mapsto xy.$






                share|cite|improve this answer












                I'd prove using the following general rules of differentiation:




                1. Chain rule $(g circ f)'(x) = g'(f(x)) circ f'(x)$;


                2. Leibniz rule $B'(x,y) cdot (h, k) = B(x, k) + B(h, y)$ for continuous bilinear functions.


                3. Derivate of product spaces $(f, g)'(x) = (f'(x), g'(x)).$



                Then, you want to differentiate $psi(x) = F(x) G(x) = (B circ (F,G)) (x),$ where $B$ is the bilinear function $(x, y) mapsto xy.$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 '18 at 4:51









                Will M.

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