Computing Frechet derivative












1












$begingroup$


Let $F:C^{([-1,1])}rightarrow C^{([-1,1])}$ defined as $F(q)(x)=q^{2}(-x)$.



I want to find the derivative of this function my current stratergy is to use the directional derivative so $dF(q)(x)h= frac{d}{depsilon}|_{epsilon=0} (q+epsilon h)^{2}(-x)=2hq(-x)$.



Is this correct of do i need to do something different?






functional-analysis frechet-derivative







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    1












    $begingroup$


    Let $F:C^{([-1,1])}rightarrow C^{([-1,1])}$ defined as $F(q)(x)=q^{2}(-x)$.



    I want to find the derivative of this function my current stratergy is to use the directional derivative so $dF(q)(x)h= frac{d}{depsilon}|_{epsilon=0} (q+epsilon h)^{2}(-x)=2hq(-x)$.



    Is this correct of do i need to do something different?






    functional-analysis frechet-derivative







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $F:C^{([-1,1])}rightarrow C^{([-1,1])}$ defined as $F(q)(x)=q^{2}(-x)$.



      I want to find the derivative of this function my current stratergy is to use the directional derivative so $dF(q)(x)h= frac{d}{depsilon}|_{epsilon=0} (q+epsilon h)^{2}(-x)=2hq(-x)$.



      Is this correct of do i need to do something different?






      functional-analysis frechet-derivative







      share|cite|improve this question











      $endgroup$




      Let $F:C^{([-1,1])}rightarrow C^{([-1,1])}$ defined as $F(q)(x)=q^{2}(-x)$.



      I want to find the derivative of this function my current stratergy is to use the directional derivative so $dF(q)(x)h= frac{d}{depsilon}|_{epsilon=0} (q+epsilon h)^{2}(-x)=2hq(-x)$.



      Is this correct of do i need to do something different?






      functional-analysis frechet-derivative




      functional-analysis frechet-derivative






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Feb 3 at 14:44







      Roger

















      asked Feb 3 at 10:17









      RogerRoger

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      847






















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          Let $mathrm{H} = mathscr{C}(mathrm{I}),$ the space of continuous real-valued functions defined on the compact interval $mathrm{I}$ the with norm of suprema. Let $s$ be the function $mathrm{H} to mathrm{H}$ given by $s(f)(x) = f(-x),$ $z$ the function $mathrm{H} to mathrm{H} times mathrm{H}$ given by $q mapsto (q, q)$ and let $B$ the function $mathrm{H} times mathrm{H} to mathrm{H}$ given by $B(f,g) = fg.$ It is easy to see $s$ and $z$ are continuous linear functions and $B$ is a continuous bilinear (with $|B| leq 1$). The function you want to differentiat is given $F = B circ z circ s.$ By the chain rule, $F'(q) = B'(s(q), s(q)) circ z circ s;$ hence, $F'(q) cdot h = B'(s(q), s(q)) cdot (s(h), s(h)) = 2 s(q) s(h),$ if you evaluate in a point $x$ this becomes $big(F'(q) cdot h big)(x) = 2q(-x)h(-x).$ Q.E.D.






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            $begingroup$

            Let $mathrm{H} = mathscr{C}(mathrm{I}),$ the space of continuous real-valued functions defined on the compact interval $mathrm{I}$ the with norm of suprema. Let $s$ be the function $mathrm{H} to mathrm{H}$ given by $s(f)(x) = f(-x),$ $z$ the function $mathrm{H} to mathrm{H} times mathrm{H}$ given by $q mapsto (q, q)$ and let $B$ the function $mathrm{H} times mathrm{H} to mathrm{H}$ given by $B(f,g) = fg.$ It is easy to see $s$ and $z$ are continuous linear functions and $B$ is a continuous bilinear (with $|B| leq 1$). The function you want to differentiat is given $F = B circ z circ s.$ By the chain rule, $F'(q) = B'(s(q), s(q)) circ z circ s;$ hence, $F'(q) cdot h = B'(s(q), s(q)) cdot (s(h), s(h)) = 2 s(q) s(h),$ if you evaluate in a point $x$ this becomes $big(F'(q) cdot h big)(x) = 2q(-x)h(-x).$ Q.E.D.






            share|cite|improve this answer









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              0












              $begingroup$

              Let $mathrm{H} = mathscr{C}(mathrm{I}),$ the space of continuous real-valued functions defined on the compact interval $mathrm{I}$ the with norm of suprema. Let $s$ be the function $mathrm{H} to mathrm{H}$ given by $s(f)(x) = f(-x),$ $z$ the function $mathrm{H} to mathrm{H} times mathrm{H}$ given by $q mapsto (q, q)$ and let $B$ the function $mathrm{H} times mathrm{H} to mathrm{H}$ given by $B(f,g) = fg.$ It is easy to see $s$ and $z$ are continuous linear functions and $B$ is a continuous bilinear (with $|B| leq 1$). The function you want to differentiat is given $F = B circ z circ s.$ By the chain rule, $F'(q) = B'(s(q), s(q)) circ z circ s;$ hence, $F'(q) cdot h = B'(s(q), s(q)) cdot (s(h), s(h)) = 2 s(q) s(h),$ if you evaluate in a point $x$ this becomes $big(F'(q) cdot h big)(x) = 2q(-x)h(-x).$ Q.E.D.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $mathrm{H} = mathscr{C}(mathrm{I}),$ the space of continuous real-valued functions defined on the compact interval $mathrm{I}$ the with norm of suprema. Let $s$ be the function $mathrm{H} to mathrm{H}$ given by $s(f)(x) = f(-x),$ $z$ the function $mathrm{H} to mathrm{H} times mathrm{H}$ given by $q mapsto (q, q)$ and let $B$ the function $mathrm{H} times mathrm{H} to mathrm{H}$ given by $B(f,g) = fg.$ It is easy to see $s$ and $z$ are continuous linear functions and $B$ is a continuous bilinear (with $|B| leq 1$). The function you want to differentiat is given $F = B circ z circ s.$ By the chain rule, $F'(q) = B'(s(q), s(q)) circ z circ s;$ hence, $F'(q) cdot h = B'(s(q), s(q)) cdot (s(h), s(h)) = 2 s(q) s(h),$ if you evaluate in a point $x$ this becomes $big(F'(q) cdot h big)(x) = 2q(-x)h(-x).$ Q.E.D.






                share|cite|improve this answer









                $endgroup$



                Let $mathrm{H} = mathscr{C}(mathrm{I}),$ the space of continuous real-valued functions defined on the compact interval $mathrm{I}$ the with norm of suprema. Let $s$ be the function $mathrm{H} to mathrm{H}$ given by $s(f)(x) = f(-x),$ $z$ the function $mathrm{H} to mathrm{H} times mathrm{H}$ given by $q mapsto (q, q)$ and let $B$ the function $mathrm{H} times mathrm{H} to mathrm{H}$ given by $B(f,g) = fg.$ It is easy to see $s$ and $z$ are continuous linear functions and $B$ is a continuous bilinear (with $|B| leq 1$). The function you want to differentiat is given $F = B circ z circ s.$ By the chain rule, $F'(q) = B'(s(q), s(q)) circ z circ s;$ hence, $F'(q) cdot h = B'(s(q), s(q)) cdot (s(h), s(h)) = 2 s(q) s(h),$ if you evaluate in a point $x$ this becomes $big(F'(q) cdot h big)(x) = 2q(-x)h(-x).$ Q.E.D.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 5 at 18:06









                Will M.Will M.

                2,915315




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