Computing Fréchet derivative of $F(f)(x) = int^{x}_{0} cos(f(t)^{2})dt, x in [0,1]$












6












$begingroup$


Let $X = mathcal{C} left( [0,1] right)$ be the Banach space of continuous functions on $[0,1]$ (with the supremum norm) and define a map $F : X rightarrow X$ by
$$F(f)(x) = int^{x}_{0} cos(f(t)^{2})dt, x in [0,1].$$



Show that $F$ is Fréchet differentiable and compute the Fréchet derivative
$DF|_{f}$ for each $f in X.$




So far I have the following.



Using the identity $cos(A+varepsilon B)=cos(A)-varepsilon Bsin(A)+mathcal{O}(varepsilon^2).$ We have
begin{align*}F(f+h)(x)
&=int^{x}_{0} cos((f(t)+h(t))^{2})dt \
&=int^{x}_{0}cos(f^2(t))-h(t)[2f(t)h(t)]sin(f^2(t))+mathcal{O}(h^2(t))dt \
&=F(f)(x)+int^{x}_{0}-2f(t)(h(t)text{sin}(f^2(t))-h^2(t)sin(f^2(t))+mathcal{O}(h^2(t))dt \
end{align*}
Let $T(h)=int^{x}_{0}-2f(t)(h(t)sin(f^2(t))dt$. I am able to show that $T(h)$ is a linear map. Then we have
$$F(f+h)(x)-F(f)(x)=T(h)+int^{x}_{0}mathcal{O}(h^2(t))dt $$
I am confused about the Big-$mathcal{O}$ notation and I do not understand what it would mean to integrate over it. Also, am I on the right track? Thank you in advance for any help provided.






real-analysis functional-analysis operator-theory asymptotics frechet-derivative







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  • 1




    $begingroup$
    You're missing some parentheses after "Let $T(h) = dots$"
    $endgroup$
    – 0x539
    Jan 3 at 14:59
















6












$begingroup$


Let $X = mathcal{C} left( [0,1] right)$ be the Banach space of continuous functions on $[0,1]$ (with the supremum norm) and define a map $F : X rightarrow X$ by
$$F(f)(x) = int^{x}_{0} cos(f(t)^{2})dt, x in [0,1].$$



Show that $F$ is Fréchet differentiable and compute the Fréchet derivative
$DF|_{f}$ for each $f in X.$




So far I have the following.



Using the identity $cos(A+varepsilon B)=cos(A)-varepsilon Bsin(A)+mathcal{O}(varepsilon^2).$ We have
begin{align*}F(f+h)(x)
&=int^{x}_{0} cos((f(t)+h(t))^{2})dt \
&=int^{x}_{0}cos(f^2(t))-h(t)[2f(t)h(t)]sin(f^2(t))+mathcal{O}(h^2(t))dt \
&=F(f)(x)+int^{x}_{0}-2f(t)(h(t)text{sin}(f^2(t))-h^2(t)sin(f^2(t))+mathcal{O}(h^2(t))dt \
end{align*}
Let $T(h)=int^{x}_{0}-2f(t)(h(t)sin(f^2(t))dt$. I am able to show that $T(h)$ is a linear map. Then we have
$$F(f+h)(x)-F(f)(x)=T(h)+int^{x}_{0}mathcal{O}(h^2(t))dt $$
I am confused about the Big-$mathcal{O}$ notation and I do not understand what it would mean to integrate over it. Also, am I on the right track? Thank you in advance for any help provided.






real-analysis functional-analysis operator-theory asymptotics frechet-derivative







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You're missing some parentheses after "Let $T(h) = dots$"
    $endgroup$
    – 0x539
    Jan 3 at 14:59














6












6








6


1



$begingroup$


Let $X = mathcal{C} left( [0,1] right)$ be the Banach space of continuous functions on $[0,1]$ (with the supremum norm) and define a map $F : X rightarrow X$ by
$$F(f)(x) = int^{x}_{0} cos(f(t)^{2})dt, x in [0,1].$$



Show that $F$ is Fréchet differentiable and compute the Fréchet derivative
$DF|_{f}$ for each $f in X.$




So far I have the following.



Using the identity $cos(A+varepsilon B)=cos(A)-varepsilon Bsin(A)+mathcal{O}(varepsilon^2).$ We have
begin{align*}F(f+h)(x)
&=int^{x}_{0} cos((f(t)+h(t))^{2})dt \
&=int^{x}_{0}cos(f^2(t))-h(t)[2f(t)h(t)]sin(f^2(t))+mathcal{O}(h^2(t))dt \
&=F(f)(x)+int^{x}_{0}-2f(t)(h(t)text{sin}(f^2(t))-h^2(t)sin(f^2(t))+mathcal{O}(h^2(t))dt \
end{align*}
Let $T(h)=int^{x}_{0}-2f(t)(h(t)sin(f^2(t))dt$. I am able to show that $T(h)$ is a linear map. Then we have
$$F(f+h)(x)-F(f)(x)=T(h)+int^{x}_{0}mathcal{O}(h^2(t))dt $$
I am confused about the Big-$mathcal{O}$ notation and I do not understand what it would mean to integrate over it. Also, am I on the right track? Thank you in advance for any help provided.






real-analysis functional-analysis operator-theory asymptotics frechet-derivative







share|cite|improve this question











$endgroup$




Let $X = mathcal{C} left( [0,1] right)$ be the Banach space of continuous functions on $[0,1]$ (with the supremum norm) and define a map $F : X rightarrow X$ by
$$F(f)(x) = int^{x}_{0} cos(f(t)^{2})dt, x in [0,1].$$



Show that $F$ is Fréchet differentiable and compute the Fréchet derivative
$DF|_{f}$ for each $f in X.$




So far I have the following.



Using the identity $cos(A+varepsilon B)=cos(A)-varepsilon Bsin(A)+mathcal{O}(varepsilon^2).$ We have
begin{align*}F(f+h)(x)
&=int^{x}_{0} cos((f(t)+h(t))^{2})dt \
&=int^{x}_{0}cos(f^2(t))-h(t)[2f(t)h(t)]sin(f^2(t))+mathcal{O}(h^2(t))dt \
&=F(f)(x)+int^{x}_{0}-2f(t)(h(t)text{sin}(f^2(t))-h^2(t)sin(f^2(t))+mathcal{O}(h^2(t))dt \
end{align*}
Let $T(h)=int^{x}_{0}-2f(t)(h(t)sin(f^2(t))dt$. I am able to show that $T(h)$ is a linear map. Then we have
$$F(f+h)(x)-F(f)(x)=T(h)+int^{x}_{0}mathcal{O}(h^2(t))dt $$
I am confused about the Big-$mathcal{O}$ notation and I do not understand what it would mean to integrate over it. Also, am I on the right track? Thank you in advance for any help provided.






real-analysis functional-analysis operator-theory asymptotics frechet-derivative




real-analysis functional-analysis operator-theory asymptotics frechet-derivative






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edited Jan 4 at 0:08







Gaby Alfonso

















asked Oct 29 '18 at 7:52









Gaby AlfonsoGaby Alfonso

693315




693315








  • 1




    $begingroup$
    You're missing some parentheses after "Let $T(h) = dots$"
    $endgroup$
    – 0x539
    Jan 3 at 14:59














  • 1




    $begingroup$
    You're missing some parentheses after "Let $T(h) = dots$"
    $endgroup$
    – 0x539
    Jan 3 at 14:59








1




1




$begingroup$
You're missing some parentheses after "Let $T(h) = dots$"
$endgroup$
– 0x539
Jan 3 at 14:59




$begingroup$
You're missing some parentheses after "Let $T(h) = dots$"
$endgroup$
– 0x539
Jan 3 at 14:59










1 Answer
1






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oldest

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2












$begingroup$

I think your derivative $T$ is correct. In my experience big-$O$ notation is useful for obtaining the solution for this kind of problems, but for actual proofs it is often easier to write down explicit inequalities. So let's start doing exactly that.



$renewcommand{epsilon}{varepsilon}$
By the error theorem for Taylor expansion $cos(A + x) = cos A - x sin A - frac12x^2 cos alpha$ for some $alpha in (A, A+x)$. Since bounding $cos$ by $1$ we get
$$
|cos(A + x) - (cos A - x sin A)| leq frac12 x^2
$$



Now let $f, h in mathcal{C}[0, 1]$ such that $|h|_infty < epsilon$ for some $epsilon > 0$. Then
$$
begin{align}
|F(f + h)(x) - F(f)(x) - T(x) | \
&hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) +2 f(t) h(t) sinleft(f(t)^2right) ,dt right|\
&hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) + left( 2 f(t) h(t) + h(t)^2 right) sinleft(f(t)^2right) \
- h(t)^2 sinleft(f(t)^2right),dtright| \
&hspace{-45mm}leq int_0^x frac12left| 2 f(t) h(t) + h(t)^2 right|^2 + left| h(t)^2 sinleft(f(t)^2right) right| ,dt \
&hspace{-45mm}leq frac12 left( 2 |f|_infty epsilon + epsilon^2 right)^2 + epsilon^2
end{align}
$$



This last bound is obviously $O(epsilon^2)$.






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    1 Answer
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    1 Answer
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    active

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    2












    $begingroup$

    I think your derivative $T$ is correct. In my experience big-$O$ notation is useful for obtaining the solution for this kind of problems, but for actual proofs it is often easier to write down explicit inequalities. So let's start doing exactly that.



    $renewcommand{epsilon}{varepsilon}$
    By the error theorem for Taylor expansion $cos(A + x) = cos A - x sin A - frac12x^2 cos alpha$ for some $alpha in (A, A+x)$. Since bounding $cos$ by $1$ we get
    $$
    |cos(A + x) - (cos A - x sin A)| leq frac12 x^2
    $$



    Now let $f, h in mathcal{C}[0, 1]$ such that $|h|_infty < epsilon$ for some $epsilon > 0$. Then
    $$
    begin{align}
    |F(f + h)(x) - F(f)(x) - T(x) | \
    &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) +2 f(t) h(t) sinleft(f(t)^2right) ,dt right|\
    &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) + left( 2 f(t) h(t) + h(t)^2 right) sinleft(f(t)^2right) \
    - h(t)^2 sinleft(f(t)^2right),dtright| \
    &hspace{-45mm}leq int_0^x frac12left| 2 f(t) h(t) + h(t)^2 right|^2 + left| h(t)^2 sinleft(f(t)^2right) right| ,dt \
    &hspace{-45mm}leq frac12 left( 2 |f|_infty epsilon + epsilon^2 right)^2 + epsilon^2
    end{align}
    $$



    This last bound is obviously $O(epsilon^2)$.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      I think your derivative $T$ is correct. In my experience big-$O$ notation is useful for obtaining the solution for this kind of problems, but for actual proofs it is often easier to write down explicit inequalities. So let's start doing exactly that.



      $renewcommand{epsilon}{varepsilon}$
      By the error theorem for Taylor expansion $cos(A + x) = cos A - x sin A - frac12x^2 cos alpha$ for some $alpha in (A, A+x)$. Since bounding $cos$ by $1$ we get
      $$
      |cos(A + x) - (cos A - x sin A)| leq frac12 x^2
      $$



      Now let $f, h in mathcal{C}[0, 1]$ such that $|h|_infty < epsilon$ for some $epsilon > 0$. Then
      $$
      begin{align}
      |F(f + h)(x) - F(f)(x) - T(x) | \
      &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) +2 f(t) h(t) sinleft(f(t)^2right) ,dt right|\
      &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) + left( 2 f(t) h(t) + h(t)^2 right) sinleft(f(t)^2right) \
      - h(t)^2 sinleft(f(t)^2right),dtright| \
      &hspace{-45mm}leq int_0^x frac12left| 2 f(t) h(t) + h(t)^2 right|^2 + left| h(t)^2 sinleft(f(t)^2right) right| ,dt \
      &hspace{-45mm}leq frac12 left( 2 |f|_infty epsilon + epsilon^2 right)^2 + epsilon^2
      end{align}
      $$



      This last bound is obviously $O(epsilon^2)$.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        I think your derivative $T$ is correct. In my experience big-$O$ notation is useful for obtaining the solution for this kind of problems, but for actual proofs it is often easier to write down explicit inequalities. So let's start doing exactly that.



        $renewcommand{epsilon}{varepsilon}$
        By the error theorem for Taylor expansion $cos(A + x) = cos A - x sin A - frac12x^2 cos alpha$ for some $alpha in (A, A+x)$. Since bounding $cos$ by $1$ we get
        $$
        |cos(A + x) - (cos A - x sin A)| leq frac12 x^2
        $$



        Now let $f, h in mathcal{C}[0, 1]$ such that $|h|_infty < epsilon$ for some $epsilon > 0$. Then
        $$
        begin{align}
        |F(f + h)(x) - F(f)(x) - T(x) | \
        &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) +2 f(t) h(t) sinleft(f(t)^2right) ,dt right|\
        &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) + left( 2 f(t) h(t) + h(t)^2 right) sinleft(f(t)^2right) \
        - h(t)^2 sinleft(f(t)^2right),dtright| \
        &hspace{-45mm}leq int_0^x frac12left| 2 f(t) h(t) + h(t)^2 right|^2 + left| h(t)^2 sinleft(f(t)^2right) right| ,dt \
        &hspace{-45mm}leq frac12 left( 2 |f|_infty epsilon + epsilon^2 right)^2 + epsilon^2
        end{align}
        $$



        This last bound is obviously $O(epsilon^2)$.






        share|cite|improve this answer











        $endgroup$



        I think your derivative $T$ is correct. In my experience big-$O$ notation is useful for obtaining the solution for this kind of problems, but for actual proofs it is often easier to write down explicit inequalities. So let's start doing exactly that.



        $renewcommand{epsilon}{varepsilon}$
        By the error theorem for Taylor expansion $cos(A + x) = cos A - x sin A - frac12x^2 cos alpha$ for some $alpha in (A, A+x)$. Since bounding $cos$ by $1$ we get
        $$
        |cos(A + x) - (cos A - x sin A)| leq frac12 x^2
        $$



        Now let $f, h in mathcal{C}[0, 1]$ such that $|h|_infty < epsilon$ for some $epsilon > 0$. Then
        $$
        begin{align}
        |F(f + h)(x) - F(f)(x) - T(x) | \
        &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) +2 f(t) h(t) sinleft(f(t)^2right) ,dt right|\
        &hspace{-45mm}= left|int_0^x cosleft((f(t)^2 + h(t)^2)right) - cosleft(f(t)^2right) + left( 2 f(t) h(t) + h(t)^2 right) sinleft(f(t)^2right) \
        - h(t)^2 sinleft(f(t)^2right),dtright| \
        &hspace{-45mm}leq int_0^x frac12left| 2 f(t) h(t) + h(t)^2 right|^2 + left| h(t)^2 sinleft(f(t)^2right) right| ,dt \
        &hspace{-45mm}leq frac12 left( 2 |f|_infty epsilon + epsilon^2 right)^2 + epsilon^2
        end{align}
        $$



        This last bound is obviously $O(epsilon^2)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 15:43

























        answered Jan 3 at 15:30









        0x5390x539

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