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Thread: Frechet Derivative

  1. #1
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    Frechet Derivative

    Consider the normed linear space, $\displaystyle (C[0,1],+,\cdot,||\cdot||_\infty)$ where $\displaystyle ||u||_\infty = \sup_{t\in[0,1]}|u(t)|$.
    Let $\displaystyle f:C[0,1] \rightarrow C[0,1]$ and define

    $\displaystyle f(x)(t) = \int_0^1 g(t,x(s))ds$

    for some kernel $\displaystyle g: \mathbb{R}^2 \rightarrow \mathbb{R} $. Compute the Frechet Derivative of f at x. State any assumptions required for g. (Hint: Use Taylor's Thorem).

    I have deduced that g must be continuous in the variable s, but that's all i can figure out. I would greatly appreciate a hint.
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  2. #2
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    May be this helps. The definition is

    $\displaystyle
    \displaystyle
    d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)-f(x)}{\epsilon}
    $

    where

    $\displaystyle
    x \; and \; h \ \in \ D.
    $


    So we have

    $\displaystyle
    \displaystyle
    d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)(t)-f(x)(t)}{\epsilon}=
    $

    $\displaystyle
    \displaystyle
    =lim_{\epsilon \to 0} \frac{1}{\epsilon}\; ( \; \int_0^1 g(t,x(s)+\epsilon h)ds-\int_0^1 g(t,x(s))ds \; )=\; \int_0^1 \frac{\partial g(t,x(s))}{\partial x} \ h \ ds \ .
    $
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    May be this helps. The definition is

    $\displaystyle
    \displaystyle
    d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)-f(x)}{\epsilon}
    $

    where

    $\displaystyle
    x \; and \; h \ \in \ D.
    $


    So we have

    $\displaystyle
    \displaystyle
    d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)(t)-f(x)(t)}{\epsilon}=
    $

    $\displaystyle
    \displaystyle
    =lim_{\epsilon \to 0} \frac{1}{\epsilon}\; ( \; \int_0^1 g(t,x(s)+\epsilon h)ds-\int_0^1 g(t,x(s))ds \; )=\; \int_0^1 \frac{\partial g(t,x(s))}{\partial x} \ h \ ds \ .
    $
    I have actually figured it out for the most part, but i am operating on the definition of the Frechet Derivative, $\displaystyle d_f(h)$, that for $\displaystyle x,h \in C[0,1]$,
    $\displaystyle \lim_{h \rightarrow 0} ||\frac{f(x+h)(t) - f(x) - d_f(h)}{h}|| = 0$

    So, i use the fact that
    $\displaystyle g(t,x+h) = g(t,x) + h(s)\frac{\partial}{\partial s} g(t,x(s_0))$
    for some $\displaystyle s_0 $.
    So,
    $\displaystyle || \int_0^1g(t,x(s)+h(s)) ds - \int_0^1 g(t,x(s))|| = ||\int_0^1\frac{\partial}{\partial s} g(t,x(s_0)) h(s) ds||$ Which is our Frechet Derivative.
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