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Math Help - Frechet Derivative

  1. #1
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    Frechet Derivative

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

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

    for some kernel 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

    <br />
\displaystyle<br />
d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)-f(x)}{\epsilon}<br />

    where

    <br />
x \;  and \; h \ \in \ D.<br />


    So we have

    <br />
\displaystyle<br />
d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)(t)-f(x)(t)}{\epsilon}=<br />

    <br />
\displaystyle<br />
=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 \ .<br />
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    May be this helps. The definition is

    <br />
\displaystyle<br />
d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)-f(x)}{\epsilon}<br />

    where

    <br />
x \;  and \; h \ \in \ D.<br />


    So we have

    <br />
\displaystyle<br />
d_h f=lim_{\epsilon \to 0} \frac{f(x+\epsilon h)(t)-f(x)(t)}{\epsilon}=<br />

    <br />
\displaystyle<br />
=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 \ .<br />
    I have actually figured it out for the most part, but i am operating on the definition of the Frechet Derivative, d_f(h), that for x,h \in C[0,1],
    \lim_{h \rightarrow 0} ||\frac{f(x+h)(t) - f(x) - d_f(h)}{h}|| = 0

    So, i use the fact that
    g(t,x+h) = g(t,x) + h(s)\frac{\partial}{\partial s} g(t,x(s_0))
    for some s_0 .
    So,
    || \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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