1. ## 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.

2. May be this helps. The definition is

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

where

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

So we have

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

$
\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 \ .
$

3. Originally Posted by zzzoak
May be this helps. The definition is

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

where

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

So we have

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

$
\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, $d_f(h)$, 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.