I have been attempting this problem for quite a while. The problem is to show that the following function $$F((f))(x)=\int_{0}^{x}f^2(t)\cos(t)dt$$ is differentiable on the normed vector space $C[0,1]$ equipped with the norm $||f||=\underset{x\in[0,1]}\max|f(x)|$, and to then compute its derivative.
I don't really know where to go from here. I know that the frechet derivative is defined as a linear operator $T:C[0,1]\rightarrow C[0,1]$ that satisfies
$$\underset{h\rightarrow0}\lim \frac{||F(f+h)-F(f)-T(h)||}{||h||}=0$$
I though that things would become clear after writing everything out, but I'm stuck right around here: $$\frac{\underset{x\in[0,1]}\max |\int_{0}^{x}[f^2(t)+2f(t)h(t)+h^2(t)]\cos(t)-f^2(t)\cos(t)dt-T(h)|}{\underset{x\in[0,1]}\max |h(t)|}$$
Do I show that this goes to zero in the limit as $h$ approaches the constant zero function? Apart from the obvious cancellation of the $f^2(t)\cos(t)$ term I don't know how to proceed. I would greatly appreciate some help here. Thank you!
First, we need to find a candidate for $T$. Using Gateaux variation, we see that \begin{align} T(f)g=&\ \frac{d}{dt}F(f+tg)\Big|_{t=0} = \frac{d}{dt}\int^x_0(f(s)+tg(s))^2\cos(s)\ ds\Big|_{t=0}\\ =&\ 2 \int^x_0 f(s)\cos(s) g(s)\ ds. \end{align} Hence it follows \begin{align} \|F(f+g)-F(f)-T(f)g\|_\infty=&\ \Big\|\int^x_0 [(f(s)+g(s))^2-f^2(s)-2f(s)g(s)]\cos(s)\ ds \Big\|_\infty\\ =&\ \Big\|\int^x_0 g^2(s)\cos(s)\ ds \Big\|_\infty \leq \| g\|_\infty^2 \Big\|\int^x_0\cos(s)\ ds \Big\|_\infty\\ \leq&\ \|g\|_\infty^2 \end{align} which means \begin{align} \frac{\|F(f+g)-F(f)-T(f)g\|_\infty}{\|g\|_\infty} \leq \|g\|_\infty \rightarrow 0 \end{align} as $\|g\|_\infty\rightarrow 0$. Thus, we see that the Frechet derivative of $F(f)$ is \begin{align} (T(f)g )(x) = 2\int^x_0 f(s)g(s)\cos(s) \ ds. \end{align}