I'd like to know if anyone could provide a general instruction to proving that a function is Fréchet differentiable. I've run into some problems where usually I can propose a form for the linear operator $D_{f,x}$, but I'm having trouble with that.
The specific case I refer in the title is:
Let $\phi: \mathbb{C}[0,1]\to\mathbb{R}$ be defined by:
$$\phi(f) = \int_{0}^{1} |f(t)|^2 \ dt.$$
Prove that it is Fréchet differentiable and compute $D_{\phi,f}$.
I've tried solving this and reached the following result:
$$D_{\phi,f} = 2\int_{0}^{1} g(z)f(z) \ dz$$
Thanks in advance
The Frechet derivative generalizes the notion of derivative to functionals. As with functions, there are some well known cases with a formula, but in general, we have to go back to the definition. The Frechet derivative of a functional $K$ at $f$ is a linear functional $L_f$ for which $K(f+g) = K(f) + L_f(g) +R(g)$ and \begin{equation*} \lim_{\|g\|\rightarrow 0} \frac{\|R(g)\|}{\|g\|} = 0 \end{equation*}
In your case, \begin{equation*} K(f) = \int_0^1 f(t)^2 dt \end{equation*} and \begin{align*} K(f+g) &= \int_0^1 (f(t)+g(t))^2 dt\\ &= \int_0^1 f(t)^2 dt + 2\int_0^1 f(t)g(t) dt + \int_0^1 g(t)^2 dt \end{align*}
So, we define $L_f$ and $R$ to be the latter two terms. Then we can verify that $|R(g)|\rightarrow 0$ faster than $\|g\|\rightarrow 0$. Hence $L_f$ is in fact the Frechet derivative.
Note To clarify a few of the comments below, the norm on $R$ is just absolute value and the norm on $g$ is the $L^{\infty}$ norm. So we have \begin{align*} \frac{\|R(g)\|}{\|g\|} &= \frac{|\int_0^1 g(t)^2 dt |}{\| g\|_{\infty}}\\ &\leq \frac{\int_0^1 \|g\|_{\infty}^2 dt }{\| g\|_{\infty}}\\ &= \| g\|_{\infty} \end{align*} which converges to zero as $\| g\|_{\infty}\rightarrow 0$.