Proof of Fréchet Differentiability - general instruction and specific problem 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
 A: 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 $\| g\|_{\infty}\rightarrow 0$.
