Frechet derivative of an operator Let an operator $T:C[a,b]\to C[a,b]$ be defined as:
\begin{equation}
(Tu)(x)=\int_{a}^{b}K(x,t)f(t,u(t))dt
\end{equation}
where $K:[a,b]\times[a,b]\to \mathbb{R}$, $f:[a,b]\times \mathbb{R}\to \mathbb{R}$ are given functions. 
How can I compute the Frechet derivative of $T$?
 A: Differentiation under integral sign. Assume:


*

*$[a, b]$ is a compact interval of $\Bbb R$;

*$\mathrm{E}_1, \mathrm{E}_2$ are normed and complete (completeness is essential as Riemann integral fails to exist for non-complete vector spaces) linear spaces;

*$\mathrm{O}$ is an open set in $\mathrm{E}_1$;

*$f:[a, b] \times \mathrm{O} \to \mathrm{E}_2$ is a continuous and
$\mathbf{D}_2f$ exists and is continuous.


Then, the function $\displaystyle \varphi(x) = \int\limits_a^b dt\ f(t, x)$ defines a continuous (resp. differentiable with continuity) function $\mathrm{O} \to \mathrm{E}_2$ and its derivative is given by $\displaystyle \mathbf{D} \varphi(x) = \int\limits_a^b dt\ \mathbf{D}_2 f(t, x).$
To your exercise. Define $\varphi:[a,b] \times \mathscr{C}[a,b] \to \Bbb R$ by $\varphi(t, u) = u(t),$ which is a continuous bilinear map, so $\dfrac{\partial \varphi}{\partial u} \cdot h = h(t).$ The function $f$ has to be assumed continuous with a continous partial derivative with respect to the second factor (or if $f$ is a continous function of $(t, y),$ then you want $\dfrac{\partial f}{\partial y}$ to be continuous also). By chain rule, if $\psi(t, u) = f(t, u(t)) = f(t, \varphi(t, u)),$ then $\dfrac{\partial \psi}{\partial u} \cdot h = \dfrac{\partial f}{\partial y} \dfrac{\partial \varphi}{\partial u} \cdot h = \dfrac{\partial f}{\partial y} h(t).$ Therefore, $$\mathbf{D}\mathrm{T}(u) \cdot h = \int\limits_a^b dt\ \mathrm{K}(\ , t) \dfrac{\partial f}{\partial y} h(t) \in \mathscr{C}[a,b]$$
is the continuous function
$$x \mapsto \int\limits_a^b dt\ \mathrm{K}(x, t) \dfrac{\partial f}{\partial y} h(t).$$
