Why this functional isn't differentiable? I saw in the article Alt, H. M. and Caffarelli, L. A. Existence and regularity for a minimum problem with free boundary. J. Reine Angew. Math., 325, (1981), 105–144. That the functional 
\begin{equation}
J(v):= \int_{\Omega}(|\nabla v|^{2} + \chi(\{v>0\})Q^2)
\end{equation}
is not differentiable. Why this functional isn't differentiable? My difficulty in this question is because  I do not know exactly what means that the functional is differentiable. So if you can explain, what is a differentiable functional, or to say in what sense this specific functional isn't differentiable I will be very grateful. Above $\chi$ denotes the characteristic function, this $\chi(A)(x) = 1 $ if $x \in A$ and $0$ if $x \notin A.$, and $Q(x)$ is non-negative and measurable.

A functional is differentiable if it has a derivative. There are two popular kinds of derivative: 


*

*Fréchet derivative

*Gâteaux derivative
Accordingly, afunctional may be Fréchet differentiable or Gâteaux differentiable, the former property being stronger than the latter. 
The functional $F(v) =  \int_\Omega |\nabla v|^2$   is Fréchet differentiable on $W^{1,2}(\Omega)$. Its derivative at point $v\in  W^{1,2}(\Omega)$ is given by 
$$
(D_v F)(\phi)=2\int_\Omega\nabla v\cdot \nabla \phi,\quad \phi\in W^{1,2}(\Omega) \tag1
$$
where you should understand that the derivative $ D_v F $ is a linear functional given by (1). The verification that $D_v F$ meets the definition of Fréchet derivative is below:
$$\begin{split}
F(v+\phi ) &=\int_\Omega |\nabla v+\nabla \phi|^2=\int_\Omega |\nabla v|^2+2 \int_\Omega\nabla v\cdot \nabla \phi + \int_\Omega |\nabla \phi|^2 \\ & = F(v)+(D_vF)(\phi)+ O(\|\phi\|^2)
\end{split}$$
However, the second term $G(v)=\int_{v>0} Q^2$ is not even Gâteaux differentiable. Let   $v\equiv 0$ and $\phi\equiv 1$. Then the restriction of $G$ to   the line $\{v+t\phi:t\in\mathbb R\}$ is
$$G(v+t\phi)=\begin{cases} \int_\Omega Q^2 \quad &\text{if } t>0
\\ 0\quad &\text{if } t\le 0\end{cases} $$ which is not even continuous, let alone differentiable.
