Is $ f(x(.)) := \int_{0}^{1} F ( x(t)) \; dt$ differentiable? Let $f : AC[0, 1] \to R$ be defined by $ f(x(.)) := \int_{0}^{1} F ( x(t)) \; dt$. Where, $AC[0, 1]$ is the set of absolutely continuous functions with the norm $W^{1,1}$, and  $F: R^n \to R$ is continuously differentiable function. My question is that, Is $f$ Fréchet differentiable? If not is it at least Gateaux differentiable? 
The motivation of this question comes from the question https://math.stackexchange.com/q/1517707/219176  . Note that in my question  no growth condition is assumed. However I believe since $[0,1]$ is compact and $F$ is locally Lipschitz, this growth condition is automatically satisfied.  
 A: From the context, here $AC[0, 1]$ should be the set of all absolutely continuous functions from $[0,1]$ to $\mathbb R^n$ with the norm $W^{1,1}$.
Note that for all $x=(x_1,\dots,x_n)\in AC[0, 1]$ we have $\|x\|_\infty:=\sup\{|x_i(t)|\colon i=1,\dots,n,0\le t\le 1\}\le2\|x\|_{1,1}$; see the lemma at the end of this answer for details. So, for any $x$ and $h$ in $AC[0, 1]$ and any $s\in[0,1]$ we have $\|x+sh\|_\infty\le2\|x\|_{1,1}+2\|h\|_{1,1}$. Because $F$ is continuously differentiable, its derivative $F'$ is uniformly continuous on any bounded set in $\mathbb R^n$. So, for any given $x\in AC[0, 1]$ we have  $F'(x(t)+sh(t))-F'(x(t))\to0$ uniformly in $s,t$ in $[0,1]$ as $\|h\|_{1,1}\to0$ (and hence $\|h\|_\infty\to0$). So, 
\begin{multline}
f(x+h)-f(x)-\int_0^1 dt\, F'(x(t))\cdot h(t) \\ 
=\int_0^1 dt\,\int_0^1 ds\,[F'(x(t)+sh(t))-F'(x(t))]\cdot h(t) \\
=o(\|h\|_\infty)=o(\|h\|_{1,1}), 
\end{multline}
where $\cdot$ denotes the dot product. 
Thus, $f$ is indeed Fréchet-differentiable, with $f'(x)(h)=\int_0^1 dt\, F'(x(t))\cdot h(t)$. 
Added:

Lemma: If $\|x\|_{1,1}\le1$ then $\|x\|_\infty\le2$. 

Proof. It is enough to prove this lemma when the dimension $n$ is $1$. Let then $m$ and $M$ denote the minimum and maximum of $x$, respectively. Suppose that $\|x\|_\infty>2$. Then without loss of generality $M>2$. On the other hand, the condition $\|x\|_{1,1}\le1$ implies $\|x'\|_1\le1$ and hence $M-m\le1$. So, $m>1$ and hence $1<\|x\|_1\le\|x\|_{1,1}\le1$. This contradiction completes the proof. $\Box$
