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. 



Is $ f(x(.)) := \int_{0}^{1} F ( x(t)) \; dt$ differentiable?
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. $\Box$


