Frechet and Gateaux derivative of integrals I am new to differential calculus on normed spaces and I struggle with some easy things. Let $f:[a,b]\times\mathbb{R}\longrightarrow\mathbb{R}$ and $g:\mathbb{R}\longrightarrow\mathbb{R}$ two continuous functions and $$F(t,u)=\int_0^u f(t,\xi)\,d\xi$$ Sexa $c$ be a fixed point, $c\in(a,b)$. I want to prove that the functions
$$
\varphi_1:u\in H_0^1(a,b)\longrightarrow \int_a^b F(t,u(t))\,dt\in\mathbb{R}
$$
and
$$
\varphi_2:u\in H_0^1(a,b)\longrightarrow \int_0^{u(c)} g(t)\,dt\in\mathbb{R}
$$
are Frechet differentiable and in $\mathcal{C}(H_0^1(a,b),\mathbb{R})$.
Let $H=H_0^1(a,b)$. So far I have been to prove


*

*$$\varphi_2(u+sv)-\varphi_2(u)=\int_{u(c)}^{u(c)+sv(c)} g(t)\,dt=g(u(c)+s_0v(c)) sv(c)
$$
by the mean value theorem of integrals. So dividing by $s$ and taking limits as $s\to 0$ I get
$$
g(u(c))v(c)
$$
and this is the Gateaux derivative. However, trying I can't prove that the limit goes to $0$ on the Frechet derivative taking this as the derivative.

*Doing some computations, I arrive at
$$
\lim_{s\to0}\int_a^b F(t,u(t)+s_0v(t))v(t)\,dt
$$
with $s_0$ between $0$ and $s$.
Can I move the limit inside? In that case, the Frechet derivative would be $$\int_a^bf(t,u(t))v(t)\,dt$$ However, as in the case above, I struggle with the Gateaux derivative
 A: We need to prove that$$
\lim_{v\rightarrow0}\frac{|\varphi_{1}(u+v)-\varphi_{1}(u)-L(u)(v)|}{\Vert
v\Vert_{H^{1}}}=0.
$$
You have
\begin{align*}
& \varphi_{1}(u+v)-\varphi_{1}(u)-L(u)(v)\\
& =\int_{a}^{b}F(t,u(t)+v(t))-F(t,u(t))-f(t,u(t))v(t)\,dt\\
& =\int_{a}^{b}\int_{u(t)}^{u(t)+v(t)}f(t,s)\,ds-f(t,u(t))v(t)\,dt\\
& =\int_{a}^{b}\int_{u(t)}^{u(t)+v(t)}[f(t,s)-f(t,u(t))]\,ds\,dt.
\end{align*}
Now let $M:=\max_{t\in\lbrack0,b]}|u(t)|+1$. Since $f$ is continuous in
$[a,b]\times\lbrack-M,M]$, it is uniformly continuous there and so given
$\varepsilon>0$, there exists $0<\delta<1$ such that $|f(t,s)-f(t,r)|\leq
\varepsilon$ for all $t\in\lbrack a,b]$ and all $s,r\in\lbrack-M,M]$ with
$|s-r|\leq\delta$. Using the fact that $\Vert v\Vert_{L^{\infty}}\leq
(b-a)^{1/2}\Vert v\Vert_{H^{1}}$ (do you know this? It's the fundamental theorem of
calculus and Holder's inequality), if $\Vert v\Vert_{H^{1}}\leq\delta/(b-a)^{1/2}$,
then $\Vert v\Vert_{L^{\infty}}\leq\delta$. Hence
\begin{align*}
& |\varphi_{1}(u+v)-\varphi_{1}(u)-L(u)(v)|\\
& \leq\left\vert \int_{a}^{b}\int_{u(t)}^{u(t)+v(t)}%
|f(t,s)-f(t,u(t))|\,ds\,dt\right\vert \\
& \leq\int_{a}^{b}\left\vert \int_{u(t)}^{u(t)+v(t)}\varepsilon
\,ds\right\vert\,dt \leq\varepsilon\int_{a}^{b}|v(t)|\,dt\leq\varepsilon
(b-a)^{1/2}\left(  \int_{a}^{b}|v(t)|^{2}\,dt\right)  ^{1/2}.
\end{align*}
It follows that
$$
\frac{|\varphi_{1}(u+v)-\varphi_{1}(u)-L(u)(v)|}{\Vert v\Vert_{H^{1}}}
\leq\varepsilon(b-a)^{1/2}\frac{\Vert v\Vert_{L^{2}}}{\Vert v\Vert_{H^{1}}
}\leq\varepsilon(b-a)^{1/2},
$$
which shows that the limit is zero.
You should use similar tricks for $\varphi_{2}$. 
