Frechet and Gateaux derivative of integrals

Question

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

1. $$\varphi_2(u+sv)-\varphi_2(u)=\int_{u(c)}^{u(c)+sv(c)} g(t)\,dt=g(u(c)+s_0v(c)) sv(c) $$ 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.

2. 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

Answer 1

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 $\varphi_{2}$.