Show that the Lagrangian $\int_I H(t,c(t),c^\prime(t)) dt$ is Frechet differentiable 
Consider an interval $I = [t_0,t_1]$ and a finite dimensional Banach space $X$. Let $U$ be an open subset of $\mathbb{R} \times X \times X$ and let  $V \subseteq \mathcal{C}^{1}(I,X)$ be the set of all curves $c:I \rightarrow X$, where $(t,c(t),c^\prime(t))$ is contained in $U$ for all $t$. Let further be
$$\lvert\lvert c \rvert\rvert_{\mathcal{C}^1(I)} := \sup_{t \in I} \lvert\lvert c(t) \rvert\rvert_X + \sup_{t \in I}\lvert\lvert c^\prime(t) \rvert\rvert_X.$$
We note that $V$ is open in $\mathcal{C}^1(I,X)$.
For a $\mathcal{C}^r$ function $H: U \rightarrow \mathbb{R}$ ($r \ge 2$) consider the functional (sometimes known as Lagrangian)
$$f: V \rightarrow \mathbb{R}, c \mapsto \int_I H(t,c(t),c^\prime(t)) dt. $$
Why is $f$ Frechet differentiable?

From the derivation of the Euler-Lagrange equation we know that the derivative must be
$$f^\prime = \frac{f_\varepsilon}{d \varepsilon} = \int_{t_0}^{t_1} \eta(t) \frac{H_\varepsilon}{\partial c_\varepsilon} + \eta^{\prime}(t) \frac{H_\varepsilon}{\partial c^\prime_\varepsilon}$$
, but using this on the definition of Frechet-differentiability only messy terms for me. I also can not find a proof for this fact online. Could you please help me?
 A: Assuming $H$ is twice continuously differentiable. By the definition of differentiability for each fixed $t$
$$H(t, x+h_1, y+h_2)-H(t, x, y)-H'_x(t, x, y)h_1-H'_y(t, x, y)h_2=g(t, h_1, h_2)$$
$$\lim_{h_1\to0,h_2\to0}\frac{|g(t, h_1, h_2)|}{||h_1||_X+||h_2||_X}=0$$
We can also write the remainder $g(t, h_1, h_2)$ in the Lagrange form explicitly. This form for a function of multiple variables will be something like this:
$$g(t, h_1, h_2)=l[h_1^TH''_{x,x}(t,x',y')h_1+2h_1^TH''_{x,y}(t,x',y')h_2+h_2^TH''_{y,y}(t,x',y')h_2]$$
where $l$ is a positive constant; $x',\ y'$ lie on the lines between $x$ and $x+h_1$, $y$ and $y+h_2$ respectively.
Taking $x=c(t),\ y=c'(t),\ h_1=\eta(t),\ h_2=\eta'(t)$ we have
$$\int_IH(t, c(t)+\eta(t),c'(t)dt+\eta'(t))-\int_IH(t, c(t),c'(t))dt-\int_IH'_x(t, c(t),c'(t))\eta(t)dt-\int_IH'_y(t, c(t),c'(t))\eta'(t)=$$
$$=\int_I\bigg[H(t, c(t)+\eta(t),c'(t)dt+\eta'(t))-H(t, c(t),c'(t))-H'_x(t, c(t),c'(t))\eta(t)-H'_y(t, c(t),c'(t))\eta'(t)\bigg]dt=\int_Ig(t, \eta(t), \eta'(t))dt$$
Now to show that $\int_IH'_x(t, c(t),c'(t))\eta(t)dt-\int_IH'_y(t, c(t),c'(t))\eta'(t)dt\ $ is the Frechet derivative, we need to prove that
$$\frac{|\int_Ig(t, \eta(t), \eta'(t))dt|}{\sup_{t\in I}||\eta(t)||_X+\sup_{t\in I}||\eta'(t)||_X}\to0,\ \text{as}\ ||\eta||_{\mathcal{C}^1(I)}\to0$$
We know that $c(t)$ and $c'(t)$ are bounded, all $\eta$ go to $0$ in the norm, so we can also bound any $\eta$ for any $t$ by some constant. This means that $c(t)+\eta(t)$ and its derivative lie in some closed ball, also $t\in I$ (closed interval). Thus we are interested in the values of $g(t, h_1, h_2)$ only on some compact set. Each element of the hessian $H''_{i,j}$ is continuous by assumption, thus bounded on our compact set by some shared constant, $|H''_{i,j}|<M$. Using the Lagrange form of $g$ and the Cauchy-Schwarz inequality, we can write
$$\frac{|\int_Ig(t, \eta(t), \eta'(t))dt|}{\sup_{t\in I}||\eta(t)||_X+\sup_{t\in I}||\eta'(t)||_X}\leq\frac{lM\int_I(||\eta(t)||^2_X+2||\eta(t)||_X||\eta'(t)||_X+||\eta'(t)||^2_X)dt}{\sup_{t\in I}||\eta(t)||_X+\sup_{t\in I}||\eta'(t)||_X}\leq$$
$$\leq \frac{(t_1-t_0)lM(\sup_{t\in I}||\eta(t)||_X+\sup_{t\in I}||\eta'(t)||_X)^2}{\sup_{t\in I}||\eta(t)||_X+\sup_{t\in I}||\eta'(t)||_X}=(t_1-t_0)lM(\sup_{t\in I}||\eta(t)||_X+\sup_{t\in I}||\eta'(t)||_X)\to0,\ \text{as}\ ||\eta||_{\mathcal{C}^1(I)}\to0$$
