My question concerns a step from the following derivation of the Euler-Lagrange equation, from the standard book on calculus of variations by Gelfand and Fomin
While I understand the general procedure, I'm not sure why the integral of the reminder goes to zero. I've see other more general derivations of the Euler-Lagrange equation but they use more machinery from calculus over Banach spaces, so I'm trying to do this specific in $\mathbb R^3$ which uses more elementary calculus methods.
Reformulating my question more precisely, let's assume $F: \mathbb R ^3 \to \mathbb R$ is $C^2$ everywhere. By Taylor's formula, $$F(a+h_1, b+h_2, c+h_3) = F(a,b,c) + \langle \nabla F(a,b,c),(h_1,h_2,h_3)\rangle + R(h_1,h_2,h_3),$$ where we have the usual inner product in $\mathbb R^3$ and $R$ is the Taylor reminder, with $R(h_1, h_2, h_3)/\|(h_1,h_2, h_3)\|_{\mathbb R^3} \to 0$ as $(h_1,h_2,h_3) \to 0$ for a norm $\|\cdot\|_{\mathbb R^3}$ of $\mathbb R^3.$
By defining the functional J : $C^1([a,b], \mathbb R) \to \mathbb R,$ $$ y \in C^1([a,b],\mathbb R) \mapsto J(y) = \int_a^b F(t,y(t),y'(t))dt$$ the idea is to show that $$D J_y(h) = \int_a^b [F_y(t,y(t),y'(t))h(t) + F_{z}(t,y(t),y'(t))h'(t)]dt $$ (where $F_y$ and $F_z$ denote the partial derivatives of $F$ with respect to the second and third coordinates) is the Frechet derivative of $J.$ Now the text is a bit vague but I believe this has to be done over the norm in $C^1([a,b], \mathbb R)$ given by $\|h\|_{C^1} = \sup_{[a,b]} |h(x)| + \sup_{[a,b]} |h'(x)|.$ What we must prove is that $$\lim_{h \to 0} \frac{J(y+h) - J(y) - DJ_y(h)}{\|h\|_{C^1}} = 0 $$ (where {$h \to 0$ means convergence on its norm). So, by applying Taylor's formula for $(0, h(t), h'(t)),$ we get $$ F(t,y(t) + h(t), y'(t)+ h'(t)) = F(t,y(t),y'(t)) + F_y(t,y(t),y'(t))h(t) + F_z(t,y(t),y'(t))h'(t) + R(0,h(t),h'(t)), \quad \text{for each } t\in[a,b] $$ integrating and rearanging, I must prove $$ \lim_{h \to 0}\frac{\int^b_a R(0,h(t),h'(t))dt}{\|h\|_{C^1}} = 0$$
While this seems to be intuitively true since at each point $R$ is "small" compared to $h,$ I'm not sure how to precisely prove this last step. The integral and the norm are complicating my analysis.
Any help is appreciated, thanks.