Here is an exercise in the calculus of variations:
Exercise 1.5 Consider the space $V=\mathcal C^0([0,1],\Bbb R)$, let $g:\Bbb R\to\Bbb R$ be a $\mathcal C^1$ function, and define the functional $J$ on $V$ by $J(y)=\int^1_0 g(y(x))\,\Bbb dx$. Show that its first variation exists and is given by the formula $\delta J|_y (\eta) = \int^1_0 g' (y(x)) \eta(x) \,\Bbb dx$.
I think I can solve it as follows: use the Taylor approximation of $g\left(y(x)+\alpha \eta (x)\right)$ to get: $$g\left(y(x)+\alpha \eta (x)\right)-g(y(x))=g'(y(x))\cdot \alpha \eta(x) +O(\alpha^2\eta(x)^2)$$
Then the first variation of $J$ becomes: $$\lim_{\alpha\to 0}\left(\frac{\displaystyle \int_0^1 \left( \alpha g'(y(x))\eta(x)+O(\alpha^2\eta^2(x))\right)\,\Bbb dx} \alpha \right)= \lim_{\alpha\to 0}\left({\int_0^1 \left( g'(y(x))\eta(x)+O(\alpha\eta^2(x))\right)\,\Bbb dx} \right)=\int_0^1 g'(y(x))\eta(x)\,\Bbb dx$$
The answer is correct, but I don’t feel entirely comfortable with the derivation. Specifically, it seems to me that I am assuming that $g(x)$ is analytic, even though this is not an assumption, since otherwise we cannot assume that it is equal to its Taylor series.
So is my use of this Taylor expansion correct for non-analytic $g(x)$, and if so why? otherwise, how would one prove the result for non-analytic $g(x)$?
If $g$ is continuously-differentiable on some interval containing a point $x_{0}$, the fundamental theorem of calculus gives \begin{align*} g(x_{0} + \alpha) &= g(x_{0}) + \int_{x_{0}}^{x_{0}+\alpha} g'(t)\, dt \\ &= g(x_{0}) + \int_{0}^{\alpha} g'(x_{0} + t)\, dt \\ &= g(x_{0}) + \int_{0}^{\alpha} [g'(x_{0}) + \underbrace{g'(x_{0} + t) - g'(x_{0})}_{o(1)}]\, dt \\ &= g(x_{0}) + g'(x_{0})\alpha + o(\alpha) \end{align*} for $|\alpha|$ small.