Assume that the Lagrangian $L$ is $C^2$, $y(x)$ is a $C^1$ curve and $\eta$ is a $C^1$ perturbation. Define $$ J (y) = \int^b_a L(x,y,y')dx. $$ If we define the first variation to be $$ \left.\delta J\right|_{y}(\eta)=\int_{a}^{b}\left(L_{y}\left(x, y(x), y^{\prime}(x)\right) \eta(x)+L_{z}\left(x, y(x), y^{\prime}(x)\right) \eta^{\prime}(x)\right) d x, $$ then it satisfies the relation $$ J(y+\alpha \eta)=J(y)+\left.\delta J\right|_{y}(\eta) \alpha+o(\alpha) \label{1}\tag{1} $$ w.r.t variable $\alpha\in \Bbb{R}$, where $o(\alpha) = \alpha \psi(\alpha)$ such that if $\alpha \to 0$ then $\psi(\alpha) \to 0$ in at a faster pace, which is obvious by definition.
The question is: even if the $o$-relation w.r.t. $\alpha$ \eqref{1} above holds, the following analogous $o$-retlation w.r.t. a norm of the $\eta$ variable $$ J(y+ \eta)=J(y)+\left.\delta J\right|_{y}(\eta) +o(\|\eta\|_{1,\infty}), \label{2}\tag{2}$$ where $\|\eta\|_{1,\infty} = \|\eta\|_\infty + \|\eta'\|_\infty$, does not hold in general. I have no idea how to show this fact.
My attempt. First, by using Taylor expansion with remainder, I wrote:
$$
L(x,y+\eta ,y'+\eta') = L(x,y,y')+ \bar{L}_y\eta + \bar{L}_z\eta'
$$
with $\bar{L}_y(x,y,\eta),\bar{L}_z(x,y,\eta') \to L_y(x,y,y'), L_z(x,y,y')$ when $\|\eta\|_\infty,\|\eta'\|_\infty \to 0$ respectively.
Then, seeking to obtain \eqref{2}, I substituted the found relation into into original formula, and got the following relation
$$
\begin{split}
|J(y+\eta)-J(y) -\delta J | & =\left| \int\big[(\bar{L}_y-L_y)(x,y,y')\big]\eta(x) + \big[(\bar{L}_z - L_z)(x,y,y')\big]\eta(x)'dx\right|\\
& \le \|\eta\|_{1,\infty}\int|(\bar{L}_y-L_y)(x,y,y')|+|(\bar{L}_z - L_z)(x,y,y')| dx
\end{split}
$$
Now I am stuck in showing that the last right hand term goes to zero when $\|\eta\|_{1,\infty} \to 0$.