In "The Calculus of Variations" by Bruce van Brunt, he says:
Let $J:C^2[x_0,x_1]\to\mathbb{R}$ be a functional of the form $$J(y)=\int_{x_0}^{x_1}f(x,y,y^\prime)dx,$$ where $f$ is a function assumed to have at least second-order continous partial derivatives with respect to $x,y$ and $y^\prime$. Given two values $y_0,y_1\in\mathbb{R}$, the fixed endpoint variational problem consists of determining the functions $y\in C^2[x_0,x_1]$ such that $y(x_0)=y_0$, $y(x_1)=y_1$, and $J$ has a local extremum in $S$ at $y\in S$. Here $$S=\{y\in C^2[x_0,x_1]\,:\,y(x_0)=y_0\mbox{ and }y(x_1)=y_1\}.$$
(...) Suppose that $J$ has a local extremum in $S$ at $y$. For definiteness, let us assume that $J$ has a local maximum at $y$. Then, there is an $\epsilon>0$ such that $J(\widehat{y})-J(y)\le 0$ for all $\widehat{y}\in S$ such that $\|\widehat{y}-y\|<\epsilon$.
However, he doesn't specify which is the norm $\|\cdot\|$. He continues saying:
For any $\widehat{y}\in S$ there is an $\eta\in C^2[x_0,x_1]$, $\eta(x_0)=0=\eta(x_1)$, such that $\widehat{y}=y+\epsilon\eta$, and for $\epsilon$ small Taylor's theorem implies that $$f(x,\widehat{y},\widehat{y}^\prime)=f(x,y,y^\prime)+\epsilon\left\{\eta\frac{\partial f}{\partial y}+\eta^\prime\frac{\partial f}{\partial y^\prime}\right\}+O(\epsilon^3).$$
Thus $$J(\widehat{y})-J(y)=\epsilon\int_{x_0}^{x_1}\left(\eta\frac{\partial f}{\partial y}+\eta^\prime\frac{\partial f}{\partial y^\prime}\right)dx+O(\epsilon^3)$$
By the last, I suppose that this norm is a type of "sup norm", but I don't know.
Can you helpme to deduce which is the norm used? That is, which is the norm from which he is defining the local maximum?