# Derivation of Euler-Lagrange equation: locally stationary in what sense?

I previously learned the smallest amount of calculus of variations from a physics course. I understand most part of the derivation of Euler-Lagrange equation except one or two line.

A derivation often starts by assuming that a function is a locally stationary point of a functional. What does it mean by locally stationary? Obviously we need topology to talk about "local", but it is insufficient. For example, the function $\Bbb R\to\Bbb R$, $x\mapsto x^3$ is locally stationary at $0$ by considering $\Bbb R$ as a normed space, but "locally stationary" is not defined if we consider $\Bbb R$ as purely a topological space.

Do we need some structure on the function space where we can do an analog of Taylor series expansion? I thought of this because I saw from the notes from the physics course with a sentence goes "there should be no linear terms". I would like to see some formal definitions of this "locally stationary" instead of informal arguments.

• When you say locally stationary, where is always a function in mind. Yes, you cannot just think of $\mathbb R$ alone. It's $X=\mathbb R$ together with a map $f :X\to \mathbb R$ where you can say a point $x\in X$ is locally stationary. – user99914 Aug 11 '17 at 12:21
• I would say that a functional $F : C^k \to \mathbb R$ is locally stationary at $f$ if $$\left. \frac{d}{d\lambda} F[f+\lambda \phi] \right|_{\lambda=0} = 0$$ for all $\phi \in C_c^k,$ where ${}_c$ stands for compact support, i.e. that $\phi=0$ outside a compact set. – md2perpe Aug 11 '17 at 15:18