# Is there any justification for a path between two fix points to have a compact support in the derivation of Euler-Lagrange Equation?

In the derivation of Euler-Lagrange equation, as given here (under the "Statement" section, with the name "Derivation of one-dimensional Euler–Lagrange equation"), at the end of the derivation, it is assumed that the arbitrary pertubation path $\eta(t)$ has compact support in order to use the fundamental lemma of calculus of variations.

However, is there any justification for this assumption? Can't a path have non-compact support? If so, why?

• Your link is not to a derivation of the Euler-Lagrange equation. There is one above it for the one-dimensional version, but the $\eta$ there is an arbitrary perturbation, and the fact that $\eta(a) = \eta(b) = 0$ simply comes from the fact that $f + \eta$ has to satisfy the same boundary value requirements as $f$. – Paul Sinclair Jun 26 '18 at 4:03
• @PaulSinclair I did not understand anything from your comment expect the first sentence. – onurcanbektas Jun 26 '18 at 5:32

The derivation takes place on the compact interval $[a,b]$. Any function whose domain is a compact space automatically has compact support.
Since the support of a function is the closure of the set where the function is not $0$, it is a closed subset of a compact space, and is therefore compact itself. The function doesn't even need to be continuous for this to be true.