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The basic form of the fundamental lemma of the calculus of variations states:

Let $f$ be a continuous function in an interval $[a,b]$ then:

1) $\forall x \in ]a,b[.f(x) = 0 $

2) $\forall \phi \in C_0^1(a,b).\int_{a}^{b} f(x)\phi(x) dx$

are equivalent.

My problem comes with $C_0^1(a,b)$, the set of smooth functions with compact support.

According to Wikipedia compact suport means that the support $supp(\phi) = \overline{\{ x \in ]a,b[:\phi(x) \neq 0\}}$ is compact. For my professor the definition of support is: $$supp(\phi) = \Big(\cup \{B \subseteq \Omega \text{ open }:\forall x \in B.\phi(x) = 0\}\Big)^c$$

Is it evident that the two notions are equivalent? (I will have to come back to the question to finish it myself but probably direct definitions give this equivalence)

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  • $\begingroup$ In your second definition, don't you have to take the complement of this set? $\endgroup$ – Bremen000 Feb 25 '18 at 13:30
  • $\begingroup$ BTW usually, in a topological space, one takes the closure of the set in the first definition and then the two definitions are the same. $\endgroup$ – Bremen000 Feb 25 '18 at 14:10
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The answer is with basic logic and topology arguments:

The first set can be written:

$\{y:\forall O \text{ open }.y \in O \implies O \cap \{x \in \Omega:\phi(x) \neq 0\} \neq \emptyset\}$

equivalently:

$\{y:\forall O \text{ open }.y \in O \implies \exists z \in O.\phi(z) \neq 0\}$

The second set can be written:

$\{y:y \notin \cup \{B \text{ open }: \forall x \in B.\phi(x) = 0\}\}$

equivalently:

$\{y:\forall B \text{ open }. y \in B \implies \lnot \forall x \in B.\phi(x) = 0\}$

equivalently:

$\{y:\forall B \text{ open }. y \in B \implies \exists x \in B.\phi(x) \neq 0\}$

Thus, the two sets are equal.

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