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We can find on the web several forms of the fundamental lemma of calculus of variation, the simplest one I could find (French wikipedia ) is: for $f\in C^1([a, b])$

$$ \int_a^b f(x) g(x) dx = 0, \quad \forall g\in C^1([a, b]), \quad g(a)=g(b)=0 \quad \Leftrightarrow \quad f(x)=0, \quad \forall x\in [a, b].$$

After studying the proof I have two questions

  • why assuming derivability for $g$ (sometimes we even see $C^\infty([a, b])$)?
  • why do we need $g(a)=g(b)=0$ (sometimes we even see $C_c([a, b])$?

It seems to me that we need neither of them. Is it because it is a lemma (and not a theorem) and have premisses coming from another context (for example to anticipate an integration by part)?

KevMoriarty
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  • Hmmmm, is $f(x)=\begin{cases} 1\quad x=c\in [a, b] \ 0 \quad \text{otherwise}\end{cases}$ not a counterexample? Or need $f$ be continuous? – user160110 Feb 22 '16 at 17:37
  • Right, $f$ need to be continuous. – KevMoriarty Feb 22 '16 at 17:51
  • g is usually the variation associated to the function f, it details how f varies from the optimal path, since f is defined at specific points a and b, the variation there must be zero – Triatticus Feb 22 '16 at 18:57
  • g needs to be differentiable because differentiability is stronger than continuity otherwise you could get complicated paths from point a to b which are not continuous – Triatticus Feb 22 '16 at 18:58

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