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The fundamental lemma of the calculus of variations, as it is most commonly defined, says that if $g(x)$ is continuous on an interval $[a,b]$ and $$\int_a^b g(x)h(x) dx = 0$$ for an arbitrary infinitely differentiable $h(x)$ s.t. $h(a) = h(b) = 0$, then $g(x) = 0$ on the entire interval.

Is a stronger version of this lemma, which removes the requirement that $h(a) = h(b) = 0$, also true?

I say stronger version because the hypothesis that "$h(x)$ is an arbitrary differentiable function" is a weaker statement than "$h(x)$ is an arbitrary differentiable function with $h(a) = h(b) = 0$", and a theorem with weaker hypotheses (and the same conclusion) is stronger. Please correct me if I am wrong in this understanding.

Blahblahblacksheep
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    You are not taking the quantifier into consideration. "$h(x)$ is an arbitrary differentiable function" is weaker than "$h(x)$ is an arbitrary differentiable function with [additional conditions]" -- therefore the set of things in the first class is larger. Therefore asking that $g$ satisfy a condition for all such functions is a stronger hypothesis in the first case. – JBL Mar 26 '23 at 01:29
  • You can always change $h(a)=h(b)=0$ to $h(a) = h(b) = h(\frac {a+b}{2})=0$, or that $h(c)=0$ for all $c\in F$, where $F$ is a finite set. Whether or not it is useful is up for debate. – Arctic Char Mar 26 '23 at 01:33
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    @JBL Oh I think I see what you mean! IF I understand correctly, $\int g(x) h(x) dx = 0$ for $h(x)$ any arbitrary function $\implies$ $\int g(x) h(x) dx = 0$ for $h(x)$ such that $h(a) = h(b) = 0$. So what I am proposing is actually a weaker theorem, which follows from the stronger theorem defined usually and hence is true? – Blahblahblacksheep Mar 26 '23 at 01:43
  • Yes, that's right. – JBL Mar 26 '23 at 01:46
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    @JBL Ah perfect thanks so much. If you want to make that an answer, I'd be happy to accept it. – Blahblahblacksheep Mar 26 '23 at 02:00
  • Done, as suggested. – JBL Mar 26 '23 at 14:35

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The condition "$h(x)$ is infinitely differentiable" is a weaker restriction on the function $h(x)$ than the condition "$h(x)$ is infinitely differentiable and $h(a) = h(b) = 0$", i.e., more functions satisfy the first condition than the second. However, the statement of the theorem is that $g(x)$ should satisfy a certain condition for each function $h$. Therefore, your version of the theorem requires that $g$ satisfy extra conditions (including those in the standard version, but also others). This makes it a stronger hypothesis on $g$ (in order to get the same conclusion), and consequently makes your version of the theorem weaker than the standard version.

JBL
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