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.
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.