In this link, the second fundamental lemma of calculus of variation (Lemma 2 in the link):
If $\alpha(x)$ is continuous in $[a, b]$, and if $$ \int_a^b \alpha(x) h^{\prime}(x) d x=0 $$ for every function $h(x) \in \mathscr{D}_1(a, b)$ such that $h(a)=h(b)=0$, then $\alpha(x)=c$ for all $x$ in $[a, b]$, where $c$ is a constant.
... is proved by constructing $h(x)$ and $c$ for a given $\alpha(x)$ function, and claims that this proves the lemma.
This doesn't make sense, since the lemma claims that the above holds for every function that is continuous & continuously differentiable within [a,b] satisfying the boundary constraints, not just a specifically constructed h(x).
Am I misunderstanding the proof?