Well, I need some help with the lemmas presented in the title. I didn't understand their demonstration as the authors pick some "adequate functions" while considering all of functions in the class $\mathcal{C}(a,b)$. For example, Lemma 1 says:
If $\alpha(x)$ is continuous in $[a,b]$, and if
$\int_{a}^{b}\alpha(x)h(x)dx=0$
for every function $h(x)\in\mathcal{C}(a,b)$ such that $h(a)=h(b)=0$, then $\alpha(x)=0$ for all $x$ in $[a,b]$.
Proof:
Suppose the function $\alpha(x)$ is also positive in some interval $[x_1,x_2]$ contained in $[a,b]$. Then $\alpha(x)$ is also positive in some interval $[x_1,x_2]$ contained in $[a,b]$. If we set
$h(x) = (x-x_1)(x_2-x)$
for $x$ in $[x_1,x_2]$ and $h(x)=0$ otherwise, then $h(x)$ obviously satisfies the conditions of the lemma. However,
$\int_{a}^{b}\alpha(x)h(x)dx=\int_{a}^{b}\alpha(x)(x-x_1)(x_2-x)dx > 0$
since the integrand is positive (except at $x_1$ and $x_2$). This contradiction proves the lemma.
The definition $h(x)=(x-x_1)(x_2-x)$ in this proof really confused me. My thought is that such proof only demonstrates the existence of this particular function $h(x)$, not considering all $h(x)$ in the class of admissible functions.
Thanks in advance.
The lemma is proved by contraposition : instead of showing $A\implies B$, you can show equivalently that $\neg B \implies \neg A$. In the contrapositive, the universal quantifier $\forall$ becomes $\exists$. Does this help?