# Bad proof in Fomin's Calculus of Variations?

I was just reading through the first few pages of Fomin's Calculus of Variations and I came across this proof, which really doesn't seem to prove the Lemma (I may be missing something though) could someone give me a second opinion and perhaps some clarification?
It goes like this;

If $\alpha(x)$ is continuous in [a,b] and if $\int_a^b \alpha(x) h'(x) dx=0$ for every function $h(x)\in D_1(a,b)$ such that h(a)=h(b)=0 then $\alpha(x)=c$ for all x in [a,b], where c is a constant.
Where $D_1(a,b)$ is the space of all once differentiable functions.

Now, here's the given proof;

Let c be the constant defined by the condition $\int_a^b (\alpha(x) - c)dx=0$ and let $h(x) = \int_a^x (\alpha(\xi) - c) d \xi$ so that h(x) automatically belongs to $D_1(a,b)$ and satisfies the conditions h(a)=h(b)=0. Then on the one hand;
$\int_a^b(\alpha(x) - c)h'(x)dx = \int_a^b\alpha(x)h'(x)dx - c (h(b)-h(a))=0$
while on the other hand;
$\int_a^b(\alpha(x)-c)h'(x)dx = \int_a^b(\alpha(x)-c)^2dx.$
It follows that $\alpha(x)-c=0$ for all x in [a,b]

It just seems to me that this only proves the lemma for one specific case and that we've used the 'then' in the proof of the theorem.. Am I wrong in thinking this?

AlephZero