# Quantifiers and contrapositives in proof of fundamental lemma of variational calc

In Gelfand and Fomin's Calculus of Variations, they present the lemma

If $$\alpha(x)$$ is continuous in $$[a, b]$$, and if $$\int_a^b \alpha (x) h(x) \textrm{d}x = 0$$ for every continuous function $$h(x)$$ such that $$h(a) = h(b) = 0$$, then $$\alpha(x) = 0$$ for all $$x \in [a, b]$$.

The authors claim that the proof that follows is by contradiction, but from what I can tell, it appears to be a proof by contrapositive---yet, I can't quite grasp it. They assume $$\alpha(x) > 0$$ on some interval $$[x_1, x_2] \subset [a, b]$$ and set $$h(x) = (x - x_1)(x - x_2)$$ in $$[x_1,x_2]$$ (but equal to zero elsewhere $$[a, b]$$), then show that the integral is also greater than zero.

In a rough first-order sense, the lemma seems to have the form $$(\forall a)(P(a) \rightarrow Q)$$, with the integral being $$P$$, the quantified variable representing all functions $$h$$ that satisfy the given conditions, and $$Q$$ being the state of $$\alpha = 0$$. But it seems to me that the authors prove $$(\exists a)(\neg Q \rightarrow \neg P(a))$$, when the contrapositive proof should have a $$\forall$$ quantifier instead.

Shouldn't the proof involve $$h$$ being literally all/any possible functions that satisfy the necessary properties? It seems like picking one specific function proves nothing at all, and I can't see why it works to just pick one function. Did I mess my first-order logic up?

• The statement is of the form $(\forall a)P(a)\to Q$, and the authors are proving the contrapositive $\lnot Q\to\lnot(\forall a)P(a)$, i.e. $\lnot Q\to(\exists a)\lnot P(a)$. Jan 30 at 23:43

The lemma is of the form $$(\forall\alpha)\big( \big( (\forall h)P(\alpha,h) \big) \implies Q(\alpha) \big)$$, and thus the contrapositive of the implication would yield $$(\forall\alpha)\big( \lnot Q(\alpha) \implies \big( (\exists h)(\lnot P(\alpha,h)) \big) \big)$$. And indeed this is what the proof demonstrates: given a suitable (but generic) $$\alpha$$, it constructs an $$h$$. [Side note: $$\lnot Q(\alpha)$$ literally says that there's a single point at which $$\alpha(x)\ne0$$, but then WLOG $$\alpha(x)>0$$ and then continuity provides an interval containing $$x$$ on which $$\alpha$$ is positive.]