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?

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    $\begingroup$ 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. $\lnot Q\to(\exists a)\lnot P(a)$. $\endgroup$ Jan 30 at 23:43

1 Answer 1


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.]

  • $\begingroup$ Ah, so an existential quantifier does actually slip in; I need to review my logic, it seems. Quite a proof structure to throw in within the first few pages of the text! $\endgroup$
    – AtomJZ
    Jan 31 at 20:31

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