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Let me preface this question by saying I'm actually a physicist, not a mathematician, so a lot of the language I see you guys using here is over my head, so if you can keep it simple, that would be preferable.

The fundamental lemma of calculus of variations states that, if $f\in C^k[a,b]$, $\int_a^b f(x) h(x) dx = 0$, and $h(a) = h(b) = 0$, then $\forall h \in C^k[a,b]$: $f(x) = 0$ $\forall x \in [a,b]$.

The proof of this usually goes something along the lines of assuming some specific form of $h(x)$, something like $h(x) = r(x)f(x)$ where $r(x) > 0$ on $(a,b)$ and $r(a) = r(b) = 0$, substituting this into the integral, and then proving that this for the integral to remain zero, this requires that $f(x) = 0$, and proof completed. My confusion is this: does this not only mean that the theorem holds for some small subset of $h$'s, namely, those fitting the form assumed to replicate the proof, rather than the general class of $h$'s proposed in the lemma above? The proof assumes that if you can demonstrate $f(x) = 0$ for some small subset of $h$'s, then it works for a larger class of $h$'s, which I have a hard time seeing as justified. What am I missing here?

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You're quoting the lemma wrong. It should be something like

Assume $f\in\mathcal C^k[a,b]$ and that for all $h\in\mathcal C^k[a,b]$ which is zero at the endpoints it holds that $\int_a^b f(x)h(x)\,dx=0$. Then $f(x)=0$ for all $x\in[a,b]$.

In other words the $\forall h$ is in the assumptions of the lemma, not the conclusion.

The fact that only a certain few $h$s are used in the proof just means that the lemma assumes more than it strictly needs to -- which is to say that it promises less than it can keep. That can't make it less true than it would be if it listed precisely those $h$ that it needed the premise to hold for.

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  • $\begingroup$ I understand that $\forall h$ is in the assumptions of the lemma, but when you conclude $f(x) = 0$ you only do so assuming certain things for $h$. In other words $f(x) = 0$ under more strict considerations than when the integral is 0. $\endgroup$
    – Joe Horne
    Jan 15, 2015 at 18:56
  • $\begingroup$ @Joe: No, at the time you conclude $f(x)=0$, there is no $h$ anymore because this conclusion is outside the scope of the quantifier $\forall h$. In order to use the theorem you need to supply it with proof that such-and-such hold for all $h$, whether or not they have the special form the proof actuall uses. $\endgroup$ Jan 15, 2015 at 18:59
  • $\begingroup$ How can the conclusion be outside the assumptions? And that's what I'm saying, we haven't proven it for all $h$, just those of the special form the proof actually uses. $\endgroup$
    – Joe Horne
    Jan 15, 2015 at 19:19
  • $\begingroup$ @JoeHorne: Proving it for all $h$ is something the user of the lemma has to do. The proof of the lemma gets to take it for granted for every $h$ because it's an assumption! $\endgroup$ Jan 15, 2015 at 19:26
  • $\begingroup$ In some sense you can view the lemma as having infinitely many different assumptions, one for each possible $h$ -- and they must all be true before the lemma yields any conclusion to you. If you're looking to use the lemma, you need to establish the each of those assumptions. If it turns out that the proof of the lemma only uses some of its assumptions, that doesn't make the conclusion any less true. It just means that the user may have done more work than he strictly needs to do. $\endgroup$ Jan 15, 2015 at 19:28

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