Talk:Fundamental lemma of calculus of variations

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This article incorporates material from PlanetMath, which is licensed under the GFDL.

Should the fact that the lemma is a necessary but not sufficient condition ( => ) for the functional extremal clearly stated?

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[edit] Proof of the principle

I think the proof is important - some students use wikipedia to help them understand what they learn in lectures better. Plenty of other pages have proofs: for example Noether's theorem, (and it is one of the main reasons I use wikipedia). This is a quite a short proof. Oh yeah, and I missed out the details of h(x) because they were written in the statement of the lemma (h ∈ C2[a,b] with h(a) = h(b) = 0)


The proof is by contradiction:

\int_a^b f(x)h(x) dx = 0, \forall \, h(x): h(a)=h(b)=0.

Assume that for some c in the interior of the interval one has f(c) = 2e > 0.

By continuity, and the intermediate value theorem, there exists a neighbourhood [c0,c1] of c within [x0,x1] on which f(x) > e. Then,

\int_{c_0}^{c_1} f(x)h(x) \, dt > e\int_{c_0}^{c_1} h(x) > 0.

That gives a contradiction: therefore the only way for the integral to be zero in general is if

f(x)=0 \forall x \in [x_0, x_1].
Your proof is still wrong. How do you know that
\int_{c_0}^{c_1} h(x) > 0.
also, how do you know that
\int_{c_0}^{c_1} f(x)h(x) \, dt > e\int_{c_0}^{c_1} h(x)?
The function h needs to be chosen such that h(a) = h(b) = 0 but more is needed. It must be non-negative, and positive only in the small interval [c0,c1]. Why does such a function exist? Things are a bit more complicated than what you wrote. Oleg Alexandrov (talk) 16:15, 29 March 2006 (UTC)
It is not hard to find such an h, but the proof above is certainly sloppy. The first inequality is just wrong, and there is no need for the Intermediate Value Theorem. -cj67

For smooth f doesn't this simple proof work? Let r be any smooth function that's 0 at a and b and positive on (a, b); for example, r = − (x − a)(x − b). Let h = rf. Then h satisfies the hypotheses, so

0 = \int_a^b f h \; dx = \int_a^b r f^2 \; dx.

But the integrand is nonnegative, so it must be identically 0. Since r is positive on (a, b), f is 0 there and hence on all of [a, b]. Joshua R. Davis 04:49, 25 March 2007 (UTC)

[edit] Definition of Functional

Functional J is a functional of Lagrangian NOT the dependent variable y ! --mcyp 11:35, 24 January 2007 (UTC)

[edit] External Links

I have removed the link to a web page that only contains a proof of the Euler-Lagrange equation, but not of the lemma. I'd guess that some sort of Hilbert space basis is needed for a proof, but I haven't actually seen any proof of this lemma... Anyone have a good book handy? --Shastra 20:10, 15 July 2006 (UTC)

[edit] Continuity assumption

71.165.190.99 just changed the hypothesis from f smooth to f continuous. Now the given proof doesn't work, because rf is not necessarily smooth. The proof can be edited to work as long as f is Ck and for all h in Ck the integral of fh is 0. That is, the smoothness on f and h must match. The given proof is nice for Wikipedia because it's so simple. The lemma can be stated more generally (see PlanetMath's version), but then the proof is more demanding. What version should we state and prove? Joshua R. Davis (talk) 20:35, 8 May 2008 (UTC)