Talk:Fundamental lemma of calculus of variations

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

Proof of the principle[edit]

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 [c_0, c_1] of c within [x_0, x_1] 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 [c_0, c_1]. 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 = r f. 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)

This is supposed to work for any f(x) and h(x). But by stating that h = r f and that r is positive on (a, b), aren't you restricting f(x) to having the same sign as h(x) on (a, b)? So this proof doesn't work. ---- Yaxy2k (talk) 06:33, 28 April 2010 (UTC)
No; I think that the proof is correct. The theorem is about a given function f, that has a special property: for all functions h, a certain integral is 0. So, if f is such a function, then the integral is 0 for any h that we care to talk about. So let's talk about the function h = r f. The integral must be 0 for that h (and infinitely many others, that we don't care about), so the proof can proceed.
In other words, I think you're reading the statement of the theorem as "for all f, for all h, (integral is 0 implies f is 0)". But the statement of the theorem is "for all f, (for all h integral is 0) implies f is 0". Mgnbar (talk) 12:42, 28 April 2010 (UTC)
Ok, I understand now. Thank you for pointing out my error! ----Yaxy2k (talk) 02:22, 29 April 2010 (UTC)

I am no expert in maths, but is there an issue with assuming h=rf? Because, looking at the conditions on r, it says that h can never be 0 except at endpoints unless f is 0 at that point. Please remove this if I am wrong. — Preceding unsigned comment added by (talk) 04:20, 15 May 2014 (UTC)

I do not see any issue. You are right that h is zero at the endpoints and wherever f is zero, and nowhere else. That is not a problem. The proof goes on to show that f and h are zero everywhere. Mgnbar (talk) 13:46, 15 May 2014 (UTC)

Definition of Functional[edit]

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

External Links[edit]

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)

Continuity assumption[edit] just changed the hypothesis from f smooth to f continuous. Now the given proof doesn't work, because r f is not necessarily smooth. The proof can be edited to work as long as f is C^k and for all h in C^k the integral of f h 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) ≤

Is the proof really more demanding? I am not sure what exactly is gained by the continuity assumption at all. It seems that what one really needs to prove in the final result is that \int_a^b f h dx exists, which is true if f,h are locally integrable on [a,b]. After the substitution h \leftarrow (x - a)(b - x)f, the integrand in the equation is strictly greater than or equal to 0. Nothing really needs to change if all of the instances of C^k are replaced with the weaker assumption of local integrability and the theorem is changed to 'f is 0 almost everywhere'. —Preceding unsigned comment added by (talk) 05:58, 21 January 2009 (UTC)

But then (if f is smooth, say, as it often is) the result is much weaker, since in your version our hypothesis is that the integral of fh is zero for all locally integrable h, while in the current version the hypothesis only involves smooth h. Algebraist 23:14, 22 March 2009 (UTC)

Small correction for correctness[edit]

Putting this in discussion part as I am not sure whether this requirement is already included by h(x)\in C^k. But I think it is neccessary to require \exists x\in[a,b]:h(x)\neq 0 i.e. h(x) cannot be constant zero for the lemma to hold. —Preceding unsigned comment added by (talk) 01:03, 16 January 2011 (UTC)

The statement of the lemma is correct as it stands. The conclusion of the lemma holds if a certain property is satisfied for every h, including the h that is identically 0 and many other hs. The lemma makes no claim about f, if the property holds only for the h that is identically 0. Mgnbar (talk) 01:57, 16 January 2011 (UTC)


I got bold and rewrote it, after some discussion on Wikipedia talk:WikiProject Mathematics#Fundamental lemma of calculus of variations. Hope it is better. You are invited to improve it further. (Really, I intend to do some more soon.)

About the proof: if you want to restore it, do so. But we should not restrict the formulations to the very weak case according to the very simple proof. Just formulate the relevant simple case and prove it, if you like.

About Section "Applications": I doubt it is correct, but for the moment I am tired; let it stay as it was, and then we'll see.

About the old refs: restore them if you like (but please keep my refs). Boris Tsirelson (talk) 20:10, 2 June 2015 (UTC)