# 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

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)

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

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 182.64.210.74 (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

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