Bad proof in Fomin's Calculus of Variations?

In summary, the conversation discusses a proof in Fomin's Calculus of Variations, which aims to prove a lemma about continuous functions. The proof involves using a particular function h(x) that satisfies certain conditions, and then showing that for this function, a specific integral must be equal to zero. The conversation also discusses the implications of this proof and raises questions about its applicability to other cases.
  • #1
genericusrnme
619
2
I was just reading through the first few pages of Fomin's Calculus of Variations and I came across this proof, which really doesn't seem to prove the Lemma (I may be missing something though) could someone give me a second opinion and perhaps some clarification?
It goes like this;

If [itex]\alpha(x)[/itex] is continuous in [a,b] and if [itex]\int_a^b \alpha(x) h'(x) dx=0[/itex] for every function [itex]h(x)\in D_1(a,b)[/itex] such that h(a)=h(b)=0 then [itex]\alpha(x)=c[/itex] for all x in [a,b], where c is a constant.
Where [itex]D_1(a,b)[/itex] is the space of all once differentiable functions.

Now, here's the given proof;

Let c be the constant defined by the condition [itex]\int_a^b (\alpha(x) - c)dx=0[/itex] and let [itex]h(x) = \int_a^x (\alpha(\xi) - c) d \xi[/itex] so that h(x) automatically belongs to [itex]D_1(a,b)[/itex] and satisfies the conditions h(a)=h(b)=0. Then on the one hand;
[itex]\int_a^b(\alpha(x) - c)h'(x)dx = \int_a^b\alpha(x)h'(x)dx - c (h(b)-h(a))=0[/itex]
while on the other hand;
[itex]\int_a^b(\alpha(x)-c)h'(x)dx = \int_a^b(\alpha(x)-c)^2dx.[/itex]
It follows that [itex]\alpha(x)-c=0[/itex] for all x in [a,b]

It just seems to me that this only proves the lemma for one specific case and that we've used the 'then' in the proof of the theorem.. Am I wrong in thinking this?

Thanks in advance! :biggrin:
 
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  • #2
The proof look OK to me. The point is that if the condition is true for EVERY function h(x), then it must be true for one PARTICULAR function h(x) that you can invent. You pick a function such that the integral is ##\int_a^b (a(x)-c)^2\,dx >= 0## and it can only be 0 if ##a(x) = c##.

You will find the same type of argument quite often in calculos of variations: if something is true for all functions meeting some condition, it must be true for the "worst case" that you can invent.

FWIW the converse argument is trivial. If ##a(x) = c##, then ##\int_a^b a(x)h'(x)\, dx = c(h(b) - h(a)) = 0##
 

1. What is Fomin's Calculus of Variations?

Fomin's Calculus of Variations is a mathematical theory that deals with finding optimal solutions to problems involving variations in functions. It is commonly used in physics and engineering to solve problems involving optimization.

2. What is considered "bad proof" in Fomin's Calculus of Variations?

Bad proof in Fomin's Calculus of Variations refers to a proof that contains logical errors, incorrect assumptions, or incomplete explanations. These types of proofs can lead to incorrect solutions and can undermine the validity of the theory.

3. Why is it important to avoid bad proof in Fomin's Calculus of Variations?

Avoiding bad proof in Fomin's Calculus of Variations is crucial because it ensures the accuracy and reliability of the theory. Bad proof can lead to incorrect solutions and can hinder progress in solving complex problems in physics and engineering.

4. How can one identify bad proof in Fomin's Calculus of Variations?

Identifying bad proof in Fomin's Calculus of Variations requires a careful examination of the proof for any logical errors, incorrect assumptions, or incomplete explanations. It is also important to compare the proof with other established proofs to ensure consistency.

5. What can be done to prevent bad proof in Fomin's Calculus of Variations?

To prevent bad proof in Fomin's Calculus of Variations, it is important to thoroughly understand the theory and its principles. Additionally, seeking feedback and constructive criticism from peers and experts can help identify and correct any potential errors in the proof.

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