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?(adsbygoogle = window.adsbygoogle || []).push({});

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Bad proof in Fomin's Calculus of Variations?

**Physics Forums | Science Articles, Homework Help, Discussion**