- #1

genericusrnme

- 614

- 2

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!