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Fundamental lemma in calculation of variation states that $\int_a^b f(x)g(x) dx =0$ for $f$ continuous in $(a, b) $ and and $g(x)$ which has compact support and is continuous in $(a,b)$ , then $f(x)=0$ $\forall x\in (a,b)$ . I found out the proof of this theorem is easy . But i have some doubts related to it if i want to extend this concept in heigher dimensions and to different class of functions .

a) Instead of interval $(a,b) $ if i take a set $\Omega \subset R^n$ , then what do i have to think about here, to prove the above theorem ?

b) Is it necessary that $g(x)$ should be infinitely differentiable and compactly supported ?

c) If $\int_a^b f(x) g'(x) =0$ with $g(x) , f(x) \in C^1[a,b]$ and $g(a)=g(b)=0$ . Then $f(x)$ is constant but does it still remain the same if i take $f(x)\in C^0[a,b]$ and $g(x) \in C_c^1[a,b]$ ?

A bit vague question : If i want to extend this idea to large class of functions what are the important things that i need to go about ? Thank you for your help .

What i am thinking about $c$ is to use mollification but the problem is it will be valid only almost everywhere, which i am not interested . is it possible after all ?

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I can't quite follow what you're asking in (c). But see my response about the other two. – coffeemath Oct 18 '12 at 2:25

It needs to be included in the theorem statement that the statement $\int_a^b f(x)g(x) dx =0$ is to hold for $every$ allowable $g$. The idea of it is to fix a particular point $x$ in the interior of the domain of $f$, and based on that, and how $f$ behaves near $x$, you then select your $g$ to have a bump nearly 1 at $x$, and tapering off quickly enough that the behavior of $f$ away from $x$ puts only a negligible amount into the integration of $f(x)g(x)$ over the region.

When I've seen it applied, $g$ is taken to be actually zero outside of a ball about the point $x$. Then the size of this ball is shrunk to zero, at the same time keeping the $g$'s each individually continuous. After all this, if $f(x)$ were not zero we'd get a contradiction in the limit.

I gather you understand what I just wrote already. If so then it seems for your question (a) it should be clear it works in $n$ dimensions, especially if you use boxes instead of balls around the $x$ for ease of notation/calculation.

For $b$ the answer is: absolutely not! We can select our $g$ ourselves to be anything we want, as long as the relevant properties of integration of the product work. Certainly continuous is enough. And I guess the compact support hypothesis is there so the product integrals exist. If the given funcion $f$ has an unbounded domain, then integrating it against another function of unbounded domain might give no value. But if $g$ has compact domain the integration of the product $f(x)g(x)$ can be done over that domain, and so will converge provided $g$ is also continuous. [It's already assumed that $f$ is continuous for this application]

The only place I ever heard talk of integrating against functions which had to be $C^\infty$ with compact support was in a quite abstract course about distribution theory, in which things like the Dirac delta function had rigorous meaning. That course was over my head at the time...

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