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The Fundamental Lemma of Calculus of Variation says that if a continuous function $f$ on an open interval $(a,b)$ satisfies the equality

$$\int_{a}^{b} f(x) h(x) = 0$$

for all compactly supported smooth functions $h$ on $(a,b)$ then $f$ is identically equal to zero.

Is it possible to weaken this statement by requiring this to only be true where $h$ is a polynomial by using Stone-Weierstrass? I think I am misunderstanding something because as far as I can tell it just follows immediately and I feel like the theorem would be stated like this instead if it were true.

JessicaK
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    Note that since the domain is open, to make sure that the integral makes sense you need to only integrate over some closed subinterval (so it is compact). A priori you cannot guarantee that the integral makes sense for a general polynomial and a continuous $f$. This is why the $h$ have compact support. – copper.hat Feb 26 '16 at 01:51
  • I am having difficulty understanding why the integral would not make sense. The smoothness of $h$ seems strong enough to remove any weird cases I could think of that would break down the argument. – JessicaK Feb 26 '16 at 02:03
  • $f$ may be unbounded. For example, $f(x) = {1 \over x-a}$ lies in $C(a,b)$. The integral is undefined for any polynomial that does not have $(x-a)$ as a factor. – copper.hat Feb 26 '16 at 02:06

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It wouldn't be a a weakening, since polynomials are not compactly supported. Notice also that you really desire boundary values to be zero, since otherwise you cannot derive the differential form of Euler-Lagrange equation (getting rid of boundary terms during integration by parts).

Lorenzo Cecchi
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