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I am doing some reading on the calculus of variations and one of the first examples uses the following theorem:

Let $f\in C[a,b]$. If $\int_a^b f(x)\varphi(x)dx=0$ for all $\varphi\in C[a,b]$, then $f\equiv 0$.

Intuitively this is clear. But I am not sure how to prove this. I was thinking of using the mean value theorem for integrals. Since $f,\varphi$ are continouus they are integrable and hence there exists $\xi\in [a,b]$ such that

$$0=\int_a^b f(x)\varphi(x)dx=f(\xi)\int_a^b\varphi(x)dx$$

Assuming $f\equiv\hspace{-1.5ex}|\hspace{1ex} 0$, we can assume $f(\xi)\neq 0$ and hence get $\int_a^b\varphi(x)dx$, but this definitely does not hold for all contiouus functions on $[a,b]$.

Thus $f\equiv 0$ on $[a,b]$.

Is this a way to argue? Or how can I prove this differently? Thanks!

Harry Peter
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dinosaur
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    Let $\varphi=f$.${}$ – David Mitra Feb 20 '14 at 12:21
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    You might find this useful http://math.stackexchange.com/questions/16831/nonzero-f-in-c0-1-for-which-int-01-fxxn-dx-0-for-all-n – Ludolila Feb 20 '14 at 12:23
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    In addition to David Mitra's comment, recall that if $g$ is continuous at a point, $a$, and $g(a)>0$, then there is some nontrivial interval surrounding $a$ where $g(x)>0$. – J126 Feb 20 '14 at 12:23
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    then note a positive function whose integral is 0, must be 0 almost everywhere, then use continuity – Lost1 Feb 20 '14 at 12:23
  • Thanks everyone! This really isn't as complicated to prove as I first thought it was. – dinosaur Feb 20 '14 at 12:26

1 Answers1

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Take $\varphi :=f$.

Then $\int_a^b f(x)^2 dx=0$

$f^2$ is continuous, positive. If it's non zero at a point $c \in [a,b]$, there's an interval $]c-\epsilon,c+\epsilon[ \subset [a,b]$ such that $f^2$ is strictly positive over $]c-\epsilon,c+\epsilon[$ . This contradicts $\int_a^b f(x)^2 dx=0$.

Hence $f^2=0=f$

Gabriel Romon
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