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Let $f\in C[0,1]$ be such that $\forall \phi\in C[0,1]$ with $\int_0^1\phi \, dx=0$,we have $\int_0^1f\phi \, dx=0$. Then $f$ is constant.

May be someone can help me proof this problem. Thanks!

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up vote 3 down vote accepted

Let $c=\int_0^1fdx$ and let $\phi=f-c$. Then $\phi\in C[0,1]$ and $\int_0^1\phi dx=0$. Therefore, $$0\le \int_0^1\phi^2 dx=\int_0^1f\phi dx-c\int_0^1\phi dx=0.$$ It follows that $\phi\equiv 0$, i.e. $f\equiv c$.

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To richard: Thank you very much! The answer is what I want to find out! – Darry Oct 27 '12 at 0:09

Here's an outline for how to proceed. If $f$ is not constant, then $f$ has a maximum value $M$ that is strictly larger than its minimum value $m$. Assume $f(x_1) = M$ and $f(x_2) = m$. Choose $\phi$ to be positive in a small neighborhood of $x_1$, negative in a small neighborhood of $x_2$, and zero everywhere else, so that $\phi$ is continuous and $\int_0^1 \phi \, dx = 0$. Then argue that $\int_0^1 f \phi \, dx > 0$. (This will dictate how small to make your neighborhoods.)

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To Michael Joyce: Thank you! Your outline makes me believe the result $f=constant$ must be right in intuition. – Darry Oct 27 '12 at 0:15

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