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Let $f$ be a continuous function on $[0,1].$ Suppose that $\int_0^1 f(x) g(x) dx = 0$ for every integrable function $g(x)$ on $[0,1].$ Prove that $f(x) \equiv 0$ on $[0,1]$

This proof is easy to write out if $g(x) \ge 0.$ If it is integrable on $[0,1]$ is that implied?

mfl
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Travis
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  • your question is not clear $f(x) \equiv !$??? – Anurag A Jun 22 '14 at 22:29
  • right, it's unclear because the way your question is stated you can pick any $g$ you want. did you mean $f \geq 0$? anyways integrability does not imply that a function is nonnegative – mm-aops Jun 22 '14 at 22:45

1 Answers1

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I don't think you need the First Mean Value Theorem for integrals to conclude that $f=0$.

Hint: Consider $g=f$.

user71352
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