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This question already has an answer here:

Let $f$ be $C^1$ in $[-\pi, \pi]$ and satisfies $\int_{-\pi}^\pi f(x)dx=0$, periodic boundary condition.

Then, prove that $\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx$.

I try to prove it by Parseval's equality(with $X_n=\sin nx$) and Schwartz inequality,

but then some constants come out. Also why condition $\int_{-\pi}^\pi f(x)dx=0$ needs?

Give some advice.

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marked as duplicate by Norbert, user99914, Etienne, Git Gud, M Turgeon Apr 28 '14 at 12:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If the condition is dropped, then the claim is false (eg constant functions) $\endgroup$ – user99914 Apr 28 '14 at 9:31
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This is known as Wirtinger's inequality. See the wikipedia article for its proof.

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Consider the function ${f(x)-f'(x)}^2$ . Integrate this function from $-\pi$ to $\pi$ . As this is a positive function this definite integral is greater than or equal to zero . Now expand the ${f(x)-f'(x)}^2$ using $(a-b)^2$ . In the question he meant that $f(x)$ is an odd function which means $f'(x)$ is an even function . The product $f(x).f'(x)$ is also an odd function and when you integrate, this the term becomes zero . Arrange the other two terms you will get the result . :)

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  • $\begingroup$ I think $f$ might not be odd? $\endgroup$ – user99914 Apr 28 '14 at 9:36
  • $\begingroup$ The condition ∫π−πf(x)dx=0 indicates that the function is an odd function . $\endgroup$ – user146334 Apr 28 '14 at 9:38
  • $\begingroup$ There are functions that are not odd and satisfies that.... $\endgroup$ – user99914 Apr 28 '14 at 9:39
  • $\begingroup$ Can you please give an example which satisfy the condition , integration from -a to +a of a function is zero and the function is not odd $\endgroup$ – user146334 Apr 28 '14 at 9:46
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    $\begingroup$ $\cos x$ on $[-\pi, \pi]$ $\endgroup$ – user99914 Apr 28 '14 at 9:59

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