# How to prove $\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx$ [duplicate]

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?

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

• If the condition is dropped, then the claim is false (eg constant functions) – user99914 Apr 28 '14 at 9:31

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 . :)
• I think $f$ might not be odd? – user99914 Apr 28 '14 at 9:36
• $\cos x$ on $[-\pi, \pi]$ – user99914 Apr 28 '14 at 9:59