5
votes
0answers
38 views

If $f$ is $2 \pi$ periodic and $\int_{0}^{2 \pi} f(t) dt=0$ then $\int_{0}^{2 \pi} (f(t))^2 dt \le \int_{0}^{2 \pi} (f'(t))^2 dt$ [duplicate]

Given $f$ a real differentiable function, $2 \pi$ periodic such that $\int_{0}^{2 \pi} f(t) dt=0$ show that: $\int_{0}^{2 \pi} (f(t))^2 dt \le \int_{0}^{2 \pi} (f'(t))^2 dt$. When does equality hold? ...
1
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1answer
81 views

Exercice on periodic function

Let $f$ be a periodic function, $\mathcal{C}^1$ on $\mathbb{R}$ such that: $$\displaystyle\int_0^{2 \pi} f(t) \, dt = 0$$ $$f(2 \pi) = f(0)$$ Prove that $$\forall t \in [0,2 \pi]: \int_0^{2 \pi} ...
1
vote
0answers
101 views

Inequalities of integrals of periodic functions

I have a function that has a shape similar to $\sin(x)^2$ (could be periodic extensions of $(x/(\pi/2))^2$ defined between $-\pi/2$ to $\pi/2$ for example). Let's call it $g(x)$. I want to show that ...