# Tagged Questions

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### 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? ...
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### An inequality on $C^1$ periodic functions

Suppose $f \in C^1(\mathbb{R})$ and $f(x + 1) = f(x) \ \forall x \in \mathbb{R}$. Show that $$||f||_{\infty} \leq \int_0^1|f| + \int_0^1|f'|.$$ I have tried using techniques in Fourier Analysis such ...
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### Integral inequality using positive and negative parts

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a measurable (with respect to the Lebesgue measure), $\pi$-periodic function which is Lebesgue integrable over $[0,\pi]$. Moreover assume that ...
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 ...
Given $f:\mathbb R\to \mathbb R^+$ continuous and periodic of period $T\geq 0$, I am asked to prove that $$\int_0^T\frac{f(x)}{f(x+\alpha)}\mathrm dx\geq T,\; \forall \alpha\in\mathbb R.$$ How does ...