# A definite integral inequality

Suppose $$f(x)$$ has continuous derivative on $$[-\pi, \pi]$$, $$\,f(-\pi)=f(\pi)\,$$ and $$\,\int_{-\pi}^{\pi}\, f(x)\, dx=0$$. Then prove that:

$$\int_{-\pi}^{\pi} [\,f'(x)]^2\, dx \ge \int_{-\pi}^{\pi} f^2(x)\, dx,$$ with the equal sign holding if and only if $$\,f(x)=A\cos x+B\sin x$$.

If $$f: [-\pi,\pi]\to\mathbb R$$ is continuously differentiable, then $$f$$ and $$f'$$ are also $$L^2$$, and hence they are expressed as $$f(x)=\sum_{k\in\mathbb Z}\hat f_k\,\mathrm{e}^{ikx} \quad \text{while}\quad f'(x)=\sum_{k\in\mathbb Z}ik\,\hat f_k\,\mathrm{e}^{ikx},$$ and we have that $$\int_{-\pi}^\pi|\,f(x)|^2\,dx=2\pi\sum_{k\in\mathbb Z}|\,\hat f_k|^2 \quad \text{while}\quad \int_{-\pi}^\pi|\,f'(x)|^2\,dx=2\pi\sum_{k\in\mathbb Z}k^2|\,\hat f_k|^2$$
If $$\int_{-\pi}^\pi f(x)\,dx=0$$, then $$\,\hat f_0=0$$, and hence $$\int_{-\pi}^\pi|\,f'(x)|^2\,dx\ge \int_{-\pi}^\pi|\,f(x)|^2\,dx,$$ with the "=" to hold only if $$\hat f_k=0$$, for all $$|k|\ne 0$$, i.e., if $$f(x)=a\cos x+b\sin x$$.