Let \(f:[0,2\pi]\to\mathbb R\) be twice differentiable and satisfy \[ \int_0^{2\pi}f(x)\,dx=0, \qquad f(2\pi)=f(0). \] Prove that \[ \int_0^{2\pi}f(x)^2\,dx \le \int_0^{2\pi}f'(x)^2\,dx. \]
Solution
For \(n\geq1\), define the real Fourier coefficients \[ a_n = \frac1\pi\int_0^{2\pi}f(x)\cos(nx)\,dx, \qquad b_n = \frac1\pi\int_0^{2\pi}f(x)\sin(nx)\,dx, \] and put \[ a_0=\frac1\pi\int_0^{2\pi}f(x)\,dx. \]
1. One has\[ a_0=0. \]
This is exactly the zero-mean hypothesis.
2. For every \(n\geq1\), the cosine and sine Fourier coefficients of\(f'\) are respectively\[ n b_n \qquad\text{and}\qquad -n a_n. \]
2.1. The cosine coefficient of \(f'\) is \(n b_n\).
Integration by parts and \(f(2\pi)=f(0)\) give \[ \begin{aligned} \frac1\pi \int_0^{2\pi}f'(x)\cos(nx)\,dx &= \frac1\pi \left[f(x)\cos(nx)\right]_0^{2\pi} + \frac n\pi \int_0^{2\pi}f(x)\sin(nx)\,dx\\ &= n b_n. \end{aligned} \]
2.2. The sine coefficient of \(f'\) is \(-n a_n\).
Again by integration by parts, \[ \begin{aligned} \frac1\pi \int_0^{2\pi}f'(x)\sin(nx)\,dx &= \frac1\pi \left[f(x)\sin(nx)\right]_0^{2\pi} - \frac n\pi \int_0^{2\pi}f(x)\cos(nx)\,dx\\ &= -n a_n, \end{aligned} \] because the boundary term vanishes.
2.3. Q.E.D.
Steps 1 and 2 prove step 2.
3. Parseval’s identity gives\[ \int_0^{2\pi}f(x)^2\,dx = \pi\sum_{n=1}^{\infty}(a_n^2+b_n^2). \]
For the real Fourier coefficients of \(f\), Parseval’s identity is \[ \int_0^{2\pi}f(x)^2\,dx = \pi \left( \frac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2+b_n^2) \right). \] Step 1 gives \(a_0=0\).
4. Parseval’s identity also gives\[ \int_0^{2\pi}f'(x)^2\,dx = \pi\sum_{n=1}^{\infty}n^2(a_n^2+b_n^2). \]
The constant Fourier coefficient of \(f'\) is \[ \frac1\pi\int_0^{2\pi}f'(x)\,dx = \frac{f(2\pi)-f(0)}{\pi} =0. \] By step 2, its \(n\)th cosine and sine coefficients are \(n b_n\) and \(-n a_n\). Parseval’s identity for \(f'\) therefore gives \[ \begin{aligned} \int_0^{2\pi}f'(x)^2\,dx &= \pi\sum_{n=1}^{\infty} \left((n b_n)^2+(-n a_n)^2\right)\\ &= \pi\sum_{n=1}^{\infty}n^2(a_n^2+b_n^2). \end{aligned} \]
5. Therefore\[ \boxed{ \int_0^{2\pi}f(x)^2\,dx \leq \int_0^{2\pi}f'(x)^2\,dx }. \]
For every \(n\geq1\), \[ n^2(a_n^2+b_n^2) \geq a_n^2+b_n^2. \] Summing this inequality and applying steps 3 and 4 gives the result.
6. Q.E.D.
Step 5 is the required inequality.