Let $f(t)$ and $g(t)$ be periodic functions on $t\in[0,2\pi]$. By using the Fourier series of the two functions, we can easily prove the inequality $$\left|\int_0^{2\pi}f(t)g'(t)dt\right|= \left|\int_0^{2\pi}f'(t)g(t)dt\right|\le \frac{1}{2}\int_0^{2\pi}[f'(t)^2+g'(t)^2]dt\text.$$

I have been trying to find a reference for this inequality because I need to use it to solve some problem. The closest I have been able to find is Pachpatte 1986, which gives $$\frac{1}{2}\int_0^{2\pi}\left[|f(t)||g'(t)|+|f'(t)||g(t)|\right]dt\le \frac{\pi}{2}\int_0^{2\pi}[f'(t)^2+g'(t)^2]dt\text.$$

The extra factor of $\pi$ is highly undesirable and the absolute values inside of the integral unnecessary for me. I can easily provide a short proof in the text, but if anybody can think of where the first inequality might appear, that would be better.

  • $\begingroup$ I think there must be a mistake in the second inequality, since $f=c$ and $|g^\prime|=1$ would give the inequality $c \leq \pi$. The first inequality is an easier inequality from this perspective, and is provable by using a standard Poincare inequality for $||f-\overline{f}||_{L^2}\leq C ||f^\prime||_{L^2}$ with the optimal constant $C$ and realizing that you can introduce the average value on the left hand side, due to periodicity of $g$, followed by Cauchy-Schwartz. $\endgroup$ – Daniel Spector Dec 11 '12 at 20:13
  • $\begingroup$ Oh yeah. The second inequality has more assumptions that I forgot to include $f(0)=g(0)=0$. $\endgroup$ – Yoav Kallus Dec 11 '12 at 20:42

Your inequality is implicit in Hurwitz's Fourier series proof of the isoperimetric inequality in the plane. See for example section 36 of Körner's Fourier Analysis or section 4.1 of Groemer's Geometric Applications of Fourier Series and Spherical Harmonics.


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