# A Wirtinger-like inequality involving two functions

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.

• 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. – Daniel Spector Dec 11 '12 at 20:13
• Oh yeah. The second inequality has more assumptions that I forgot to include $f(0)=g(0)=0$. – Yoav Kallus Dec 11 '12 at 20:42