A Wirtinger-like inequality involving two functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:01:22Z http://mathoverflow.net/feeds/question/116106 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116106/a-wirtinger-like-inequality-involving-two-functions A Wirtinger-like inequality involving two functions Yoav Kallus 2012-12-11T18:25:13Z 2012-12-11T19:00:34Z <p>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 $$\left|\int_0^{2\pi}f(t)g'(t)dt\right|= \left|\int_0^{2\pi}f'(t)g(t)dt\right|\le \frac{1}{2}\int_0^{2\pi}[f'(t)^2+g'(t)^2]dt\text.$$</p> <p>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 <a href="http://www.sciencedirect.com/science/article/pii/0022247X86902283" rel="nofollow">Pachpatte 1986</a>, which gives $$\frac{1}{2}\int_0^{2\pi}\left[|f(t)||g'(t)|+|f'(t)||g(t)|\right]dt\le \frac{\pi}{2}\int_0^{2\pi}[f'(t)^2+g'(t)^2]dt\text.$$</p> <p>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.</p> http://mathoverflow.net/questions/116106/a-wirtinger-like-inequality-involving-two-functions/116108#116108 Answer by Mark Meckes for A Wirtinger-like inequality involving two functions Mark Meckes 2012-12-11T19:00:34Z 2012-12-11T19:00:34Z <p>Your inequality is implicit in Hurwitz's Fourier series proof of the isoperimetric inequality in the plane. See for example section 36 of K&ouml;rner's <em>Fourier Analysis</em> or section 4.1 of Groemer's <em>Geometric Applications of Fourier Series and Spherical Harmonics</em>.</p>