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Dec 11, 2012 at 20:42 comment added Yoav Kallus Oh yeah. The second inequality has more assumptions that I forgot to include $f(0)=g(0)=0$.
Dec 11, 2012 at 20:13 comment added Daniel Spector 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.
Dec 11, 2012 at 19:21 vote accept Yoav Kallus
Dec 11, 2012 at 19:00 answer added Mark Meckes timeline score: 6
Dec 11, 2012 at 18:25 history asked Yoav Kallus CC BY-SA 3.0