I am trying to prove the following inequality: let $f,g:S^1\to R$ (here $S^1$ is the unit circle parametrized by arc-length) be differentiable and have zero mean. Then $$ 4\pi \int f(t) g(t)\, dt \le \left(\int \sqrt{f^2(t) + g'^2(t) }\, dt \right)^2. $$ The inequality is easy to show if the $4\pi$ on the left-hand side is replaced by $1$. The $4\pi$ seems to be related to an isoperimetric inequality. An equality is obtained for $f(t) = g(t) = \sin t$.
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$\begingroup$ Do you perhaps want to have $f'^2$ on the right instead of $f^2$? $\endgroup$– Aleksei KulikovCommented Nov 21 at 13:54
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$\begingroup$ "The inequality is easy to show if the $4\pi$ on the left-hand side is replaced by $1$." -- How is this done? $\endgroup$– Iosif PinelisCommented Nov 21 at 14:27
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1$\begingroup$ @IosifPinelis Cauchy--Schwarz + Wirtinger's inequality, for example. $\endgroup$– Aleksei KulikovCommented Nov 21 at 14:40
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$\begingroup$ @AlekseiKulikov Not clear to me how you use Wirtinger. On the RHS you have an $L^1$-norm. I guess it is $f$, not $f'$. $\endgroup$– Giorgio MetafuneCommented Nov 21 at 19:52
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$\begingroup$ @GiorgioMetafune you are absolutely right, it is indeed $L^1$-norm. I see how to get some constant with a bit of suffering, but not sure if I can get $1$... $\endgroup$– Aleksei KulikovCommented Nov 21 at 19:56
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1 Answer
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Since $f$ has zero mean, we have $f=F'$ for a continuous $F$ on the circle. Then for the curve $\gamma(t)=(F(t), g(t)) $ the integral $\int \sqrt{F'^2+g'^2}$ is the length, and the integral $\int fg=\int gdF$ is the oriented enclosed area. Thus it is a variant of isoperimetric inequality. It may be deduced from the usual isoperimetric, for example, by the following steps:
approximate $F', g'$ uniformly by piecewise constant functions (or trigonometric polynomials if you prefer) with zero mean;
now our curve is a finite union of simple loops;
apply isoperimetric inequality for each loop and sum up. It remains to use the bound "the sum of squares does not exceed the square of the sum" for the lengths of loops.
answered Nov 22 at 4:20
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$\begingroup$ Thanks! This seems to nail it indeed, $\endgroup$ Commented Nov 22 at 14:07