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Note: We view the sphere $S^1$ as $[0,1]$ with the endpoints identified, and equip it with its usual addition structure, and Lebesgue measure.

Question: Does there exist an absolute constant $C > 0$ such that for all $L^1$ functions $f: S^1 \to \mathbb R$,

$$\sup_{t \in S^1} \int_{S^1} |f(x + t) - f(x)| \, dx \geq C\int_{S^1} \left|f(x) - \left(\int_{S_1} f(y) \, dy\right) \right| \,dx\text{?}$$

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1 Answer 1

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Using the condition $\int_{S^1}dx=1$ and Jensen's inequality, we have $$\sup_{t\in S^1} \int_{S^1}dx\,|f(x + t)-f(x)| \ge\int_{S^1}dt\, \int_{S^1}dx\,|f(x + t)-f(x)| \\ =\int_{S^1}dx\, \int_{S^1}dt\,|f(x)-f(x + t)| \ge\int_{S^1}dx\, \Big|f(x)-\int_{S^1}dt\,f(x + t)\Big| \\ =\int_{S^1}dx\, \Big|f(x)-\int_{S^1}dy\,f(y)\Big|. $$ So, your conjectured inequality holds with $C=1$. $\quad\Box$

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    $\begingroup$ $+1$, is $C=1$ optimal? $\endgroup$ Commented Aug 18, 2023 at 1:33
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    $\begingroup$ @mathworker21 : This is a good question. I don't think that $C=1$ is optimal. Do you have an idea about the optimal value? $\endgroup$ Commented Aug 18, 2023 at 1:39
  • $\begingroup$ @NateRiver It's the triangle inequality for integrals: $$\int_{S^1} \left| f(x) - \int_{S^1} f(x+t)dt \right| dx \le \int_{S^1} \int_{S^1} |f(x)-f(x+t)| dt dx.$$ $\endgroup$ Commented Aug 18, 2023 at 1:52
  • $\begingroup$ Oh.. so it is haha $\endgroup$
    – Nate River
    Commented Aug 18, 2023 at 1:52
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    $\begingroup$ The best constant is 1. This is seen best for measures: just take a $\delta$ at any point. For any $t$, the total variation of the difference between $\delta$ and $\delta_t$ is 2 as well as that of $\delta-1$. For functions, one approximates the $\delta$ with $\epsilon^{-1} \chi (0, \epsilon)$. $\endgroup$ Commented Aug 19, 2023 at 15:57

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