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 $f: S^1 \to \mathbb R$,

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


1 Answer 1


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 $C=1$. $\quad\Box$

  • 2
    $\begingroup$ $+1$, is $C=1$ optimal? $\endgroup$ Commented Aug 18, 2023 at 1:33
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.