# The maximal difference between a function and translates of itself

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

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

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 $$C=1$$. $$\quad\Box$$
• $+1$, is $C=1$ optimal? Commented Aug 18, 2023 at 1:33
• @mathworker21 : This is a good question. I don't think that $C=1$ is optimal. Do you have an idea about the optimal value? Commented Aug 18, 2023 at 1:39
• @NateRiver It's the triangle inequality for integrals: $$\int_{S^1} \left| f(x) - \int_{S^1} f(x+t)dt \right| dx \le \int_{S^1} \int_{S^1} |f(x)-f(x+t)| dt dx.$$ Commented Aug 18, 2023 at 1:52
• 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 $\epsilon^{-1} \chi (0, \epsilon)$. Commented Aug 19, 2023 at 15:57