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

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

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

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

Let $f: S^1 \to \mathbb R$ be an $L^1$ function, where 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

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

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

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