Deriving an inequality for the integral of maximum indicator functions under measure-preserving transformations

Question

Let's denote the measure space by $(X, \mathcal{B}, \mu)$ and the measure-preserving transformation by $T: X \to X$. Let $A \in \mathcal{B}$ be a measurable set with $0 < \mu(A) < \infty$. Let $0 < t < 1$ and $n \ge 1$. For any $s > 0$, we set $Y_s := \left\{x \in X: \sum_{i=1}^{n} 1_A \circ T^i(x) \ge s\right\}.$

We aim to show:

$$ \frac{1}{n+1} \int_X \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu \ge \frac{(1-t)\mu(A)}{1 + \sup\left\{s > 0: \mu\left(A \cap Y_s\right) \ge t \mu(A)\right\}}. $$

Approach:

Let us assume $s^* = \sup\left\{s > 0: \mu\left(A \cap Y_s\right) \ge t \mu(A)\right\}$. We aim to show that:

$$ \frac{1}{n+1} \int_A \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu \ge \frac{(1-t)\mu(A)}{1 + s^*}. $$

Now, if we take $f = \sum_{i=1}^{n} 1_A \circ T^i$, then we have $\frac{f}{n+1} \le \max_{0 \le k \le n} 1_A \circ T^k$. It follows that:

$$ \frac{1}{n+1} \int_A f \, d\mu \le \int_A \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu. $$

By the definition of $Y_s$,

$$ Y_s = \left\{x \in X: \sum_{i=1}^{n} 1_A \circ T^i(x) \ge s\right\}. $$ and since $s^*$ is the supremum, we have,

$$ \int_A \left(\sum_{i=1}^{n} 1_A \circ T^i\right) \, d\mu \ge \int_{A \cap Y_{s^*}} s^* \, d\mu \ge s^* \cdot \mu(A \cap Y_{s^*}). $$

By our assumption, $\mu(A \cap Y_{s^*}) \ge t \mu(A)$. Hence,

$$ \int_A \left(\sum_{i=1}^{n} 1_A \circ T^i\right) \, d\mu \ge s^* \cdot t \cdot \mu(A). $$

Therefore,

$$ \frac{1}{n+1} \int_A \left(\sum_{i=1}^{n} 1_A \circ T^i\right) \, d\mu \ge \frac{s^* \cdot t \cdot \mu(A)}{n+1}. $$

Combining the results, we obtain:

$$ \int_A \left(\max_{0 \le k \le n} 1_A \circ T^k\right) \, d\mu \ge \frac{s^* \cdot t \cdot \mu(A)}{n+1}. $$ At this point, I am unsure how to proceed further. Could you please assist me in solving the remaining steps? I appreciate your time and effort. Thank you.

Answer 1

Fix $t \in (0, 1)$. Let $s > 0$ be such that $\mu(A \cap Y_s) \leq t \mu(A)$.

Then $(1-t) \mu(A) = \mu(A) - t\mu(A) \leq \mu(A) - \mu(A \cap Y_s) = \mu(A \cap Y_s^C)$ and therefore $$ (n+1)(1-t)\mu(A) \leq (n+1)\mu(A \cap Y_s^C) = \sum_{k = 0}^{n} \int_{X} \, \mathbb{1}_{A \cap Y_s^C} \, \mathrm{d}\mu = \sum_{k = 0}^{n} \int_{X} \, \mathbb{1}_{A \cap Y_s^C} \circ T^k \, \mathrm{d}\mu\\ = \int_{X} \, \left(\sum_{k = 0}^{n} (\mathbb{1}_{A} \circ T^k)(\mathbb{1}_{Y_s^C} \circ T^k) \right)\, \mathrm{d}\mu \leq (1+s) \int_{X} \, \left(\max_{0 \leq k \leq n} \mathbb{1}_{A} \circ T^k \right) \, \mathrm{d}\mu. \label{1}\tag{1} $$ The last inequality follows from $$ \sum_{k = 0}^{n} (\mathbb{1}_{A} \circ T^k)(\mathbb{1}_{Y_s^C} \circ T^k) \leq (1+s) \mathbb{1}_{\exists 0 \leq k \leq n \colon T^k \in A} = (1+s) \left(\max_{0 \leq k \leq n} \mathbb{1}_{A} \circ T^k \right) \label{2}\tag{2} $$ This is in turn true by the following argument:

If $T^j \in A$ for $0 \leq j \leq n$, then $$ \mathbb{1}_{Y_s^C} \circ T^j = \mathbb{1}_{\sum_{k = 1}^{n} (\mathbb{1}_A \circ T^k) \circ T^j < s} = \mathbb{1}_{\sum_{k = j + 1}^{n+j} \mathbb{1}_A \circ T^k < s} \leq \mathbb{1}_{\sum_{k = j + 1}^{n} \mathbb{1}_A \circ T^k < s}. $$ Now if you let $j = \min\{0 \leq l \leq n \,\vert\, T^l \in A \cap Y_s^C\}$ (in the case that this minimum doesn't exist, \eqref{2} is clearly true), it actually is $$ \sum_{k = 0}^{n} (\mathbb{1}_{A} \circ T^k)(\mathbb{1}_{Y_s^C} \circ T^k) = \sum_{k = j}^{n} (\mathbb{1}_{A} \circ T^k)(\mathbb{1}_{Y_s^C} \circ T^k)\\ \leq (\mathbb{1}_{A} \circ T^j)(\mathbb{1}_{Y_s^C} \circ T^j) + \sum_{k = j + 1}^{n} (\mathbb{1}_{A} \circ T^k) \leq (\mathbb{1}_{A} \circ T^j)\mathbb{1}_{\sum_{k = j + 1}^{n} \mathbb{1}_A \circ T^k < s} + \sum_{k = j + 1}^{n} (\mathbb{1}_{A} \circ T^k)\\ \leq 1+s. $$

Now let us go back again to proving your proposed inequality. If $s^* := \sup\{s > 0 \,\vert\, \mu(A \cap Y_s) \geq t \mu(A)\}$ is $+\infty$, the inequality is obviously true, so let us assume $s^* <+\infty$. Then $\mu(A \cap Y_s) < t\mu(A)$ for all $s > s*$ and therefore \eqref{1} holds. By letting $s > s^*$ decrease to $s^*$, we get \eqref{1} for $s^*$ instead of $s$ and this is your proposed inequality in a slightly rearranged form.