Show that the Infimum is bounded below (with prob. 1)

Question

I have a sample of $n$ i.i.d. "observations" of a random process $X(t)$ where $t\in [a,b]=T$.

I want to show that there exists a constant $0<D<\infty$ such that $$\lim_{n\to \infty} P(\inf_{t\in T} \frac{1}{n} \sum_{i=1}^n X_i(t)^2 \geq D a_n) = 1,$$ where $a_n>0$ is a sequence with $a_n \to 0$, as $n \to \infty.$

what i know:

1. I know that for all $t\in [a,b]$: \begin{equation} 0< D_1 a_n \leq E(X(t)^2) \leq D_2 a_n <\infty \end{equation} for some constants $D_1$ and $D_2$ (i.e. $\inf$ and $\sup$ of $E(X(t)^2)$ can be bounded.)
2. Moreover I know that there exist a constant $0<D_3<\infty$ such that for some sequence $\tilde{a}_n$ with $\tilde{a}_n/a_n \to 0$ as $n \to \infty$: $$\lim_{n\to \infty}P(\sup_{t\in T}|\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)|\leq D_3 \tilde{a_n}) = 1.$$

How can I use these results to show the assertion?

Usually I would apply the triangle inequality to get

$$|\frac{1}{n}\sum_{i=1}^n X_i(t)^2 | = |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2) + E(X_i(t)^2)| \leq |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| + E(X_i(t)^2).$$

and hence $$\inf_t \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \leq \sup_{t} |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| + D_2 a_n.$$ Obivously this attemp then goes in the wrong direction.

edit: the attemp was obviously very wrong. we can't conclude $$\inf_t \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \leq \sup_{t} |\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| + D_2 a_n$$ from the triangle inequality.

However: I think I found a solution (see the answer below ): It turns out (unexpected) that Assertion (2.) was too weak: I updated it.

Answer 1

Proof of the Statement:

From Assertion (2.) we have, as $n \to \infty$. ($\forall t$ meaning $\forall t \in T$) :
$$P(|\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| \leq D_3 \tilde{a}_n \, \forall t) \to 1.$$

But since $X \leq |X|$ implies $P(X\leq c) \geq P(|X| \leq c)$, we have: $$P(|\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| \leq D_3 \tilde{a}_n \, \forall t) \leq P(E(X_i(t)^2) - \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \leq D_3 \tilde{a}_n \, \forall t)$$ Now manipulating the last expression and using from (1.) the fact that $E(X_i(t)^2) \geq D_1 a_n$ we find $$P(E(X_i(t)^2) - \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \leq D_3 \tilde{a}_n \, \forall t)\leq P(\frac{1}{n}\sum_{i=1}^n X_i(t)^2 \geq D_1 a_n - D_3 \tilde{a}_n \, \forall t).$$

Since $\tilde{a}_n/a_n \to 0$, for $n$ large enough there now exists a constant $0<D < D_1$ such that $D_1 a_n - D_3 \tilde{a}_n\geq D a_n$ and hence we have (at least for large enough $n$):

$$P(\frac{1}{n}\sum_{i=1}^n X_i(t)^2 \geq D_1 a_n - D_3 \tilde{a}_n \, \forall t) \leq P(\frac{1}{n}\sum_{i=1}^n X_i(t)^2 \geq D\tilde{a}_n \, \forall t).$$ But $$P(\frac{1}{n}\sum_{i=1}^n X_i(t)^2 \geq D a_n \, \forall t) = P(\inf_{t\in T} \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \geq D a_n ).$$

To summarize: for large enough $n$ we have that there exists a constant $D$ such that $$P(\sup_{t\in T}|\frac{1}{n}\sum_{i=1}^n X_i(t)^2 - E(X_i(t)^2)| \leq D_3 \tilde{a}_n) \leq P(\inf_{t\in T} \frac{1}{n}\sum_{i=1}^n X_i(t)^2 \geq D a_n )$$ From which, together with Assertion (2.), the statement immediately follows.