I was reading around some books on probability and saw that most moment inequalities are for independent random variables/martingales. That lead me to wonder if the following problem can be solved. Suppose $Y_1,Y_2,...$ are dependent random variables, such that $\frac{1}{n}\sum\limits_{k=0}^{n-1}\mathbb{E}[Y_k^2]\to \sigma^2>0$ (so that $\frac{\mathbb{E}[Y_n^2]}{n}\to 0$). Is it possible to show that $\frac{1}{n}\mathbb{E}[\max\limits_{k-0,...,n} Y_k^2]\to 0$ as $n\to\infty$? Would we perhaps require some extra conditions (a Lindeberg type condition, for example?) Or maybe I am just overthinking this and it is trivial.
Thanks in advance! Ian
Edit: Because I see no harm in adding it, suppose in addition that the Lindeberg condition is satisfied. That is, $\frac{1}{n}\sum\limits_{k=0}^{n-1}\mathbb{E}[Y_k^2\mathrm{1}_{|Y_k|\geq\epsilon\sqrt{n}}]\to 0$ for any $\epsilon>0$.
Assume $Z_1,Z_2,\dots$ are nonnegative random variables. Let $$M_n=\max\{ Z_j:j \le n\}$$
Assume that there exists $\alpha\in (0,1]$ such that for any $\epsilon>0$, $$ \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^n E[Z_j{\bf 1}_{\{Z_j > \epsilon n^\alpha\}}]=0.$$
Then $\lim_{n\to\infty} \frac{1}{n} E[M_n]=0$.
Indeed,
$$M_n {\bf 1}_{\{M_n > \epsilon n^{\alpha}\}} \le \sum_{j=1}^n Z_j{\bf 1}_{\{Z_j > \epsilon n^{\alpha}\}},$$
because if the maximum is above a certain level, then at least one of the random variables has to be above that level, and the maximum coincides with one of the random variables. Also,
$$E [M_n ] \le \epsilon n ^{\alpha} + E [ M_n {\bf 1}_{\{M_n > \epsilon n^{\alpha}\}}].$$
(first term on RHS is upper bound for expectation of $M_n$ on the event $\{M_n\le \epsilon n^{\alpha}\}$).
Therefore,
$$ \frac{1}{n} E[M_n] \le \epsilon n^{\alpha-1} +\frac{1}{n} \sum_{j=1}^n E[Z_j {\bf 1}_{\{Z_j > \epsilon n^{\alpha}\}}],$$
so
$$\limsup_{n\to\infty} \frac{1}{n} E[M_n] \le \epsilon.$$
Since $\epsilon>0$ is arbitrary, the result follows.