There is an interesting problem that has been resisting my efforts for a while. Assume that $\{X_n: n = 1, 2, \ldots\} $ is a sequence of uniformly integrable random variables. I would like to show that $$ \lim_n \frac{1}{n}\cdot E\left[\sup_{1\le i \le n} |X_i|\,\right] = 0. $$
Of course $\lim_n \frac{1}{n} \sup_{1\le i \le n} E[|X_i| ] = 0, $ but the limit above, albeit it might very well be as easy, I do not seem to find the way to establish its validity....
I wonder if anyone has an easy proof for this fact. Thank you.
Maurice
Notice that for a fixed $R$, $$\max_{1\leqslant i\leqslant n}|X_i|\leqslant\max_{1\leqslant i\leqslant n}|X_i|\chi\{|X_i|\lt R\}+\max_{1\leqslant i\leqslant n}|X_i|\chi\{|X_i|\geqslant R\}\leqslant R+\max_{1\leqslant i\leqslant n}|X_i|\chi\{|X_i|\geqslant R\},$$ hence, taking the expectation and using the inequality $$\mathbb E\left[\max_{1\leqslant i\leqslant n}Y_i\right]\leqslant n\max_{1\leqslant i\leqslant n}\mathbb E[Y_i]$$ valid for non-negative random variables $(Y_i)_{1\leqslant i\leqslant n}$, we obtain $$\mathbb E\left[\max_{1\leqslant i\leqslant n}|X_i|\right]\leqslant R+ n\max_{1\leqslant i\leqslant n}\mathbb E\left[|X_i|\chi\{|X_i|\geqslant R\}\right].$$ Dividing by $n$, we obtain for each $R$ $$\limsup_{n\to \infty}\frac 1n\mathbb E\left[\max_{1\leqslant i\leqslant n}|X_i|\right]\leqslant \sup_{i\geqslant 1}\mathbb E\left[|X_i|\chi\{|X_i|\geqslant R\}\right]$$ and we conclude using the assumption of uniform integrability.