I do an exercise from a textbook, where one has to show for $(X_n)_{n\in\mathbb{N}}$ iid. having finite second moment that $$\frac{1}{\sqrt{n}}\max_{1\le i\le n}|X_i|\rightarrow 0$$ in $L^1$. For this I have proved something which should allow me to prove convergence. I proved that $nP(|X_1|\ge \varepsilon\sqrt{n})\rightarrow 0 $ for all $\varepsilon>0$.
Then I am instructed to use the following $$E\left [\frac{1}{\sqrt{n}}\max_{1\le i\le n}|X_i|\right ]=E\left [\frac{1}{\sqrt{n}}\max_{1\le i\le n}|X_i|\left ( 1_{\max |X_i|<\varepsilon \sqrt{n}}+ 1_{\max |X_i|\ge\varepsilon \sqrt{n}}\right )\right ]$$
I continue this with using CS ineqality for the second term (denote $M_n:=\frac{1}{\sqrt{n}}\max_{1\le i\le n}|X_i|$)
$$\le E\left [M_n 1_{\max |X_i|<\varepsilon \sqrt{n}}\right ]+\sqrt{E\left [ M_n^2\right ] E\left[1_{\max |X_i|<\varepsilon \sqrt{n}}\right ]}\\ \le \varepsilon + \sqrt{E\left [ M_n^2\right ] nP(|X_1|\ge\varepsilon\sqrt{n})}$$
My problem is, that I cannot finish, because I now face the exact same problem as before, I have to bound $E[M_n]$ and even worse, $E[M_n^2]$.
As you pointed out it is enough to show that $E[M_n^2]$ is bounded: $$ E[M_n^2] = E\left[\frac{1}{n} \max_{1\leq i \leq n} X_i^2\right] \leq E\left[ \frac{1}{n}\sum_{i=1}^n X_i^2 \right]= E[X_1^2]. $$