Let $X_1, \cdots, X_n$ be i.i.d. random variables with distribution $P$. Let $g$ be a measurable function with $P g = 0$ and $P g^2 = 1$. Show that $\max_{1 \leq i \leq n}|g(X_i)| = o_p (\sqrt{n})$.
I read this claim in a proof but couldn't figure it out myself.
EDIT: $P g = 0$ and $P g^2 = 1$ mean: $E[g(X_1)] = 0$ and $E[g(X)^2] = 1$. $o_p(\cdot)$ is small oh pee notation: $X_n = o_p(Y_n)$ means that $X_n = Y_n R_n$ and $R_n \stackrel{P}{\to} 0$.
Let $G_i=g(X_i)$. Let $\epsilon>0$. Note $$ \sum_{n=1}^\infty P(|G_n|/n^{1/2}>\epsilon)=\sum_nP(G_n^2/\epsilon^2>n)\le\int_0^\infty P(G_1^2/\epsilon^2>t)\,dt=E[G_1^2/\epsilon^2]=1/\epsilon^2<\infty $$ Therefore, the second Borel Cantelli lemma implies $P(|G_n/n^{1/2}|>\epsilon\text{ infinitely often})=0$. This holds for all $\epsilon>0$, allowing you to conclude $|G_n|/\sqrt{n}\to 0$ almost surely.
You can then use the following fact: if $a_n$ is any sequence such that $a_n/\sqrt{n}\to 0$, then it follows $\frac1{\sqrt n}\max_{1\le i \le n}a_i \to 0$. Thus, you can conclude $\frac1{\sqrt n}\max_{i\le n} |G_i|\to 0$ almost surely, and therefore also in probability.