Probability- $L^2 \implies \max$ tail inequality

Question

I wonder if anyone can give a hint about whether the following is correct or wrong.

If given a probability measure $P$, $D \in L^2(P)$ is a function. Is it true that $$ \max_{1\leq i\leq n} D(X_i) = o_p(n^{1/2}) $$

Answer 1

Yes, it is true if $(D(X_i))_{i=1}^\infty$ is an i.i.d. sequence of $L^2$-random variables. As a hint, estimate $$ \sum_{i=1}^\infty \Bbb P\left(|D(X_i)|^2\ge \epsilon^2 i\right)=\Bbb E[|D(X_1)|^2]/\epsilon^2 + O(1) $$ and deduce that $$ \frac{\max_{i\le n}|D(X_i)|}{n^{1/2}}\xrightarrow{n\to\infty}_p 0. $$