I apologize in advance if this question is too basic, but I've received no response on Math Stack Exchange, so perhaps it is more appropriate here:
Let $X_n$ be a square integrable martingale with $\mathbb{E}\lbrack X_n^2\rbrack=O(n)$. Is it true that $X_n/n$ tends to $0$ almost surely?
Note that if one demanded instead that $\mathbb{E}\lbrack X_n^2\rbrack =O(n^{1-a})$ for some $a>0$, then the claim would follow for any random variables from Markov's Inequality and the Borel Cantelli Lemma.
The answer is yes, and it is based on an idea by Prokhorov (cf. e.g. Theorem 10 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). We have $EX_n^2\le Cn$ for some real $C>0$ and all natural $n$. For natural $s$, let \begin{equation} T_s:=\max_{2^s\le n<2^{s+1}}\frac{|X_n-X_{2^s}|}{2^s}. \end{equation} Then, by [Doob's martingale inequality], for any real $t>0$ \begin{equation} P(T_s\ge t)\le\frac{E(X_{2^{s+1}}-X_{2^s})^2}{(2^s t)^2} \le\frac{EX_{2^{s+1}}^2}{(2^s t)^2}\le\frac{C2^{s+1}}{(2^s t)^2}=\frac{2C}{2^s t^2}, \end{equation} so that $\sum_{s=1}^\infty P(T_s\ge t)<\infty$. So, by the Borel--Cantelli lemma, $T_s\to0$ almost surely (a.s.) as $s\to\infty$. Therefore, for any natural $n$ and $r$ such that $2^r\le n<2^{r+1}$ (so that $r=r_n:=\lfloor\log_2 n\rfloor$), one has \begin{equation} \frac{|X_n-X_1|}n\le2 \frac{|X_n-X_1|}{2^{r+1}} \le2\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s\to0 \tag{1} \end{equation} a.s., since $\sum_{s=0}^r 2^s<2^{r+1}$; cf. e.g. Lemma 9 in Section 3 of Ch. IX in [Petrov, V. V., Sums of independent random variables, Springer-Verlag, 1975]). Thus, $\frac{X_n}n\to0$ a.s., as desired.
Details on $(1)$: Since $T_s\to0$ a.s., without loss of generality for each real $\epsilon>0$ there is a natural-valued random variable $R_\epsilon$ such that for any natural $s$ the event $s>R_\epsilon$ implies $T_s\le\epsilon$. Therefore,
\begin{equation}
\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s
\le\frac1{2^{r+1}}\,\sum_{s=0}^{R_\epsilon} 2^s T_s+\frac1{2^{r+1}}\,\sum_{s=R_\epsilon+1}^r 2^s \epsilon
\le\frac1{2^{r+1}}\,\sum_{s=0}^{R_\epsilon} 2^s T_s+\epsilon.
\end{equation}
So, $\limsup_{r\to\infty}\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s\le\epsilon$, for any $\epsilon>0$, and hence $\lim_{r\to\infty}\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s=0$.
