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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 $\lim_{r\to\infty}\frac1{2^{r+1}}\,\sum_{s=0}^r 2^s T_s=0$.
