Let $\{Y_n\}_{n \ge 0}$ be a martingale with $ \mathbb{E}[Y_n^2] < \infty $ for all $n$. Show that $$ \sup_{n \ge 0} \mathbb{E}[Y_n^2] < \infty \Leftrightarrow \sum_{n \ge 0} \mathbb{E} [(Y_{n + 1} - Y_n)^2] < \infty. $$
My attempt: I think I was able to solve it for "$\Rightarrow$": For any $n \in \mathbb{N}$, $$ (Y_n)^2 = (Y_0 + (Y_1 - Y_0) + ... + (Y_n - Y_{n - 1}))^2 $$ Then setting $D_m = Y_m - Y_{m - 1}$, we get
$$ (Y_n)^2 = Y_0^2 + \sum_{m = 1}^n D_m^2 + 2 \sum_{1 \le r < m \le n} D_rD_m $$ and taking the expected value
$$ \mathbb{E}[Y_n^2] = \mathbb{E}[Y^2_0] + \sum_m \mathbb{E}[D_m^2] + 2 \sum_{r < m} \mathbb{E}[D_rD_m]. $$
For $1 \le r \le m$, $$ \mathbb{E}[D_rD_m] = \mathbb{E}[\mathbb{E}(D_rD_m | \mathcal{F}_{r - 1})] = \mathbb{E}[D_m \mathbb{E}(D_r | \mathcal{F}_{r - 1})] = 0, $$
since $D_m$ is independent of $\mathcal{F}_{r - 1}$ and
$$ \mathbb{E}(D_r|\mathcal{F}_{r - 1}) = 0. $$ Since $\{Y_n\}_{n \ge 0}$ is bounded in $L^2$, i.e. $\sup_{n} \mathbb{E}[Y_n^2] < \infty$, we have that $\sum_{m = 1}^\infty \mathbb{E}[D_m^2]$ converges.
Is this correct reasoning? Also I am not sure how to solve the statement assuming "$\Leftarrow$" condition. Are there some hints or ideas I can use to solve it?
The correct reasoning would have been: For $1 \le r < m$, $$ \mathbb{E}[D_rD_m] = \mathbb{E}[\mathbb{E}(D_rD_m | \mathcal{F}_r)] = \mathbb{E}[D_r \mathbb{E}(D_m | \mathcal{F}_r)] = 0, $$
since $D_r$ is $\mathcal{F}_r$-measurable and
$$ \mathbb{E}(D_m|\mathcal{F}_r) = 0. $$ Thus, $$ \mathbb{E}[Y_n^2] = \mathbb{E}[Y^2_0] + \sum_{m=1}^n \mathbb{E}[D_m^2], $$ this being bounded iff $$\sum_{m=1}^ \infty{E}[D_m^2]<\infty.$$ So you were very close to the solution, indeed.