I am baffled with this question:
Let $(B_t)$ be a standard Brownian motion. For any $n \in \mathbb{N}$, let $(f_n)$ be a sequence of functions defined by
$$ f_n(x) = \left\{ \begin{array}{lr} |x| & , & \, &|x| \geq \frac{1}{n}\\ \frac{n}{2} x^2 + \frac{1}{2n} & , & \, & |x| < \frac{1}{n}. \end{array} \right. $$
Assuming that $(f_n)$ satisfies $$ M^{f_n}_t := f_n (B_t) - f_n(0) - \int_0^t \frac{1}{2} f_n''(B_s) \,ds \quad \text{ is a martingale} \quad \quad \text{ (A)}$$
and
$$ \mathbb{E}\bigg[ |M^{f_n}_t |^2 \bigg] = \int_0^t \mathbb{E} \bigg( |f_n' (B_s) |^2 \bigg) \,ds, \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{(B)} $$ show that there exists a continuous martingale $(M_t)$ such that $$ \mathbb{E} \bigg[ \sup_{0 \leq s \leq t} |M^{f_n}_s - M_s |^2 \bigg] \rightarrow 0 \quad \text{ as } n \rightarrow \infty$$ and hence $$ \int_0^t \bigg( n \mathbf{1}_{ \{ |B_s| < \frac{1}{n} \} } \bigg) \,ds \rightarrow |B_t| - M_t \quad \text{as } n \rightarrow \infty \quad \text{ a.s.}$$
$\mathbf{\text{My progress:}}$
Assuming the existence of $(M_t)$, I applied the Doob's $L^2$- inequality to conclude that $$ \mathbb{E} \bigg[ \sup_{0 \leq s \leq t} |M^{f_n}_s - M_s |^2 \bigg] \leq 2 \mathbb{E} \bigg[ |M^{f_n}_t - M_t |^2 \bigg].$$ Then, by (A) and (B), I get (for $n$ large, since $f_n$ converges to $|x|$ uniformly) $$ \mathbb{E} \bigg[ |M^{f_n}_t - M_t |^2 \bigg] \leq \frac{1}{2} +t -\frac{1}{n} + \mathbb{E} \bigg[ M^2_t + M_t- 2 M_t |B_t| \bigg]. $$
How can I proceed from here to prove the two facts? Any suggestions?
Hint: Show that $(M^{f_n})_{n \geq 0}$ is a Cauchy-sequence with respect to the norm
$$\|M\| := \left(\mathbb{E}\left( \sup_{s \leq t} |M_s|^2 \right) \right)^{\frac{1}{2}}.$$
Since the space of continous martingales endowed with $\|\cdot\|$ is complete, this proves the claim. In order to show that $(M_t^{f_n})_{n \geq 0}$ is a Cauchy sequence, use that $f_n-f_m$ does also satisfy (A), (B) for all $n,m \in \mathbb{N}$.